Crossover formula of potassium oxide.
1 answer:
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Answer:
ΔH = - 2020.57 kJ/mol
Explanation:
Given that :
mass of propanol = 1.685 g
the molar molar mass = 60 g/mol
Thus; the number of moles = mass/molar mass
= 1.685 g/60 g/mol
= 0.028 g/mol
However ;
ΔH = heat capacity C × Δ T
Given that:
The temperature increases from 298.00 K to 302.16 K.
Then ;
Δ T = 302.16 K - 298.00 K
Δ T = 4.16 K
heat capacity C = 13.60 kJ/K
∴
ΔH = 13.60 kJ/K × 4.16 K
ΔH = 56.576 kJ
The equation of the given reaction can be represented as :
Thus for 0.028 mol of heat liberated; ΔH = 56.576 kJ
For 1 mole of heat liberated now:
ΔH = 56.576 kJ/0.028 mol
ΔH = 2020.57 kJ/mol
SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;
ΔH = - 2020.57 kJ/mol
% composition of ethanol = 34.51%
% composition of water = 65.49%
<h3>What is density?</h3>A material's density is defined as its mass per unit volume.
Given data:
The density of ethanol = 0.7890 g/mL
The density of water = 0.9982 g/mL
The density of mixture = 0.926 g/mL
Let the % composition of ethanol = x
Let the % composition of water = 100-x
Now density of the mixture
%
Hence,
% composition of ethanol = 34.51%
% composition of water = 65.49%
Learn more about the density here:
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