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:
0.0498 mol
Explanation:
Number of moles = concentration in mol/L × volume in L
Concentration = 1 M = 1 mol/L
Volume = 49.8 mL = 49.8/1000 = 0.0498 L
Number of moles = 1×0.0498 = 0.0498 mol
Answer: sorry I’m late but it is 11 electrons
Explanation:
Answer:
Explanation:
The pressure is constant, so we can use Charles' Law to calculate the volume.
Data:
V₁ = 693 mL; T₁ = 45 °C
V₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert temperature to kelvins
T₁ = ( 45 + 273.15) = 318.15 K
T₂ = (150 + 273.15) = 423.15 K
(b) Calculate the volume
Answer:
Glucose and oxygen are required for cellular respiration. As the law of conversation states, in a biochemical reaction, mass is conserved. The mass of hydrogen in the glucose is therefore conserved in the water molecules products.
