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⁻
Rock paper scissor and that is on Albert Einstein , he made that up because he is so freaking smart so there you go have a wonderful day
Its A because I did it on connexus
The initial temperature of the metal = 35 °C
<h3>Further explanation</h3>
Heat can be formulated :
Q = m . c . ΔT
Q = heat, J
c = specific heat, J/g C
ΔT = temperature, °C
m = 20 g
c = 5 J/(g°C)
Q = 500 J
T₁ = 40 C
the initial temperature :
1. C : Mg(CN)2
2. B : N2O5
3. D : Ti(ClO4)3
4. A : Ni(NO3)3
5. D : Cobalt (III) Acetate
6. B : Nickel(II) sulfate
7. C : Dinitrogen Tetrafluoride
8. A : Phosphorus pentachloride<em />
9. C : Metallic <em>(<!> This is the only one I'm not 100% sure of)</em>
10. A : Ionic
11. C : Metallic
12. B : Covalent
