Answer:
Kc = 3.72 × 10⁶
Explanation:
Let's consider the following reaction:
NH₄HS(g) ⇄ NH₃(g) + H₂S(g)
At equilibrium, we have the following concentrations:
[NH₄HS] = 0.196 M (assuming a 1 L flask)
[NH₃] = 9.56 × 10² M
[H₂S] = 7.62 × 10² M
We can replace this data in the Kc expression.
![Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2} \times 7.62 \times 10^{2}}{0.196} =3.72 \times 10^{6}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNH_%7B3%7D%5D%20%5Ctimes%20%5BH_%7B2%7DS%5D%20%7D%7B%5BNH_%7B4%7DHS%5D%7D%20%3D%5Cfrac%7B9.56%20%5Ctimes%2010%5E%7B2%7D%20%20%5Ctimes%207.62%20%20%5Ctimes%2010%5E%7B2%7D%7D%7B0.196%7D%20%3D3.72%20%5Ctimes%2010%5E%7B6%7D)
Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol
Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.
W are given a chemical reaction:



To calculate the enthalpy change, we use the formula:

This is the amount of energy released when 0.1326 grams of sample was burned.
So, energy released when 1 gram of sample was burned is = 
Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

Answer: grams=0.048g, ounces=0.0017oz, 0.00011lb
Explanation:
Stoichiometry
48 mg x 1 g
÷ 1000 mg = 0.048 g
48 mg x 1 g x 16 oz
÷ 1000 mg ÷ 453.6 g = 0.0017 oz
48 mg x 1 g x 1 lb
÷ 1000 mg ÷ 453.6 g = 0.00011 lb