__All About Me__

Hi, my name is **Katherine** I am from Edinburgh and am currently in my 2^{nd} 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

__All About Me__

Hi, my name is **Katherine** I am from Edinburgh and am currently in my 2^{nd} 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

No DBS Check

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

C_{2}H_{5}OH + 3 O_{2} --> 2 CO_{2} + 3 H_{2}O

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 CO _{2}**

gfm CO_{2} = (1 x 12) + (2 x 16) = 44 g/mol

gfm C_{2}H_{5}OH = (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 CO_{2} from the data that we have.

C_{2}H_{5}OH + 3 O_{2} --> 2 CO_{2} + 3 H_{2}O

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 dioxide**as these can be directly compared.

C_{2}H_{5}OH + 3 O_{2} --> 2 CO_{2} + 3 H_{2}O

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 CO_{2}.

C_{2}H_{5}OH + 3 O_{2} --> 2 CO_{2} + 3 H_{2}O

n = moles of C_{2}H_{5}OH 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

C_{2}H_{5}OH + 3 O_{2} --> 2 CO_{2} + 3 H_{2}O

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.

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

C_{2}H_{5}OH + 3 O_{2} --> 2 CO_{2} + 3 H_{2}O

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 CO _{2}**

gfm CO_{2} = (1 x 12) + (2 x 16) = 44 g/mol

gfm C_{2}H_{5}OH = (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 CO_{2} from the data that we have.

C_{2}H_{5}OH + 3 O_{2} --> 2 CO_{2} + 3 H_{2}O

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 dioxide**as these can be directly compared.

C_{2}H_{5}OH + 3 O_{2} --> 2 CO_{2} + 3 H_{2}O

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 CO_{2}.

C_{2}H_{5}OH + 3 O_{2} --> 2 CO_{2} + 3 H_{2}O

n = moles of C_{2}H_{5}OH 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

C_{2}H_{5}OH + 3 O_{2} --> 2 CO_{2} + 3 H_{2}O

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.

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__

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__