**About Me:**

I'm a third year student at the University of Manchester studying for my masters degree in Chemistry. The main purpose of my tutorials, aside from allowing you to pass your exams with a great mark, is to get you interested in Maths, Chemistry or science in general. I've helped people to understand both maths and science since my GCSE days, with several different peers coming to me for help understanding something after a particularly challenging lesson, or even just a quick refresher of the year once exams loomed closer. I've had experience teaching in classrooms in the past, as i have done both work experience and volunteer work at local primary schools.

**The Sessions:**

**The primary focus of our sessions is to help you. **Your particular learning style is important in teaching you all the different concepts we will be teaching you. If you learn by doing example questions and applying what you know, we could go through those questions step by step. If however, you prefer to learn the theory in depth, which could even include concepts slightly more advanced than strictly needed if they help you understand something better than just stating this fact is true and expecting you to accept it and move on. **The choice is entirely yours.**

**What you need to do:**

If you decide you'd like to book a session or just ask a few more questions about me in particular, simply contact me through the website (Book a 'Meet a Tutor Session' or WebMail). Please include your exam board, and if there's any particular subjects you are struggling with.

I look forward to hearing from you!

**About Me:**

I'm a third year student at the University of Manchester studying for my masters degree in Chemistry. The main purpose of my tutorials, aside from allowing you to pass your exams with a great mark, is to get you interested in Maths, Chemistry or science in general. I've helped people to understand both maths and science since my GCSE days, with several different peers coming to me for help understanding something after a particularly challenging lesson, or even just a quick refresher of the year once exams loomed closer. I've had experience teaching in classrooms in the past, as i have done both work experience and volunteer work at local primary schools.

**The Sessions:**

**The primary focus of our sessions is to help you. **Your particular learning style is important in teaching you all the different concepts we will be teaching you. If you learn by doing example questions and applying what you know, we could go through those questions step by step. If however, you prefer to learn the theory in depth, which could even include concepts slightly more advanced than strictly needed if they help you understand something better than just stating this fact is true and expecting you to accept it and move on. **The choice is entirely yours.**

**What you need to do:**

If you decide you'd like to book a session or just ask a few more questions about me in particular, simply contact me through the website (Book a 'Meet a Tutor Session' or WebMail). Please include your exam board, and if there's any particular subjects you are struggling with.

I look forward to hearing from you!

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.

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Nitrogen has 5 valence electrons in its outer shell. Each Hydrogen has 1 valence electron in its outer shell, so the 3 Hydrogens donate 3 electrons in total. This gives 8 valence electrons in total. We can then divide this by 2 to calculate the number of electron pairs around the central nitrogen atom. As there are 4 valence electron pairs around the nitrogen, but only 3 Hydrogens, then we have 1 pair of electrons left over. (As a covalent bond, N-H in this example, requires 1 electron pair.) This left over pair of electrons is said to be a lone pair.

Each pair of electrons will repel the other pair of electrons, so the electron pairs will try and be as far away as possible from each other. As there are 4 pairs of electrons, we could say that the angle between them would be 109.5^{o}. However, the lone pair that was left over actually repels more than the other pairs of electrons, so the actual angle between the hydrogens would be slightly less than this at ~107^{o}.

The shape of the NH_{3} is referred to as trigonal bypramidal, as while the electron pairs are arranged tetrahedrally, the lone pair is considered to be invisible and so the NH_{3} does not look tetrahedral.

Nitrogen has 5 valence electrons in its outer shell. Each Hydrogen has 1 valence electron in its outer shell, so the 3 Hydrogens donate 3 electrons in total. This gives 8 valence electrons in total. We can then divide this by 2 to calculate the number of electron pairs around the central nitrogen atom. As there are 4 valence electron pairs around the nitrogen, but only 3 Hydrogens, then we have 1 pair of electrons left over. (As a covalent bond, N-H in this example, requires 1 electron pair.) This left over pair of electrons is said to be a lone pair.

Each pair of electrons will repel the other pair of electrons, so the electron pairs will try and be as far away as possible from each other. As there are 4 pairs of electrons, we could say that the angle between them would be 109.5^{o}. However, the lone pair that was left over actually repels more than the other pairs of electrons, so the actual angle between the hydrogens would be slightly less than this at ~107^{o}.

The shape of the NH_{3} is referred to as trigonal bypramidal, as while the electron pairs are arranged tetrahedrally, the lone pair is considered to be invisible and so the NH_{3} does not look tetrahedral.

On the periodic table, Magnesium has the symbol _{12}Mg^{24}, where 24 is the mass number and 12 is the atomic number. The atomic number is the number of protons found inside an atom, while the mass number tells us how many neutrons and protons are found in an atom. As the atomic number is 12, we can see we have 12 protons. To calculate the number of neutrons, we do 24-12, which gives us 12. Therefore, the mass number is 24 because Magnesium has 12 protons and 12 neutrons.

On the periodic table, Magnesium has the symbol _{12}Mg^{24}, where 24 is the mass number and 12 is the atomic number. The atomic number is the number of protons found inside an atom, while the mass number tells us how many neutrons and protons are found in an atom. As the atomic number is 12, we can see we have 12 protons. To calculate the number of neutrons, we do 24-12, which gives us 12. Therefore, the mass number is 24 because Magnesium has 12 protons and 12 neutrons.

As the expression contains 4 separate terms that are added together, we can treat this as 4 separate differentiations. The general rule for differentiation for a general term bx^n is (b*n)x^{(n-1)}. For example, the first differentiation in this question is for x^{3}, which would lead to (3*1)x^{(3-1)}, as b is just 1, which can be simplified to 3x^{2}. Carrying on, we can find the following differentiations, which are:

2x^{2} goes to 4x

6x goes to 6

5 goes to 0

The final differentiation is 0 due to the fact that while it is written in the question as 5, we can actually think of it as 5x^{0}. This is due to the fact that anything to the power of 0 is 1, leaving just 5. Using the general rule again, this would give the result of (5*0)x^{(}^{0-1)} = 0.

The final result is therefore dy/dx = 3x^{2}+4x+6.

As the expression contains 4 separate terms that are added together, we can treat this as 4 separate differentiations. The general rule for differentiation for a general term bx^n is (b*n)x^{(n-1)}. For example, the first differentiation in this question is for x^{3}, which would lead to (3*1)x^{(3-1)}, as b is just 1, which can be simplified to 3x^{2}. Carrying on, we can find the following differentiations, which are:

2x^{2} goes to 4x

6x goes to 6

5 goes to 0

The final differentiation is 0 due to the fact that while it is written in the question as 5, we can actually think of it as 5x^{0}. This is due to the fact that anything to the power of 0 is 1, leaving just 5. Using the general rule again, this would give the result of (5*0)x^{(}^{0-1)} = 0.

The final result is therefore dy/dx = 3x^{2}+4x+6.