Can someone please answer this question for me.
1 answer:
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Answer:
The temperature change from the combustion of the glucose is 6.097°C.
Explanation:
Benzoic acid;
Enthaply of combustion of benzoic acid = 3,228 kJ/mol
Mass of benzoic acid = 0.570 g
Moles of benzoic acid =
Energy released by 0.004667 moles of benzoic acid on combustion:
Heat capacity of the calorimeter = C
Change in temperature of the calorimeter = ΔT = 2.053°C
Glucose:
Enthaply of combustion of glucose= 2,780 kJ/mol.
Mass of glucose=2.900 g
Moles of glucose =
Energy released by the 0.016097 moles of calorimeter combustion:
Heat capacity of the calorimeter = C (calculated above)
Change in temperature of the calorimeter on combustion of glucose = ΔT'
The temperature change from the combustion of the glucose is 6.097°C.
Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:
a) moles of butane
b) moles of oxygen
According to stoichiometry :
2 moles of butane require 13 moles of
Thus 0.09 moles of butane will require = of
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.