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:
It kinda helps but not really
Thanks for trying anyway doe!
Explanation:
Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>
Answer:
D. the organism is able to survive and reproduce more than others in it's environment
Explanation:
It will only go extinct if it can't survive or reproduce.
