The given chemical reaction is:
Δ∑BE(reactants)-∑BE(products)
= {(941 kJ/mol) + (3 * 242 kJ/mol)} -[{2*(3*200 kJ/mol)}]
= 467 kJ/mol
Calculating the change in heat when 85.3 g chlorine reacts in the above reaction:
Moles of chlorine =
= 1.20 mol
Heat change when 1.20 mol chlorine reacts
=
Answer:
Explanation:
Hello,
Based on the stoichiometry, the produced moles of carbon dioxide are computed via:
Now, by using the ideal gas law and subsequently solving for the volume we get:
Best regards.
Answer:
69.8%
Explanation:
k = 0.00813 sec-1
t = 44.1 s
What percentage of the compound has decomposed?
We can obtain this using the formular;
[A] = [A]o e^(-kt)
where;
Final Concentration = [A]
Initial concentration = [A]o
[A] / [A]o = e^(-kt)
[A] / [A]o = e^(-0.00813*44.1)
[A] / [A]o = e^(-0.359)
[A] / [A]o = 0.698
Percentage decomposed = Final Concentration / Initial concentration * 100
Percentage decomposed = [A] / [A]o * 100%
Percentage decomposed = 0.698 * 100 = 69.8%
Answer:
The answer would be Triple Covalent!
Explanation:
Answer:
Explanation:
Calcium carbonate decomposes at high temperatures to give calcium oxide and carbon
dioxide as shown below.
CaCO3(s) = CaO(s) + CO2(g)
The Kp for this reaction is 1.16 at 800°C. A 5.00 L vessel containing 10.0 g of CaCO3(s)
was evacuated to remove the air, sealed, and then heated to 800°C. Ignoring the volume
occupied by the solid, what will overall mass percent of carbon in the solid once equilibrium is reached?