Answer:
uh yeah I think so technically
Answer:
665 g
Explanation:
Let's consider the following thermochemical equation.
2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol
According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:
-1.00 × 10⁴ kJ × (8 mol CO₂/-5,314 kJ) = 15.1 mol CO₂
The molar mass of CO₂ is 44.01 g/mol. The mass corresponding to 15.1 moles is:
15.1 mol × 44.01 g/mol = 665 g
PH = -log([H+])
[H+] = 10^(-pH)
[H+] = 10^(-9)
[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.
[OH-] = Kw/[H+] = (1.0*10^-14)/(1*10^-9) = 1.0*10^-5
The concentration of OH- ions is 1.0*10^-5 M.
Answer:
0.456 M
Explanation:
Step 1: Write the balanced neutralization equation
HNO₂ + KOH ⇒ KNO₂ + H₂O
Step 2: Calculate the reacting moles of KOH
9.26 mL of 1.235 M KOH react.
0.00926 L × 1.235 mol/L = 0.0114 mol
Step 3: Calculate the reacting moles of HNO₂
The molar ratio of HNO₂ to KOH is 1:1. The reacting moles of HNO₂ are 1/1 × 0.0114 mol = 0.0114 mol.
Step 4: Calculate the initial concentration of HNO₂
0.0114 moles of HNO₂ are in 25.0 mL of solution.
[HNO₂] = 0.0114 mol / 0.0250 L = 0.456 M