The heat of combustion of butane , c4h10 , is 2874 kj/mol. what is the heat of combustion of butane, in kj per gram?(provide the
2 answers:
Answer : 49.45 kJ per gram.
Explanation:
Heat of combustion is the amount of heat released on complete combustion of 1 mole of substance.
Given :
Amount of heat released on combustion of butane = 2874 kJ/mol
According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number of particles.
1 mole of weighs = 58.12 g
Thus we can say:
58.12 g of on combustion releases = 2874 kJ
Thus 1 g of on combustion releases =
Thus the heat of combustion of butane, in kJ per gram is 49.45.
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Solution :
Given :
Amount of anserine solution = 0.200 M
pH value is = 7.20
Preparation of 0.04 M solution of anserine from the 0.2 M solution.
0.2 M x = 0.04 M x 1000 ml
= 200 ml
So the 200 ml of 0.2 M anserine solution is required to prepare0.04 M of anserine.
0.1 M x = 0.04 x 1000 ml
= 400 ml
Therefore, 400 ml of HCl is needed.
Answer:
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