A resistor has a voltage Across it of 3kv and a current through it of 3m A .
1 answer:
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Answer :
(a) 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.
(b) There is no other element present in the compound.
Explanation :
(a) Now we have to determine the masses of C and H in the sample.
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
where, 'x' and 'y' are the subscripts of Carbon and hydrogen respectively.
We are given:
Mass of
Mass of
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 3.381 g of carbon dioxide, of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in 0.692 g of water, of hydrogen will be contained.
Thus, 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.
(b) Now we have to determine the compound contain any other elements or not.
Mass carbon + Mass of hydrogen = 0.922 g + 0.0769 g = 0.999 g ≈ 1 g
This means that there is no other element present in the compound.
Answer: 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require = 1 mole of
Thus 0.061 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 1 mole of give = 1 mole of
Thus 0.061 moles of give =
of
Mass of
Thus 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.