In a closed system, If a gas is transported to a container with double the volume of the previous container, the gas was held in
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Answer : The molar heat of solution of KBr is 19.9 kJ/mol
Explanation :
Mass of KBr = 7.00 g
Molar mass of KBr = 119 g/mole
Heat capacity = 2.72 kJ/K
Change in temperature = 0.430 K
First we have to calculate the moles of KBr.
Now we have to calculate the heat of the reaction.
where,
q = amount of heat = ?
= heat capacity =
= change in temperature = 0.430 K
Now put all the given values in the above formula, we get:
Now we have to calculate the molar heat of solution of KBr.
where,
n = number of moles of KBr
Therefore, the molar heat of solution of KBr is 19.9 kJ/mol
<u>Answer:</u> The total volume of the gaseous products is 1044.29 L
<u>Explanation:</u>
We are given:
Volume of butane = 116 L
At STP:
22.4 L of volume is occupied by 1 mole of a gas
So, 116 L of volume will be occupied by = of butane
The chemical equation for the combustion of butane follows:
- <u>For carbon dioxide:</u>
By Stoichiometry of the reaction:
2 moles of butane produces 8 moles of carbon dioxide
So, 5.18 moles of butane will produce = of carbon dioxide
Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L
- <u>For water vapor:</u>
By Stoichiometry of the reaction:
2 moles of butane produces 10 moles of water vapor
So, 5.18 moles of butane will produce = of water vapor
Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L
Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L
Hence, the total volume of the gaseous products is 1044.29 L