Haroon A. GCSE Maths tutor, GCSE Chemistry tutor, GCSE Physics tutor,...

Haroon A.

Currently unavailable: until 01/06/2016

Degree: Chemical Engineering with Industrial experience (Other) - Manchester University

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About me

I am currently studying Chemical Engineering at the University of Manchester. The reason I have chosen to adopt engineering as my profession is because engineers make a difference. Engineers are able to take the knowledge derived from research and apply it in the real world to make it safer and more comfortable for the people. I aim to help in the development of sustainable and environmentally friendly technologies that benefit the world. I strive for excellence and I am always on the lookout for my weaknesses so that I can improve and grow.

Personal approach towards teaching:

Helping students find interest in their subjects, communicating core concepts, providing extra online learning resources, and most importantly, helping students develop self-study skills which become essential as they transition to their higher level education. 

 

Subjects offered

SubjectLevelMy prices
Maths GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Physics GCSE £18 /hr

Qualifications

QualificationLevelGrade
AccountingA-LevelA*
ChemistryA-LevelA*
MathematicsA-LevelA
PhysicsA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: until

01/06/2016

Questions Haroon has answered

Balance the equation for the reduction of MnO4- to Mn2+

Balancing equations is usually fairly simple. However some of them involve several steps. You may have come across balanced equations in data booklets that look intimidating. The balanced equation for reduction of Mn7+ to Mn2+ is one such equation. Initially one might write: Mn7+ + 5e- --> ...

Balancing equations is usually fairly simple. However some of them involve several steps.

You may have come across balanced equations in data booklets that look intimidating. The balanced equation for reduction of Mn7+ to Mn2+ is one such equation.

Initially one might write:

Mn7+ + 5e- --> Mn2+

Although technically balanced (since the ox state of Mn in MnO4- is +7), this equation does not represent the full reaction that takes place which involves H2O molecules and H+ ions.

The fully balanced equation is:

MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

To balance this, the following steps must be followed:

Step 1: Write only what's given. 

MnO4- --> Mn2+

Step 2: Balance all atoms except for H and O.

MnO4- --> Mn2+ (Mn atoms are already balanced; one on each side).

Step 3: Balance Oxygen atome by adding H2O to the side where more oxygen atoms are needed. At this stage, we have 4 O atoms on the left hand side and need 4 on the RHS. One water molecule contains one O atom, so we need 4 water molecules.

Therefore, MnO4- --> Mn2+ + 4H2O

Notice that now we have 4 O atoms on each side but 8 H atoms on the RHS.

Step 4: Balance H atoms by adding the required number of H+ ions to the side that is short of H atoms. At this stage we have (4 x 2) 8 H atoms on the RHS and none on the LHS.

Therefore, MnO4- + 8H+ --> Mn2+ + 4H2O

Step 5: Balance the charges by adding an electron, e-. At this stage, the LHS has a  (-1 +8)  +7 charge. The RHS has a +2 charge. If we add 5e- to the LHS, the charge becomes 7 - 5 = +2

Therefore, MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

By the end of step 5, we have obtained the fully balanced equation.

You can balance any equation using these steps, however, there is a slight adjustment that has to be made to step 4 sometimes.

In acidic solutions, to balance H atoms you just add H+ to the side lacking H atoms but in a basic solution, there is a negligible amount of H+ present. Instead, OH- is abundant. In this case, you add H2O to the side lacking H atom(s) and a OH- to the opposite side. The net effect is that you end up adding 1 H atom to the side that lacks a H atom. If a side lacks 'n' number of H atoms, add 'n' number of H2O molecules to that side and 'n' number of OHions to  the opposite side.

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10 months ago

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The polynomial x^3 - 2*x + a, where x is a constant is denoted by p(x). It is given that x+2 is a factor of p(x). Find a

Before solving this question, lets look at a second order polynomial. Let the polynomial x^2 +5*x +4 be f(x) Using factorisation, we can write f(x) as (x+1)(x+4) If you were to expand (x+1)(x+4) you'd get x^2 +5*x +4. To solve f(x) = 0, we can write (x+1)(x+4) = 0 and then set each one of the...

Before solving this question, lets look at a second order polynomial.

Let the polynomial x^2 +5*x +4 be f(x)

Using factorisation, we can write f(x) as (x+1)(x+4)

If you were to expand (x+1)(x+4) you'd get x^2 +5*x +4.

To solve f(x) = 0, we can write (x+1)(x+4) = 0 and then set each one of the factors to zero.

x+1 = 0 --> x = - 1 and x+4 = 0 --> x = - 4

Check that: f(-1) = 0 and that f(-4) = 0

We can conclude that,

- setting the factors of a polynomial f(x) to zero gives the roots of the equation f(x) = 0

- plugging the roots (values of x) back in the polynomial expression will lead to f(x) equalling zero.

Coming back to the question, we know that the factor of polynomial p(x) is (x+2).

Setting x+ 2 equal to zero gives x = - 2

This is a one of the three roots/solutions of the equation p(x) = 0 and plugging it into the polynomial expression should give zero,

i.e, p(-2) = 0

p(-2) = -2^3 - 2*(-2) + a = 0

        -8 + 4 +a = 0

        -4 + a = 0

         a = 4

- setting factor of p(x) to zero gives root of the equation p(x)=0

-Plugging root into p(x) expression will satisfy p(x)=0

        

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10 months ago

278 views
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