Answer:
V = 0.356 L
Explanation:
In this case, we need to use the following expression:
M = n/V (1)
Where:
M: molarity of solution (mol/L or M)
n: moles of solute (moles)
V: Volume of solution (Liters)
From these expression, we can solve for V:
V = n/M (2)
Now, replacing the given data we can solve V:
V = 8.9 / 25
V = 0.356 L
Answer:
4.23.
Explanation:
<em>∵ pH = - log[H⁺].</em>
<em>For weak acids:</em>
∵ [H⁺] = √(ka)(c).
∴ [H⁺] = √(3.5 × 10⁻⁸)(0.10 M) = 5.92 x 10⁻⁵.
∴ pH = - log[H⁺] = - log(5.92 x 10⁻⁵) = 4.2279 ≅ 4.23.
Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
<em />
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>
