Daniel D. GCSE Chemistry tutor, A Level Chemistry tutor, GCSE Maths t...

Daniel D.

£20 - £22 /hr

Currently unavailable: for regular students

Studying: Natural Sciences Physical (Bachelors) - Cambridge University

| 1 completed tutorial

Contact Daniel

About me

About me:

I am currently studying Physical Natural Sciences at Queens’ College Cambridge and next year I am going into second year.

Particles, atoms and molecules are the building blocks of our world and from an understanding of these we can answer many questions, which is why I love Science so much!

I have tutored four students already in a face to face environment, in a range of subjects with varying levels of difficulty. Therefore, I have already gained the necessary key skills of adaptability, communication and understanding of others. I have also completed over 100 hours of volunteering helping all ages from newly born babies in orphanages to the elderly in nursing homes.

The Sessions:

During the session we can work out together what you do understand and don’t understand and from then on I can work with you to fill any gaps you have in your knowledge.

Through diagrams, words and equations I will explain thoroughly what you need to know with a range of different teaching techniques to suit your learning style.  

I want to instil my enthusiasm for science into you - I hope you enjoy every minute of our sessions.

What next?

If you have any questions, please send me a ‘Webmail’ or book a “Meet the tutor session’! Please remember to tell me what exam board and level you are doing so I can brush up on the specification.

I look forward to meeting you!

About me:

I am currently studying Physical Natural Sciences at Queens’ College Cambridge and next year I am going into second year.

Particles, atoms and molecules are the building blocks of our world and from an understanding of these we can answer many questions, which is why I love Science so much!

I have tutored four students already in a face to face environment, in a range of subjects with varying levels of difficulty. Therefore, I have already gained the necessary key skills of adaptability, communication and understanding of others. I have also completed over 100 hours of volunteering helping all ages from newly born babies in orphanages to the elderly in nursing homes.

The Sessions:

During the session we can work out together what you do understand and don’t understand and from then on I can work with you to fill any gaps you have in your knowledge.

Through diagrams, words and equations I will explain thoroughly what you need to know with a range of different teaching techniques to suit your learning style.  

I want to instil my enthusiasm for science into you - I hope you enjoy every minute of our sessions.

What next?

If you have any questions, please send me a ‘Webmail’ or book a “Meet the tutor session’! Please remember to tell me what exam board and level you are doing so I can brush up on the specification.

I look forward to meeting you!

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A*
ChemistryA-level (A2)A*
PhysicsA-level (A2)A*

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
ChemistryA Level£22 /hr
Further Mathematics A Level£22 /hr
MathsA Level£22 /hr
PhysicsA Level£22 /hr
ChemistryGCSE£20 /hr
MathsGCSE£20 /hr
PhysicsGCSE£20 /hr

Questions Daniel has answered

Describe two different test tube reactions to identify the following organic compounds: propanal and benzoic acid.

There are several different answers one could make and an exam question would just require one for each. 

1) Propanal - from the ending "al" this is an aldehyde.

        - Add Tollens' reagent. Positive result: Silver mirror formed.

        - Add Fehling's / Benedict Solution. Positive result: Red precipitate formed.

        - Acidified potassium dichromate. Positive result: Orange to green (due to oxidation). 

2) Benzoic acid - You probably have not come across this in the lab at school. Nevertheless from the name you can work out that it is a carboxylic acid. 

        - Add Sodium carbonate/ Sodium hydrogen carbonate. Positive result: Effervescence is observed. 

There are several different answers one could make and an exam question would just require one for each. 

1) Propanal - from the ending "al" this is an aldehyde.

        - Add Tollens' reagent. Positive result: Silver mirror formed.

        - Add Fehling's / Benedict Solution. Positive result: Red precipitate formed.

        - Acidified potassium dichromate. Positive result: Orange to green (due to oxidation). 

2) Benzoic acid - You probably have not come across this in the lab at school. Nevertheless from the name you can work out that it is a carboxylic acid. 

        - Add Sodium carbonate/ Sodium hydrogen carbonate. Positive result: Effervescence is observed. 

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1 year ago

902 views

Solve the simultaneous equations y = 2x-3 and x^2 +y^2 = 2

Let the first equation be equation 1 and the second equation be equation 2.

Firstly, you must substitute equation 1 into equation 2. This is because there are currently two unknown values in each equation (x and y) and therefore you must eliminate one of them so that you just have x's or y's in a singe equation. 

Therefore, lets sub in equation 1 into equation 2. But first we must square:

From equation 1: y2 = (2x-3)2 = (2x-3)(2x-3) = 4x2 -12x +9

Then substituting this into equation 2:  x2 +y2  = 2 

                                                             x2 +4x2-12x+9 =2

                                                             5x2 -12x+9 =2

Subtracting 2 from both sides of the equation: 

                                                             5x2 -12x + 7 = 0

Now factorising the quadratic equation:

                                                             (5x -7)(x-1) = 0

Now to solve for x: either of the brackets need to equal 0 so that the left hand side of the equation equals 0. 

Therefore, 5x-7 = 0, which with some simple rearrangement gives rise to x = 7/5

                 x-1 = 0, which with some simple rearrangement gives rise to x = 1

Please note we have two values for x; this is because we have a quadratic equation (one which has powers of 2 in it). As we have two values for x it means we have two corresponding values for y. To find these values substitute the values of x you have just found back into equation 1:

From equation 1: When x = 7/5, y = -1/5

                            When x = 1, y = -1 

And that's the final answer!

Let the first equation be equation 1 and the second equation be equation 2.

Firstly, you must substitute equation 1 into equation 2. This is because there are currently two unknown values in each equation (x and y) and therefore you must eliminate one of them so that you just have x's or y's in a singe equation. 

Therefore, lets sub in equation 1 into equation 2. But first we must square:

From equation 1: y2 = (2x-3)2 = (2x-3)(2x-3) = 4x2 -12x +9

Then substituting this into equation 2:  x2 +y2  = 2 

                                                             x2 +4x2-12x+9 =2

                                                             5x2 -12x+9 =2

Subtracting 2 from both sides of the equation: 

                                                             5x2 -12x + 7 = 0

Now factorising the quadratic equation:

                                                             (5x -7)(x-1) = 0

Now to solve for x: either of the brackets need to equal 0 so that the left hand side of the equation equals 0. 

Therefore, 5x-7 = 0, which with some simple rearrangement gives rise to x = 7/5

                 x-1 = 0, which with some simple rearrangement gives rise to x = 1

Please note we have two values for x; this is because we have a quadratic equation (one which has powers of 2 in it). As we have two values for x it means we have two corresponding values for y. To find these values substitute the values of x you have just found back into equation 1:

From equation 1: When x = 7/5, y = -1/5

                            When x = 1, y = -1 

And that's the final answer!

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1 year ago

1982 views

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