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Ed W.

£40 /hr

Medicine (Bachelors) - Edinburgh University

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This tutor is also part of our Schools Programme. They are trusted by teachers to deliver high-quality 1:1 tuition that complements the school curriculum.

319 completed lessons

About me

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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Ratings & Reviews

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Ayman Parent from Dublin 16

18 Mar

My son has been doing Chemistry and Biology A levels with Ed. Fantastic teacher, very patient with excellent use of lesson time. Fully recommend him.

SK
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Shula Parent from Devizes

25 Feb

Ed continues to be a fantastic tutor to my son. J was so happy after his last maths lesson as he finally understood a particular topic after trying for weeks.

MK
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Mina Student

18 Nov, 2018

Great lesson, well explained - thanks a lot Ed!

RK
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Rishabh Parent from Hartlepool

17 May, 2018

Ed is very helpful, my son is doing well in Biology, online method is really good.

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Qualifications

SubjectQualificationGrade
BiologyA-level (A2)A*
ChemistryA-level (A2)A*
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A

General Availability

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12 - 5pm

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Subjects offered

SubjectQualificationPrice
BiologyA Level£40 /hr
ChemistryA Level£40 /hr
BiologyGCSE£40 /hr
ChemistryGCSE£40 /hr
Further MathematicsGCSE£40 /hr
MathsGCSE£40 /hr
Medical School PreparationMentoring£40 /hr
Personal StatementsMentoring£40 /hr

Questions Ed has answered

A bromoalkane contains 34.9% carbon and 6.60% hydrogen by mass. The rest of the mass is made up by bromine. What is the empirical formula of this molecule?

A 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 C4H9BrA 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 C4H9Br

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2 years ago

1396 views

The first four terms of an arithmetic sequence are: 3, 10, 17, 24. What is an expression, in terms of n, for the nth term?

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.

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2 years ago

4301 views

Why is it sometimes necessary to conserve a plant species ex situ?

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.

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2 years ago

1015 views

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