The number of electrons that can be held in the second orbit are 8
Answer:
5 moles of electrons
Explanation:
The balance equation is as follow,
<span> 5 Ag</span>⁺ + Mn⁺²<span> + 4 H</span>₂O →<span> 5 Ag + MnO</span>₄⁻<span> + 8 H</span>⁺
Reduction of Ag:
Ag⁺ + 1 e⁻ → Ag
Or,
5 Ag⁺ + 5 e⁻ → 5 Ag
Oxidation of Mn:
Mn⁺² → MnO₄⁻ + 5 e⁻
Result:
Hence 5 moles of Ag⁺ accepts 5 electrons from 1 mole of Mn⁺².
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol
Find the molar mass
2Na = 23 * 2 = 46 grams
1S = 32 * 1 = 32 grams
O4 = 16 * 4 = 64 grams
Total = 142 grams / mol
Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???
given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams.
85.2 are in a 2 L solution that has a concentration of 0.6 mol/L
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