Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
Answer:
36.2 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 8.6 atm
- Initial temperature of the gas (T₁): 38°C
- Final pressure of the gas (P₂): 1.0 atm (standard pressure)
- Final temperature of the gas (T₂): ?
Step 2: Convert T₁ to Kelvin
We will use the following expression.
K = °C +273.15
K = 38 °C +273.15 = 311 K
Step 3: Calculate T₂
We will use Gay Lussac's law.
P₁/T₁ = P₂/T₂
T₂ = P₂ × T₁/P₁
T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K
Answer:
4.92 L
Explanation:
Rearrange ideal gas law and solve.
Change C to K.
- Hope that helps! Please let me know if you need further explanation.
Answer:
A. 32.06 g/mol
Explanation:
The molar mass units are always g/mol