Answer:
Molarity = moles ÷ liters
to get moles of NaBr divide grams of NaBr by its molar mass (mass of Na + mass of Bromine)
Na = 22.989769
Br = 79.904
molar mass of NaBr = 102.893769
6.6g ÷ 102.893769 = 0.064143826 moles of NaBr
0.064143826 moles ÷ 0.60 liters = 0.1069 molar concentration or 11 %
Answer:
Definitely add more points
Explanation:
I'm saying this because this is worth way more points
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
36.49 gm using law of equivalent proportion
