Es Muy Tutor Services - (c) 2011 - 2017 and onwards until the end of time!(Designed for browser resolution 1024 x 768 - but bigger is better! - hold down [Ctrl] and roll your mouse’s centre scroll wheel)
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Welcome to Es Muy Tutor ServicesThe home of educational support for one-to-one delivery in the home or in classroom groups. Es Muy Tutor Services is conscious of the fact that today’s teachers are stretched to the limit, class sizes are almost always high and there often isn’t time to ensure that every student has the attention they deserve. Es Muy Tutor Servicescan give that time to your child, teen or young adult ensuring them of 100% attention and explanation. Click on the menu link above to send an email for more information.
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(Oh...by the way....while you are browsing round the site, don’t forget to look out for the links to and from each page, such as the battery icons and so on. I have tried to make the site enjoyable, a little bit light hearted but at the same time capable of delivering quality materials. Please drop me a line if you like, I’m always open to good suggestions)
02 Sep 2017
Es Muy Tutor Services - Trouble With Font Rendering
Throughout this site I have used Buxton Sketch (the font you’re looking at now) as the predominant font. In certain places I have used an alternative called “squeaky chalk” but it isn’t a particularly easy font to use throughout a document such as this, as a result I hunted round the Internet and found Buxton.The software I use to create the website is first class, but since an upgrade I have had problems with Buxton being rendered correctly in any browser other than Internet Explorer version 11 and a proprietary URL database application with its own internal browser. The rendering problem was reported to the developers of the software, they did not appear to have any issues with it, however Microsoft Edge for example, cannot render the font properly.I must stress that I am not stating this is a fault with any of the applications that can’t display the font properly, but it seems to be a problem that I cannot overcome. In the fullness of time I will probably have to replace the font, however the site is into three figures (pages) now so it will not be a particularly simple task to do this, and I can’t say whether or not it’ll be started anytime soon (if at all).As a result I would ask you to please bear with me, and utilise something like Internet Explorer if the font appears to be messy, and the words appear to be smashed together (as per the image on the right, the one on the left shows the font as it should be)I hope that these issues do not spoil your experience on the site, please now click your browser’s “back” button or click on the home button above to continue your browsing experience of the ESMUY website. (Don’t forget, you can make the page bigger or smaller on-screen by using your [ctrl] key with your mouse scroll wheel).Many thanks - Martyn
Now…for interest sake, these are the web browsers I have found so far that will render the font (at least on my computer anyway)1. Microsoft Internet Explorer2. Comodo Icedragon3. Jordysoft Advanced URL Database’s Internal Browser4. Download and install the ttf font >> click here <<I have no affiliations here, this is simply a list of what I’ve found that works for me.
Es Muy Tutor Services - News
27/08/2017 - the concept of a news page and never really hit me until now, but it’s GCSE time here in the UK and as I’m going into the teaching profession, it may be useful to comment occasionally on anything newsworthy in that arena.With that in mind, the usual consternation over questions has arisen, the new 9-1 GCSE marking scheme for Mathematics and English has come into play, with the other subjects coming online in 2018. The Edexcel maths paper had a particularly nasty question on it >> click here << which I have explained. In fact, it probably wasn’t so nasty, just that it did require the very mentally demanding conjugation of the cosine rule twice, using surds and on top of all this, under examination conditions.So this is where the news will be - enjoy!
Es Muy Tutor Services - About
History and a bit about me too......The philosophy of Es Muy Tutor Services is quite simple and it goes back many years. I was fortunate enough to experience very little difficulty with Mathematics and Sciences whilst studying at Comprehensive / Secondary level. Inspired by some good teachers it led to a deeply ingrained passion for study which I have managed to maintain for many years.I studied at the University of Aston between 1979 and 1982 obtaining a Lower Second Class honours degree in Chemistry, but being at a somewhat loose end I decided to attend local educational establishments in Dudley and Halesowen, West Midlands, UK and obtain City and Guilds accreditation in Visual Basic Programming, “C” Programming, Network Installation and Maintenance and an Institute for the Management of Information Systems Applications Programming Certificate. It still wasn’t enough and so in 1999 I contacted the Open University and embarked on a seven year part time Bachelors Degree in IT and Computing, graduating in 2007 with Upper Second Class honours.But, still hungry for knowledge I completed a 2 year certificate in Mathematics with the Open University in the UK and I recently completed a 2 year certificate in Spanish (which is where Es Muy originated). Es Muy Tutor Services offers the benefits of a tutor with honours degrees in Chemistry and IT, Diplomas in IT and Computing, Certification in Mathematics and Spanish to a level equivalent to that of “A” Level and a continued passion to study. As of April/May 2011 I now hold the PTLLS qualification at Level 4 and this confirms my competency in the area of tuition. Also, as of 19 August 2016 I became a qualified Teaching Assistant QCF Level 3In 2017 I will start teacher training, and hopefully by the end of the one year course I will be a fully qualified Secondary School Science Teacher, teaching (in the UK) the KS3 to KS5 age groups. (In the USA this might equate to 7th grade onwards).Previously a serving member of the UK police service, I have come to the end of a very demanding, yet interesting career. In my time I have had many different experiences, and have encountered many people, both young and old who would clearly benefit from the services of an organisation like Es Muy Tutor Services. For more information please drop me an email or feedback from the menu.Martyn J CoxBSc(Hons) Chem (Aston) BSc(Hons) IT&Comp (Open) Dip Comp (Open) Dip IT (Open) Cert Maths (Open) Cert Spanish (Open) PTLLS MBCS
Es Muy Tutor Services - Here’s Engelbart
Introducing “Engelbart the Mouse”Well.......The Wise Old Owl is going to need a bit of help now that the site is growing, so we now introduce Engelbart the Mouse, his trusty old friend who will appear from time to time on these pages.Generally when you see Engelbart......click YOUR mouse on him (OR simply rest your mouse cursor on him for a short hint or tip) and he will give you some information, or possibly take you to another place where information (concerning the subject matter on the page he was on) will be.Hopefully you won’t see him that often 8-)If you see my digging friend here (or a page is ‘coned off’), the chances are that you are on a page that isn’t finished, you’ll have to bear with me on that, and when you revisit the page at some point later you may find that my digging friend has gone in which case the page can be regarded as complete.
Es Muy Tutor Services - Resources
This page will hold links to areas of documented topics (ie: discussed in greater depth but as you’re already on that page I’ve decided to leave the image without a link, so no point in clicking it (but I bet you will anyway!))This link will take you to an index page where you can choose from a selection of worked examples, eg: if you don’t want a full document, just a refresher on how to do a certain problem.
Free Toys !!
PLEASE NOTE - These pages are my own work, and my interpretation of the material that they cover. Most of them are written in a high level ‘overview’ style so that the reader can get a grasp of the subject and possibly read further (however occasionally I have gone into depth). They are NOT intended to be definitive and are aimed exclusively at computer / maths / science novices - thank you.NB: I have put a lot of time and effort into these pages, but of course I have no control over who uses them, however I would like to know who you are, please drop me a line and tell me a bit about yourself.
HANDOUT TYPE STUFF
HANDOUT TYPE STUFF
Some sections on this website will cover areas where the end objective, for example how to work at square roots, can lend itself quite nicely to being computerised. As well as being keen on mathematics, science and information technology I am also a keen (albeit amateur) computer programmer with experience in “C” and “Visual Basic”.
Es Muy Tutor Services - Downloadables
The small free downloads here have all been written using Visual Basic Dot Net, they are independent applications, so they do not need to be “installed” in the conventional sense, they are not dependent on any library files, OCX, DLLs or anything like that over and above those that you already have installed on your computer. What this means is that there should be no interference with the way your computer runs, if you choose to download and run these. Of course the Windows registry may receive slight updates and the occasional ‘temporary’ file may be created.These are offered free, but of course I can’t be responsible for how they are used. If you download one and use it then I have to take this as an unconditional acceptance of the risks that may be involved. I’m not trying to scare you off, I’m just being honest!
Click on the owl to download a small program which will show you how square roots are calculated using the “repeated estimate”method.
Click on the owl to download a small program which will show you how to approximate broadband download times depending on file size and line speed.
Click on the owl to download a small program which will compute, where possible the roots of a quadratic equation.
Click on the owl to download a small program which will demonstrate long multiplication using the popular secondary school “grid method”.
Teach yourself BINARY with this little desktop clock !
Passwords with a Black Country twang eh ?? - click on the owl “mah mon / wench!”
Perhaps the simplest note pad in the world, but it works, and its FREE!!
Es Muy Tutor Services - Worked Examples Index
This page is a one-off dedicated to “units of measurement”, entities that have their place in mathematics, science and ICT so I couldn’t really categorise it other than neutrally.
Es Muy Tutor Services - Units of Measurement
This page will talk about units of measurement. Whether it’s in mathematics, science, computing or any other scientific field of study you will at some point perform a calculation of some nature. This calculation could be something simple like moles per litre in a chemistry setting, or metres per second in physics and mathematics. In any event, and indeed you will probably find in tests, getting the numerical part of the answer right is only part of the question and carries only part of the marks. It is also just as important to make sure that your units are expressed properly, for example is quite obvious that you can’t measure weight in metres per second, nor can you measure volume in millimetres (it would be cubic millimetres of course).On this page we will take a look at a few examples and hopefully demystify units for you (see also the physics based text on this subject)DistanceThere are many units for distance, but this site will concentrate on the SI unit for distance which is the metre, the metre is defined as “the distance travelled by light in a specified fraction of a second”, that fraction being as follows:This follows simply because the speed of light in a vacuum is 299, 792,458 m/s or approximately 186,322 mi./s. there have been many definitions of “metre” but this is the latest one adopted in the 1980s. With distance of course we are thinking in one dimension, from A to B, very often (but of course not always) in a straight line but we now start to look at areas which is in 2 dimensions and will see that there will be a slight adjustment of the way we represent the unit, it will still be something to do with metres but there will be a slight twist to it.AreaWith area we are looking at two-dimensions, for example if we were looking at the area of a patio we would be discussing its length and its width (some people refer to this as breadth) so we can’t have metres as the unit of measurement as this would become confusing.
Take a look at the rectangle on the right, we are given its dimensions as 14 m x 9 m, that is it is a two-dimensional object with a width of 14 m and a depth of 9 m, so what is its area?Well, the area is the width multiplied by the depth and we do this in 2 stages:1. Multiply the numbers together to get a numerical result, then2. Multiply the units together as if we were doing a small algebra question.
The calculation on the left has been broken down into several parts, it shows that 14×9 is hundred and 26 if you follow the red, purple and yellow linesbut also shows “metres times metres equal metres squared” if you follow the orange lines.So you can see that it is quite useful to break the calculation down into the 2 pieces to ensure that not only do you get the right numerical answer but also the correct units for that answer.
VolumeQuestion - “what do we get if we add another dimension and move from two-dimensional to three-dimensional?”Answer - “we enter three-dimensional space and we move from square metres to cubic metres:Let us assume that our rectangle was indeed the markings out for the excavation required for petrol tanks in a new filling station, the excavation has now taken place to a depth of 7 m. The volume of earth removed is shown to be 882, but 882 what?.Follow the calculation above in the same style of the annotated one and you will see that “metres times metres times metres” becomes metres cubed, or cubic metres which, as shown is “m” raised to the power of 3.What about “per”?When you see “per” the units are going to be a little bit more complicated because you are going to have to combine units, as in “meters per second” or “kilometres per litre”. it’s no more complicated really to work them out provided you take your time, follow the examples I’m going to give you now and hopefully this will start to make a bit more sense.At the time of writing this page (February 2016) we are quite lucky in the United Kingdom to be able to buy petrol at a reasonable price of 99p per litre (We used to buy our fuel in gallons, and I believe in the United States of America that is the unit still used, but we have adopted European units “SI”).99p for every litre of fuel would be written this way on paper:If you think about it, it does make sense. Consider a fraction, for example 3/5……the lower number being the denominator identifies what we have, and the numerator being the number on top tells us how many of them we have. Another way of looking at this is to regard our fraction 3/5 as “3 per 5”, or “3 out of every 5”. Compare the expression “3 per 5” with “pence per litre” and you should be able to work out where this is going. Consider another example, another fraction, 1/5 which is presently the base rate for income tax in the United Kingdom, usually quoted as 20%. This can be expressed again as “1 per 5” or “1 out of every 5” meaning that for every £5 I earn the government taxes me £1.Let us go back to our petrol cost, I go to the petrol station and put into my car 39.26 L of petrol, the receipt from the pump tells me that this has cost me £38.87. What is the cost in p per litre of the petrol then?Note that I have had to convert the cost from pounds to p to make the calculation correct, had I have left the cost in pounds my result would have been pounds per litre not p per litre.What is this “l to the power of minus 1 then!” ??It’s just a way of expressing per litre without having to use fractions, if you study algebra you will see that we deal with reciprocal numbers (that’s a number inverted, for example the reciprocal of 5 is 1/5 and the reciprocal of 4/9 is 9/4 and so on). Also “ordinary numbers” can also be dealt with in this way, for example take a look at the numbers listed below:
Follow the diagram on the left down from the top, 2 to the 4th power is 16, similarly 2 to the power of 3 is 8 right down to 2 to the power of zero which is accepted mathematically to be one.When we get to negative powers we can see that “2 raised the power of -1” is in fact the same as “1 over 2 raised the power of +1” so it follows that a number raised to a negative power is the same as the reciprocal of that number raised to the positive power, this also applies to letters so that in algebra (as shown in red at the bottom) “x raised to the power of -1” is the same as “one over x”and this can quite logically be extended to other units, such as in our case - litres.Hopefully you have been able to follow this example, there are some really weird and wonderful mathematical and scientific properties in the world with very strange units indeed, we will take a look at some of the simpler ones such as speed, acceleration and momentum, then we will take a look at density and finish off with a really strange one indeed, pressure.
Let’s take a look at speed, measured in metres per second:We can see that speed is measured in metres per second which is the distance in metres divided by the number of seconds taken to cover that distance. In our example we have covered a distance of 25 m in 5 seconds, this works out at 5 m/s and I have shown the units in the index notation with the “raised to the power of -1” applicable to the seconds.When an object accelerates, it changes its speed usually at a fixed rate. An object travelling at 4 m/s changes its speed in one second to 8 m/s and then a second later it is travelling at 12 m/s, finally after the 4th second the object is travelling at 16 m/s…… what was its rate of acceleration?We can see that the object changes its speed by 4 m/s every second, 4, 8, 12 then 16 we can reasonably confidently predict that after 5 seconds the object will be travelling at 20 m/s and then 24, 28, 32 and so on. The object speed changes at 4 m/s every second. The acceleration of the body is therefore measured in metres per second per second, this is shown like this:
Momentum - the momentum that an object possesses is the product of its mass multiplied by its velocity, that is its mass in kilograms multiplied by its velocity in metres per second:Understandably this property is measured in “kilogram metres per second”. The momentum of an object can be regarded as a measure of how hard the object would be to stop from uniform motion, the more momentum object had then the harder it would be to stop it. If someone threw a tennis ball at you at 25 m/s (about 56 miles an hour) you would probably be able to stop it by catching it although your hand might sting a little bit, would you want to try to stop a cannonball at the same speed? - No I didn’t think you would, because you have already ascertained (whether you know it or not) that a cannonball would have far more momentum than a tennis ball!
Pressure - this is the last unit that I am going to describe. Pressure is defined as the effect of a force measured across a given area, force is usually measured in Newtons, with 1 N being defined as the force required to accelerate an object weighing 1 Kg at a rate of 1 m/s per second per second, the standard equation “force equals mass times acceleration” or F=MA gives an idea of the units of force:With one Newton being measured as “1 kg metre per second per second” as shown at the end of the equation directly above this line. Since pressure is a measure of force divided by area we can come up with the following equation:with “1 Kg metre per second per second” being known as “1 Pascal” or 1 Pa..I am now going to bring this page on “units of measurement” to a close. I hope that it has explained a little bit about the importance of establishing the correct units for your calculations. There are many texts available on the Internet which you could refer to for a more detailed explanation.
This question appeared on a Scottish Higher mathematics paper in 2015 and was the subject of controversy. Quite a ‘tasking’ question for the 16 year age group.Part (a) is quite straightforward, consider that if the crocodile does not travel on land, he must swim directly across to the zebra in a diagonal line, i.e.: the value of ‘x’ above is 20.Substituting x = 20 into the equation given:
This leads to:
Note that the second part of the equation disappears as it equates to 0. The first result is therefore:
Remember that the first answer of 104.4 is in TENTHS of a second so it needs to be divided by 10 before you present your solution.Similarly, the second bit of part (a) asks for the time taken if the crocodile swims the shortest distance, this is obviously when x = 0 as the shortest swim MUST be directly across the river:
This time the second part of the expression has a value, so we need to remember to include it in the answer, again in TENTHS of a second:
Not much of a difference between 10.44s and 11.0s but I suppose that a hungry crocodile might want to try to reduce his time taken!!Now….This is where the tricky bit starts. The question does in fact SAY that there IS an optimum value for x where the time taken is minimised, i.e.: a certain value for x where the time is potentially lower than 10.44s and 11.0s (but it could be the lower of the two, we don’t know yet)….but how to establish this requires the use of Differential Calculus, and this is where the consternation arose.First of all, plot the function to see where, if any, the curve produces a minimum value:
The equation / function shows a stationary point where the value for y reaches a minimum value, approximately 98 TENTHS of a second, or 9.8s at a value of x = approximately 8 metres.The issue here is that if there isn’t the option to draw a graph, we need to use calculus to arrive at these values. The calculus required is not complex, but a good knowledge of the rules of differentiation is required. The local minimum will occur where the value for x in the first derivative is zero.
Stating the equation again, we will now differentiate it.
We have two functions of x and a constant (5) viz:
The three functions are coloured and the differentiation rules to apply are shown, now let’s do it.Expand the green function to make it 80-4x, we can then differentiate it and ‘put it to one side for later’
We have performed the simpler bit first, but we have to remember to include this -4 later on. Now let’s look at the ‘function of a function’:
This is a ‘function of a function’ because we have 36+x2 as the first, and the ‘raising to the power of ½’ as the second.Now, the chain rule states:
Let u = 36+x2
And y = (u) 1/2
Substituting back for x:
Remembering our chain rule we now have:
Remember the constant 5?.......well the constant multiple rule states that:
So, since the first part of the equation that we started to differentiate was:
Then the derivative becomes 5 times what it was by virtue of the Constant Multiple Rule, that is, the derivative of the red and yellow bits above becomes:
Remember the -4 from way back?....Our final derived expression is this:
Now we need to establish a value for x when this expression is equal to zero, this will optimise x and therefore lead to the minimised time T.If:
Squaring both sides and further expansion gives us:
Rearranging:The optimum value for x is therefore 8m, and if we reinsert this into the original expression we arrive at the minimum time:Therefore:The crocodile can get to the zebra almost a full second faster if he swims to a point P = 8m and then makes the rest of the journey on land.
Es Muy Tutor Services - Hannah’s Sweets
This question appeared on an English GCSE paper in 2015 and was also the subject of complaints from many students. Not so much the question as the “probability” in it would have been covered in the curriculum, I think it was the seemingly disjointed nature of the question that caused the confusion.
I think that the part of the question which caused the most confusion was that it initially looks like a probability question but at the very end you are given a quadratic equation! What you are not asked to do is solve the equation for ‘n’ but this could be a trap that most people would fall into, after all when you are given a quadratic it is usual to try to find the roots of it.A good understanding of the rules of probability is essential here, not only for the obvious reason (to get the marks) but to be able to start the question at all you have to be able to see that the question is almost entirely probability.
Let us suppose that we have five balls, red white blue yellow and green. What would be the probability of selecting the green ball?We know that probability is defined as:
In other words the probability of selecting the green ball is one divided by the number of possible outcomes which of course is five. Therefore the probability of drawing out the green ball is one in five or 1/5. The point that I am trying to reinforce here is that generally in probability questions we are given fixed numbers, i.e. “a bag contains twelve” or “a box contains twenty-five” but in this case we are not told exactly how many sweets are in the bag, in this case we are told simply that there are “n” which could of course be any amount.We are told that six of the sweets are orange and we can then deduce from that that the remainder (which we are told are yellow) is therefore n - 6 although this little bit of information turns out to be of no use whatsoever.What is the probability of Hannah choosing a single orange sweet..............?
If you stop and study this small equation you will see that it is in fact correct, there are six ways in which we can pull out an orange sweet from a total of ‘n’ sweets. We are then told that Hannah eats the sweet, so we know for a fact that this particular sweet is not returned to the bag.Hannah then selects, by chance, another orange sweet from those left in the bag, and how many sweets are left in the bag now?Take a look at this:
I have called the probability of selecting the second orange sweet “orange 2” and if you stop and think about the expression you will realise that there are five ways now of selecting an orange sweet simply because of the fact that there are now only five orange sweets in the bag. We knew at the beginning that we had ‘n’ sweets so this time round we must have one less, i.e. n -1Summarising to this point we have the following:
Now, we refer back to the question on we are told that the probability of these two events occurring sequentially is one third. Using the rules of probability (dependent events) we arrive at the following expression:
It is this expression that the student would have needed to arrive at, and then a little bit of algebraic manipulation would lead to the quadratic given. This solution would attract full marks because, as I said earlier, the question does not ask for the value of ‘n’ to be calculated.Let us multiply out the left-hand side of the expression:
Now let us get rid of the fraction on the right-hand side by multiplying both sides by three:And for this expression to be correct, the numerator and denominator must be the same .............if this is the case then we can write the following:
Now if we subtract 90 from both sides:
We have now completed the question because we have shown that the quadratic expression is in fact correct. (Should your curiosity get the better of you and you cannot resist the temptation to find out the value of ‘n’ it is in fact 10).
Es Muy Tutor Services - Completing The Square
Completing the Square is a method used to show standard quadratic expressions in terms of a complete square, that is to show this:
As an expression in this format:
Where ‘a’and ‘b’ are real numbers (and in many cases, integers)
The way that we do this is as follows:
1. Halve the coefficient of the ‘x’ term, in this case - 8, half of which is - 4.2. Write down the expression in its square form and expand it
But note that we now have +16 whereas we started with +2. We need to subtract 14 to get to the right answer, therefore in COMPLETED SQUARE FORM the result is......
Let’s take a look at another one....
Note that we start with +40, now as before, halve the coefficient of ‘x’ and square the expression, then expand it out to give a second quadratic expression:
Well, we started off with +40 but now have +25, to make up the shortfall we ADD 15:
Prove this to yourself by expanding out the expression:
Try these 4 for yourself......I have put the answers right at the bottom of this page. Don’t look unless you are completely stuck.
Es Muy Tutor Services - Make The Subject Of....
As we have seen before in algebra, an equation is an expression split by an equality such that one side equals the other.The above expression is perhaps as simple as it gets, clearly “x” does indeed “equal one” for the expression to qualify as an equation.But what happens when we have more complicated expressions, and we need to find the value of “x”?Generally, there are several ways to solve these issues:
Collecting ‘like terms’FactorisationGraphing
Collection of ‘like terms’ is the easiest for simpler expressions, and is the preferred method for what this tutorial is aimed at:
Rearranging expressions initially in terms of one variable, in terms of anotherThen rearranging formulae in the same way
In the equation above, we know that ‘x’ has a certain value, and experience in this sort of algebra will lead you to be able to deduce this “in your head” without having to resort to the structure I’m about to go through. The value for the variable ‘x’ above is 6 but let’s see how, and why.Remember - anything you do to ONE side of an equation MUST be replicated on the other side to maintain the equality.Step 1 - subtract 4 from each side:Step 2 - divide both sides by 2 to get the variable ‘x’ on it’s own:This one was quite easy, let’s look at somthing a bit tougher.
Not quite so easy, the “like terms” are on both sides of the equality. We need to remedy this before we can attempt to solve for ‘x’. We do this in just the same way as before, by making changes to each side of the equality (identical changes that is) until we can collect the “like terms” on one side of the quality with the constant values on the other side, and therefore solve the equality for the variable concerned, in this case ‘x’:Step 1 - subtract 8x from both sides:Step 2 - subtract 6 from both sidesStep 3 - Tidy up both sides:We could at this point multiply the whole thing by -1 simply to make both sides positive, this isn’t really necessary and the next step is to divide both sides by 5 which leads us to our final answer:
Of course, it is not always going to be the case that you are given the equality straight off. You may be given a set of circumstances or a scenario in which you create your own equality and then go on to solve it.
Look at the diagram on the right. Let us suppose that in the question we are told that the perimeter of the triangle was equal to the perimeter of the rectangle. The difficulty we face here is the fact that we have two variables, ‘x’ and ‘y’ so we cannot establish a numerical value for ‘x’, only value in terms of ‘y’. This is a little bit different to the ones you have attempted up until this point, but the principles are exactly the same.
Given what we already know, we can establish an equality like the one shown below:
(If you’re not sure why this is the case, pause at this point and study the diagram for a moment). What we need to do now first of all is get rid of the brackets because all they are doing at the moment is confusing the issue. once we have got rid of the brackets we simply add up the terms until we are left with a simpler equality which we then manipulate according to the rules.As I said at the beginning, we cannot establish a numerical value for ‘x’ because we are not given a numerical value for the second variable ‘y’, in these situations, the best that we can achieve is to express our target variable in terms of the other variables in the expression as shown.Now, it may be that we are given no numbers at all, just variables and these can become quite tricky. For instance in the example below we could be asked to re-state the expression in terms of any of the variables ‘a’ ‘b’ or ‘c’. For our purposes we will restate in terms of ‘a’
Step 1 - we will collect our ‘a’ terms by adding ‘a’ to both sides of the equality:
Step 2 - we will subtract ‘c’ from both sides of the equality:
Step 3 - factorise both sides of the equality:
Step 4 - divide both sides by (b+1):
It is this penultimate step that sometimes leads to confusion, sometimes it is not obvious that the expressions need to be factorised to completely isolate the variable required. Once this is established as in step 3 above, the rest of the operation is quite straightforward as in step 4.
Es Muy Tutor Services - Sequences (Compound Interest)
This is an example of sequences, but what I’ve tried to do in this particular question is put it into an “real-life context” concerning savings accounts and the methods used to calculate the interest on them. The sequence formula that I’m going to use in this question is a variant of the standard geometric progression formula:This formula tells us that the ‘nth’ term of the sequence is found by multiplying the first term by the common difference raised to the appropriate power which is usually the number of the term requested less one ** (study the equation and you should see this).The question is this:I have invested £2000 in a savings account where the money has been locked away for ten years, this means that I cannot (except in an emergency) make any withdrawals from it. The interest rate is 1% and is paid, that is added to the principal sum, every year. After the first year I was notified that my money had earned £20 in interest, and at the end of the second year the balance stood at £2040.20. At the end of year three my balance was £2060.602 (which of course in financial terms would be rounded off to 60p).Question one - what would my balance be at the end of year ten?Question two - assuming I then immediately reinvested for a further ten years, what would the balance be at the end of the twentieth year?Let us first of all create our sequence terms:
** Note that this is one of those rare occasions where the sequence terms will start from index 0 not index 1, therefore our balance at the end of the first year (a subscript 1) will be 2020 pounds as given, at the end of year two (a subscript 2) 2040 pounds and 20p and so on. Because we have started the index notations at zero we will not raise the common difference to the power of (n-1) but it will be just simply ‘n’ when we come to evaluate further terms.You may be wondering where the value 1.01 comes from?, well this is the common difference and there are two ways of working it out. firstly if you divide each term by its predecessor you will arrive at the value of 1.01 because this is in fact interest at 1% compounded, alternatively we know that if the interest rate is 1% (or 0.01) we have to multiply the principal sums by 1.01.if you study the formation of the first three terms (a subscripts 1,2 3) you will see that each one has been arrived at by multiplying the previous one by the common difference, that is 2000 multiplied by 1.01 gives us the first term, 2020 multiplied by 1.01 gives us the second term and finally 2040.20 multiplied by 1.01 gives us our third term 2060.602 which of course will be rounded to 2 decimal places in a financial situation.In other words the closed form for our sequence (or “formula”)is this:
Now we can start to evaluate the values of certain terms as the question asks. For example at the end of the first ten year period, what would be the balance? Well..... substitute n=10 into the equation given where a0 is 2000 as the question states.
Not a particularly exciting amount of money for a ten year investment, but with such a low interest rate it’s to be expected. If we further apply the equation substituting n=20 to satisfy the question part two we will see that the situation hasn’t really improved very much:
If we plug the equation into something like Microsoft Excel we can see what the intervening years would be as well.
All things considered a pretty poor return for twenty years saving!Having said that, for the purposes of this exercise we are only interested in the figures but it just goes to show that it is a very good idea to look around for the best investment deal if you have money you can lock away like this.I hope that you enjoyed this question :-)(NB: the method used for this calculation was to apply the interest at the end of each year, if you apply the interest at other time frames the results may differ slightly).
Es Muy Tutor Services - Expanding Algebraic Expressions
A typical activity in GCSE mathematics, particularly at the Year 10 (age 14-15) stage is to have to expand and factorise algebra. Now this is a “love hate” relationship because you have to keep an eye on sign changes and in the early stages take everything “step by step” which isn’t always easy to do (if you’re like me and you want to dive in head first).Even though these aren’t really that hard, the “step by step” approach is highly recommended until you’re a bit more skilled, let’s take a look at one:
What we have to do here is “collect like terms”, that is, get all of the ‘x’ values on one side of the equation, and the rest of the stuff on the other side. We do this by adding and subtracting certain elements until we have completed the “collection”, remembering that ANYTHING done to one side of an equation MUST be done to the other to maintain the equality.In this case, the value of x = 6, we’ll now have a look at how to show it.Step 1 - Let’s subtract 2 from both sides:
Can you see what I have done?, we have “lost” the “-2” from the right side, now we can “lose” the 2x by subtracting it from BOTH sides.Step 2 - Subtract 2x from both sides:What we’ve done is simply take away 2x from both sides, the right hand side becomes 0 because 2x-2x=0 and the left side becomes x-6 since 3x-2x = xYou should be able to see that if x-6=0 then x=6 as we said at the start, OK?
Let’s have another go OK ? - see what you think of this one:In fairness, although it looks scary, it’s no harder than the first one, but you must take care and work each step carefully:Step 1 - Expand the parentheses (posh word for brackets):Well, you might be thinking “thats all well and good, but HOW did he get that?”, OK well we take it in steps:1. 3 x 4 = 12 and 3 x (-2x) = -6x...this sorts out the left side, then2. 6 x (x) = 6x and 6 x (-6) = -36...this sorts out the right side.Now we are at the stage that we were at in Example 1, I was a bit sneaky here and put you back a step to see how you’d get on....you did great!Let’s collect our “like” terms now......Step 1 - Add 36 to both sides:Step 2 - Subtract 6x from both sides:Step 3 - Add 12x to both sides:It’s always a good idea to put your new-found value for x back into the equation you were given at the start to see if it works, in this case we end up with the expression -12 = -12 which is indeed correct.
Now let’s just work through another, then I will leave one for you to look at.Once again we have to expand the brackets and then gather all of our “like” terms to the one side, so first of all we will perform the expansion.Step 1 - add 33 to both sides:Step 2 - Subtract 4x from both sides:Step 3 - Divide both sides by 5 and rearrange:As you become more and more adept at these sorts of problems, you might start to skip some of the steps, but while you are getting used to them I highly recommend that you take it step-by-step.
Try these for yourself, the answers are at the bottom of this page, but I haven’t shown the steps.Question 1:Question 2:I have been a bit sneaky with Q2.....you cannot have a numerical value for ‘x’ because there are two variables, ‘x’ and ‘y’. All you can do here is express the result for ‘y’ in terms of an expression containing ‘x’ (and vice-versa).Have a go anyway !!
Question 1 Question 2
Es Muy Tutor Services - Simultaneous Equations
An old favourite in GCSE level algebra is the solving of equations involving two variables by “Elimination” (there are other ways), that is....permutation of one or both of the equations until one or the other can be eliminated, then the solution to the variable that is left can be used to identify the other one.There are a few ways that you can end up with Simultaneous Equations, either by being given them, OR being given a scenario from which you must create and THEN solve the equation pair. This page will go through a few examples.
This very easy example above shows two basic equations, just by studying these you can probably work out that x = 6, y =2 but let’s go through the process to see how we get there:Step 1 - it is usual to label the equations, so we’ll restate them:It might not be obvious yet, why we “number” them.....but it makes the permutations easier to follow in later (harder) examples.Step 2 - We need to “eliminate” a variable, it doesn’t matter which one we choose, but it can be seen that ‘y’ can be eliminated IF we ADD thequations together, we ALWAYS state our intentions (this is where the numbering comes in handy):(1) + (2) gives:Now that we have a value for one of our variables, we “substitute” it into one of our original equations, usually choosing the simplest one. In this case it hardly matters but for the sake of followng the steps we say:Step 3 - Substituting for ‘x’ in (2)Using the principal of “doing to one side what we do to the other to maintain the equality” we now:Step 4 - Subtract 6 from BOTH sides:Step 5 - Multiply both sides by -1:Ideally you should now put BOTH values into the equation of your choice to make sure you have the right solution, of course this depends on time.Lets take a look at a slightly more complex example now.
Numbering is definitely going to come in handy here. We have to decide which variable to eliminate, but again it really doesn’t matter, logically though you would choose the easiest one but in this case there’s not much to choose from, however I’ll go for ‘y’.(1) x 3(2) x 2Now we can eliminate ‘y’ by adding equations (3) and (4) together:Now choose the simplest equation to substitute back the new found value for ‘x’, equation ( 1 ) is as good as any I think:
Now it could be that you are given a scenario, not a direct set of equations, and from the scenario you would be expected to create your own equations and then solve them. An example of this could be:Last week I bought some fruit. I paid £3.31 for 3 apples and 7 oranges, this week I had to buy more because they were so nice. I bought 4 apples and 11 oranges and it cost me £5.03 this time.How much does an individual apple and orange cost ?Well, let’s start by calling an apple = x and an orange = y, so last week I bought 3x apples and 7y oranges, costing me 331. This week I bought 4x apples and 11y oranges, costing me 503. Now......let’s create our equations:So far, so good ?Right, let’s eliminate a variable......’x’ this time I think......(1) x 4(2) x 3(4)-(3)If we substitute our value for ‘y (37) into, say, equation (1) we can see that:ALWAYS WRITE A CONCLUSION LINE TO THE QUESTION...... “....therefore an apple costs 24p and an orange costs 37p”Simultaneous Equations are good fun, they can also be evaluated using MATRICES (Pronounced “May-TRIH-seas” not “Mattresses”) (this will come later)OK....over to you....show that the values for ‘x’ and ‘y’ are correct by solving these sets of Simultaneous Equations:
Es Muy Tutor Services - Simultaneous Equations (Matrices)
On the previous page I stated......”An old favourite in GCSE level algebra is the solving of equations involving two variables by “Elimination” (there are other ways)” and now it’s time to look at the Matrix Method.This particular “worked example page” is actually being written before the section on matrices, so I won’t be going through the mechanics of matrix algebra, I will simply be using the rules so if this looks complicated or confusing you have to bear with me because I don’t want to have to reinvent the wheel here by going through all of the “ins and outs” of matrix algebra. Suffice it to say that in this particular “worked example” I will assume that you understand the expression “determinant of a matrix” and how to calculate it.First of all let’s take a look at a fairly standard pair of simultaneous equations:A fairly straightforward set of basic equations as I’m sure you’ll agree, and if we were going to use the “elimination” method we would have to get rid of X or Y, note also that I haven’t numbered them “1” and “2” because using matrices to solve them, this is not necessary.
What we need to do is construct our matrices from the equations given (once again I will just place the matrices here and assume that you understand where they came from).if you use the rules of multiplication of matrices you will easily get back to the source equation. When we are looking at simultaneous equations in matrix form it is usual to refer to what we have put above in the following format:Where, I’m sure you can see the A refers to the matrix with the numbers 2 3 3 and -2 inside it and the X refers to the 2 element matrix ‘x’ and ‘y’, finally the B refers to the 2 element matrix with the numbers 17 and 6.Given AX = B we can multiply both sides by the inverse of A (denoted A to the power of -1), provided this exists, to give: But as you will see when you study matrices, if you multiply a matrix by its own inverse you end up with a thing called the identity matrix, and the special feature about the identity matrix is that if you multiply it by matrix, you leave the value of that matrix unchanged.Effectively, the meaning of the last 3 lines leads us to this:You must understand that I’ve missed out a lot of the rules of matrices in arriving at this expression because, as I said earlier, we have not yet covered matrices. When we do, you will find that there are a lot of idiosyncrasies, certain matrices cannot be multiplied together and certain matrices do not have an inverse.OK we can leave the mechanics of matrix algebra to that particular section, let’s just crack on with solving the set of simultaneous equations:We can see the matrix A, we need to calculate the inverse of the matrix and we do this by calculating the determinant of A and then multiply this by a slight rearrangement of matrix A. Symbolically this is shown as follows:In our particular case the “reciprocal determinant” of A is - 1/13 (minus one over thirteen) and a suitably transposed matrix A is as shown:Stop and study this for a short while, you will see that “ad-cb” = -13 ((2 x -2)-(3 x 3) = -4 -9 = -13) . When we multiply the reciprocal of this by our transformed matrix A we arrive at the rather unwieldy looking expression involving fractional components of 13.We now multiply our inverse matrix A to the minus one by the matrix B which represented the numerical right-hand sides of the simultaneous equations we started with, that is 17 and 6.Finally, remembering that the matrix at the extreme right-hand side does in fact represent 2 improper fractions in 13th’s which can be simplified to 4 and 3 respectively.Therefore the solutions to our simultaneous equations are.... “x = 4” and “y = 3”Another example will follow now, but it will simply be worked out. Follow the steps as above to verify that it’s correct:
We now need to find the INVERSE of A (A to power -1)
And finally, as we should always do......QUOTE the answer........”So x = 6 and y = -2”I personally think that Matrices are a thing of beauty, not necessarily the ideal choice for Simultaneous Equations but a challenge well worthy of attention. Have a go at a few more, using matrices and see if you can show that the solutions are as given.
Es Muy Tutor Services - Positive and Negative Numbers
A source of confusion for many students at early Secondary School level, is just how exactly to keep the signs in order. Getting a sign wrong at almost any point in a calculation will most certainly produce the wrong answer at the end, with almost no possibility whatsoever of finding out where it went wrong!
It is probably obvious to you that the above equation is correct. 1-1 is indeed zero but where are the signs? We can see the “minus sign” in front of the number one telling us that this is a negative number, or a subtraction that we are dealing with but what about the sign in front of the first number one? Is there a sign anyway?Well in fact yes there is. The first number one is in fact +1 but we choose not to show the sign when it is a plus, only if it is a minus sign. So, technically speaking the above equation should have been written like this:Already it is starting to look a little bit confusing, but if you break it down from left to right you’ll see where I’m going: 1. Our first number is +1, our second number is also +12. We want to take our second number away from our first number3. Working on the principle above, that we don’t show the signs of positive numbers, if we, in our imagination, take away the plus signs and the brackets which are only there for convenience anyway, we will be left with the first equation at the very top of the page.let’s take a look at our next example, this time our first number is still +1 but our second number is -1, that is, it is a negative number this time,so how do we deal with this?When we see a negative sign inside a bracket, with a negative sign outside the bracket, this becomes an addition. Brackets in mathematics usually mean multiplication so we can simply say that “a negative times a negative equals a positive”. Based on this then, equation above is not correct because the answer indeed is 2 not 0.Let’s take a look at another one:The only thing that has changed here is the numbers themselves, the principle remains unaltered. “Minus” -3 becomes “+3” and so the answer is 5+3 which is indeed 8. Remember that in all of the examples so far, any number which is naturally positive has not had a sign shown, it is inferred by the absence of the sign that the number is in fact positive.In this example, our first number is negative already and we are taking away from it a further 3 (that is, if you think about is we are adding a -3 to -9) in the 2nd part of the example we encounter our sign change because as we said in the purple bit above “a negative times a negative equals a positive” and so what we are doing is adding +3 to-9 which does in fact give us -6 as the answer (see the red bits above to see where the +3 comes from).Multiplying positive and negative numbers togetherWhen we come to multiply numbers with different signs together we have to take into account the sign that each number possesses, whether it is a “shown negative” number or a “not shown positive” number. Hopefully a few examples will make this clearer:The first example simply highlights what I’ve been saying before, that: “a negative times a negative equals a positive”But the 2nd 3rd and 4th examples go like this:“a negative times a positive equals a negative”“a positive times a negative equals negative”“a positive times a positive equals a positive”
This equation is wrong, read on to see why the answer isn’t 0
Dividing positive and negative numbersWithout going into too many examples it will probably be sufficient at this point just to give you examples and the rules, and let you study them for yourself:
“a positive divided by a positive equals a positive”
“a positive divided by a negative equals a negative”
“a negative divided by a positive equals a negative”
“a negative divided by a negative equals a positive”
Es Muy Tutor Services - B.O.D.M.A.S.
Basic calculations have to be done in a certain order for the correct answer to be arrived at. There are many rules in mathematics concerning “orders of hierarchy”, “commutative” operations and so forth, but one of the most fundamental rules, or set of rules, is brought together under the umbrella of B.O.D.M.A.S. (You may see this referred to also as B.I.D.M.A.S. here in the UK or P.E.M.D.A.S. in the USA)BODMAS specifies the order in which certain operations in a calculation MUST be carried out to obtain the correct answer.
Brackets MUST be executed first, irrespective of what is inside OR the hierarchy that might have applied outside the brackets (BODMAS within BODMAS).
Orders of Numbers, or Powers are processed next, if they were inside brackets they would be processed as part of that operation anyway (BODMAS again).
Division and Multiplication have the SAME status in the hierarchy, so they are executed from LEFT to RIGHT, ie: in the ORDER they are encountered.
Addition and Subtraction have the SAME staus in the hierarchy, so they are executed from LEFT to RIGHT, ie: in the ORDER they are encountered.
Let us take a look at a few examples, the first ones are basic but the later ones were run through a computer based maths system to confirm their accuracy.
Using the rules of BODMAS and especially the fact that Addition and Subtraction are equally weighted and performed in the “natural” order, ie: from left to right we conculde that the answer is 2.
Using the rules of BODMAS we must perform the ORDER calculation first, so the question then becomes (answer also shown):
How should we set about solving this example?, well just follow the rules and you will see the answer unravel in front of you. I have put a series of “stepped equations” below to show you the stages.
Now follow the same logic to confirm that the answer is indeed 137. The brackets evaluate to 72, then 10x 13=130, 130 / 2 =65 = 65 which when added to 72 becomes 137.
Es Muy Tutor Services - Square Roots (Manually)
When I bought my first calculator way back in 1973 it was expensive (for the time) and was only a “four function” machine. It could add subtract, multiply and divide and if my memory serves me correctly it would also calculate percentages. What it didn’t do and what didn’t come along till a bit later was the facility to calculate square roots.
I was taught at school how to calculate square roots manually, and it isn’t particularly difficult. This particular page will teach you how to do it, ideally with the aid of a calculator (but try to avoid the temptation of cheating using the square root key :-)).The method is quite simple, repeated averaging of 2 estimates of the square root of the number until the estimates coincide in value.
Let’s take a look at an example, we will start off with a number which we know is an exact square root, that is there is no fractional part, it is a “perfect square”:
Question: The square root of 289 is what?Step 1 : Make a reasonable estimate, let’s say 14 and divide 14 into 289 using full calculator accuracy:Guess 1 = 14Answer 1 = 20.642857142857142857142857142857We call the first number “Guess 1” and when we divide “Guess 1” into the target number we arrive at the huge number above which we call “Answer 1”. We take “Guess 1” and “Answer 1” and we add them together and divide them by 2, that is... we take the average.Average of “Guess 1” and “Answer 1” is therefore:This long answer 17.3... becomes our new “Guess 2”and we divide it into the target number 289 once again which gives us this answer (”Answer 2”):Guess 2= 17.321428571428571428571428571429Answer 2= 16.684536082474226804123711340206We take our “Guess 2” 17.3... and our “Answer 2” 16.6...... number and we average them once again, to make yet another “Guess 3”This “Guess 3” is divided once again using full calculator accuracy into our target number 289 producing 16.997…(”Answer 3”)
If we carry on, creating new guesses, then creating new answers and subsequently new averages we will see that the numbers do eventually start to converge. You can see that at Guess 4 and certainly Guess 5 we are converging quite nicely onto our square root of 289 which is in fact 17.
Not particularly difficult although you must agree it is a little bit fiddly, there are many ways to combat this for example you could build the equations into a Microsoft Excel spreadsheet and use that, or do what I’ve done and write a small application that does it all for you.Look at the picture (left click on it to make it larger). You will see the small application which I have, rather predictably, called “Square Rooter”, if you click on the download link you can download a zip file which contains the small executable for your own use. For those people who want to know it was written using Visual Basic.net 2015 community edition with no custom controls so it doesn’t need to be installed, it can just be extracted from the archive and run.
Es Muy Tutor Services - Trigonometry Tree Example
This worked example concerns a man standing 30m away from the base of a tree in a park:Looking at the top of the tree requires the man to angle his line of sight to 30 degrees as shown.Question Part 1 - Given that the angle between the tree and the ground is a right angle, approximate the height of the tree, in metres, to one decimal place (click on the image for a bigger view).
This is in fact a right triangle, and a Pythagoras theorem question. It may help (if you have time that is) to create a small diagram emphasising the mathematics rather than the aesthetics. We can now see that we have a 30-60-90 triangle, and a fairly basic problem in Pythagorean trigonometry. you should be able to see that the value of B is the value that we are interested in, and we are given the value of A and in this particular instance all of the internal angles. To evaluate the value of the height of the tree B we use the trigonometric “sine rule”:
Now if you study the diagram in conjunction with the the expression above you will see that it is the angle opposite the lettered side that is labelled with the lowercase version of that letter, i.e. the 90° angle is c because the side opposite is C, similarly the 60° angle shown at the top of the tree is a because the distance from the tree to the man is A which is given as 30 m. What we need to do is establish which pair of the equalities above (that is out of the 3 of them) we have sufficient values to use.Upon inspection you should be able to see that we have side A with corresponding angle a, and we have angle b so we need to do is establish the value for the height B. We do this by utilising a part of the expression as shown below:All we have to do now is “plug in the values” of the variables we have and we will be able to calculate the value of B (note that we will be working in degrees not radians and we start working out the sines of the angles concerned).Substituting our known values leads us to this expression:Remember that we always make a final quotation at the end of any answer “therefore to one decimal place the height of the tree is 17.3 m” (incidentally you can see that I rounded off the denominator of the fraction to 5 decimal places, you should always work to full calculator accuracy until you’re ready to quote your final answer.
Question Part 2 - the man now walks forward in a perfectly straight line towards the base of the tree. He continues to walk forward until his angle of elevation to look towards the top of the tree has now doubled to 60°.How far has the man walked, that is how far away from his original starting position is he now standing?
Once again, you will find that a sketch, or drawing of the situation will help you to comprehend what you are being asked to find. from the diagram below we can see that the angle of elevation is indeed 60° as we are given and marked upon the diagram is of course now our newly found height of 17.3 m.We need to calculate exactly how far away from the base of the tree the man now stands and take this away from the original 30 m, this will give us the answer to the question we are being asked.Once again we use the “sine rule” and we select a portion of it as shown:We have a value for B = 17.3, we have a value for angle b of 60° and by virtue of the fact that all of the internal angles of a triangle add up to 90° we can establish that the value of our angle a is 30°.We “plug in” these values to establish a value for our distance (which I have I suppose rather confusingly also called A, so I will call it “new A” from now on).Once again you can see that I have rounded to 5 decimal places the denominator showing the value for sine 60°, you shouldn’t of course do this until you are ready to present an answer. To round off the answer to this particular question we can see that the answer is 9.988 which is to all intents and purposes 10 m.As the man was 30 m away when he started and is now 10 m away we can say that he has in fact walked 20 m which is the answer expected.
Es Muy Tutor Services - The Congruent Parallelograms
On the face of it, a tricky-looking question, but it does in fact boil down to a couple of fairly straightforward applications of the cosine rule. Here is the wording of the question, read it in conjunction with the diagram on the left.
ABCDEF is a hexagon, made up of two congruent parallelograms, ABEF and BCDE.Angle ABC is 30° as shown, and the points P and Q lie on the lines AF and CD such that BP equals BQ equals 10 (arbitrary units). AB equals BC which equals X.Prove that:
On the face of it, this question appears to just jump out of thin air, and it does rely on some prior knowledge of angles, and surd representations of them.Lets make a start:You should be able to see the relationship to the cosine rule as I have annotated it, study this with the diagram to make sure that you are happy before we go on.
We are trying to work out the value of the side AC but in terms of ‘x’ as we don’t have enough numbers to come to an actual value for it. We need to now tidy this expression to arrive at a suitable representation of AC squared….With a little bit of mathematical dexterity we come to the following:This is the crafty bit, because it does rely on you knowing that the cosine of 30° is “root 3 divided by 2”. Interestingly the answer to the question, that is a bit you’re being asked to prove contains a surd like expression, so this would have been a clue.We now reach a turning point because AC = PQ from the diagram above, and letting the angle PBQ=THETA we have:Now we need to rearrange his expression in terms of cos THETA (angle PBQ):Therefore:First of all let’s tidy up the left-hand side and the right-hand side of the above equation, to get rid of the fraction and to factorse the right-hand side:From now on, it is a case of simple arithmetic manipulation of the equation, but there are still pitfalls here so you need still to be very cautious:Multiplying throughout by -1 and reversing right and left sides gives us the following final result:
Es Muy Tutor Services - The Canal Lock Example
This is a photograph of a typical British canal lock, I have annotated it with approximate length and width, both of which will be used in this example.Question part one - given the dimensions as shown on the photograph, what is the surface area of the top of the lock?This is a fairly straightforward one, the answer is of course the width multiplied by the length:
Remember of course that we are dealing with lengths measured in “metres” so of course areas will be in “metres squared”. A useful way to remember this of course is to perform the calculation first of all with just the numbers, and then perform the calculation with just the units. I will show you the result of this but the accurate determination of the units of the final measurement in a calculation problem could be subject to a document on its own.Question part 2 - although this can’t be shown, accept the statement that the lock is 5 m deep. with this in mind what is the volume of water held by the lock when it is full as shown?We are dealing with a cuboid, the dimensions of which are 8 m, 2 1/2 m and 5 m. To obtain the volume we multiply the 3 values together.
in just the same way that we multiplied the numbers and the values separately in the previous example, I have done exactly the same this time to show you that the answer we require is 100 but it is in fact 100 m³. When we deal with linear distance we talk about metres, when we deal with two-dimensional areas we talk about metres squared and when we deal with volumes (three-dimensions) we deal with metres cubed.Question part 3 - although this varies slightly, the average density of water is approximately 1 g/cm³, (one gram per cubic centimetre) that is 1000 kg/m³ (one thousand kilograms per cubic metre).Armed with this information, calculate the weight of the water in the lock when it is full.
Now we know that we have 100 m³ of water because we calculated this in the 2nd part of the question. we are told that one cubic metre weighs 1000 kg so we should be able to see quite easily that the weight of the water in our lock is this:That is.....100,000 Kg or 100T....... Now I did mention before that the determination of the correct units could become a subject on its own, and the think the above might explain the reason for that to you. Density is measured in kilograms per cubic metre which is depicted as kilograms over cubic metres and volume is depicted in cubic metres. When we multiply density by volume we are left with mass because density equals mass over volume.
Question part 4 - canal lock gates are notoriously leaky, and this one is no exception. Given that the leakage rate of this particular lock has been calculated to be 30 Litres of water per minute, how long will it be before the lock reaches half capacity (assuming it is not topped up at all in any way)?We hit a snag here, we haven’t encountered litres yet, we have been talking in kilograms. However we know that the density of water for the purposes of our question is 1000 kg/m³. Accepting the fact that there are 1000 litres in a cubic metre (that is 1000 cm³ per litre, and we have 1000 litres in a cubic metre, that is we have 1,000,000 cm³ in our cubic metre) we can establish that the weight of a litre of water is 1 kg (See Question part 3’s wording).Now we have that extra bit of information, we can see that the lock loses 30 kg of water every minute (remember we have just established above that 1 kg equals 1 L). Previously we worked out that we have 100,000 kg of water in a full lock and we need to know how quickly we can lose 50,000 kg of it because that will take us to half capacity. The number of minutes is therefore the amount of water we lose divided by the rate (in kilograms) at which we lose it:PLEASE NOTE THAT ‘m’ ABOVE IS REFERRING TO MINUTES IN THIS CONTEXT, NOT METRES (SORRY ABOUT THAT!)You can see I’m sure, that this is another one of those nasty examples where the units could go horribly wrong. We lose 50,000 kg at a rate of 30 kg per minute which would normally be written as KG over M, but the alternative way to write it is KG multiplied by M raised to the power of -1At this stage I’m simply going to ask you to accept what I’ve said about units, hopefully in the not too distant future I will be able to produce a “walk-through” of units and how to calculate them because in mathematics and indeed physics there are some terrifying combinations so it’s always a good idea to be able to do this.Returning to the problem in hand, we have established that the minutes value is 1667, so we finalise our question by making the following statement:The lock will reach half capacity in approximately 1667 minutes, or 27.8 hours. If you consider that some of these canal lock gates have been standing there for years, slowly decomposing under the weight of the water they are holding back you will probably accept that the calculation of leakage rate can’t be particularly accurate so in this particular case it would probably be acceptable to say “the lock would be half empty after about 28 hours”, (of course if this was a homework or examination question you would have to quote to the accuracy you are asked for).
Es Muy Tutor Services - ICT Index
This page will hold an index of worked examples. More often than not, these pages will require you to scroll down, depending on the length and complexity of the solution explanation.
Es Muy Tutor Services - Science (Index)
This page will hold an index of worked examples. More often than not, these pages will require you to scroll down, depending on the length and complexity of the solution explanation.
Es Muy Tutor Services - Chemistry Index
This page will hold an index of worked examples. More often than not, these pages will require you to scroll down, depending on the length and complexity of the solution explanation.
This page will hold an index of worked examples. More often than not, these pages will require you to scroll down, depending on the length and complexity of the solution explanation.
Es Muy Tutor Services - FS Index
This page will hold an index of worked examples. More often than not, these pages will require you to scroll down, depending on the length and complexity of the solution explanation.
The functional skill set of pages will hold content which is intended to be very basic and in many cases will be explanatory rather than tutorial. What I’m trying to achieve here is to explain “real life” situations without becoming too technical or theoretical (although some technicalities and theories will creep in occasionally this is inevitable).Remember some of these topics you may already be able to deal with, there are many people however who I hope will benefit from these pages.
Functional Skills Level Examples in Mathematics
Functional Skills Level Examples in Science
Functional Skills Level Examples in IT and Computing
For FUNCTIONAL SKILLS LEVEL examples follow these small icon links
I’ve no doubt that we have all been to the supermarket probably in the not too distant past and we have seen special offers. What makes these offer special? Are these offers actually special? To be absolutely certain that these bargains are what they seem to be requires a basic knowledge of arithmetic to see how many of a particular smaller purchase will be obtained by buying larger bulk items and is it cost-effective to do it?Many supermarkets now will quote you a cost per, for example, hundred grams so that you can work out for yourself if the larger item is more cost-effective than buying the smaller item, for example does it make more sense to buy 1 kg of coffee at £13.50 or should we stick to our 250 g jars which are priced at £3.55 each?One thing is for certain, not only do you need to be able to work out which is the bargain are, you need to also be aware of why certain things are reduced, are they approaching their sell by date? If you buy lots of them will you use them before the sell by date? because if you end up throwing things away then that’s not cost-effective at all.Let’s take a look at some examples:
Question - “My local supermarket, ‘Tescbury’s’ sells washing-up liquid in 1 L bottles for 99p. Recently they had in a stock of 5 L cartons (catering overstock) and were selling them off at £4.50 each (their RRP or recommended retail price was previously £7.95 but they’ve been reduced to £4.50 for a quick sale). Would I be better off sticking to the smaller bottles or would it be more cost-effective to buy one of the larger ones which would of course last 5 times as long?Let’s do a little bit of arithmetic.........Our normal wash-up liquid would cost us 99p so if we bought 5 of them we would end up spending £4.95, but the equivalent 5 L is actually £4.50 so although it is only a small saving of 45p it is still a saving and therefore a worthwhile purchase. If you consider that the recommended retail prices £7.95 then it is clearly a bargain because once it goes back up to that price you would actually be paying £1.59 a litre as opposed to your normal 99p
Question - a very tasty biscuit and caramel finger coated in chocolate usually retails at 39p per pack, each pack contains 2 biscuits and the weight of the pack is 75 g. Recently the supermarkets have offered “20 packs” where each biscuit is wrapped individually, therefore there are only 20 in the pack, the pack weighs 600 g and is currently on sale for £4.05. Let us work on the assumption that if you bought the larger pack you wouldn’t exceed the sell by date, so is it cost-effective to buy the larger pack or is it better just to buy the smaller ones?Once again, let’s do a little bit of arithmetic........... first of all let’s take a look at the packs:
Small (2 Pack)
Large (20 Pack)
On the face of it we can see that the large multi pack containing 20 biscuits becomes more expensive to buy because not only are the biscuits individually more expensive, they are also considerably lighter. Usually we would resort to a ‘pence per gram’ or ‘ pence per 100 g’ calculation to be absolutely certain.Well if the first calculation didn’t swing it for you, then the second one should. The multipack is not value for money because you are paying two thirds of a penny per gram as opposed to just over a half a penny per gram, and if we take this up to “pence per hundred grams” we can see that the smaller packs will cost you 52p per hundred grams whereas the larger packs will actually cost you 67 1/2p per 100 g, a significant increase and enough to put me (at least, and hopefully you too) off the larger packs.
We will now take a look at our 3rd and final example, but could carry on producing examples ad infinitum but of course this is impracticable and once you have seen a couple you should be able to apply the principles to most others. Let’s take a look at buying cheese:Question - our favourite supermarket sells cheese for £1.95 per 227 g block (that’s 8oz, half a pound to old people like me!), but a few weeks ago they had some 1 kg blocks (that’s just under 2 1/4 pounds, again to old people like me!) which they were selling for £6.99 reduced from £8.99 while stocks last.Assuming that cheese is something you eat a lot of and the bulk purchase would be sensible, i.e. it’s not going to take long to eat it so it won’t go bad and have to be thrown away, which would be the bargain buy?Let’s take a look at the arithmetic, but this time we will adopt the “p per 100 g” system that many shops have now taken on.We can work out that in our smaller block the cheese costs 85.9p per hundred grams, because we work it out as a fraction of the whole amount, i.e. 100 g is “’one hundred’ ‘two hundred and twenty sevenths’” of £1.95 which works out at 85.9p. Note the “p per 100g” can also be written as I’ve shown it above in red (if you look at some of the more complicated mathematics on this site you will see this “index notation” quite a lot).Now let’s take a look at the “p per gram” for the bulk buy:When you are a little bit more skilled at these sorts of calculations you will be able to see that 100 g is a 10th of a kilogram and so all we have to do is divide the price of 1 kg by 10 to get the price per hundred grams, but I decided to show it this time as a full calculation. It can be seen quite clearly that the “bulk” purchase would be the better one by 16p per 100g, a not insignificant saving. Would the cheese have been the same bargain at its previous price of £8.99?Well, let’s redo the calculation above replacing 699 with 899:At the previously high price of £8.99 the cheese would have been 4p per hundred grams more expensive than buying it in the smaller blocks, so the smaller blocks would have been more cost-effective until the price reduction. The 1 kg block at the high price would cost you 40p more than buying just under 5 of the smaller blocks (to make up the thousand grams).
Es Muy Tutor Services - Area and Volume
Sometimes it is necessary to work out areas, for example if you want to re-slab your patio, re-turf your lawn or have any carpet down in one of your rooms you will need to have some idea how to work out roughly how much you need, whether it is slabs turf or carpet.This page will talk about some basic areas, and the units of measurement involved.
The picture on the left shows a fairly simple slab area, let us assume that the slabs we are using for this particular project are 30 cm x 45 cm as shown. It is quite usual to receive a quotation from a supplier, or to find a price by supplier being quoted in square metres, so a good idea would be to approximate how many of these slabs we would be getting in one square metre.Given that the metre is 100 cm and a square metre is therefore 100 cm x 100 cm we would need to work out how many “30s” and how many “45s” are in a hundred and to use this result to establish roughly how many slabs we would get per square metre.
Let me explain this seemingly complicated-looking expression above. The red section is simply telling me the size of a slab in square metres, that is 30 cm times 45 cm which is 1350 cm² but as there are 10,000 cm² in a square metre (as shown by the dark green section) we then have to divide 10,000 x 1350 which works out at roughly 7.4. (as shown in the blue section). So to all intents and purposes we are going to get just about 7 1/2 slabs per square metre. these calculations can never be very accurate, and when you are looking at actually buying and laying your slabs you must take into account odd sizes, breakages and suchlike so from a “working” perspective you may expect 5 or 6 usable slabs from your square metre.As you’ve probably already worked out by now, slabs are not cheap, but the “upside” is that a slabbed area will last a very long time.Let us now do a calculation based on a fictitious patio that we want to slab. Let us suppose that our patio is shaped like the capital letter L and has the dimensions as shown in the diagram below.
Now as if by magic, purely by coincidence of course this is our patio with dimensions that look suspiciously convenient given the choice of slab. Indeed if only such a perfect environment existed I’m sure the more people would get involved in this sort of decoration. Anyway we can see that our patio is 8 slabs by 5 slabs but is missing a section 4 slabs by 2 slabs.What we would normally do is simply count the number of slabs we want, in this particular case 40-8 = 32 but life isn’t always this simple so as to give us the perfect measurements.We worked out earlier that we will get roughly 7 1/2 slabs per square metre so based on this we would need to order at least:
4.3 m²Practically you will never ask for 4.3 m², remember that this calculation has to only be a guideline because you can’t be specific and accurate in these sort of matters.In this particular example you would just order the 32 slabs because the dimensions have been deliberately calculated so that there’ll be no cutting involved.
Very often such things are quoted as a price per square metre, but also you will find that many garden centres will let you buy per slab, in this particular case because you know you need exactly 32 this would be the better option, otherwise you would probably have to sit down and study your requirements probably buying 5 m² to allow for cutting and breakages.Just on a final note before we leave this particular example, you will see that the layout in the diagram does not match the layout in the picture. To replicate the pattern in the picture will almost certainly result in some form of cutting for every other row, this would probably add to the cost, but having said that the pattern would look far more attractive.
OK well that’s the back garden sorted but what about the front garden? Let us suppose my front garden is perfectly square measuring 6 m x 6 m excluding the border walls shown in blue) and in the middle of it I have a nice flowerbed which is also 4 m², this leaves me with a border 1 m wide around the flowerbed. I want to put down some coloured gravel to a depth of about 5 cm (to us older people at about 2 inches). What I need to do is work out how many cubic metres of gravel I need to buy.So how do we actually go about doing this?well we can see from the diagram that each grey square is 1 m x 1 m or 1 m² in area. We have 2 ways in which we can calculate the area of all of the grey slabs:1. Add up the number of slabs, in this case we have 20 which is therefore 20 m², or2. Calculate the area that would be taken up if the whole garden was slabbed and subtract from that the area of the garden part itself, that is the flowerbed with the 4 shrubs and tree as shown.
So now we know that we need 20 m² of our decorative chippings or gravel, but earlier on in the question we spoke about cubic metres? Well in fact we do now need to calculate a volume which will come out in cubic metres because we want to layer our decorative gravel to a depth of 5 cm which is 0.05 m. We multiply our original 20 m² by 0.05 to find the volume of decorative gravel required in cubic metres:We have now established that to decoratively coat our border will require about 1 m³ of gravel, now I say “about” because it is unlikely that you will coat the relevant area to an exact depth of 5 cm..... Remember that these examples are guidelines on how to do these jobs but much of the mathematics can only be approximate.
Es Muy Tutor Services - FS Examples in Chemistry
Es Muy Tutor Services - FS Examples in IT / ICT
Es Muy Tutor Services - IT and Computing (Index)
Miscellaneous IT and Computing Topics
Miscellaneous Email Information
Various bits on IT under the subject headings shown.
Es Muy Tutor Services - Microsoft Access
An Introduction to Databases and Introducing Microsoft Access
Es Muy Tutor Services - Microsoft Excel
An Introduction to Microsoft Excel
Using the Microsoft Excel = SUM() Function
Microsoft Excel - Data Formatting
Microsoft Excel - Graphs and Charts
Microsoft Excel - Inserting Rows and Columns
Microsoft Excel - A Few Other Useful Functions
Microsoft Excel - Basic Introduction to Macros
Es Muy Tutor Services - Microsoft PowerPoint
Microsoft PowerPoint - An Introduction (Basic Overview)
Microsoft PowerPoint - Pictures, Themes, Animations and Transitions
Es Muy Tutor Services - Microsoft Word
An Introduction to Microsoft Word
Microsoft Word - Text Entry and Formatting
Microsoft Word - Tables and Text Boxes
Microsoft Word - Inserting Pictures
Es Muy Tutor Services - Using Email
An Introduction to Email (Overview)
An Introduction to Email using Yahoo! Mail
Es Muy Tutor Services - Miscellaneous IT Stuff
Bits and Bytes
Colour and Bit Depth
A Bit About Hard Disk Drives
A Bit About Motherboards
Overview of the Parts of a Personal Computer
The Basics of Computer Viruses
Es Muy Tutor Services - Mathematics (Index)
Various bits on Maths under the subject headings shown.
Starting Algebra“Why should I learn Algebra? It’s not like I’m ever going to go there!”A standard joke by many comedians, algebra is probably one of the most misunderstood areas of mathematics yet it has a beautifully pure logic attached to it. Algebra is about “unknowns” or more correctly, “variables”.The name “algebra” comes from the Arabic al-jabr which means “reunion of broken parts” and in its elementary form uses letters in the place of numbers to represent the above mentioned variables.You have probably already met algebra without in fact knowing it, if you have used any formula where you have been required to enter numbers to obtain an end result then it is almost certain that your formula could be properly regarded as an “algebraic expression”.Take for example the simple equation that we all learn in infant school when we first start arithmetic:We have no variables here, “1” has a fixed value as does “2”, the values never change, that is they are constant and so logically they are referred to as “constants”, whereas variables above represented by letters can have literally any value.Consider the first algebraic restatement of the above expression:Of course we know straightaway that the value of ‘x’ is one, but how did we get there? Take a look at the “= sign”:For any equation to retain its “equality” both sides must represent the same value, in other words they must be equal. In the picture shown if the two blue weights were taken away from the left-hand side then to retain the balance two of the grey weights would have to be taken away from the right hand side.This leads to our first rule “anything done to one side of the equation must be replicated on the other side if the balance is to be maintained”So, going back to our equation, we can obtain the value of ‘x’ by doing the same thing to both sides…that is…take 1 away:We removed 1 from each side to reveal a value for ‘x’Let us look at a slightly more difficult example, but one that we can solve just as easily:OK, not exactly a ‘one and one’ type of equation but to solve it involves only two simple steps:1.Add 7 to each side:2.Divide both sides by 2:Simple enough?Effectively you are “rearranging the equation to get the variable quantity ‘x’ on its own on one side”, and as you get better at this you will find that you can perform the steps to do this automatically without having to write them down individually, and in some simple cases in your head!OK what about this one?Well…just do the same as before:1.Add 12 to each side 2. Now multiply each side by -1 3. Divide each side by 6
It is a good practice to ‘substitute for x’, that is, put your new found value for ‘x’ back into the equation to see if it is right.
Es Muy Tutor Services - Basic Algebra
More than one variable?This can present a problem, take two variables ‘x’ and ‘y’:This means “x times y equals 12” but how do we work out the values of x and y? Well as things stand we can’t, because ‘x’ and ‘y’ could have infinite variable values, for example x=1, y=12 or x = -24, y = -½ and so on.We can usually work out ‘representative’ values for mixed variables when they are on opposing sides of the equality:Use the method above, divide both sides by 2 to leave:Or divide both sides by 5:But these answers are only representations of each variable in terms of the other, to be able to find out the values of two separate variables, you would need two equations and would have to solve them simultaneously (this sort of action will be seen in another section, when we do indeed start to look at “simultaneous equations”).We will now look a little bit more at these variables, and at ways in which we can manipulate them using the four basic operators and powers.The Four Basic OperatorsWe are now going to look at addition subtraction multiplication and division using algebra.AdditionWhen we think of addition, usually it involves something like this:This is in fact the addition of constants, 3 never changes, 7 never changes and so the answer of 10 can never change, whereas if we look at the equation below:The situation is somewhat different. 10 is constant and can never change but X and Y can have an infinite range of values respectively, provided of course that they complement each other by adding up to 10:Although I have only shown four examples, I am surYou will agree that there are literally infinite values for X and Y which, when added together, will produce the number 10.CoefficientsOK, so what if we have more than one ‘X’?This means “four times X” but because X is a variable we can’t write the result as another entity (such as 10, because we wouldn’t leave ‘3+7’ as it is, we’d say ‘10’).The number ‘4’ in this case is called the ‘Coefficient of X’Can we add these together? – Well provided that we have ‘like terms’ then yes, we can:We can collect ‘like terms’ and simplify, but if we have different variables we can’t simplify:It can’t go any further, the answer is 4x+8y – end of!It can’t go any further, the answer is 4x+8y – end of!SubtractionSimilarly we can subtract ‘like terms’:But this one has to stay ‘as it is’If we have values for x and y then they are not variables anymore and can then be added / subtracted etc.
Es Muy Tutor Services - Basic Algebra
MultiplicationA slightly tidier operation than addition or subtraction because the end result can usually be simplified or “tidied up”:The answer to this little problem is “three times X times four times Y”. Now we can multiply the constants together to make 12 but what can we do with X and Y? Well, all we can do is put them together like this:In mathematics when you see sets of parentheses back-to-back it means multiply, so in the above example 3X will be multiplied by 4Y and it will be this representation that we use in the rest of this document. As can be seen 12XY is the answer and cannot be simplified any further.Try substituting some values for X and Y and you will see that the above equation is in fact correct.DivisionIn this simple fraction, 8X is referred to as the numerator and 2X is the denominator. If you recall above 8X means “eight times X” and of course 2X means “two times X”. Because we have X above and below the ‘fraction line’ or ‘Vinculum’ they cancel each other out, leaving just the contact values, the result of the expression then becomes four:Of course the situation is not the same here:We can’t have a ‘hard and fast’ answer such as 4, because we still have a variable amount ‘X’ that has not been ‘cancelled out’ this time.PowersMultiplicationWe will leave this particular document with a brief discussion of powers.If we return briefly to our expression:But this time we have only X values:This is correct but wouldn’t be left that way, instead when we multiply a variable by itself we denote this as a representation of a ‘power of that variable’Where the ‘superscript two’ denotes “squared”, we have been able to tidy up a little bit by showing the resultant value of X in its “power” format. Phonetically the above equation would be read out like this:“Three ecks times four ecks equals twelve ecks squared”Let us move on:Okay, this one is a little bit more complicated because it involves a squared power before we have even worked everything out. 3×5 as you have probably gathered is quite easy to work out as 15:But what do we do with the X values? Well, one way to look at it is if we expand the left-hand side of the equation:By counting the number of X’s that we end up with we do in fact come to the right answer, but of course we would not write it this way, instead we would choose another power of X known as “cubed”:“Three ecks squared times five ecks equals fifteen ecks cubed”Before I let you into a little secret, let us just take a look at one more example:Just as we did previously, we can work this out by expansion:Again the constant ‘coefficients’ come to 35 but what about the X values? Well add them up:“Seven ecks cubed times five ecks squared equals thirty five ecks to the fifth”The secret? - What we are doing is adding up the powers or the index of each value of X, in multiplication of variables raised to powers we add the powers, so the equation above could just as easily have been written like this:You can see, I’m sure, the advantage of being able to simply add up the powers rather than having to keep expanding each indexed power of X, for example imagine having to work this expression out by expansion:In summary:
Es Muy Tutor Services - Basic Algebra
DivisionThe rules for division are quite similar to those for multiplication, but in this case we do not add the index powers, we take one away from the other, that is, we subtract.We will stick with our first equation but this time instead of a multiplier it will become a divisor:Well, in this case because we are dealing with dissimilar variables we cannot go any further, but take a look at this example:Remembering from our discussion on index / powers or indices these expressions are the same although the index of one is not normally shown:So, in this case if we take the lower index power (that is the one on the denominator) away from the index power of the numerator we end up with zero, in other words X vanishes:Once again, we could expand our powers of X and simply ‘cancel down’ to see what we have left at the end of that procedure, for example have a look at this one:Note that it is also OK to call this but generally with fractions it is accepted to leave them as they are, even if they are ‘top heavy’ or ‘improper’.Note: The subtraction MUST always be from the ‘top to bottom’, i.e.: the power of X on the denominator is taken from the power of X on the numerator, or ‘top power minus bottom power’:With multiplication of index powers, involving addition it doesn’t matter:But division of powers, involving subtraction, does matter:In summary:
Es Muy Tutor Services - Algebraic Manipulation
Algebraic EquationsAs we have seen before in algebra, an equation is an expression split by an equality such that one side equals the other.The above expression is perhaps as simple as it gets, clearly “x” does indeed “equal one” for the expression to qualify as an equation.But what happens when we have more complicated expressions, and we need to find the value of “x”?Generally, there are several ways to solve these issues:Collecting ‘like terms’FactorisationGraphingCollection of ‘like terms’ is the easiest for simpler expressions:ExampleSolve for ‘x’ This isn’t so bad, because the ‘like terms’ are already ‘collected’ on the left side of the equation, so what we need to do now is a little bit of ‘left side’ –v- ‘right side’ manipulation:Subtract 4 from both sidesDivide both sides by 2And here we have our solution.Now things are going to get a bit tougher….look at this one:Solve for ‘x’Not as easy huh?, well as you get better you will be able to use the ‘bring it across and change the sign’ method when collecting ‘like terms’Bring across and change its sign:Take across and change its sign:Tidy up:Divide both sides by 5This takes some getting used to, and the benefits of checking back with your new found value for ‘x’ can’t be stated strongly enough.Until you get more skilled, the expression above would best be sorted by our ‘do the same on both sides’ strategy:1.Add 34 to both sides:2.Subtract from both sides:3.Divide both sides by 5:Effectively the methods are the same, but the second method takes it a bit slower and allows you to see each step.
Es Muy Tutor Services - Algebraic Manipulation
OK these can get tough, and if you are running for a GCSE in Maths you will need to crack these, as they can be a great source of marks in an exam.Look at the picture below:The triangle and the rectangle have the same perimeter, so we need to restate the problem and THEN make ‘x’ the subject.So if the perimeter of the triangle = perimeter of the rectangle we are saying that:OK?Let’s do some tidying up, first remove the parentheses (they weren’t really needed but useful to demonstrate that 2x+1 was the whole of the side).So we haveLet’s have a go at a tougher one now……Rearrange the expression to make ‘a’ the subject:Arguably this is about as nasty as a GCSE examiner would make it, but who knows??Anyway….Step 1 – restate the problemStep 2 – Collect the ‘a’ terms on the left (add ‘a’ to both sides…remember?)Step 3 – Collect the ‘c’ terms on the right (subtract ‘c’ from both sides)Step 4 – Now it gets naughty, you have to bring in some factorisation to release the ‘multiplied a’ and ‘b’ terms….see if you follow this (by expanding back out)Step 5 – Got that?.....OK divide both sides by (b+1)
Es Muy Tutor Services - Powers, Indices and Standard Form
Powers, Indices and Standard FormWhen we look at a number, we are actually looking at more than just a number, it is an entity representing a value which occupies a place on the number line.The picture above, depicts a small section of the natural number line. The origin is set at zero with all natural positive numbers to the right and all natural negative numbers to the left. To the right of number one we have number two and to the right of number two we have number three and so on, similarly to the left of -1 we have -2 and to the left of -2 we have -3 and so on.From the point of origin, set at zero, the number line extends to the right and to the left infinitely and between each number we have an infinite number of fractional/decimal values so that for example exactly midway between zero and one we have the fraction ½ or the decimal value 0.5So, what do we mean by Powers? - It would probably be useful at this point to bring in the correct terminology, “Exponentiation” which is defined as being a mathematical operation involving two components:What we have here, is a number ‘a’ “raised to the power of ‘n’”Let us take a look at an example:In this particular case our base and the exponent have the same value, that is the value 2 but the subtle difference is that the larger 2 is the base which is being raised to the power of the exponent so effectively they have different jobs.In this particular instance 2 raised to the power of 2 is 2×2 which is of course 4.What about this……Well, if I was to say that this is 2 x 2 x 2 = 8 can you see a pattern forming already?OK……formally we can say that:“A number raised to a positive power (exponent) is that number multiplied by itself as many times (less 1) as the exponent value”So 23 is in fact 2 multiplied by itself a further 3 - 1 = 2 times.Right, you have to be aware that if you are told that 23 is 2 multiplied by itself 3 times this isn’t correct, because that would be 2 x 2 x 2 x 2 = 16.Accepted practice is to exclude the first instance of the base and to say that:
Es Muy Tutor Services - Powers, Indices and Standard Form
OK here are a few more examples before we reach our first general rule:The general rule for positive powers:There are a few special cases to remember:a2 is also referred to as ‘Ay Squared’, similarly a3 is referred to as ‘Ay Cubed’ but from then on we usually say that an is “Ay to the Enth” or just “Ay to the Enn” .What happens if the exponent is 1 or 0? – Perhaps we should talk about these cases before we go further back in the number line (into the negative numbers).OK…….think about the general rule, an = ‘a’ times ‘a’ up to n-1 times, well in this case we have n=1 so n-1 is zero, in other words we don’t multiply ‘a’ by itself at all (well, we do….but its zero times) so effectively ‘a’ stay as it is:General rule: So…..if n = 1 then this means “2 multiplied by itself zero times, or just left alone = 2”And what about when n = 0? Well this is a simple statement of arithmetic fact that: ANY number raised to the zero power is 1.So now that we have established this, what happens when the number ‘n’ is negative? For example what if n was ‘minus one’?Well, the rule here is that any number ‘a’ raised to a negative power ‘-n’ is the same as the reciprocal of ‘a’ raised to the positive equivalent of ‘n’:So:This is quite straightforward, we have looked at positive integer powers and negative integer powers:
Numbers“We use numbers to count”“A number is a mathematical entity used in counting”Just two of the possible ways to describe numbers, but of course numbers are far more than this.Numbers can be classified into many different subheadings. We have:Natural numbers, real numbers rational numbers, irrational numbers, integers, fractions, even and odd numbers and prime numbers. This list could go on but for the benefit of this particular part of the book I don't want to make it too complicated.The first thing that we should do is briefly define each of the expressions that has just been listed above. I will start with (because I'm not going to do this in any particular order) "Real Numbers".Real NumbersThese are the numbers that would naturally take a place on the number line:‘real numbers’ are also regarded as the ‘measuring numbers’, therefore we are in fact talking about the decimal numbers, used in day-to-day calculations.Natural NumbersThese are perhaps the most familiar numbers of all, the numbers that we count with and it wasn’t until the 1800s that the number 0 was even considered to be a number and the purposes of counting, and so the sequence started with one.Perhaps now is a good time to introduce the next set of numbers, integers.IntegersThese are what you would call a “whole numbers”, written without fractional parts or decimal parts. For example the numbers 1, 26, 241, 4027 etc are integers because they carry no fractional component whereas numbers such as 23.7, 248.964 etc are not integers but are in fact examples of “real numbers”.Numbers that can be expressed as fractions would be part of the family of “rational numbers”.Rational NumbersA rational number, is a number that can be shown as a fraction with integers on the numerator and on the denominator, for example:Is an example of a rational number.Now this may seem contradictory but integers belong to the family of rational numbers because, although they aren’t shown that way normally, an integer can be shown as a fraction when it is shown as “divided by 1”I’m only going to mention three more types of numbers; I want to keep this particular part simple. We have even numbers and odd numbers, the former are the ones that are divisible by 2 without there being anything left over and of course the odd numbers are the ones that aren’t!Prime NumbersFinally we have prime numbers, perhaps the most “quirky” of all the numbers, these are the ones that only divide by themselves or 1. For example I have listed below all of the prime numbers between 1 and 100:2, 3, 5 ,7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97however, if a number can be divided evenly it is called a composite number, being formally defined as a number which has at least one other positive divisor other than itself and one, so for example whereas the number 17 is a prime number, the number 18 is a composite number because it divides by 1, 2, 3, 6 and 9 (indeed, being next door neighbour to a prime number but having so many divisors itself you could in fact say that the number 18 is indeed a very composite number :-)).As I said before I don’t intend to mention any other types of number, suffice it to say that there are very many other types of number such as complex numbers, perfect numbers etc.So, what can we do with numbers? Well we can add them together, subtract one from the other, multiply them together or divide one by the other. There are of course many other operations we can square them cube them, raise them to powers, turn them into fractions, turn them into decimals, change the bases (that is change the number that the groups are based on, more on that later) and so on, and so forth.For the first part of this section I’m going to concentrate on the 4 basic mathematical functions of addition, subtraction, multiplication and division. Further on into this section I will talk about bases (for example base 10, otherwise known as the decimal or denary system, base 2 which is also known as the binary system, base 16 which is hexadecimal (and a bit more complicated than some of the others)).
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The Four Basic Mathematical FunctionsWe deal with these every day of our lives, whether we are adding up shopping bill items, taking our weekly bill away from our allocated budget, working out how much a certain number of items will cost us given the price of one of them or splitting up our wages/salary into equal chunks to last us for the month. Since we were small children, we have done this in one way or another and probably not given very much thought to it (except in maths class when we probably had very little choice :-))This part of the section is going to go over these 4 functions and explain how to do it.The diagram below shows the symbolic representations of the 4 functions were going to talk about. Top left is a red plus sign representing addition, directly below it is a blue minus sign representing subtraction, to the right of the plus sign is the green sign representing division (as you can see it represents a fraction if you think about it) and finally below that in purple is the cross which represents multiplication.Since we work in “tens”, that is we use the decimal system, my explanation of these 4 functions will also be a description of the decimal system. Other bases will be discussed separately.Simply put, we count in tens, we start at 1 (it is very rare that we say that we have nothing of something so that is why we wouldn’t start at 0 although the number 0 does feature prominently in the decimal system as a placeholder, more on that later) and we proceed through 2, 3, 4…9. When we reach 9 we have to start on “another 10” so we indicate this by representing our “unit account” as being reset to 0 and then our tens (that is our completed batches of 10) are incremented by one, and we start again at 1 through 9 for our next batch of 10.It might be simpler to think of it like this. Imagine having a tray, not unlike an egg box, with 10 holes, each of which can hold one item. We pick up our first item and place it in any empty hole in the tray. We now have 9 free holes. We pick up our next item and place it again in any free hole. Now we have 8 free holes.We continue to pick up items and place them into free holes until there are no free holes left, whereupon our tray is full and if we wish to continue collecting items we need another tray (that is we have a completed 10, but if we have more items than 10 we need to start on a second 10 until we have completed counting items.Assume that we have more than 10 items, so we pick up another blue tray and start filling it up in the same way that we did the last one, but having reached 10 on the second tray we find we still have more items:We keep filling up trays until we run out of items to put in them, let us assume that we filled up four trays and we have 3 items left over:We have 4 completed trays with 10 in each one and 3 left over:So in fact we have 43 items.What about if we had actually filled up 10 trays before we found that we had 3 leftover? Well this time of course we would have 10×10 which is 100, plus the remaining 3 which is 103.What this is in fact leading us to is the fact that we ordinarily count in tens, and when we fill up a 10 we start on another one, until we’ve filled up 10 of them, and we call a group of ten 10s = one “100”.When we reach 10 groups of 100, we call this 1000, and 10 of these is called 10,000. In our primary school education we are taught “hundreds, tens and units” and this forms the basis of our mathematical learning.Earlier on, we said we had 4 trays full with 3 leftover, the top line of the table above shows as this is a number, i.e. four tens, and 3 units. A little bit later on we suggested that we might have 10 full trays, and 3 left over. Now this time we won’t write 10 in the tens column, because we already said that when we reached ten 10s we call this 100 so we put one in the hundreds column and 3 in the units column. We don’t have any ungrouped trays of 10 lying about so this time we put a 0 in the tens.With me so far?.....good 8-)
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Addition = Adding Up NumbersWe call the process of totalling items “addition” and we use this to group together a number of items which we can represent by one number called “the total”.For example, if we have 3 of something and we then receive another 4 of something, we can at these together to make a total of 7 of something:So for example, if initially we had 3 books but then we buy a further 4, we can say that in total we have 7 books:We show this operation of “addition” by the simple mathematical expression shown below:Where the plus symbol represents addition and the equal symbol represents the equality, in other words (as you will see much later in the book) the information on the one side of the = is equal to the information on the other, that is indeed 3 + 4 does equal 7!Because a book is a whole item, that is you are unlikely to buy a fraction of a book, we are in fact dealing with “integer addition” which is the simplest addition we can have. Things start to become a little bit trickier when we are looking at negative numbers, or numbers with decimal or fractional parts. We will of course come to these in due course.We are taught in primary school to add numbers together by putting one on top of the other with the + at the side, and a line underneath the bottom number where we would write our answer (in fact when I was at primary school we were taught to put two lines at the bottom to make a box without 2 of the sides, and the answer would be written between them):In this example we are simply adding 3 to 5 to make a total of 8, as these are all “units” they sit underneath each other because the answer does not exceed 9 we don’t have to start on the next column to the left which would be the tens. In the next example we will create an addition which will go above 10 and I will show you what we have to do then.Of course you’ve probably worked out that HTU means “hundreds, tens and units”Okay, let’s look at another example where the answer will put us into the tens column as well:This one is a bit more complicated and I have drawn a few lines all over it. We perform additions using the units first then we move across to the tens when our number of units is bigger than 9, and likewise when our number of tens is bigger than 9 we move across to the hundreds, and so on.5+7 = 12 but we can’t simply write 12 in the box, because 12 represents one 10 and 2 units, so what we have to do is write the units in the box under the units column and write 10 just outside the box underneath under the tens column (we do this because we have to add it to the tens and putting it underneath the tens outside the box means that we won’t forget it) I have coloured it red to make it standout, and also to indicate that it is a by-product of the addition of the units column which has a red line through it. Taking this 10 and putting it outside the box ready to be added to the tens is a process called “carrying” and this particular example means we have added 2 to the units box but we have carried the 10.Going back to the top of the sum, and now moving across to the tens column, you can see that we have one 10 but below it there is nothing to add, so we prepare to move it into the answer box, however we have carried 10 from our previous operation so we add it to the 1 making 2. There is nothing in the hundreds column so we have in fact finished our sum and can see that 15+7 is indeed 22.The next example now to show you will involve some hundreds, but the principle is exactly the same, we start with units, then we move on to tens, then finally we move on to hundreds, remembering to “carry” where necessary from one column to the next. Here is our next example which will involve some hundreds, but the principle is exactly the same, we start with units, then we move on to tens, then finally we move on to hundreds, remembering to “carry” where necessary from one column to the next. Here is our next example:This time we have “carried” twice, and the carried numbers are coloured blue and red to indicate which columns they came from. In just the same way as before we first of all add up the units column:5+6 = 11We put the unit into the units column (below the number 6 in 96) and we carry the 10 which is coloured in red underneath outside the box and under the tens column. Next, we add up the tens column:1+ 9 = 10Remember though we have one extra 10 carried from the previous operation, so we add this to our answer here which makes it 11. Once again we put the one in the tens column, and we carry the hundred (which is coloured in blue) and put it outside the box underneath the hundreds column.Finally, we add up the hundreds column, we can see that there is nothing below the 2 so we get ready to place the 2 directly into the answer box in the hundreds column, but remember that when we added up the tens we ended up with 100 leftover which we carried (in blue) so we add this to the 2 making 3.Answer therefore is 215+96 = 311
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For the final time, I’m going to go through another example but this time I’m going to include tens of thousands as well, so we are going to jump from hundreds tens and units straight to “ten thousands, thousands, hundreds tens and units!”.Let’s start by adding together 5+6, this of course is 11 so we put the one in the box (the rightmost number 1 in the answer box) and we carry the 10 (as a number 1, coloured red because it came from the units column).Move into the tens column, 1+8 = 9, but we have to add the ‘red 10’ which makes the total in the tens column also 10, so we put a 0 in the answer box to the left of the 1 and we carry the hundred, coloured in blue because it came from the tens column. We now move into the hundreds and we add 2 and 9 together making 11, we add the ‘blue 100’ from the previous addition to make 12, we then put the 2 in the answer box and we carry the one, this time coloured green because it came from the hundreds. Our last but one step is to look the thousand column where we add 4 and 9 to make 13, but don’t forget we carried the ‘green 1000’ from the previous addition so we add one to 13 making 14, put the 4 in the answer box and we carry the one this time in purple to show that it came from the thousands column. Our final step is to bring the 6 down, and add to it the one ‘10000 in purple’ making 7 and we have now arrived at our answer, that 64,215+9986 does indeed come to 74,201.As you can probably see, addition isn’t that difficult provided you line things up in proper columns and take your time, not forgetting every time you ‘carry’ you have to add it on to the subsequent addition. Of course nowadays you would probably just use a calculator to do this sort of thing, and indeed you could use a calculator is to prove that you’ve actually done the manual addition correctly. In fact I would be inclined to do just that, have lots of practice in addition, and where possible check your answers with the calculator. It is fair to say that calculators have made us lazy, so these basic skills are very worthwhile (after all you might find yourself with a calculator or a mobile phone with a flat battery one day!).In this “addition” section we have “carried” when the addition of each column has exceeded 10, in the next section that we are going to look at “subtraction” we will find that we have to use in some cases a not dissimilar action which we will call (for reasons which might become obvious later) “borrowing”, but like with all good “borrowing” we also pay back quite quickly.Subtraction = Taking One Number Away from AnotherThe opposite to “addition” is “subtraction” where we remove the value of one number from another to arrive at a new result which of course is (normally) less than we started with. “Subtraction” is all around us, income tax, mortgages, indeed any debt that we have involves the subtraction of money from our pockets leaving us with less than that we started with. Subtraction is yet again another skill which is worth mastering so that you can for example make sure that your monthly budget is on target by taking your routine and daily expenses away from the money you have in the bank. Of course a calculator is probably more critical in subtraction and you would probably default to a calculator straightaway as I believe most people would, but learning to do it on paper with a pen is a desirable skill which we will start to go through now.Our first example will be very simple, 15 take away 4 equal 11. The first part involves the “taking away” of 4 from 5. If you line up for example 5 oranges and take 4 of them away it is quite obvious you are left with just one so in this case would put the one in the answer box (the rightmost number 1 of the two shown). The second operation would be to take something away from the 10, but as you can see there is nothing to take away from it, so we bring it down into the answer box unchanged.In each individual column, where the number at the bottom is smaller than the one at the top, the subtraction is always quite straightforward. For example the above 5-4 is one, but what about if the 5 and 4 had been reversed?This time the subtraction isn’t quite so straightforward, because in simplistic terms you can’t take 5 away from 4 (later on in the book you will find that in fact you can, but your answer will be negative and so is not to confuse the matter at the moment we won’t bother with those).Consider for a moment if we want to buy something, but we have not got the money, what do we do? Well… We can consider a loan from the bank so in fact we “borrow” and we do exactly the same in this subtraction. Study the question, we need to borrow 10 from the tens column to add to our 4, so that we can take away 5 from it:This time, because 4 is too small to do the subtraction I’ve borrowed a 10 from the tens column and so the tens column has had to be reduced by 1 to 0. I’ve placed the “borrowed” 10 near to the 4 making it 14, I have then taken awaysthe 5 leaving 9, and this is our answer because we have nothing left in the tens column to do anything with.The next example I’m going to go through will contain some hundreds, and will be a little bit more complicated than this first one but by the same token I hope it will make the processes a little bit clearer to you.l to do the subtraction I’ve borrowed a 10 from the tens column and so the tens column has had to be reduced by 1 to 0. I’ve placed the “borrowed” 10 near to the 4 making it 14, I have then taken away the 5 leaving 9, and this is our answer because we have nothing left in the tens column to do anything with.The next example I’m going to go through will contain some hundreds, and will be a little bit more complicated than this first one but by the same token I hope it will make the processes a little bit clearer to you.
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In this question we take away 187 from 455, leaving us with 268. There will be a couple of instances of “borrowing” during this question, but as always, we will take it step by step.First of all look at the units, we can see straight away that we cannot ordinarily take 7 away from 5 so straight away we have to consider “borrowing” a 10 from the five 10s we have in the top line (the 50 component of 455).We “borrow” 10 from the 50 in the 455 component of the question, making our units 15, from which we take away 7 leaving us with 8. The 5 in ‘455’ representing the tens is reduced by 1, temporarily making the number 445 as shown below:Study the diagram in conjunction with the last few lines if you’re not sure where this has all come from.If you’re satisfied to move on then you will see that 15 - 7 is indeed 8 and our five 10s has been reduced four 10s because we borrowed one.Moving into the tens column we now need to subtract 8 from 4 and once again (ordinarily) we cannot do this without borrowing 100 from the hundreds column. We do this by borrowing 100 from the 400 and reducing it to 300, then we make our 4 in the tens column into a 14, we subtract 8 from it leaving 6 which were put in the answer box. Study the diagram for a while, you will see that it is in fact correct and that we now only have three 100s to play with not 4 because we borrowed 1.The remaining step is quite simple, looking at the hundreds column 3 minus 1 equals 2 so the answer that we place in the answer box to the left of the 6 will be 2, giving us our final answer of 268Addition and Subtraction with Negative NumbersAt the very beginning of the section on subtraction I did say “you can’t take 5 away from 4” and at the very beginning of your primary school education it would be left at that, however, I want to go through negative numbers very briefly to show you that you can in fact do this but you would be left with a negative answer.We understand that:But what would happen if we took (in the case of the second equation) 2 away from 1?, Well to answer this will take a look at the number line which we will see more of later.The number line starts at 0 which is called the origin and moves infinitely to the right along the positive scale and infinitely to the left along the negative scale. The first increment to the right of the origin is the positive number 1 and the first increment to the left of the origin is the negative number 1. Between each increment in the line shown above there is an infinite number of further increments representing the decimals/fractions so for example halfway between 0 and positive 1 we would have positive 0.5 (positive one half ) and halfway between the origin and -1 would have -0.5 (negative one half).So, how do we add negative numbers? Well, before we answer that let’s just quickly recap on adding positive numbers by using the number line above:Consider the very simple addition:We know that this is correct, but let’s show it on the number line:
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What we’ve actually done here is start at 2, then we have moved to the right 5 times in increments of 1 to arrive at our answer of 7, so what we’ve actually done as this:2 to 3, 3 to 4, 4 to 5, 5 to 6 and 6 to 7And if you count them, you’ll see that we have made 5 “jumps”Now let’s take a look at this:The answer is correct, at -3 because if you think about it, it can’t be 7 because that was the result of our previous example when we added +5, this time we are adding -5. Have a think about it for a minute, if “adding a positive number” means that we move to the right then it makes sense that “adding a negative number” means that we have to move to the left, but of course (in this case) with the same number of “jumps”:Addition:Adding a + moves us to the rightAdding a – moves us to the leftWhen we consider subtracting negative numbers, the method is slightly different.You can see that above I’ve written two general rules for addition , the rules for subtraction slightly different to this and what I am going to do now, by taking a slightly different approach, is state the rules and use the next page or so to prove them.Subtraction:Subtracting a + moves us to the leftSubtracting a – moves us to the rightWell, this looks confusing? maybe…..but bear with it!Consider the expression:Well, what on earth could the answer be to this one? Once again, let’s go back to the number line:Now I know what you’re thinking, “this is just the same diagram that we used above when we worked out 2+5”… And you would be correct! Look at the second rule of subtraction above, where I stated that “subtracting a negative moves us to the right” so if we start with 2 and we are taking away -5 then strange as this might appear, we jump 5 places to the right not the left, therefore our answer is indeed 7. Let us show it is an equation:What this is telling us is that if we subtract a negative number we effectively add what would be the positive version of that number to the first number:Study these examples and hopefully this will become clear:Now….let’s take a look at the first rule of subtraction now which states that “subtracting a positive moves us to the left”
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This time we are taking away a positive number from a number which is in fact smaller, so the result will be negative in this particular case.We start at 4, and following the rule “subtracting a positive moves us to the left” we jump to the left 5 times arriving at an end result of -1, and in fact this is our answer:Study these examples and just like before hopefully this will become clear:Multiplication and DivisionWhen we were at school, one of the first things we learnt was the “Times Tables”, from 2×2 right up to 12×12, but in some cases up to 20×20 and we were meant to memorise these so that if we were asked for example what 8×4 was we could immediately respond 32. In this particular section I will deal with multiplication and division separately, starting with multiplication, looking at an old system that will probably a little bit familiar with called “long multiplication”Before I do any of that, I have reproduced below a grid of the “Times Tables” up to 20×20 simply for ease of reference (click on it to enlarge, click on it once more to return to smaller size):The way to use the table is to select the number from the top row in yellow that you wish to multiply by the number in the left-hand column in yellow, where the row and column intersect, there is your answer. Just out of interest you will see that some columns have intersection points shaded in green because these are the first 20 squares.You can of course if you wish, simply use this table when you need to multiply numbers up to 20×20 but you will need a grasp of “long multiplication” if you wish to multiply larger numbers.MultiplicationLet’s take a look at an example of multiplying 2 numbers together which are not on this grid.Well, this one looks particularly beastly and as you can see I have retained the hundreds tens and units column headers but this time I have added thousands and tens of thousands (Th and TTh) because when we start looking multiplication we are really starting to hit some big numbers. Now there are two ways of starting this calculation and it would appear that the more modern way is to multiply the units first followed by the tens and if necessary followed by the hundreds and thousands and so on.The only difference of course is that the sub totals will appear in the answer box in a different order other than that the process is pretty much the same. Okay, starting with the units column we need to evaluate 5×7.Using our table we can see that 5 times 7 is 35 which of course is 3 tens and 5 units, so we put the 5 in the answer box underneath the 7 in the units column and, just like in addition we “carry” the 3 across to the tens column:<< Our interim answer is now starting to look like this.We stick with the 7 on the bottom line, but this time we multiply by the 5 on the top line under the tens, coincidently the answer again is 35 but remember that we carried 3 from the previous operation on the units, so we must add this to the 35 to make 38. We put the 8 to the left of the 5 in the answer box so it sits in the tens column and we carry the 3. Our final step in this particular part of the calculation is to now multiply the 7 by the 4 making 28. Not forgetting the 3 we carried from the previous operation 28 becomes 31, we put the one in the hundreds column underneath the 4 in the answer box and because we have no further operations to perform on the 455 line we put the 3 next to it underneath the thousands column as its first entry.What we’ve actually got here is the first part of the answer which is 455×7= 3185.
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What we need to do now is work out the second part which is 455×80 and then add the two sub totals together to make the final sum.Because the steps are almost identical in this second part, I am not going to go through them all except for the first step which is a little bit different.Because we have completed our operations involving the units on the bottom line (that is on the 7 in 87) what we now need to do is to multiply the 8 (of the 87) by each element in 455 and on the second row, below where it says 3185 write out the subtitle of this particular second step:The red 0 beneath the 5 in the units column the answer box is a placeholder, telling us that first of all we have finished with the units and that secondly our answers in the second step must be displaced one place to the left because we are now dealing with numbers which are 10 times bigger.Look at the diagram and you can see that the red 0 is in place and that 8×5 which is 40 has been calculated, the 0 has been placed under the 8 in the tens column of the answer box and the 4 (which represents hundreds) has been placed outside the answer box underneath the hundreds column ready to be added to the next step.As I said above, I’m going to now complete the sum, so what you need to do is study the next picture carefully to see if you can work out where the steps came from.455×87= 39,585.The 44 in the bottom but one box is just the “carryovers” from the intermediate steps in the calculation, the most important part is using your knowledge of addition to make sure that you correctly add up 3185 and 36,400 to get to the right answer, which of course in this case as above is 39,585.The next picture is the same sort, but “annotated” with some useful information for you to bear in mind when you’re doing sums of this type.DivisionJust as multiplication involved “carrying”, division will involve “borrowing” and out of the two (multiplication and division) division is considered to be the hardest.Example 1We lay out the multiplication using hundreds tens and units, and if necessary thousands tens of thousands or even higher but with division the layout appears differently.In some books you will find the upright bar is curved inwards like a right-hand parenthesis, and in some books the 2 bars won’t even be there, instead there will be a box. I prefer this “upright and horizontal bar” method because it is the way that I was taught to do long division many years ago. Unlike in multiplication where we start with units, in long division we approach the problem number (85 in this case) from left to right irrespective of its size. Okay, let’s see how many fives there are in 85:Step 1: working from the left, we first of all ask ourselves “how many fives in 8?”There is one 5 in 8, so we enter a one above the 8 (shown in green) and we subtract 5 from the 8 leaving a remainder of 3 which we write underneath the 8 as shown.We now bring down the 5, making the 3 into 35.Step 2: again working from the left, we now ask ourselves “how many fives in 35?” . If you’re not sure of the answer, refer to the grid a few pages back, find 35 in the grid and with number 5 itself as either the side or top header, move sideways or upwards and you will find that there are in fact 7 fives in 35.We put 7 above the 5 as shown in green and we take the 35 from 35 leaving 0.Because we have nothing left, we have in fact finished and the answer is 85 ÷ 5= 17.That was a fairly straightforward example, in fact if you look at the grid of the 20×20 times table you could’ve just simply read the answer straight off.
Es Muy Tutor Services - Science (Index)
Various bits on Science under the subject headings shown.
Matter in the universe exists in one of three common states, as shown in the diagram above. Although scientifically there are 5 states of matter, the 3 mentioned above and also plasma and Bose- Einstein condensates it isn’t my intention to go into that sort of depth in these pages, so what I will do here is confined the discussion to the first 3 which are of course the commonest.In solids, particles making up the solid are packed together very tightly, and so they can’t move around very much. Apart from a little bit of vibration because of the energy they possess, these particles are effectively imprisoned. Solids have a definite shape and volume and do not take the shape of the container that they are in (that is they don’t do this naturally, if they have been made that way then that’s another thing of course, but generally they retain their own shape). Because of this resistance to move, solids are usually very hard and although many solids can be hammered into shape (this is called “malleable”) and in some cases of metals can be drawn out into thin wires (this is called “ductile”) they generally retain the shape they hold.Liquids on the other hand will take on the shape of the vessel that they are in. Like solids, they are not compressible and if you look at the diagram above you will see that in liquids it would not be uncommon to see them in clumps, but generally they flow more freely than solids (which don’t of course!). The fact that liquids can’t be easily compressed, coupled with the fact that they readily take on the shape of the container that they are held in, means that liquids are useful in hydraulic systems where we can take advantage of this “non-compressibility” to force liquids through pipes, for example in the hydraulic lifting platforms of industrial plant and simpler systems such as car brakes.It should be mentioned at this stage that solids and liquids, and the particles making them up, possess a certain amount of vibrational (kinetic) energy. In the case of solids this is nowhere near enough for them to break free, and in the case of liquids they can move around to a greater extent but forces between particles tend to hold them, or attract them together, overpowering the kinetic energy that the particles possess and making sure that they don’t separate easily. This is not the case with gases.Particles making up gases have a terrific amount of kinetic energy by comparison, and so can break free from the weaker forces which tend to bind liquids together. In fact, the particles making up a gas hold so much kinetic energy that they can completely overcome any intermolecular forces which might be trying to hold them together. As a result a gas will expand to fill up all of the available space with relatively large distances between each atom or molecule.Gases can be turned back into liquids by number of processes, cooling down the gas to a point below its boiling point will cause it to condense into a liquid, and further cooling down to a point below the freezing point will cause it to solidify. Gases can also be turning to liquid by compression but this is often accompanied by a reduction in temperature.
Es Muy Tutor Services - Atoms and Molecules (1)
Atoms and Molecules
Atoms are the fundamental unit of matter, although they are themselves made up of smaller, subatomic particles any further description of that is way beyond the intention of this particular page. For our purposes ( here, anyway) we will regard the atom as the smallest unit, from which everything else in the universe is made.Essentially, an atom is made up of 3 fundamental particles: protons, neutrons and electrons. The picture below is a representation of the currently accepted perception of what an atom looks like. There have been many interpretations over the years, atoms being regarded as solid spheres with the electrons embedded inside (this is known as the plum pudding model) but scientists have decided to settle on the model shown here as there is much scientific research and evidence to support it.Taking this as our accepted model, the nucleus shown in red and grey consists of protons and neutrons, with the exception of the element hydrogen which contains only one single proton (again, not strictly true because there are other types of hydrogen, known as “isotopes” which do contain neutrons as well as protons in their respective nuclei).Let us assume that the proton is red, a proton carries one unit of positive charge and the grey neutron is electrically neutral (as its name would suggest). Balancing out the net positive charge of the nucleus are the (in this particular diagram) blue electrons, considerably lighter in weight than protons and neutrons but carrying a unit negative charge. In a neutral atom (not an ion that is) the number of protons and number of electrons are the same. It is the number of protons in the nucleus that determines the particular element.II previously mentioned isotopes, these are (if you like) different versions of the same element, but with differing numbers of neutrons in the nucleus. Remember that if the number of protons changes, so does the element, and if the number of electrons changes then you end up with a positively, or negatively charged ion (depending upon whether the number of electrons increases or decreases) unless the number of protons also changes, in which case (like i’ve already said) the element changes too.Most elements have more than one isotope, as I mentioned above hydrogen has 3 (that is, 3 that are well known, there are approximately 7 others which have been created in the laboratory but these needn’t concern us here). Protium, Deuterium and Tritium respectively, of which protium is the commonest.That’s a rough, whistlestop tour of what an atom is made up of, now let’s go quickly take a look at molecules……..>> click here <<
Es Muy Tutor Services - Atoms and Molecules (2)
Atoms and Molecules
Chemistry would be pretty uninteresting if atoms just hung around, not doing anything. Fortunately for us atoms love to combine to form units known as molecules. Perhaps the simplest molecule is that molecule produced when 2 atoms of hydrogen get together:
Two hydrogen atoms, each containing one proton and one electron combine together to form a hydrogen molecule which consists of 2 protons and 2 electrons. Now I can hear you say to yourself “but if it’s got two protons how can it be hydrogen?”, well…..this is because we are talking about molecules now not atoms. Hydrogen, in this case is said to be “diatomic” and this is how our friend hydrogen is normally found when it is not bound in compounds.When hydrogen is first formed, it is monatomic (like on the left-hand side above) but very quickly to atoms will join together to form a molecule. You may hear of the expression “nascent” hydrogen, which refers to hydrogen (and some other elements) in its newborn monatomic state.Atoms contain electrons, and these orbit around the nucleus at certain levels of energy, best thought of as “shells” which can be visualised as perhaps a series of concentric circles of increasing size, and they like to have a certain number of these in their outer shell, usually 8 electrons as in the case of the element Neon, but of course even by pooling resources a hydrogen molecule can only come up with 2, so the hydrogen molecule is satisfied with 2 electrons for its stable state. Hydrogen molecules achieve this age by sharing their single atomic electron to make a pair of electrons and by doing this each hydrogen atom feels as if it owns both, but of course it doesn’t, this feeling is achieved by sharing. This “sharing” of electrons forms what is called a covalent bond.There will be more information on chemical bonding later on, but suffice it to say at this point that atoms can either share electrons, or donate electrons depending on the type of bond being formed.
Atoms combine together to form bonds, this is the basis of the construction of molecules of any size from the component parts. Atoms bond by sharing or donating electrons, and it is the way that this is done that determines many of the properties of the resulting compound.Essentially we have 4 types of chemical bond :1. Covalent2. Ionic3. Coordinate (or Dative)4. Metallicif you click on each name in turn, you will be taken to a page dedicated to a brief discussion of that particular type of bond. Essentially though, bond types 1 and 3 perform their activities by the sharing of electrons (however in the case of a coordinate bond the shared electrons are all provided by one of the atoms), and in the cases of 2 and 4 the bonding is effected by the complete transfer of one or more electrons from a donor atom to a recipient atom creating positively charged and negatively charged entities which then bind together by electrostatic attraction (in the case of metallic bonding the donated electron joins with others to form a “sea” of delocalised electrons which, by similar electrostatic forces, help to bind the positively charged metal ions together).Please note that these pages will be very basic, and brief indeed, the subject of chemical bonding can on its own fill many books and my intention here is to literally just give you a brief overview of the types of bonds that can form between atoms.
On each page you will see a blue paperclip, once you have read that page, click on the clip and it will bring you back to this page.
Es Muy Tutor Services - Covalent Bonding
In general, non metal atoms form molecules by bonding covalently. They do this by “sharing” electrons, so that each atom ends up in the belief that it has a full outer shell of electrons. Having a full outer shell is also known as the “inert gas configuration” as the elements helium, neon, argon, krypton, xenon and radon have this full outer shell configuration. This means that they have no desire to lose, or try to gain, an extra electron or electrons to complete their outer shell (because it is already complete).As a result of this, atoms with this configuration are generally quite unreactive.The diagram below is a hypothetical molecule of one of the halogens, group 7 elements with 7 electrons in their outer shell but wishing to obtain the 8th one. They do this by sharing electrons.. If you look at the diagram you can see that the shared electrons give each atom in the molecule the feeling that they have a full outer shell:
The “dot and cross” representation is quite universally accepted in most science books to show the shared pair of electrons. We can assume therefore that the above representation could be that of a molecule of chlorine gas, because a chlorine atom has the electronic configuration 1s2 2s2 2p6 3s2 3p5 or[Ne] 3s2 3p5In other words a chlorine atom has 17 electrons in total,, 7 of which are in the “valence shell”, in this particular example at quantum level 3 (in other words, the 3s and 3p orbitals), but when bonded covalently with another chlorine atom the outer shell at quantum level 3 appears to be complete (8 electrons) in each atom, hence the bond is formed quite strongly and the chlorine molecule exists diatomically.It should be noted that in the above electronic configuration of chlorine, I have also used the shorthand version on the right-hand side. This is indicating that the 7 electron valence shellis built on the “core” of the inert gas neon. This is a convenient method of shorthand particularly when the atoms start to get very large.
Es Muy Tutor Services - Ionic Bonding
Ionic Bonding generally takes place between elements in groups 1 and 2, and groups 6 and 7, for example the element sodium will react with the element chlorine by transfer of the single, outermost electron of the sodium atom to the chlorine atom, in the latter case completing the “octet”but producing (by reduction) a negatively charged ion and in the former case the sodium atom is oxidised to a positively charged sodium ion.
The sodium atom initially has 11 electrons, and the chlorine atom has 17. As we said before the objective is to achieve the “inert gas configuration” by losing or gaining one or more electrons. In this particular example can be seen that the outermost electron of the sodium atom (the 3s1 electron at quantum level 3) is completely transferred to the vacancy in the incomplete octet at the same quantum level (this is in fact a vacancy in one of the 3p orbitals for those who wish to dig a little bit deeper).As a result of this complete electronic transfer, the sodium atom is oxidised to a positively charged sodium ion, and the chlorine atom is reduced to a negatively charged chloride ion. Now we have a completely ionic (ionised) situation, and as opposites attract the sodium ion and the chloride ion will attract each other, forming a very strong ionic bond.The structure of solid sodium chloride is a cubic lattice, and in a single grain of salt there will be millions, if not billions of sodium and chloride ions, electrostatically attracted to each other.
Es Muy Tutor Services - Coordinate (Dative) Bonding
Coordinate (Dative) Bonding
Coordinate Bonding a.k.a. Dative Bonding is a type of covalent bond, but in this case all of the bonding electrons are provided by one of the atoms concerned. The most often quoted example of this is in the donation of the lone pair of electrons from the nitrogen atom of an ammonia molecule to a proton, to form the ammonium ion.In a similar fashion, the hydroxonium ion is formed by the donation of a lone pair of electrons from an oxygen atom to a proton in the following mechanism: H20 + H+ = H3O+I suppose that one of the ways to remember what happens in a dative bond is to look at it this way….. if you’re a guy and you’re out on a date….. you will probably end up paying for everything!
Es Muy Tutor Services - Metallic Bonding
In Metallic Bonding, there is a mutual attraction between the positive metallic ions and the delocalised valence electrons forming the lattice structure. Of course not all of the electrons of the element are delocalised, just the valence electrons leaving behind a positively charged ion which is bonded to its neighbour by the attraction of the delocalised valence electrons.In most, if not all chemistry books, you will hear this referred to as “a sea of electrons” and in my diagram shown to the left I have attempted to show this in a far greater scale than most chemistry textbooks do. The blue positively charged spheres are the metal ions and the smaller red dots formed the “sea” of electrons.It is as a result of this delocalisation that metals are able to conduct electricity, the delocalised nature of the electrons means that they can pass completely throughout the lattice, and of course this delocalisation means that no particular electron belongs to any particular ion.Metallic bonding is extremely powerful, this accounts for the relatively high melting and boiling points of most metals, in addition it accounts for other properties such as malleability (the ability to be beaten into thin sheets) and ductility (the ability to be drawn out into thin wires such as copper) as well as electrical conductivity.Delocalisation of electrons will be seen again when we come to look at organic chemistry, in particular the structure of the benzene ring.
Es Muy Tutor Services - Inorganic Chemistry
Es Muy Tutor Services - Organic Chemistry
Es Muy Tutor Services - Relative Atomic Mass etc
If you take a look at the picture to the left, you will see that a chlorine atom has 17 protons (as per the number shown at the bottom left). The number at the top left would normally be the atomic mass (which is the sum of protons and neutrons in the nucleus) but we cannot have half of a neutron, so where does the “.5” come from in this particular example?35.5 represents an average value, based on the fact that chlorine exists in a number of isotopes (an isotope being the same element, but with differing numbers of neutrons in the nucleus. If the number of protons were to change, so would the element).So how do we come to the value of 35.5? Well what we do, is take a look at the “relative abundances” of the different isotopes of the element. In the case of chlorine there are 2 isotopes, both having 17 protons of course but in one case the number of neutrons is 18 and in the other case the number of neutrons is 20. With this in mind you can see that the atomic masses would be 35 and 37 respectively. We need to come to some sort of agreement as to how we are going to represent this in the periodic table as it is clearly not practicable to show every isotope in the table (when you consider that elements such as tungsten have many more isotopes).We perform a calculation based on the relative abundance of the isotope as a percentage of the whole in relation to the particular atomic mass of that isotope, and how we do it is by a fairly straightforward formula which I will show you now (the funny looking E is actually “sigma” which means the “sum of”):
In the case of chlorine there are 2 isotopes, chlorine-35 and chlorine-37 with relative abundances of 75% and 25% respectively. If we enter these values into the above equation we will arrive at the value of 35.5. This is shown in the equation below and is true for any element with more than one isotope, that is it can be applied in the same way to arrive at the relative atomic mass of the element concerned.We need to also take account of a the concept called the “relative formula mass” which is simply the sum of all of the relative atomic masses of the atoms making up the compound, so for example if we considered nitric acid we will take hydrogen, nitrogen and three lots of oxygen, summing up 1+14+ (16×3) which of course makes sixty-three which is the relative formula mass for nitric acid. The formula for relative formula mass is as follows:So, going back to our initial example of nitric acid (and a new one, copper sulphate) we will have this:Percentage MassThis is a little bit tricky to work out. If we need to identify for example the percentage of a particular element in the compound, we use the following formula:So, let us suppose that we need to find out the percentage of oxygen in a sample of copper hydroxide, the first thing we need to do is to obtain the relative atomic masses of the elements concerned which are of course copper, oxygen and hydrogen.Cu = 63, O = 16 and H = 1, and from the formula of copper hydroxide we can see that we have two atoms of oxygen.Given that the relative formula mass is ninety-seven as shown, and we can see that we have two oxygen atoms which will add together to make thirty-two we can now work out the percentage mass using the above expression:Of course, the questions may be more complicated than this may require a little bit more effort, take for example the question below:Q. A mixture contains 40% by mass of the element magnesium. The only compound in the mixture containing magnesium is magnesium oxide. Calculate the mass of magnesium oxide in 45 g of the compound.A. The first thing we need to do is obtain the relative atomic masses of the elements concerned, so for example magnesium equals twenty-four, and oxygen equals sixteen giving a relative molecular mass for magnesium oxide of forty. In numbers:Mg = 24, O = 16 and MgO = 40.We are told that the compound contains 40% by mass of magnesium, and so we can say that 45 g of the compound will contain 40% of this element, that is:The relative molecular mass of magnesium oxide is also given as 40, so the percentage mass of magnesium in magnesium oxide is simply this:Now what I think you should do at this point is to go back over what we`ve already done so far. We we have established that our 45 g sample contains 18 g of magnesium, but the magnesium itself only accounts for 60% of the sample, therefore the remaining 40% must be the oxygen which makes up magnesium oxide. Therefore if 18 g accounts for 60% we have the following calculation to complete:
Es Muy Tutor Services - The Mole and Molarity
Yes, quite predictably I chose this little rodent cartoon to represent this particular topic. Almost certainly the first thing anybody would think of, the mole in the context of chemistry and science is a completely different thing.A “mole” of a substance (atoms or molecules) is the molecular weight of that substance (atoms or molecules) expressed in grams. So, for example a mole of the element carbon would be 12 g because the accepted relative atomic mass of that element is 12.This is not confined to atoms, compounds can also be described as having a molarity based on their relative molecular mass so for example the molecule of water (H2O) with a molecular mass of 18 would have a mass of 18 g per mole, similarly sulphuric acid (H2 S04) would have a mass of 98.1 g per mole.This can lead to some quite interesting mathematics because we now bring into the discussion the Avogadro number or Avogadro constant which is this:
This rather large number, usually taken to 2 places of 6.02 (plus the exponent part) is the number of atoms, molecules, or items per mole of the substance. So in our 12 g of carbon, 18 g of water or 98.1 g of sulphuric acid there will be this number of atoms or molecules. It is a simple case of basic mathematics to work out how many particles (for want of a better expression) would be in a mole or part mole:Q. How many atoms in 0.023 moles of xenon gas?A. And just out of interest, as the relative atomic mass of xenon is 131.3 we can see that 0.023 moles would weigh approximately 3.02 g
So now we can see that “the mole” tells us how many particles will exist in a mass of the substance equal to the relative molecular mass in grams. now to other expressions you’ve probably heard of are molarity and molality.Molarity - this is the concentration of a solution of the substance dissolved in a solvent and made up to 1 LMolality - this is the concentration of a solution of the substance but this time dissolved in a specific mass of the solvent, usually 1 kgLet’s take a quick look at a couple of molarity calculations, just to get out and read some of the (it has to be said, relatively basic mathematics) involved.
Q. How many moles of lithium chloride would be contained in a 25 cm³ solution at a concentration of 1.2 moles per litre?A. This is a fairly standard and basic calculation based on the formula below:So you can see is a fairly basic calculation of placing 25 where the volume is, and replacing the “concentration in moles per litre” with 1.2,, this will give us the following result:Don’t be fooled that all calculations are going to be this easy, they can get quite tricky, although usually the “tricky part” is in the setting up of the calculation, the actual execution of the calculation is usually quite straightforward.
Q. What mass of sodium hydroxide, in grams, will be needed to produce 50 cm³ of a 2M solution?A. This one isn’t quite as straightforward, because you need a little bit more information, read the question once more and you will see that you are being asked for a mass in grams.. Previously we have talked about “mass in grams” and relative molecular mass to work out the definition of mole, so it is clear that we are going to need the relative molecular mass of sodium hydroxide.Take a look at the periodic table and you will see the relative atomic mass of sodium is 23, oxygen is 16 and hydrogen is 1, so the relative molecular mass of sodium hydroxide must be 40 - NaOH = (Na = 23) +( O = 16) + (H =1) = 40)So a molar solution of sodium hydroxide would require 40 g, it therefore stands to reason that a 2 molar solution require 80 g but of course this is in 1 L (1000 cm³ or 1 dm cubed), in our question we only require 50 cm³ so the calculation will be this one:
Q. A solution of 4.08 g of compound X dissolved in pure water makes up a solution of 100 cm³ at a strength of 1.2M. Given this information calculate the relative molecular mass of compound X.A. See….. I told you they could get harder, but don’t panic because it isn’t as complicated as it might appear.We know that molarity involves solutions measured at usually 1000 cm³, but we are given 100 cm³ so the first thing that we need to do is work out how many grams of compound X would be in 1 L. This is fairly straightforward, it is simply 40.8 g (a simple power of 10 in this particular case).If our solution was one molar, we would have arrived at our answer of 40.8 but unfortunately the solution we have is 1.2M. We arrive at our relative molecular mass thus:An alternative way to work out the calculation, which with hindsight is probably simpler would be to ask ourselves this question “if 4.08 g of the compound makes 100 cm³ at 1.2M then how many moles does this volume contain?”.The answer is quite straightforward, it is 100/1000 of 1.2 = 0.12moles.Therefore, if 4.08 g = 0.12 moles then we need to divide both sides by 0.12 to find out how many grams in 1 mole, that is the relative molecular mass:
Es Muy Tutor Services - The Mole and Molarity
Ok, we will now take a look at some more calculations involving moles, and we already know that a mole is the relative atomic mass or relative molecular mass of the element or compound expressed in grams, so for example one mole of carbon will be 12 g, one mole of hydrogen bromide will be 81 g and one mole of carbon dioxide would be 44g.The law of conservation of mass states that in a chemical reaction atoms cannot be created or destroyed, they may be “swapped around a bit” but they can`t just appear or disappear. So basically this is telling us that the mass of the product must be equal to the sum of the masses of the reactants, in other words we lose nothing.With this in mind, let`s start taking a look at a few calculations.Q. Calculate the mass of hydrogen that reacts with 56 g of nitrogen to create 68 g of ammonia.A. As is always likely to be the case, the first thing we should do is show the balanced equation for the reaction. In this particular case the production of ammonia from hydrogen and nitrogen follows a reversible reaction utilising the Haber Process, but for the sake of this question we will pretend it just goes the one-way:We can see from this balanced equation that one mole of nitrogen reacts with three moles of hydrogen to produce 2 moles of ammonia. we can therefore say that one mole of nitrogen is 28 g and this would react with 3 moles of hydrogen which would be 6 g to create 34 g of ammonia.It is probably no coincidence that the 56 g of nitrogen equates to 2 moles, a “doubling up” of amounts which should tell you that 56 g of nitrogen will in fact react with 12 g of hydrogen to produce the given 68 g of ammonia.Let’s take a look at another example:Q. How many moles of water will be created if two moles of methane gas is burned in excess oxygen?A. First of all, we must define ‘excess’ as being enough of a particular reactant to ensure that the reaction goes to completion, so in this case there will be more than enough oxygen to make sure that the methane is completely burned.Let us start, as is usually the case, by writing out a balanced equation for the reaction we are interested in, that is the total burning of methane in oxygen to produce carbon dioxide and water:Looking at the equation we can see that 1 mole of methane will react with 2 moles of oxygen to produce 2 moles of water and 1 mole of carbon dioxide, so you can simply deduce that doubling the amount of methane to the given 2 moles will double the quantity of water produced, ergo 4 moles.We can also, given the correct information, deduce the balanced equation if we know the masses of reactants and products formed, to do this we will also need to know the relative atomic masses and relative molecular masses of the components concerned.Q. 4.6 g of sodium reacts with 1.6 g of oxygen to form 6.2 g of sodium oxide, given the relative atomic mass of sodium as 23 and that of oxygen as 16, derive the balanced equation for this reaction.A. This is a calculation that we must do in stages, the first parties to calculate the molar quantities of the reagents and products. The relative molecular mass of sodium oxide is 62.Step 1 - calculate the molar masses of sodium, oxygen and sodium oxide:Step 2 - divide these molar quantities by the lowest molar quantity to obtain the ratios, that is we divide 0.2 by 0.05, then we divide 0.05 by 0.05 and finally we divide 0.1 by 0.05 to obtain our third ratio:Step 3 - we now substitute back our ratios to the elements concerned, that is the ratio 4 belongs to sodium, 1 belongs to oxygen and 2 belongs to the product sodium oxide. Our ballast equation is therefore:
Es Muy Tutor Services - The Mole and Molarity
These questions concerning deducing the balanced equation can be quite tricky, they can also carry a lot of marks so they are well worth getting your head round, and with this in mind, as if by magic is another one:Q. 1.2g of a hydrocarbon X is burnt completely in 4.48 g of oxygen to produce 3.52 g of carbon dioxide and 2.16 g of water, given this information deduce the balanced equation for the burning of this hydrocarbon.Mr (X) = 30, Mr (O2) = 32, Mr (H2O) = 18 and Mr (CO2) = 44A. we are given the relative atomic masses of the reactants and the products, we are also given the starting and finishing amounts, that is we are given the amount of hydrocarbon and oxygen (the reactants) and the amounts of the products (carbon dioxide and water) so we can start to look at molar ratios.The number of moles of each reactants or product is given below:This gives us the mole amounts of the reactants and products, but what we need to do now is evaluate the ratios, we do this as before by choosing the lowest value of the four and dividing all results into it to obtain integer ratio numbers:the lowest value is of course 0.04:You should be able to see from this that 1 mole of the hydrocarbon ‘X’ reacts with 3 1/2 moles of oxygen to produce 3 moles of water and 2 moles of carbon dioxide, the balanced equation can therefore be written like this:You should note that it is perfectly acceptable to show fractional amounts of any reagent, or product, so for example it is perfectly okay to say “3 ½ O2”. You may be tempted to double everything up to get rid of the fraction, this is also perfectly okay but not necessary.So can we identify the hydrocarbon from this equation?….. Well in fact we can, if you take a look at the equation you can see that two atoms of carbon exist on the right-hand side and six atoms of hydrogen exist on the right-hand side. We know that a hydrocarbon only consists of carbon and hydrogen anyway, so we can quickly deduce that this particular hydrocarbon ‘X’ is in fact Ethane, as ‘X’ must have provided both carbons and all six hydrogens:
Es Muy Tutor Services - The Ideal Gas Equation
Produced by combining 3 other laws concerning the behaviour of gases, Boyle’s law, Charles’s law and Gay-Lussac’s law, the ideal gas equation or otherwise known as the ideal gas law has the following representation:Each of the components of this law are defined thus:
pressure measured in Pascal (Pa)
volume measured in cubic metres (m^3)
number of moles of the gas (number)
the universal gas constant measured in joules per kelvin per mole
temperature measured in Kelvin)
In questions involving this law, you will be given 4 out of the 5 values and be asked to calculate the 5th. Chemistry examiners have universal imaginations so your best keyed to expect the most unusual, complex -looking calculation possible. All you have to remember is to transpose the equation accordingly so that it ends up in terms of the unknown quantity (whatever that is) and then insert the values you are given, but: Always make sure that your units are as shown, you may need to make some conversions if you are given different units or powers such as kilo, mega et cetera.For example you may be given kilopascals, or degrees Celsius, or decimetres cubed. If you miss these are simply enter the figures that you have been given into the equation it is likely that your answer will be severely compromised (i.e. wrong).Let us now take a look at a couple of examples:
Q1. How many moles of hydrogen gas would be contained in 0.06 m³ at a temperature of 283 K and 50 kPa?A1. Taking the universal gas constant “R” as 8.31 J per kelvin per mole (as in the definition/description above) you should be able to see that you have now been given 4 out of the 5 items and you are required to manipulate the gas equation in terms of ‘n’.So: Although we have now manipulated the equation and are ready to enter the values that we’ve been given, you must always check that the units are compatible, that is the units you’ve been given the question are compatible with those that the equation expects in this particular case we have to convert kPa to Pa, by multiplying the given value by 1000. Once we have done this we are now ready to plug in the values into the equation:
Q2. Given the pressure of 110,000 Pascal, 0.05 moles of hydrogen gas occupies 1200 cm³. What is the temperature in degrees Celsius?A2. We can see that we have quite a few conversions to do before we can go anywhere near the equation. First of all we need to convert the volume into cubic metres and we do this by dividing 1200 by 1,000,000 giving us a value of 0.0012 m³. We do have another conversion but that is of course the temperature, the aim of the question is to calculate that first in degrees Kelvin.Rearranging the ideal gas equation in terms of temperature gives us the following:Plugging our known values into this equation gives us the following result:If we now subtract 273 from this result we arrive at a temperature in degrees Celsius, which is 45°C
Q3. 10 moles of oxygen gas at a temperature of 298 K holds a volume of 0.1 m³ - calculate the pressure in kPascal to 3 significant figures.A3. Once again we need to keep an eye on our units, remembering to make any corrections or conversions before or after the calculation to make sure that we present the answer in the form required.The first thing we have to do of course is rearrange the ideal gas equation so that it is presented now in terms of pressure:We can see that the values that we’ve been given for the other variables are already in the appropriate units, so we have no actual conversions to make (at this stage anyway). We “plug-in” our given values and arrive at the following expression:An interesting ‘aside’ to this calculation would be to see what the pressure would be at a lower temperature. The pressure given of 298 K is in fact the standard to signify “room temperature” but the standard in terms of STPis in fact 0°C or 273 K. Keeping all of the other values exactly the same we will now swap into the equation this lower temperature to see what effect it will have on the pressure:We can see that as we lower the temperature, the pressure also decreases. If you think about it, this makes common sense because as the temperature decreases, so does the vibrational energy of the atoms or molecules making up the gas. In other words as the temperature drops the atomic/molecular vibration slows down. At 0° Kelvin the pressure will also be 0 (if you look at the equation) because this is the temperature at which (theoretically) all motion ceases and as it is molecular motion that causes the molecules to crash into the walls of any container (and therefore exert a pressure upon the container) it stands to reason that if the molecules aren’t moving they won’t exert any pressure at all.
Es Muy Tutor Services - Balancing Chemical Equations
It is a fundamental law of the sciences that matter cannot be created or destroyed. what you start off with in a chemical reaction has to be what you end up with, although it may be in a different format, bonded differently forming different compounds or it may simply have changed state or physical shape/appearance. The fact of the matter is, if you start off with X atoms of an element, you have to end up with the same number of atoms but they may be in a different format (that is forming different compounds to the ones they started off in). So sayeth the law of conservation of mass.
There is a certain skill to balancing equations, and although there are many calculators available on the Internet which will give you the answers it is likely that you will not be privy to such a calculator when you’re sitting in front of your examination paper, so the best thing you can do is learn how to do this “the hard way” (but in fairness it’s not really that hard although some can be tricky).Let us take a look at the fairly straightforward equation shown above. This tells us that nitric acid reacts with sodium hydroxide to produce sodium nitrate and water. To make sure the equation is balanced we have to look at each atom on the left-hand side and make sure that it is represented on the right-hand side in the same numbers.On the left-hand side we have 2 hydrogen, 1 nitrogen, 4 oxygen and 1 sodium atom(s)on the right-hand side we have 1 sodium, 1 nitrogen, 4 oxygen and 2 hydrogen atom(s)Quite clearly this equation is already balanced, but of course we not always this lucky. let’s take a look at another example, and this time I have chosen a reaction where the products and reactants are not balanced straightaway.the reaction between sulphuric acid and potassium hydroxide produces potassium sulphate and water, because this is a classic “acid plus base equals salt plus water” type of reaction. You should be able to see straight away that the potassium atoms are not balanced because we only have one K on the left-hand side but on the right-hand side we have 2. Let’s tally up our atoms to see exactly how bad the situation is:On the left-hand side we have 3 hydrogen, 1 sulphur, 5 oxygen and 1 potassium atom(s)On the right-hand side we have 2 hydrogen, 1 sulphur, 5 oxygen and 2 potassium atom(s)We are missing some hydrogen and some potassium, so we need to “juggle” with a few numbers to make this all work out. Because we have 2 potassium on the right-hand side, better start by making the potassium hydroxide on the left-hand side 2 molecules instead of one:On the face of it we think we might have cracked this, because we now have 2 potassium atoms on the left and on the right, but in doing this have we “upset the apple cart” elsewhere?:On the left-hand side we now have 4 hydrogen, 1 sulphur, 6 oxygen and 2 potassium atom(s)On the right-hand side we now have 2 hydrogen, 1 sulphur, 5 oxygen and 2 potassium atom(s)This doesn’t seem to be going very well does it? previously we were okay with oxygen, but now we appear to have messed up our numbers there. Okay let’s carry on:If you take a look at the deficit, we had 2 too many hydrogen atoms on the left-hand side, and 1 too many oxygen atoms on the left-hand side otherwise everything else was okay. Now we know basically that 2 hydrogens and 1 oxygen will go together to make a water molecule, but we already had one of those….. But there’s nothing to say we can’t have another one so if we add a further water molecule to the right-hand side as shown above we find that our equation now balances:On the left-hand side we now have 4 hydrogen, 1 sulphur, 6 oxygen and 2 potassium atom(s)} - success!On the right-hand side we now have 4 hydrogen, 1 sulphur, 6 oxygen and to potassium atom(s)} - everything now balances as we expectedNow we will take a look at a slightly more complicated one:Straight away we can see that in the reaction between sodium carbonate and hydrochloric acid, producing sodium chloride, water and carbon dioxide the equation is already off-balance because we have 2 sodium atoms on the left-hand side but only 1 on the right. Obviously, as part of our balancing exercise will have to put that right..So:On the left-hand side we have 2 sodium, 1 carbon, 3 oxygen, 1 hydrogen and 1 chlorine atom(s)On the right-hand side we have 1 sodium, 1 chlorine, 2 hydrogen and 3 oxygen atom(s)We know straightaway that we have to increase the number of sodium atoms, but if we double the number of NaCl molecules by adding 2 to the right-hand side we throw off the number of chlorine atoms on the left-hand side. In this case let us try to double up straightaway the sodium chloride and hydrochloric acid molecules in one go:Let us now count up our “atomic inventory” to see where this has led us to:On the left-hand side we now have 2 sodium, 1 carbon, 3 oxygen, 2 hydrogen and 2 chlorine atom(s)} - success!On the right-hand side we now have 2 sodium, 2 chlorine, 2 hydrogen, 3 oxygen and 1 carbon atom(s)} - as before, everything now balances correctlyFinally then, to round off this particular set of demonstrations, let us take a look at the burning of butane gas in excess oxygen (when we say excess of anything we mean that the reaction will be taken to completion):The burning of an alkane hydrocarbon produces carbon dioxide and water only. This is the reaction (the unbalanced version):This is the balanced version - I want you to try to study this and work out for yourself how I have arrived at this
Es Muy Tutor Services - Ionic Equations
This is another exercise in the production and balancing of chemical equations, but the subject of “ionic equations” refers to those processes taking place in solution. It is only the reacting particles and the products formed that we are interested in, other items will be deleted from our final equation.We must also remember to show our “state symbols” in the equations, this is something that has not been added yet to this website but it is important in many facets of chemistry to show the state of the reactants and of the products in any equations that we might create.Let us use our sodium carbonate and hydrochloric acid example from one of our previous pages. As you can see (above) I have replicated the balanced equation, showing where additions needed to be made to make it balance. What I haven’t done here is show the state symbols so, I’m going to do that now:I think you can probably work out that “AQ” means aqueous, “L” means liquid and “G” means gas. One that isn’t shown here would of course be “S” which unsurprisingly would mean solid.What we need to do next is to show the “ionised” state of our reactants and products:You should note that we only “show as ionised” those products and reactants in aqueous (aq) state. The water and carbon dioxide are liquid and gas respectively and so we don’t need to show these. Now we are not interested in those items that appear on both sides of the “=” sign…..all we need to know is what’s left:Now let’s take a look at another, slightly more complicated example:The reaction between sodium phosphate and calcium chloride producing sodium chloride and calcium phosphate proceeds according to the following equation:It should be fairly obvious to you by now that this equation is not balanced, because we have 3 sodium atoms on the left and only 1 on the right, in addition we only have 1 calcium atom on the left but we have 3 on the right so there is a little bit of work to be done before we can go any further.The balanced version of the equation is this one:Notice that this time one of the products is a precipitate, that is it does not stay in solution and so does not need to be shown in its ionic form. Let us now “decompose” our ionic reactants as per the example that we went through previously:And just as we did previously, we delete those entities which exist on each side of the equality sign, this leads us to our completed ionic equation which is shown below:
Es Muy Tutor Services - Calculate Mass of Reaction Product
When chemical reactions take place, one or more products of those reactions will be produced. Sometimes it is necessary to be able to calculate how much of a particular product is produced from a certain amount of starting reactants. In another page I will talk about “atom economy” which is away in which chemists calculate the efficiency (both in a chemical and financial context) of the reaction, but in this particular page I will just talk about calculating masses of products.Let’s take a look at our first reaction, the burning of a certain amount of iron in air.Q. What mass of iron (iii) oxide will be produced upon burning 28 g of iron in air?A. The first thing we should do every time is to write down the balanced equation for the reaction that we are looking at, in this particular case we are looking at the oxidation of iron:From this balanced equation we can see that 2 moles of iron reacts with one and a half moles of oxygen to produce one mole of iron oxide. The atomic weight of iron is 56 and we are told that our starting amount was 28 g. We can work out the number of moles of iron this way:You should be able to see, by looking back at the equation that 2 moles of iron produces ultimately one mole of the oxide, i.e. we have a 2-to-1 ratio, it therefore stands to sense that 0.5 moles of iron will produce 0.25 moles of the oxide.What we need to do now is work out how many grams of iron oxide this represents. We know that one mole of any element or compound is the relative atomic mass/relative molecular mass expressed in grams so what we need to do now is work out the relative molecular mass of iron oxide.Using a periodic table you will be able to work out the following:Fe = 56, O = 16 so the molecular mass of iron oxide is (2 x 56) + (3 x 16) = 160.Since one mole of iron oxide would weigh 160 g we can work out that 0.25 moles would weigh 40 g, and this is the answer that we require.Q. 3.4 g of zinc metal was dissolved in excess hydrochloric acid producing zinc chloride and hydrogen. Calculate the mass of zinc chloride produced.A. Fir1st of all….our balanced equation:From the equation we can see that 1 mole of zinc will react with 2 moles of hydrochloric acid to produce 1 mole of zinc chloride and 1 mole of hydrogen gas. Given the fact that we started with 3.4 g of zinc we can work out exactly how many moles this was.We can see that one mole of zinc produces 1 mole of zinc chloride, in other words the reaction is 1:1The reaction therefore produces 0.052 moles zinc chlorideThe molecular weight of zinc chloride is 65.4 + (2 x 35.5) = 136.4 and so the mass of zinc chloride produced in this reaction is 0.052 x 136.4 = 7.1 g ( 2d.p.)Q. A sample of ethene gas is burned in excess oxygen producing carbon dioxide and water. At the end of the reaction it is found that 15 g of water has been produced. What was the mass of ethene gas at the start of the reaction?A. Once again, the first thing that we do is write out a balanced reaction:(it is worth noting here that sometimes ethene can be referred to as ethylene, but this is an old name and the IUPAC convention prefers ethene)We can see from our balanced equation that one mole of ethene will react to produce 2 moles of water. Since we end up with 15 g of water, the first thing that we need to do is to calculate how many moles that actually is:The molecular mass of water is 18………Note that the dot above the number 3 means “recurring” in other words the 3 moves to the right to infinity but we normally round to 2 or 3 decimal places.because we have this amount of water, now measured in moles,, we know that we must’ve started with half of that amount of moles of the gas because of the 1 to 2 ratio of the reactant to product. We therefore started with 0.415 moles of ethene.The molecular mass of ethene is (2 x 12) + (4 x 1) = 28 so the mass in grams must have been 0.415 x 28 = 11.7 (1 dp).Q. Hexane is a straight chain hydrocarbon with the formula C (6) H (14) and is a liquid which boils at approximately 342K. A sample of Hexane is cracked in a chemical process at 400K and the pressure of 100 kPa to produce butane and ethene gases. The mass of butane produced is 3g. Taking the gas constant ‘R’ to be 8.31 J per kelvin per mole calculate the initial volume of hexane gas.A. As is always the case, we produce our balanced equation to show the reaction involved:We can see from the reaction that one mole of hexane breaks down into one mole of butane and one mole of ethene. We are told in the question that the mass of butane formed is 3g so we need to work out what this is in moles.There is a reason why I have left so many decimal places, you should always use full calculator accuracy until you arrive at your final answer, which you will then present in the format asked for by the question, i.e. to a certain number of significant figures or decimal places. If you “round” too early you will affect the accuracy of your final answer.Taking into account the equation, we can see that the “cracking” process produces the same number of moles of butane and ethene, and these must have come from the same number of moles of hexane. We now know the number of moles of hexane,, and the gas constant, temperature and pressure so we can now calculate the volume of our starting material hexane according to the “ideal gas law”:This is approximately 1719 cm³It should be remembered in all calculations - “NEVER ‘round’ ‘till the answer is found!”
Es Muy Tutor Services - Titrations (Volumetric Analysis)
You will find that there is an incredible amount of mathematics involved in chemistry, and indeed other sciences such as physics and biology. It is important that you are able to manipulate data provided, this is especially relevant in the case of titrations, formerly known as volumetric analysis. Volumetric analysis involves adding a substance of a known concentration in a known amount, to a known amount of a substance whose concentration is not known, enabling the unknown concentration to be calculated according to the amounts combined and the chemical reactions taking place.The first example we will look at is an acid/base neutralisation reaction where a known quantity of a known strength base is used to neutralise a known quantity of an unknown strength acid:Q1. 40 cm³ of 0.25 molar potassium hydroxide solution is used to neutralise 22 cm³ of dilute nitric acid. Using this information, calculated the molarity of the nitric acid.Q2. As you are probably aware by now, the first thing that we need to do is write out a balanced chemical equation to represent the neutralisation which takes place.We can see that 1 mole of potassium hydroxide will react with 1 mole of nitric acid to produce 1 mole of potassium nitrate and 1 mole of water. We are told that our potassium hydroxide solution has a strength of 0.25 moles per litre, we are also told that we were required to use 40 cm³ of it, so what we need to establish now is exactly how many moles of potassium hydroxide we in fact used.Because this reaction occurs on a one-to-one ratio, we can see that as well as having used 0.01 moles of potassium hydroxide we must also have used 0.01 moles of nitric acid.So our 22 cm³ of nitric acid contained 0.01 moles, the molarity of the nitric acid solution therefore must have been:It is possible of course that the examiners may give you the strengths of the solutions but one of the quantities will not be provided, it is therefore your problem to calculate the unknown amount. I will run through another one-to-one ratio equation/calculation to show you how this would be achieved.Q2. 37.3 cm³ of 0.42 molar potassium hydroxide solution is titrated against a solution of 1.1 molar ethanoic acid. The neutralisation occurred after the addition of X cubic centimetres of the acid. Using this information calculate the value of X in cubic centimetres and quote your answer to one decimal place.A2. Predictably, we must first write out of the equation (balanced) for the reaction:Working from left to right we can see that potassium hydroxide reacts with ethanoic acid (sometimes referred to as acetic acid) to produce potassium ethanoate (potassium acetate) and water. It can be seen that one mole of potassium hydroxide reacts with one mole of the acid, hence the equation is one-to-one and already balanced.We are told that we have used 37.3 cm³ of 0.42 molar potassium hydroxide, the number of moles in this amount of solution is as follows:The reaction is one-to-one, so it is understood that the number of moles of acid neutralised is also 0.015666.The strength of the acid is given as 1.1 molar therefore we must use the following volume in cubic centimetres in this reaction:
Es Muy Tutor Services - Molecular Formul(as)(ae)
We know that compounds are made up of atoms and molecules, for example we probably can write out the molecular formulae, for example of the first few alkanes but did you know that there is a bit more to it than that?Let us take the example one of the simpler organic carboxylic acids, one we are all familiar with as we probably all using regularly with its ionic companion NaCl when we eat our .chips from the chip shop. I am of course talking about ethanoic acid, or some people might recall it, acetic acid.The molecular formula for this acid is generally written this way:A quick “recce” will tell you that this is an organic compound with 2 carbon atoms, 4 hydrogen atoms and 2 oxygen atoms, so another way of writing it could be this way:Although it’s probably unlikely that you would recognise it as ethanoic acid from this particular representation of its molecular formula.You may have heard of the expression “empirical formula” and you may be asked to produce one in an exam. The empirical formula of a substance or compound is shown as in the second example above but with the lowest possible ratio of atoms, so for example our second representation above would become:it is highly unlikely that you will be able to establish what the compound is from its empirical formula, unless you are given some other information besides. The reason I say this is because the sugar glucose has the same empirical formula as shown above, but I’m sure you’ll agree that ethanoic acid and glucose are widely dissimilar substances, and their empirical formula is probably where the similarities end!So how else can we display molecules?…. will of course there is a clue in the word “display”, because there is also such a thing as a displayed formula which is a molecular formula which is spread out on the page to give a sort of visual representation of the way the atoms are bonded together, so for example using our acid above we may consider the “displayed formula” to be something like this:Although this is somewhat closer to reality, we are still not quite there. The atoms are indeed bonded to each other in the way that I have shown, that is the 3 hydrogen atoms are bonded singly to the carbon which is itself singly bonded to its neighbouring carbon, the first oxygen atom is double bonded to the carbon and the second one is singly bonded and in turn bonded to a single hydrogen atom (this particular hydrogen atom, bonded singly to the oxygen is the one which is readily given up to create many “ethanoic acid derived compounds” such as salts and esters.Perhaps the favourite (indeed my favourite anyway) molecular representation is the three-dimensional representation which can be created by the using a ball and stick modelling set, or a suitable piece of software which can render three-dimensional molecules. This particular acid, rendered using the molecular drawing package ChemDoodle is shown below:It is customary to occasionally refrain from showing the hydrogen atoms, usually because there are so many of them, and use what is called a “skeletal” formula. This becomes very relevant when we are dealing with large molecules such as those containing one or more ring structures and multiple bonds. To show every single hydrogen atom would make the molecule very difficult to understand and hugely difficult to draw in the first place, so we tend to omit the hydrogen atoms but we must still show other atoms and functional groups.Let us take for example the below structural formula:Well, this doesn’t tell us a great deal but think about what I said with regard to showing hydrogen atoms and other groups. The junction at the point where each line touches the next represents a carbon atom, we know that carbon is tetravalent and so we would expect to find hydrogen atoms sitting at each intersection (three hydrogen atoms on each of the carbons at the end of the chain and two on each intermediate carbon. We know that the can’t be any other atoms involved or they would have been shown.Let’s put those hydrogens into place now, and they should give us a bit of an idea of the molecule’s identity:This looks very much like…….Otherwise known as…….This is octane, a major constituent of petrol. Even though this is quite a simple hydrocarbon molecule (they do get far more complicated, believe me!) You can see that it would be quite tedious to have to write in all of those hydrogens, even in this simple case there are 18 of them. This is why the skeletal formula is so popular when we are dealing with very complicated structures.
Es Muy Tutor Services - Enthalpy, Entropy & Free Energy
This particular section could get quite long, it is a particularly advanced section as the topics that will be discussed are usually found on the GCSE A-level syllabus here in the United Kingdom. We are going to talk about Enthalpy, Entropy and free energy. The first two subjects will talk about changes during a chemical reaction and the third talks about determining the feasibility of the reaction, i.e. whether it will take place or not.Before we start talking about these, it will be as well to give definitions of what they are:Enthalpy - this is a measurement of the energy in a thermodynamic system, it is the thermodynamic quantity equivalent to the total heat content of the system. There are many different types of enthalpy, such as enthalpy of combustion, enthalpy of formation and these are all to do with energetic changes in the system during, or as a result of, a chemical reaction.Enthalpy is measured in kilojoules per mole.Entropy - this is usually defined as the “degree of disorder” in the system, that is the more disordered the system is the greater the entropy. Entropy is measured in Joules per Kelvin per mole.Free Energy - the energy that can be converted into work at a constant temperature and pressure throughout a system.Free Energy is measured in kilojoules per mole.We will first talk about enthalpy, which is basically a measurement of energy. When chemical reactions occur there will be a chang in energy, and in thermohemistry and energetic terms this is known as an “enthalpy change”. Enthalpy change is usually denoted as a triangle with a H next to it, known as “Delta H” and is the heat transferred in the reaction measured at a constant pressure. Let’s take a look at a couple of symbols now so that we can get our heads around this before I carry on to the more complicated bits:Enthalpy changes are measured in kilojoules per mole, and energy changes can be positive or negative. If, during a chemical reaction energy is taken in (meaning that the products of the reaction contain more energy than the reactants) then the enthalpy change will be positive and this is said to be an endothermic reaction.In any reaction where the enthalpy change is negative, that is the products contain less energy than the reactants, then for this to be the case energy must have been given out to the surroundings, in such cases this is said to be an exothermic reaction.So where does all this energy come from? Well….chemical reactions involve the making and breaking of bonds between atoms in molecules. The overall change in enthalpy depends therefore on the bonds made (the products) and the bonds broken (the reactants). If the energy involved in breaking bonds is more than that involved in making them, then the reaction will need to take in energy from the environment and will therefore be endothermic . If the reverse applies, and the energy involved in breaking the bonds is less than that given out when new bonds are made, there will be an excess of energy which will be given out to the environment in an exothermic reaction.We talk about bond enthalpy, in days gone by (in older chemistry textbooks) you would have seen this referred to as bond energy, and despite the fact that the name is changed, the meaning is still the same. Atoms in molecules are held together by very strong covalent bonds and it takes energy to break these bonds. Bond enthalpy is always positive as energy is needed to be applied to break the bond, i.e. bond breaking is an endothermic process.Chemistry textbooks usually have tables of bond enthalpy values, but these will vary depending on the environment of the bond, or in other words the energy required to break a particular bond will depend on the environment that that particular bond finds itself in.For example, let us consider the water molecule which contains two O-H bonds, you would be forgiven for thinking that the energy to break them both would be identical because the molecule is symmetrical, but once one of the bond is broken it is slightly easier to break the second one because when the first hydrogen leaves the molecule, it does so as a proton, leaving behind its electron and what remains is a hydroxide ion. This hydroxide ion contains an overall negative charge, and electron repulsion makes the remaining OH bond a little bit less stable than the first one was:Picture the water molecule on the left-hand side. During the breaking of the right-hand OH bond, the hydrogen atom’s electron is transferred to the remaining hydroxide ion. Electron repulsion within the remaining OH one means that this one will break easier than the first. This accounts for the fact that the first bond requires 492 kJ per mole whereas the second bond requires only 428 kJ per mole. You will therefore often find an average, or ‘mean’ value quoted, which in this particular case is 460 kJ per mole but this may vary between texts, as there are many other types of OH bond and values quoted for the ‘mean’ will include these other examples (such as OH on alcohols, carboxylic acids etc).(In fact, in the chemistry data book by Stark and Wallace, second edition dated 1982 (ISBN - 0-7195 - 3951 - X ) the average bond enthalpy for the OH bond is in fact 463 kJ per mole).There are different types of “Delta H” or standard enthalpies, there is the standard enthalpy change of formation, combustion and reaction and these will be defined on the following page.
Es Muy Tutor Services - Enthalpy, Entropy & Free Energy
The standard enthalpy change of formation - is the enthalpy change when one mole of a compound is formed from its elements in their standard states, under standard conditions.The standard enthalpy change of combustion - is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions, with all reactants and products in their standard states.The standard enthalpy change of reaction - is the enthalpy change when a reaction occurs in the molar quantity shown in the chemical equation, under standard conditions with all reactants and products in their standard states.
We spoke previously about making bonds and breaking bonds, and considering that energy cannot be created nor destroyed it stands to reason that the energy you put into making bonds and the energy that you put into breaking them will determine the Exo or Endo thermic nature of the reaction. To try to demystify this last sentence, let’s take a look at a simple example:let us take a look at the standard enthalpy of combustion of but-1-ene:This alkene burns in oxygen to produce carbon dioxide and water, according to the following balanced equation.This will involve the making, and breaking of many bonds. Firstly you should be able to see that it will involve the breaking of 8 C-H bonds, 2 C-C bonds and 1 C=C double bond, and not forgetting the oxygen to be used it will involve the breaking of 6 O=O double bonds.The bonds that will be made will be those putting together the carbon dioxide and water products, now the bonding in carbon dioxide involves 2 double bonded C=O associations per molecule of carbon dioxide and 2 single bonded O-H associations per molecule of water. In total we will produce 8 C=O and 8 O-H bonds in our products. The difference between the energy required to break all of the bonds and the energy given out when the new bonds are formed will determine whether the reaction is endothermic or exothermic (we would of course expect it to be exothermic).In any good data book you will be able to obtain the mean bond enthalpies, those that we require for this particular reaction given below:We now need to do a little bit of adding up to work out how much energy is required to break all of the bonds that are needed to be broken in this reaction.8 C-H bonds = 8 x 413}2 C-C bonds = 2 x 347} a total of 7598 kJ per mole1 C=C bonds = 612}6 O=O bonds = 6 x 498}Now we take a look at the bonds that are made:8 C=O bonds = 8 x 805} a total of 10,152 kJ per mole8 O-H bonds = 8 x 464}The total enthalpy change is the difference between the energy absorbed to break the bonds and the energy released in the making of the new bonds:And as we expected, a quite strongly exothermic reaction.
Es Muy Tutor Services - Reaction Rates and Arrhenius Eqn
This particular set of pages will be quite advanced, taken mainly from the UK A-level syllabus we will be talking about reaction rates, reaction constants and the Arrhenius equation.So what do we mean by reaction rates? Well quite logically and as you would probably have expected it is all to do with how fast a reaction takes place, this is governed by many factors such as concentration of reagents, activation energies and temperatures to name but a few.The rate of reaction can be determined by the speed at which the reactants are used up and/or the products are formed. There are many ways to determine this, depending on what the product actually is, for example if the product (or one of the products) is a gas then we can determine the rate of the reaction by capturing the gas using a device such as a gas syringe and taking volume readings at fixed intervals of time. Alternatively we can take a reading of the mass of the reactants plus the reaction vessel at a time equal zero, and by taking regular readings of the mass we can determine the rate of the reaction (if for example a gas is being given off such as hydrogen or carbon dioxide we can take readings of the decreasing mass of the reaction vessel at fixed intervals of time).When this set of pages eventually reaches the Arrhenius equation we will look at reaction rates and rate constants, but before that we will take a look at a typical laboratory type experiment involving sulphuric acid and sodium carbonate but instead of measuring the rate of the reaction, we will, by looking at some recorded data, take a look at the effect that the reaction has on the concentration of the acid because by definition if the acid is being used as a reagent its concentration will decrease as the reaction progresses.This question is taken from the UK A-level syllabus, and although it involves a considerable amount of mathematics, this is not particularly complicated maths and we will touch upon the ideal gas equation during this calculation. If you want to refresh your memory on the ideal gas equation click on the link >> here << but remember to come back to this page by using your web browser’s “back” button.OK, assuming that you did flick back.. Welcome back 8-)Q. 100 cm³ of 1M sulphuric acid is reacted with 15 g of sodium carbonate. The gas given off is carbon dioxide and the volume of carbon dioxide is measured at fifteen second intervals up to and including seventy-five seconds. Our task is to calculate the concentration of sulphuric acid at each measurement point.A. As is always the case, the first thing we need is a balanced reaction:We can see straight away that one mole of sodium carbonate will react with one mole of sulphuric acid to produce one mole of sodium sulphate, one mole of carbon dioxide and one mole of water.We are told, that at the start, we have 100 cm³ of 1M sulphuric acid, so it should be quite easy to deduce that we do in fact start with 0.1 moles of sulphuric acid as we have a tenth of a litre of a one molar solution.At fifteen seconds - the reaction has produced 3.2 cm³ of carbon dioxide gas. Unless you are told otherwise, you must assume that the pressure and temperature of the reaction are set at RTP, which is 100 kPa and 298 K. How many moles of carbon dioxide gas have we produced at this point? This is where the ideal gas equation comes in, because we have all of the necessary information to work out ‘n’:Given the pressure of 100 kPa, the volume of 3.2 cm³, the gas constant of 8.31 J per kelvin per mole and a temperature of 298 K we can convert the information we have into suitable format (remember what I always talk about --> Units) to calculate the number of moles of carbon dioxide created at this time point:Because the reaction is one-to-one in as much as one mole of carbon dioxide will be produced from one mole of sulphuric acid, we conclude that if we have created the above amount of carbon dioxide after the first fifteen seconds then we have to use the same amount of sulphuric acid to do this. So the number of moles of sulphuric acid left after fifteen seconds is this:But remember that we started off with 100 cm³ of acid so the actual molarity of the acid would be ten times this amount or 0.9987 MThis therefore, is in molarity of the acid after fifteen seconds.On the next page will continue with the calculation by working out the molarity of the acid at the next time point of thirty seconds.
Es Muy Tutor Services - Reaction Rates and Arrhenius Eqn
At thirty seconds - we would have generated 6.51 cm³ of carbon dioxide gas. We now use the ideal gas equation once again to work out the number of moles of carbon dioxide and hence how much more of the sulphuric acid has gone in the reaction.It must be remembered that this is the number of moles of carbon dioxide that has been created since the start of the reaction, not since the last time reading of 15 seconds. So it stands to common sense that this will also represent the total reduction in the strength of the acid.The number of moles of sulphuric acid therefore at 30 seconds:And since we initially started off with 100 cm³ we need to multiply this result by 10 to get the actual molarity of the acid at t = 30s Which is of course 0.9974 Mthe same calculation will be applied at 45 seconds, 60 seconds and finally at 75 seconds producing acid strengths of 0.996 M, 0.995 M and finally 0.994 M.We can see therefore, quite understandably, that as the reaction progresses, the acid will become weaker and weaker.These types of problems are not difficult when it comes to the mathematics, but you need to approach them logically and make sure you keep your wits about you, also making sure that you do not “round” your calculator results until the very end, if you “round” too early, you may compromise your answers.Just to make sure that we got this, let us take a look at another example:Q. the reaction of zinc sulphide with hydrochloric acid proceeds according to the equation below, to produce aqueous zinc chloride and hydrogen sulphide gas. This time the reaction was tracked by taking mass readings of the reaction vessel plus contents at 10 second intervals up to 60 seconds, with the following results: The balanced equation is as follows:From the balanced equation we can see that one mole of zinc sulphide will react with two moles of hydrochloric acid to produce one mole of aqueous zinc chloride plus one mole of hydrogen sulphide gas (a reaction which must be performed in a fume cupboard, because of the highly toxic nature and revolting smell of the gas, that of rotten eggs).We are also given a starting value for the amount of acid, we have 250 cm³ of 1.5 M hydrochloric acid.A. Looking at our table of results, we can see that at the 10 second stage we have lost 0.6 g and this must be from the production of hydrogen sulphide gas which will escape from the vessel. The relative molecular mass of hydrogen sulphide is 34, so that would be 34 g of hydrogen sulphide gas in one mole. We have in fact only produced 0.6 g, therefore the number of moles of gas produced would be this:Now, if we look back at our balanced reaction, we should be able to understand that this number of moles of hydrogen sulphide gas required twice the number of moles of hydrochloric acid to produce it, since the ratio of hydrochloric acid to hydrogen sulphide gas is 2 to 1.Therefore the number of moles of hydrochloric acid “lost” or “used” up until this time point (10 seconds) would be 0.03529411765 moles.We started with 250 cm³ of 1.5 M hydrochloric acid which would contain 0.375 moles, so the number of moles of acid left at 10 seconds would simply be this:But this is of course in 250 cm³, therefore the molar strength of the acid at 10 seconds would be four times this amount = 1.36 M (two decimal places).Remembering that our readings indicate a mass loss from zero, not a difference mass loss (in other words not from 0 to 10, 10 to 20, 20 to 30 and so on) so next reading 20 seconds shows a loss of 1.18 g overall. Following the same calculation this evaluates to 0. 03470588235 moles of hydrogen sulphide gas used, with a commensurate reduction in hydrochloric acid moles of twice this value. The number of moles of hydrochloric acid left after 20 seconds would be:Finally we must remember that we started off with 250 cm³ of hydrochloric acid, therefore the molarity of the acid that we now have would be four times the above value or 1.22 M quoted to 2 decimal places.Now, following the same calculation for 30 seconds, 40 seconds, 50 seconds and 60 seconds we arrive at molarities for the acid of 1.08, 0.944, 0.810 and 0.677 M respectively.In the next couple of pages we will now start to look at the rate equation, which talks about “orders of reaction” and the effects of doubling, trebling or even more the concentrations of one or more of the reactants present, and the effect (iif any) on the speed (rate) of the reaction. The mathematics involved in this section is not difficult, but can be little bit fiddly, so you will need to keep your wits about you, especially when you are working with units as these can become quite tricky.
Es Muy Tutor Services - Reaction Rates and Arrhenius Eqn