Katherine M. GCSE Chemistry tutor, A Level Chemistry tutor, GCSE Biol...

Katherine M.

Currently unavailable: for regular students

Degree: Chemistry (Masters) - Edinburgh University

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About me

All About Me

Hi, my name is Katherine I am from Edinburgh and am currently in my 2nd year studying Medicinal and Biological Chemistry, also in my home town, at the University of Edinburgh.  I have always had a real passion for science and maths which has only strengthened at university.  I am very enthusiastic about the subjects I study and hope to pass on that passion to my students.

I am very patient and friendly.  I have been a Young Leader with Girl Guiding UK since 2011, so I have a lot of experience in leading a variety of sessions with people as young as 5 years old.  This has given me the ability to cater to individual learning and to encourage young people to follow their own interests.  My sister has frequently come to me for help with particular topics she has been struggling with at school and general advice on exam techniques, I always enjoy being able to help someone else.

The Tutorials

During a tutorial, you are in control of what we cover.  In science and in maths it is good to make sure there is a basic understanding of the topic first before going over exam questions, so I would suggest that we focus on this.

I will tailor my explanation of concepts in the simplest way possible using whatever method works best for each individual (words, diagrams, analogies etc.) until you are confident enough that you could explain it to me or anyone else. 

I hope to make these sessions fun!  They will be interactive, so won’t feel like just another hour of school!  Science and maths can be tricky ones to crack but if made enjoyable it is 100 times easier to pick them up!  Don’t be afraid to ask any questions no matter how silly you think they are.

How to get in touch

If you have any questions, or would like to meet me prior to booking a tutorial, send me a ‘WebMail’ or book a ‘Meet the Tutor Session’ through MyTutorWeb.

Thank you for your interest and I look forward to hearing from you soon!

Katherine

Subjects offered

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

Qualifications

QualificationLevelGrade
ChemistryAdvanced HigherA
BiologyAdvanced HigherA
MusicHigherA
MathsHigherA
EnglishHigherA
FrenchHigherA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

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Questions Katherine has answered

What mass of carbon dioxide is produced from burning 100 grams of ethanol in oxygen according to this reaction: C2H5OH + 3 O2 --> 2 CO2 + 3 H2O

When approaching a question like this I always find it easier to write out the equation with the known quantities written underneath each individual compound.  This allows us to easily check whether the equation is balanced                                    C2H5OH          +         3 O2     ...

When approaching a question like this I always find it easier to write out the equation with the known quantities written underneath each individual compound.  This allows us to easily check whether the equation is balanced

                                   C2H5OH          +         3 O2     -->       2 CO2                         +          3 H2O

We can see that this equation is balanced as it has 2 carbons, 6 hydrogens, and 7 oxygens on each side.

We are trying to work out the mass of CO2 and we already know what mass of ethanol (C2H5OH) we have.  We can also work out the gram formula mass of each compound we are interested in. 

gfm CO2 = (1 x 12) + (2 x 16) = 44 g/mol

gfm C2H5OH = (2 x 12) + (6 x 1) + (1 x 16) = 46 g/mol

Next we need to decide what formula we can use to figure out the mass of CO2 from the data that we have. 

                                   C2H5OH          +         3 O2     -->       2 CO2                         +          3 H2O

                       m = 100 g                                            m = ?

                       gfm = 46 g/mol                                gfm = 44 g/mol

Using n = m/gfm we can figure the number of moles of ethanol.  We can use this to work out the number of moles of carbon dioxideas these can be directly compared. 

                                  C2H5OH          +         3 O2     -->       2 CO2                         +          3 H2O

                      n = m/gfm                      

                      = 100/46                         

                      =2.174 moles                 

The mole ratio of ethanol to carbon dioxide is 1:2. So by multiplying our number by 2 we get the number of moles of CO2.

                                    C2H5OH          +         3 O2     -->       2 CO2                         +          3 H2O

                n = moles of C2H5OH x 2

           = 2.174 x 2

           = 4.349 moles

Now that we have the number of moles of carbon dioxide we can figure out the mass produced by using the formula m = n x gfm

                                  C2H5OH          +         3 O2     -->       2 CO2                         +          3 H2O

                                                                                    m = n x gfm                       

                                                                                        = 4.349 x 44                 

                                                                                        = 191.3 g

Therefore, the mass of carbon dioxide is produced from burning 100 grams of ethanol in oxygen is 191.3g.

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8 months ago

457 views

Q: How to solve the simultaneous equations 3x+2y=7 and 5x+y=14

When solving simultaneous equations you are aiming to use both equations together to find out the value of x and y for values that with ‘fit’ in each equation. 3x + 2y = 7 5x + y = 14 If we subtract these equations from each other in a way that we end up eliminating y, when can find the val...

When solving simultaneous equations you are aiming to use both equations together to find out the value of x and y for values that with ‘fit’ in each equation.

3x + 2y = 7

5x + y = 14

If we subtract these equations from each other in a way that we end up eliminating y, when can find the value of x. The way we do this is to multiply equation 2 by 2, remembering to multiply both sides of the equation. This gives

2. 10x + 2y = 28

Now, when we subtract equation 1 from equation 2 we get

(10x – 3x) + (2y - 2y) = 28 – 7

7x = 21

x = 3

Now we have found the value of x we can substitute it back into equation 1 to find the value of y.

3x + 2y = 7 and x = 3

9 + 2y = 7

2y = -2

y = -1

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8 months ago

213 views
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