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About me

About Me:

Hi! I am currently a third-year student studying Chemistry at the University of Warwick; and have had previous tutor experience at my secondary school. I take a patient approach to tutoring and believe that a lot of the concepts they teach in schools can be broken down further to simple points. 

I specialise in Maths  and Chemistry tutoring - but please feel free to message me with questions of any nature.

Tutorials:

If you're interested please do send a message, with which subject you're doing and what exam board. Before tutorials please do message me with a given topic you're finding challenging; or if its for exam preparation, I can find some challenging questions from a variety of topics. Often, particularly in Chemistry the reasoning beyond some of the answers goes beyond the scope of the A Level course - if you're keen an planning on studying Chemistry at Uni, I'll gladly explain some of the principles in full, if not - we can just work on getting you the right answers consistently :) 

In science, understanding is everything, and once you learn basics everything else can seem trivial! 

Subjects offered

SubjectLevelMy prices
Chemistry A Level £20 /hr
Maths A Level £20 /hr
Chemistry GCSE £18 /hr
Maths GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
ChemistryA-LevelA
Further MathematicsA-LevelB
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CRB/DBS Standard

No

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No

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Ratings and reviews

5from 3 customer reviews

Harry (Student) November 23 2016

excellent, very clear and helpful.

Michaela (Parent) November 30 2016

Harry (Student) November 30 2016

Questions Martin has answered

Proton NMR Made Easier

Proton or 1H NMR can look complicated but with practice; there are techniques that can make the peak assignment a little easier. 1. Make sure you count the proton environments correctly, if the molecule displays a form of symmetry you may count an identical environment! 2. If there is a multi...

Proton or 1H NMR can look complicated but with practice; there are techniques that can make the peak assignment a little easier.

1. Make sure you count the proton environments correctly, if the molecule displays a form of symmetry you may count an identical environment!

2. If there is a multiplet at 7.2 ppm - this is always indicative of a benzene like ring.

3. Electronegative groups (N, O, Cl, F) cause de-shielding, moving protons in these environments further down the spectrum, use this as an indicator to find which hydrogens are those near functional groups. In contrast, branching alkane groups CH2 and chain end CH3 will mostly have quite low chemical shift.

4. The (n+1) rule can be used if you have proteins in a similar environment. The peak for a proton environment will split if a carbon adjacent to the one the proton you're observing is attached to protons in a different environment - This will cause the peak to split into (n+1) little peaks, where n is the number of protons on neighbouring carbon.

5. Often NMR questions will have other analysis such as MS in the same question. If your molecule you propose based on your NMR isnt compatible with the MS, try again - always look for differences in 14 m/z values for alkane chains etc.

These questions are worth a lot of marks so take your time.

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1 month ago

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What is a stationary point on a curve? How do I calculate the co-ordinates of a stationary point?

A stationary point simply means a point in a curve where the gradient is equal to 0. For example, in the June 2015 C3 Paper it is asked: Find the exact values of the coordinates of the stationary points of the curve. The curve function is f(x) = 6lnx + x^2 - 8x + 3 To calculate the gradient, ...

A stationary point simply means a point in a curve where the gradient is equal to 0.

For example, in the June 2015 C3 Paper it is asked:

Find the exact values of the coordinates of the stationary points of the curve.

The curve function is f(x) = 6lnx + x^2 - 8x + 3
To calculate the gradient, we need to differentiate, as the gradient can also be represented as the change in y in respect to the change in x, or in other words dy/dx.

dy/dx = 6/x + 2x - 8

Where dy/dx = 0 is where the stationary point will be, 6/x + 2x - 8 = 0; multiplying all by x will give a quadratic: 6 + 2x^2 - 8x = 0, which can then be factorised: (2x-2)(x-3) = 0

Solving this x = 1 or x= 3. Calculating y from the original function gives y = -4 and y = 6ln3 - 12, giving the co-ordinates (1,-4) and (3,6ln3 - 12)

These questions are often worth a substantial amount of marks.

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1 month ago

69 views
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