Answer:
The answer to your question is 0.34 M
Explanation:
Data
[Sr(OH)₂] = ?
Volume of Sr(OH)₂ = 35.6 ml or 0.0356 l
[HBr] = 0.0445 M
Volume of HBr = 0.549 l
Balanced chemical reaction
2HBr + Sr(OH)₂ ⇒ SrBr₂ + 2H₂O
Process
1.- Calculate the moles of HBr
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = 0.0445 x 0.549
= 0.0244
2.- Calculate the moles of Sr(OH)₂ using the coefficients of the balanced reaction
2 moles of HBr ----------------------- 1 mol of Sr(OH)₂
0.0244 moles ----------------------- x
x = (0.0244 x 1) / 2
x = 0.0122 moles of Sr(OH)₂
3.- Calculate the concentration of Sr(OH)₂
Molarity = 0.0122/ 0.0356
-Simplification
Molarity = 0.34
We are given the rate law, so we can substitute the given values for the rate constant and the concentrations of the reactants to solve for the rate of reaction. Since rate = k [NH4+][NO2-]:
rate = (2.7 x 10^-4 / M-s)(0.050 M)(0.25 M) = 3.375 x 10^-6 M/s
Answer:
-252.5 kJ/mol = ΔH H2O(g)
Explanation:
ΔH Fe2O3 = -825.5kJ/mol
ΔH H2 = 0kJ/mol
ΔH Fe = 0kJ/mol
Based on Hess's law, ΔH of a reaction is the sum of ΔH of products - ΔH of reactants. For the reaction:
Fe2O3(s) + 3 H2(g) →2Fe(s) + 3 H2O(g)
ΔHr = 67.9kJ/mol = 3*ΔH H2O + 2*ΔHFe - (ΔH Fe2O3 + 3*Δ H2)
67.9kJ/mol = 3*ΔH H2O + 2*0kJ/mol - (ΔH -825.5kJ/mol + 3*Δ H2)
67.9 = 3*ΔH H2O(g) + 825.5kJ/mol
-757.6kJ/mol = 3*ΔH H2O(g)
<h3>-252.5 kJ/mol = ΔH H2O(g)</h3>