solve the following:i)What volume of oxygen is required to burn

Subject

Chemistry

Class

ICSE Class 10

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 Multiple Choice QuestionsLong Answer Type

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51.


solve the following:
i)What volume of oxygen is required to burn completely 90 dm3 of butane under similar conditions of temperature and pressure?

2C4O10 +13O2 ------> 8CO2 +10H2O

ii)The vapour density of a gas is 8. What would be the volume occupied by 24.0g of the gas at STP.

iii)A vessel contains X number of molecules of hydrogen gas at certain temperature and pressure. How many molecules of nitrogen gas would be present in the same vessel under the same condition of temperature and pressure?


left parenthesis straight i right parenthesis space 2 straight C subscript 4 straight H subscript 10 space space plus 13 straight O subscript 2 space space rightwards arrow space 8 CO subscript 2 space space plus space 10 space straight H subscript 2 straight O
space space space
2 space Volume space of space butane space equals space 90 space dm cubed
1 space Volume space of space butane space equals space 45 space dm cubed

therefore comma

Volume space of space straight O subscript 2 space required space equals 13 space straight x space 45
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 585 space space dm cubed

left parenthesis ii right parenthesis space Molecular space Weight space space equals space 2 space space straight x space VD space left parenthesis vapour space density right parenthesis space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 16 straight g

Volume space of space 16 space straight g space gas space equals 22.4 space straight L

Volume space of space 1 space straight g space space gas space equals space fraction numerator space 22.4 space over denominator 16 end fraction

Volume space of space 24 space straight g space gas space space equals fraction numerator 22.4 over denominator 16 end fraction space straight x space 24
space space space equals 33.6 space litres

iii right parenthesis space It space will space contain space straight X space number space of space molecules.
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52.

O2 is evolved by heating KClO3 using MnO2 as a catalyst

 2 KlO subscript 3 space rightwards arrow with MnO subscript 2 space space space space on top space 2 KCl space plus space 3 straight O subscript 2

i)  Calculate the mass of KClO3 required to produce 6.72 litres of O2 at STP.

[atomic masses of K =39, Cl=35.5, O=16]

ii)  Calculate the number of moles of oxygen present in the above volume and also the number of molecules.

iii) Calculate the volume occupied by 0.01 mole of CO2 at STP
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53.

Copper sulphate solution is electrolyzed using copper electrodes

Study the diagram given below and answer the question that follows:



(i) Which electrodes to your left or right is known as the oxidising electrode and why?

(ii) Write the equation representing the reaction that occurs.

(iii)  State two appropriate observation for the above electrolysis reaction.

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