# Chemistry C7 Calculation Questions And How To Answer Them

Some practice questions on how to do certian calculations and questions mentioned.

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## Question One

i : A student has a stock solution of nitric acid containing 63g in each dm(3)

She used this to make up 250cm(3) of a standard solution containing 6.3g in each dm(3)

Describe how she makes up this solution.

:1dm(3) = 1000cm(3)

Answer: The Student Takes 100cm(3) from her stock solution this them gives her 6.3g/dm(3) therefore she adds 150cm(2) of distilled water.

Why?: She adds distilled water as it is adding no chemicals to her solution and it is pure. This method also ensures they get an exact amount, always look for a relationship between the original Dm and the aspired one, this will make the workings easier.

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## Question One: Part B

ii : The studen uses the average of her titration results, 28.2cm(3), as a best estitmate of the volume of nitric acid used.

Show by calculation that the mass of nitric acid in 28.2cm(3) of the standard solution is 0.178g.

Answer: If there is 6.3g in 1000dm(3) then whats in 28.2cm(3). This question is quite hard at first thought, however the calculation to answer it is very simple.

We begin with the 6.3g we want to know whats in 1dm(3), therefore we divide it.

6.3/1000

We want 28.2dm(3) though so we continue by multiplying the answer by the required amount.

6.3/1000 X 28.2

This gives our answer 0.178g proving the question correct.

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## Question Two

The combustion of bioethanol can be represented by this equation.

C2H5OH + 3O2 ---> 2CO2 + 3H2O

Octane, C8H18, is one of the hydrocarbons in petrol. The combustion of octane can be represted by this equation.

C8H18 + 121/2O2 ---> 8CO2 + 9H2

Burning 1.0g of bioethanol preduced 1.9g of carbon dioxide.

Burning Octane produces about 60% more carbon dioxide than the same mas of bioethanol.

Show that this is true by calculating the mass of carbon dioxide produced when 1.0g of octane burns and the percentge increase in carbon dioxide produced compared to biothanol. (relative atomic masses: C=12 H=1 O=16)

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C8H18 -> 8CO

114 --> 352

If i have 114g i will get 352g, this can be repeated for any of grams, tonnes, kilo's.

1g= 352/114 = 3.08g

Mass of carbon dioxide = 3.1grams

We also have to work out the exact percentage increase...

Octane makes 3.08g

Bioethonal make 1.9g

3.08-1.9= 1.18

(1.18/1.9) X 100 = 62%

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## Question Three

i : The relitive formula mass of hydrochloric acid is 36.5.

Work out the relitive formula mass (RFM) of magnesium hydroxide, Mg(OH)2

(relitive atomic masses: H=1, Mg=24, O=16)

24+24=58

ii : Work out the mass of hydrochloric acid in 23.5cm(3) of the hydrochloric acid solution used in the titrations.

40g in 1000dm(3)

40

-------  X 23.5 =0.94g

1000

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## Question Three

iii : Use the neutralization equation below to work out the mass of magnesium hydroxide that reacts with this mass of hydrochloric acid.

Mg(OH)2 + 2HCl --> MgCl2 + 2H2

This average mass of magnesium hydroxide in each tablet.

Mg(OH)2  with 2HCl

58g reacts with 73g

58/73 X 0.94 = 0.74

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## Question Four

Ethanoic acid and sodium hydroxide react according to this equation.

CH(3)COOH + NaOH ---> CH(3)COONa +H(2)O

Gemma used 25cm(3) of Vinegar for each titration.

The average of the results from Gemma's afternoon titrations is 12.5cm(3).

Use this average, and the concentration of sodium hydroxide you gave in the previous question to calculate the mass of ethanoic acid in each dm(3) of vinegar.

(relative formula masses: CH(3)COONa = 60, NaOH = 40)

How much is in 12.5cm(3)? 4/1000X12.5=0.05g

4g/1000cm(3) 60g reacts with 40g