Answer: I think it’s the first one
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 :]
to convert from g NaOH to mol NaOH. = 1.48 g NaOH are needed to neutralize the acid.
The answer is 23.5° but I guess 23° is closest
KCl and PbCl2 both are salts having the same white color, however, potassium salts are soluble in water while lead salts are not.
This means that KCl is soluble in water while PbCl2 is not.
So, to distinguish between them, add the same amount of each salt in a beakers containing water (each salt in a separate beaker of course), ans shake the beaker or steer it.
The salt that dissolves in water would be KCl while the salt that doesn't dissolve in water would be PbCl2.