Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
Answer:
mi juego favorito probablemente sería Mine craft cuando era más joven, aunque sigue siendo uno de mis juegos favoritos
Explanation:
Explanations:
<u>Question</u> <u>1:</u> Lithium in 20.00+ g is C. or D., but 25.00+ g is D. which means this is the correct option.
I am unsure of <u>Question</u> <u>2</u>. I don't think it is mole though.
<u>Question</u> <u>3:</u> Boron in 25.00-30.00 g is B. or D., but 25.00 g would be C.
<u>Question</u> <u>4:</u> 2.393 x 1024 atoms of Oxygen is 63.58 mole O. I don't know for sure, but I think this is correct.
<u><em>I am NOT professional. There is a chance I am incorrect. Please reply to me if I've made a mistake.</em></u>
Answer:
2.28 × 10^-3 mol/L
Explanation:
The equation for the equilibrium is
CN^- + H2O ⇌ HCN + OH^-
Ka = 4.9 × 10^-10
KaKb = Kw
4.9 × 10^-10 Kb = 1.00 × 10^-14
Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5
Now, we can set up an ICE table
CN^- + H2O ⇌ HCN + OH^-
I/(mol/L) 0.255 0 0
C/(mol/L) -x +x +x
E/(mol/L) 0.255 - x x x
Ka = x^2/(0.255 - x) = 2.05 × 10^-5
Check for negligibility
0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255
x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6
x = sqrt(5.20 × 10^-6) = 2.28 × 10^-3
[OH^-] = x mol/L = 2.28 × 10^-3 mol/L
Answer:
b is ur answer the temputer does increase
Explanation:
