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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Answer: 0.20 M
Explanation:
According to the dilution law,
where,
= molarity of stock solution = 1.40 M
= volume of stock solution = 72.0 ml
= molarity of diluted solution = m
= volume of diluted solution = 248 ml
Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.
According to the dilution law,
where,
= molarity of stock solution = 0.41 M
= volume of stock solution = 124 ml
= molarity of diluted solution = m
= volume of diluted solution = (124 +133) ml = 257 ml
Thus the final concentration of the solution is 0.20 M.