The mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g
<h3>Calculating mass </h3>
From the question we are to calculate the mass of NaHCO₃ (sodium bicarbonate) used in the experiment
From the given information
Mass of empty evaporating dish = 46.233g
Mass of evaporating dish + Sodium bicarbonate = 48.230g
∴ Mass of sodium bicarbonate (NaHCO₃) = [Mass of evaporating dish + Sodium bicarbonate] - [Mass of empty evaporating dish]
Mass of sodium bicarbonate (NaHCO₃) = 48.230g - 46.233g
Mass of sodium bicarbonate (NaHCO₃) = 1.997 g
Hence, the mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g
Learn more on Calculating mass here: brainly.com/question/15268826
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⁻
55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g
A. a mixture is a combination of two or more substances in which each substance retains its own properties.
