Andreas O. GCSE Chemistry tutor, GCSE Maths tutor, GCSE Physics tutor

Andreas O.

Unavailable

Medicinal and Biological Chemistry (MChem) (Masters) - Edinburgh University

MyTutor guarantee

New to the site and yet to acquire customer reviews. We personally interview all our tutors so if you’re not satisfied, lets us know within 48 hours and we’ll refund you.

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.

About me

Hey there!! My name is Andreas and I study Medicinal and Biological Chemistry (MChem) at University of Edinburgh (UoE). I consider myself passionate in the field of Science and Maths and at the same time I try to learn more and become better like you do!! I’m patient, friendly and very approachable which are features that a good tutor MUST have!! My genuine enthusiasm about Science and the fact that I enjoy helping people drove me to become an online tutor and who knows maybe yours as well!! J

Even though my degree is focusing on Chemistry, my knowledge on other sciences and maths is not limited. On the contrary, I was awarded for the Highest International Subject Mark for International GCSE Mathematics for scoring full marks in the exams.

The sessions will be based on your needs and focus on understanding the key concepts, explaining anything in a clear and simple way and mainly solving examples and exercises.

Hey there!! My name is Andreas and I study Medicinal and Biological Chemistry (MChem) at University of Edinburgh (UoE). I consider myself passionate in the field of Science and Maths and at the same time I try to learn more and become better like you do!! I’m patient, friendly and very approachable which are features that a good tutor MUST have!! My genuine enthusiasm about Science and the fact that I enjoy helping people drove me to become an online tutor and who knows maybe yours as well!! J

Even though my degree is focusing on Chemistry, my knowledge on other sciences and maths is not limited. On the contrary, I was awarded for the Highest International Subject Mark for International GCSE Mathematics for scoring full marks in the exams.

The sessions will be based on your needs and focus on understanding the key concepts, explaining anything in a clear and simple way and mainly solving examples and exercises.

Show more

Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

No DBS Icon

No DBS Check

Qualifications

SubjectQualificationGrade
ChemistryA-level (A2)A
Pure MathematicsA-level (A2)A
PhysicsA-level (A2)A
BiologyA-level (A2)A
GreekA-level (A2)A*

General Availability

Pre 12pm12-5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
BiologyA Level£20 /hr
ChemistryA Level£20 /hr
MathsA Level£20 /hr
BiologyGCSE£18 /hr
ChemistryGCSE£18 /hr
MathsGCSE£18 /hr
PhysicsGCSE£18 /hr

Questions Andreas has answered

Given 4x+7y=25 and 2x+5y=17, identify x and y by solving the simultaneous equations

First thing we have to do is to eliminate one of the variables, either x or y (it doesn’t matter which one). In order to achieve that, we aim to have the same co-efficient (number in front of x or y) of a variable in both equations.

In this example, we can observe that no co-efficient is the same so we have to do some manipulation first before eliminating a variable!

Multiplying 2x+5y=17 by 2, give us 4x+10y=34 (Remember to multiply by 2 all the numbers)

Now that we have the same x co-efficient in both equations we can eliminate the x variable by subtracting one equation from the other:

    4x + 10y = 34

–  4x + 7y = 25

3y=9

 y=3

Hence, we can substitute y in one of the initial equations e.g. 2x+5y=17

2x+5(3) = 17

2x=17-15

2x=2

x=1

Therefore, the variables are x=1 and y=3

First thing we have to do is to eliminate one of the variables, either x or y (it doesn’t matter which one). In order to achieve that, we aim to have the same co-efficient (number in front of x or y) of a variable in both equations.

In this example, we can observe that no co-efficient is the same so we have to do some manipulation first before eliminating a variable!

Multiplying 2x+5y=17 by 2, give us 4x+10y=34 (Remember to multiply by 2 all the numbers)

Now that we have the same x co-efficient in both equations we can eliminate the x variable by subtracting one equation from the other:

    4x + 10y = 34

–  4x + 7y = 25

3y=9

 y=3

Hence, we can substitute y in one of the initial equations e.g. 2x+5y=17

2x+5(3) = 17

2x=17-15

2x=2

x=1

Therefore, the variables are x=1 and y=3

Show more

2 years ago

1043 views

What is the volume of carbon dioxide released at room temperature and pressure when 6.2 g of copper carbonate reacts with excess dilute sulfuric acid?

Background Information

This is a neutralisation reaction between an acid (sulfuric acid) and a base (copper carbonate). When a carbonate reacts with an acid, the products are a salt, carbon dioxide and water.

carbonate + acid --> salt + carbon dioxide + water

ATTENTION:       If you are NOT familiarised with these reactions, I advise first to revise and then try to solve the question.

Example

First thing to do when we tackle problem like this one is to write down the balanced equation (if it is not given)

CuCO3 + H2SO4 --> CuSO4 + CO2 + H2O

RFM (CuCO3) = 124 g mol-1

Number of moles of CuCO3 = 6.2 g / 124 g mol-1 = 0.05 mol

For every 1 mole of CuCO3, 1 mole of CO2 is produced (Using the stoichiometry of the equation)

Number of moles of CO2 = 0.05 mol

Volume of CO2 gas released = 0.05 mol * 24 dm3 = 1.2 dm3 = 12000 cm3

Background Information

This is a neutralisation reaction between an acid (sulfuric acid) and a base (copper carbonate). When a carbonate reacts with an acid, the products are a salt, carbon dioxide and water.

carbonate + acid --> salt + carbon dioxide + water

ATTENTION:       If you are NOT familiarised with these reactions, I advise first to revise and then try to solve the question.

Example

First thing to do when we tackle problem like this one is to write down the balanced equation (if it is not given)

CuCO3 + H2SO4 --> CuSO4 + CO2 + H2O

RFM (CuCO3) = 124 g mol-1

Number of moles of CuCO3 = 6.2 g / 124 g mol-1 = 0.05 mol

For every 1 mole of CuCO3, 1 mole of CO2 is produced (Using the stoichiometry of the equation)

Number of moles of CO2 = 0.05 mol

Volume of CO2 gas released = 0.05 mol * 24 dm3 = 1.2 dm3 = 12000 cm3

Show more

2 years ago

982 views

Send Andreas a message

A Free Video Meeting is a great next step. Just ask Andreas below!


Send message

How do we connect with a tutor?

Where are they based?

How much does tuition cost?

How do Online Lessons work?

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok