C3 Energy Calculations And Titrations.

Question 0

If 475825J was produced by burning 0.9572548163 grams of energy, now would you calculate the energy produced per gram?

Answer 0

You would do 475825 divided by 0.9572548163

Question 1

What is the formula used to calculate the energy realised by a certain amount of fuel?

Answer 1

Energy released = mass(grams) X temperature change X the specific heat capacity of the liquid. (For water in chemistry it is 4.2)

Question 2

If 0.2 grams of fuel with a relative atomic mass of 46 produced a 5.25 KJ of energy, what is the energy released per mole?

Answer 2

(5.25 X 46) divided by 0.2

= 1207.5 KJ / mole.

Question 3

In titration calculations what is the formula triangle that produces concentration?

Answer 3

No. of moles

Concentration X volume.

Question 4

What is the formula triangle used to calculate mass used in titration calculations?

Answer 4

Mass

No. of moles X relative atomic mass

Question 5

What is the end point in a titration reaction?

Answer 5

The point when acid and alkali have reacted completely.

Question 6

How do you carry out a titration reaction?

Answer 6

Wash a pipettes with distilled water and then the alkali.
Measure out a known volume of alkali and put in a conical flask.
Add a few drops of indicator.
Wash a burette with distilled water then acid.
Fill the burette with acid and note the reading.
Add acid to the conical flask and swirl the flask continually.
Stop when it has been neutralised and note the reading.
Repeat three times and calculate an average.

Question 7

Find the concentration of a solution containing 40g of sonium hydroxide in 500 cm³.

Answer 7

Number of moles. Mass ÷ RFM = 40÷40=1
Concentration = moles ÷ volume = 1÷500=0.002
2md/dm³

Question 8

What mass of potassium sulphate (K”2”SO”4”)is there in 250 cm³ of a solution of a concentration of 1 mol/dm³ solution.

Answer 8

RFM of potassium sulphate = 174
Number of moles =(250÷1) ÷1000 = 0.25 moles. (You divide by 1000 because we are in decilitres)
0.25 = mass ÷ 174
174 X 0.25= 43.5

Question 9

25 cm³ of sodium hydroxide of unknown concentration reacted with 20 cm³ of 0.5mol/dm³ sulphuric acid. What was the concentration of the NaOH?

Answer 9

Sodium hydroxide = NaOH sulphuric acid = H”2”SO”4”
20÷1000 = 0.02 ( we do this to get to decilitres)
0.5 X 0.02 =0.01 ( this is the number of moles of sulphuric acid)
Write out the balanced equation and see one mole of acid gives two of alkali so we have 0.02 of sodium hydroxide.
C=moles ÷ volume.
0.02 ÷ 0.025 = 0.8 mol/dm³