Hi! I'm Ed and I'm currently studying Medicine at the University of Edinburgh. I tutor at GCSE and A Level for Biology, Chemistry and Maths, and at GCSE level for Further Maths. These are subjects I'm very comfortable in and enjoy. With my university course I am continuing to study biology and chemistry, and I have previous experience in assissting a school maths class. I'm also happy to do personal mentoring for anyone hoping to apply to medical school. With only one in ten applicants receiving a place nationwide I know how tough getting into med school is and would love to help you with this by sharing my tips & tricks for anything from personal statements to interview prep to UKCAT. As a person I'd like to think I'm enthusiastic and engaging, while still maintaining a professional style to my sessions. I am very keen on finding out which way of learning suits you best and can help you tailor this to answering exam questions as best as possible.

Hi! I'm Ed and I'm currently studying Medicine at the University of Edinburgh. I tutor at GCSE and A Level for Biology, Chemistry and Maths, and at GCSE level for Further Maths. These are subjects I'm very comfortable in and enjoy. With my university course I am continuing to study biology and chemistry, and I have previous experience in assissting a school maths class. I'm also happy to do personal mentoring for anyone hoping to apply to medical school. With only one in ten applicants receiving a place nationwide I know how tough getting into med school is and would love to help you with this by sharing my tips & tricks for anything from personal statements to interview prep to UKCAT. As a person I'd like to think I'm enthusiastic and engaging, while still maintaining a professional style to my sessions. I am very keen on finding out which way of learning suits you best and can help you tailor this to answering exam questions as best as possible.

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29/03/2016A bromoalkane is a molecule which contains bromine, carbon and hydrogen (and nothing else) and has no carbon-carbon double bonds. An empirical formula gives the simplest ratio of atoms in a molecule. Here's how I'd go about tackling this question:1) Work out the % bromine by mass: 100 - 34.9 - 6.60 = 58.5%, so this is the % mass as we are told the rest of the mass comes from bromine.2) Work out the number of moles of C, H and Br: moles of C = mass/Ar = 34.9/12 = 2.91 (we can use "% mass" as "mass" because we are simply working out a ratio of the atoms relative to each other). Similarly, moles of H = 6.60/1 = 6.60. Moles of Br = 58.5/79.9 = 0.732.3) Work out the whole number ratio of the C:H:Br. You can do this by dividing all the moles by the value for the smallest numeber of moles (i.e. 0.732 moles of bromine). So 2.91 moles of C becomes 3.98 (because 2.91/0.732 = 3.98); 6.60 moles of H becomes 9.02; 0.732 moles of Br becomes 1. As these values are all very close to whole numbers we can simply round them to the nearest, so the ratio of C:H:Br is 4:9:14) You can conclude that the empirical formula of the bromoalkane is C_{4}H_{9}BrA bromoalkane is a molecule which contains bromine, carbon and hydrogen (and nothing else) and has no carbon-carbon double bonds. An empirical formula gives the simplest ratio of atoms in a molecule. Here's how I'd go about tackling this question:1) Work out the % bromine by mass: 100 - 34.9 - 6.60 = 58.5%, so this is the % mass as we are told the rest of the mass comes from bromine.2) Work out the number of moles of C, H and Br: moles of C = mass/Ar = 34.9/12 = 2.91 (we can use "% mass" as "mass" because we are simply working out a ratio of the atoms relative to each other). Similarly, moles of H = 6.60/1 = 6.60. Moles of Br = 58.5/79.9 = 0.732.3) Work out the whole number ratio of the C:H:Br. You can do this by dividing all the moles by the value for the smallest numeber of moles (i.e. 0.732 moles of bromine). So 2.91 moles of C becomes 3.98 (because 2.91/0.732 = 3.98); 6.60 moles of H becomes 9.02; 0.732 moles of Br becomes 1. As these values are all very close to whole numbers we can simply round them to the nearest, so the ratio of C:H:Br is 4:9:14) You can conclude that the empirical formula of the bromoalkane is C_{4}H_{9}Br

Before we start thinking about an expression for this sequence, we need to figure out what's going on. An arithmetic sequence means that there's a certain number being added on each time. By working out the difference between these numbers we can see that here the number being added is 7 (for example 10-3=7). So now we need to write our expression. This is just a way of showing someone what a sequence is without having to write it all out for them. It will be of the form d.n+a (try remembering it using DNA from science), where "d" and "a" are what you need to find out. The first one is easy: "d" stands for difference and we've already worked out that is 7.To work out "a", siimply substract the difference, 7, from the first number in the sequence (called the first "term"), which is 3. 3-7=-4, so "a" is -4.You can conclude that the expression is 7n-4.Before we start thinking about an expression for this sequence, we need to figure out what's going on. An arithmetic sequence means that there's a certain number being added on each time. By working out the difference between these numbers we can see that here the number being added is 7 (for example 10-3=7). So now we need to write our expression. This is just a way of showing someone what a sequence is without having to write it all out for them. It will be of the form d.n+a (try remembering it using DNA from science), where "d" and "a" are what you need to find out. The first one is easy: "d" stands for difference and we've already worked out that is 7.To work out "a", siimply substract the difference, 7, from the first number in the sequence (called the first "term"), which is 3. 3-7=-4, so "a" is -4.You can conclude that the expression is 7n-4.

This question is all about conservation. While many students dislike ecology because it may seem a bit wishy-washy and not really "proper science", it is actually very important as we need to understand our relationships with other species to prevent degradation of our environemnt which has led to many species' extinction.There are many very good answers to this question. Conserving a plant species "ex situ" simply means outside its natural environment. Reasons you could give for this may include: their natural habitat has been lost due to human activities (e.g. deforestation), the population of the plant species is at a dangerously low level, ex situ allows protection from predators and disease, and so on.It is useful to learn a few simple reasons like this so in the exam you can just easily write them down.This question is all about conservation. While many students dislike ecology because it may seem a bit wishy-washy and not really "proper science", it is actually very important as we need to understand our relationships with other species to prevent degradation of our environemnt which has led to many species' extinction.There are many very good answers to this question. Conserving a plant species "ex situ" simply means outside its natural environment. Reasons you could give for this may include: their natural habitat has been lost due to human activities (e.g. deforestation), the population of the plant species is at a dangerously low level, ex situ allows protection from predators and disease, and so on.It is useful to learn a few simple reasons like this so in the exam you can just easily write them down.