Hayden T. GCSE Chemistry tutor, A Level Chemistry tutor, GCSE Maths t...

Hayden T.

Currently unavailable: for new students

Degree: MChem Chemistry (Masters) - Warwick University

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About me

About me:

As a second year student at the University of Warwick studying an integrated Masters in Chemistry, I have always had a keen interest in all the sciences and maths, and enjoy these subjects at all levels of education. I aim to pass this enthusiasm onto you, providing an atmosphere that can help you learn and achieve your potential.

I am friendly, approachable and patient. I am looking forward to passing on the knowledge I have gained through my degree and other qualifications – the world of science is simply amazing! With the right explanations, connections can be made through all the sciences and the possibilities are endless!

The tutorials:

Each tutorial is 55 mins long - as much, or as little, as you want can be covered in this time. Whether you would like to start from the foundations of a concept or build on top of your current knowledge, I will tailor the tutorials around you! The pace, topics and learning styles are all based on you – this is the advantage of a one-to-one tutorial.

Many different techniques will be used to aid your learning – simple explanations, diagrams, graphs, analogies and practice.  These will help you reach your potential and achieve higher grades. In an exam, and in real life, you will need to know how to apply this knowledge and not just know all the information – I intend to develop those skills alongside passing on the knowledge.

Once concepts have been covered and firmly grasped, I aim for you to be confident enough to explain the topic to me – this way you will be ready to tackle any questions on the topic. I can also help you work through and explain past paper questions, providing you with an example of how to approach questions and apply your knowledge. As an alternative, feel free to try them and bring them to me to explain concepts you don’t understand and would like clarification.

What next?

Contact me through the mailing system if you have any questions, or arrange a ‘meet the tutor’ session through this website.

Before tutorials it would benefit both of us if you send me information regarding your exam board, level of education (GCSE or A-level) and topic area/specific questions you require more help with; this will enable me to tailor my tuition to your needs.

I look forward to hearing from you!

Subjects offered

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

Qualifications

QualificationLevelGrade
ChemistryA-LevelA*
MathematicsA-LevelA*
BiologyA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

13/06/2014

Currently unavailable: for new students

General Availability

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Ratings and reviews

5from 12 customer reviews

Emily (Student) November 10 2016

Really helpful in explaining difficult questions :)

Izzy (Student) April 27 2016

very helpful! I sent some homework before the lesson and he came prepared with the answers and step by step how to get them. I've always struggled with maths in chemistry, and I'm finally starting to understand it!

Ridwaan (Student) April 26 2016

Literally the best chemistry tutor you can find on the MTW website. He's in his 1st year at university and achieved an A* in the same exam board I am currently doing (OCR A Chemistry). He understands my thoughts, he is passionate tutoring, he understands where you are coming from. 10/10, Will definitely want tutoring again from Hayden. Kind Regards, Ridwaan

Emily (Student) December 2 2016

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Questions Hayden has answered

Calculate the percentage, by mass, of carbon in sodium hydrogencarbonate (NaHCO3)

When faced with this style of problem there are steps to correctly answer the quesion: Step 1 - identify the atomic mass of carbon Step 2 - calculate the atomic mass of the compound Step 3 - work out the percentage, by mass, of carbon in the compound Method: Step 1: To  calculate the atomi...

When faced with this style of problem there are steps to correctly answer the quesion:

Step 1 - identify the atomic mass of carbon

Step 2 - calculate the atomic mass of the compound

Step 3 - work out the percentage, by mass, of carbon in the compound

Method:

Step 1:

To  calculate the atomic mass, a periodic table is required. The atomic mass of each element is located at the top of the element, displayed as a number. In this case carbon has an atomic mass of 12 g/mol.

Step 2:

The same principles in step 1 are required and applied to the remaining atoms in the compound - sodium hydrogen carbonate. The atoms in the compound are sodium (Na), hydrogen (H), carbon (C) and oxygen (O). Looking at the value of atomic mass on the periodic table, the values of atomic mass are as follows:

Na = 23 g/mol

H = 1 g/mol

C = 12 g/mol

O = 16 g/mol

There is 1 Na atom, 1 H atom, 1 C atom and 3 O atoms in the compound. Now add these values together to get the atomic mass of the compound as such:

23+1+12+(3*16) = 84 g/mol

Step 3:

Now the percentage, by mass, needs to be calculated. The general rule for calculating a percentage is:

percentage = (value/total)*100%

Therefore by applying this rule for our problem we find that

Percentage = (12/84)*100% = 14.3 % (to 1 d.p)

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8 months ago

656 views

What are isotopes and how do they differ from each other?

An isotope is a different form of the same element. They differ from each other by the number of neutrons, however they have the same number of protons and electrons. This results in a different atomic mass. Lets use an example - carbon: Carbon (C) can form different isotopes, with the most a...

An isotope is a different form of the same element. They differ from each other by the number of neutrons, however they have the same number of protons and electrons. This results in a different atomic mass.

Lets use an example - carbon:

Carbon (C) can form different isotopes, with the most abundant being C-12 and C-13.

C has 6 protons (stated under the element on a periodic table) and an equal number of electrons to balance the charge. C-12 has 6 neutrons and C-13 has 7 neutrons, because adding the number of neutrons and protons determines the atomic mass of the isotope (electrons have negligable mass).

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8 months ago

330 views
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