Answer:
+5
Explanation:
You use the rules for oxidation numbers to calculate that the oxidation number of N is +5.
Answer:
1.125 moles of H2S
Explanation:
From 2HCl(aq) + ZnS(s) ---------> H2S (g) + ZnCl2 (aq)
From the reaction equation;
2 moles of HCl produces 1 mole of H2S
Therefore 2.25 moles of HCl will produce 2.25 ×1/2 = 1.125 moles of H2S
Recall that it was explicitly stated in the question that ZnS is the reactant in excess. This implies that HCl is the limiting reactant and controls the amount of product obtained.
Answer is: Ka for propanoic acid is 6,57·10⁻⁵.
Chemical reaction: C₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).
n(C₂H₅COOH) = 0,04 mol.
V(C₂H₅COOH) = 750 mL = 0,75 L.
c(C₂H₅COOH) = 0,04 mol ÷ 0,75 L.
c(C₂H₅COOH) = 0,053 mol/L = 0,053 M.
[C₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [</span>C₂H₅COO⁻] ·
[H₃O⁺] / [C₂H₅COOH].
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10⁻⁵.
Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
