Answer:
V₁ = 208.3 mL
Explanation:
Given data:
Initial molarity of HCl = 6.0 M
Final volume = 500 mL
Final molarity = 2.5 M
Volume of initial solution required = ?
Solution:
Formula:
M₁V₁ = M₂V₂
Now we will put the values in formula.
6.0 M × V₁ = 2.5 M ×500 mL
6.0 M × V₁ = 1250 M.mL
V₁ = 1250 M.mL / 6.0 M
V₁ = 208.3 mL
Answer is: 153.52 grams of hypobromous acid <span>must be added.
</span>Chemical dissociation: HBrO ⇄ H⁺ + BrO⁻.
pH = 4.25.
pH = -log[H⁺].
[H⁺] = 10∧(-pH).
[H⁺] = 10∧(-4.25).
[H⁺] = [BrO⁻] = 5.62·10⁻⁵ M.
Ka = [H⁺] · [BrO⁻] / [HBrO].
2.00·10⁻⁹ = (5.62·10⁻⁵ M)² / [HBrO].
[HBrO] = 3.16·10⁻⁹ M² / 2.00·10⁻⁹.
[HBrO] = 1.58 M.
m(HBrO) = n(HBrO) · M(HBrO).
m(HBrO) = 1.58 mol · 96.91 g/mol.
m(HBrO) = 153.52 g.
Answer:
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2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Mno4- (aq) + Cl- (aq) → Mn2+ + Cl2 (g)
Required
Half reaction
Solution
1. Add coefficient(equalizing atoms in reaction)
2. Adding H₂O on the O-deficient side.
3. Adding H⁺ on the H-deficient side.
4. Adding e⁻
5. Equalizing the number of electrons and sum the all the half-reaction
1.MnO₄⁻(aq) = Mn²⁺(aq) reduction
2.MnO4(aq) = Mn2+(aq) + 4H2O
3. MnO4-(aq) + 8H+ = Mn2+(aq) + 4H2O
4. MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O
Cl⁻(aq) = Cl₂(g) oxidation
1. 2Cl-(aq) = Cl2 (g)
2-3 none
4. 2Cl-(aq) = Cl2 (g) + 2e-
5.
MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O x2
2Cl-(aq) = Cl2 (g) + 2e- x5
2MnO4-(aq) + 16H+ + 10e- = 2Mn2+(aq) + 8H2O
10Cl-(aq) = 5Cl2 (g) + 10e-
<em>2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)</em>