Answer:
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Explanation:
2Ag⁺(aq) + Mg(s)→ 2Ag(s) + Mg²⁺ (aq)
<h3>Further explanation</h3>
Given
Standard cell notation:
Mg(s) | Mg2+ (aq) || Ag+(aq)| Ag(s)
Required
a balanced redox reaction
Solution
At the cathode the reduction reaction occurs, the anode oxidation reaction occurs
In reaction:
Ag⁺ + Mg → Ag + Mg²⁺
half-reactions
- at the cathode (reduction reaction)
Ag⁺ (aq) + e⁻ ---> Ag (s) x2
2Ag⁺ (aq) + 2e⁻ ---> 2Ag (s)
- at the anode (oxidation reaction)
Mg (s) → Mg²⁺ (aq) + 2e−
a balanced cell reaction
<em>2Ag⁺(aq) + Mg(s)→ 2Ag(s) + Mg²⁺ (aq)
</em>
Answer:
39.72 g
Explanation:
Given data:
Mass of CsF = 15.2 g
Mass of XeF₆ = 260 g
Mass of Cs[XeF₇] = ?
Solution:
Chemical reaction:
CsF + XeF₆ → Cs[XeF₇]
Number of moles of CsF:
Number of moles = mass/ molar mass
Number of moles = 15.2 g/151.9 g/mol
Number of moles = 0.1 mol
Number of moles of XeF₆ :
Number of moles = mass/ molar mass
Number of moles = 260 g/245.28 g/mol
Number of moles = 1.06 mol
Now we will compare the moles of Cs[XeF₇] with both reactants.
CsF : Cs[XeF₇]
1 : 1
0.1 : 0.1
XeF₆ : Cs[XeF₇]
1 : 1
1.06 ; 1.06
Number of moles of Cs[XeF₇] produce by CsF are less so it will limiting reactant and limit the yield of Cs[XeF₇].
Mass of Cs[XeF₇]:
Mass = number of moles × molar mass
Mass = 0.1 mol × 397.2 g/mol
Mass = 39.72 g
Explanation:
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