Answer:
The answer to your question: 0.7 M
Explanation:
Data
V of KOH = 90 ml
[KOH] = ?
V H2SO4 = 21.2 ml
[H2SO4] = 1.5 M
2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)
Molarity = moles / volume
moles of H₂SO₄ = (1.5) (21.2)
= 31.8
2 moles of KOH -------------- 1 mol of H₂SO₄
x -------------- 31.8 mol of H₂SO₄
x = (31.8)(2) / 1
x = 63.8 moles of KOH
Molarity = 63.8 / 90
= 0.7 M
**Visible confusion** no problem
Answer:
Kc =![\frac{[8.326x10-3]^{1} }{[1.113x10-2]^{1}[1.490x10-2]^{1} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B8.326x10-3%5D%5E%7B1%7D%20%7D%7B%5B1.113x10-2%5D%5E%7B1%7D%5B1.490x10-2%5D%5E%7B1%7D%20%20%7D)
Kc = 50.2059
Explanation:
1. Balance the equation
2. Use the Kc formula
Remember that pure substances, like H2 are not included on the Kc formula
According to ideal gas equation, we know for 1 mole of gas: PV=RT
where P = pressure, T = temperature, R = gas constant, V= volume
If '1' and '2' indicates initial and final experimental conditions, we have

Given that: V1 = 100.0 kPa, T1 = 100.0 K, V1 = 2.0 m3, T2 = 400 K, P2 = 200.0 kPa
∴ on rearranging above eq., we get V2 =

∴ V2 = 4 m3