The decomposition of calcium carbonate, CaCO3(s) --> CaO(s) + CO2(g), has the following values for free energy and enthalpy at 25.0°C. G = 130.5 kJ/mol. H = 178.3 kJ/mol.
Answer:
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Explanation:
Answers:
See below
Step-by-step explanation:
1. Most food energy
(a) Pringles
Heat from Pringles + heat absorbed by water = 0
m₁ΔH + m₂CΔT = 0
1.984ΔH + 100 × 4.184 × 18 = 0
1.984ΔH + 7530 = 0
ΔH = -7530/1.984 = -3800 J/g
(b) Cheetos
0.884ΔH + 418.4 × 13 = 0
ΔH = -5400/0.884 = -6200 J/g
Cheetos give you more food energy per gram.
(c) Snickers
Food energy = 215 Cal/28 g × 4184 J/1 Cal = 32 000 J/g
The food energy from Cheetos is much less than that from a Snickers bar
2. Experimental uncertainty
The experimental values are almost certainly too low.
Your burning food is heating up the air around it, so much of the heat of combustion is lost to the atmosphere.
3. Percent efficiency
Experimental food energy = 3800 J/g
Actual food energy = 150 Cal/28 g × 4184 J/1 Cal = 22 000 J/g
% Efficiency = Experimental value/Actual value × 100 %
= 3800/22 000 × 100 %
= 17 %
Answer:
The answer is explained below:
Explanation:
Given the chemical equation:
N2H4(g)+H2(g)→2NH3(g)
The standard enthalpy of formation is given by the following formula:
ΔH^0 rxn = ∑ B reactants - ∑ product
-187.78 kJ/mol = [( 1x B N-N) + (1 x B N-H ) + (1 x b H-H)] - [ 6 x B N-H]
-187.78 kJ/mol = [( 1x B N-N) +(4 x 391 kJ/mol) + (1 x 436 kJ/mol)] - [ 6x 391 kJ/mol ]
-187.78 kJ/mol = B N-N + (1564 + 436 - 2346) kJ/mol
B = 158.22 kJ/mol
So, in this case the enthalpy of N-N bond is 158 kJ/mol
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