Must contain: 6 protons, 6 electrons and 12 neutrons.
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:
oxygen is an element because it is a pure substance which cannot be split into simpler substances by chemical means
Answer:
The original volume of the gas is 0.001 mL
Explanation:
This easy excersise can be solved by the law for gases, about pressure and volume; the volume of a gas is inversely proportional to the pressure it exerts.
We can propose the rule by this formula:
P₁ / V₁ = P₂ / V₂
We replace data given: 1.50 atm / V₁ = 0.50 atm / 750 mL
As the rule says, that volume is inversely proportional, and the pressure was decreased, volume must be lower than 750 mL.
1.5atm / (0.5 atm / 750mL) = V₁
V₁ = 0.001 mL
