Kiera D. GCSE Chemistry tutor, A Level Chemistry tutor, GCSE Biology ...

Kiera D.

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Degree: Chemistry (Masters) - Durham University

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About me

About me 

Hello! I'm Kiera, a second year Chemist undertaking a Master's degree at the University of Durham. Besides exploring the fascination and complexity of Chemistry, my degree involves many mathematical topics and links with Biology thus giving me a thorough understanding of my specialist subjects. I'm a friendly, enthusiastic student offering productive tutorials to help students achieve their full potential at GCSE and A-level. I want my students to feel relaxed and not afraid to ask any questions- there is no such thing as a silly question!

Our Tutorials 

As a Chemist, I participate in weekly interactive tutorials at university which involve group work, discussion, exam question analysis, independent work and demonstrations. With this experience I find it is important for my students to fully understand the basic concepts in order for successful application to exam questions, practical work or homework. 

During your tutorials we will cover any material you would like, whether it be a detailed explanation, simple refresher or purely exam practise. Having sat years of exams, I understand the precise phrasing and key points examiners look for to award marks. Developing model answers together for common exam questions will ensure no marks are wasted! The delivery and the pace of our tutorials will be tailored to your preference, allowing you to get the most out of our sessions. 

Please don't hesitate to get in touch, simply drop me a message and I’ll happily answer any questions. Look forward to hearing from you! 

Subjects offered

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

Qualifications

QualificationLevelGrade
Mathematics A-LevelA
Chemistry A-LevelA*
Biology A-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for new students

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Ratings and reviews

4.8from 4 customer reviews

Dushanthi (Parent) November 30 2016

Minoshi (Student) November 23 2016

Minoshi (Student) November 20 2016

Dushanthi (Parent) November 20 2016

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Questions Kiera has answered

How to formulate and balance a redox equation under acidic conditions

In a reaction between  MnO4- and Fe2+ we are told MnO4- goes to Mn2+ and Fe2+  goes to Fe3+. We then need to derive the redox equation for the reaction occuring.  First we need to idenfity which species are being reduced and which are oxidised. We can use ' OIL RIG'  when referring to electron...

In a reaction between  MnO4- and Fe2+ we are told MnO4- goes to Mn2+ and Fe2+ goes to Fe3+. We then need to derive the redox equation for the reaction occuring. 

First we need to idenfity which species are being reduced and which are oxidised. We can use ' OIL RIG'  when referring to electrons to help us.

Oxidation

Is

Loss of electrons

Reduction 

Is 

Gain of electrons

MnO4- (where Mn has a +7 oxidation state) gains electrons, hence is reduced, to Mn2+ and Fe2+ loses electrons, hence oxidised, to Fe3+

Starting with the reduction half equation,       

1. Write out the reactant and product                                     MnO4-                 ------>      Mn2+

2. Use H2O to balance the oxygen                                        MnO4-                 ------>      Mn2+ + 4H2O

3. Use H+ (as under acidic conditions) to then

balance the other side of the equation                                   MnO4-  + 8H+                ------>      Mn2+ + 4H2​O

4. Add electrons to balance the charges                               MnO4-  +  8H+ +  5e-        ------>      Mn2+ + 4H2​O

For the oxidation half equation,

1. Write out the reactant and product                                    Fe2+      ------->      Fe3+

2. Balance the equation using electrons                               Fe2+   ------->      Fe3+ + e-

To combine the reduction and oxidation half equations we need to have the same number of electrons. Multiply the oxidation half equation by 5 to give us  5Fe2+   ------->      5Fe3+ + 5e-

Then add the two half equations together, canceling any like terms (the electrons) from both sides. The overall redox equation is therefore 

5Fe2+ + MnO4-  +  8H+     ------>    5Fe3+ +  Mn2+ + 4H2​O 

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1 month ago

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