The primary colors are blue, red, and yellow.
The reactant is what you begin with.
The product is what you end up with (so the answer is B)
B) the number of electrons
Answer:
Cd is oxidized during the discharge of the battery
Explanation:
Based on the reaction:
2 NiOOH + Cd + 2H₂O → 2Ni(OH)₂ + Cd(OH)₂
And knowing Oxygen and hydrogen never change its charge, we must to find oxidation state of Ni and Cd before and after the reaction:
<em>Ni:</em>
In NiOOH: 2 O = -2*2 = -4 + 1H = +1, = -4 + 1 = -3. And as the molecule is neutral, Ni is 3+
In Ni(OH)₂: OH = -1. As there are 2 OH = -2. That means Ni is +2
The Ni is gaining one electron, that means is been reduced
<em>Cd:</em>
Cd before reaction is as pure solid with oxidation state = 0
Cd after the reaction is as Cd(OH)₂: 2 OH = -2. That means Cd is +2
The Cd is loosing 2 electrons, that means is the species that is oxidized.
Answer:
ΔG = -52.9 kJ/mol
Explanation:
Step 1: Data given
Temperature = 298 K
All species have a partial pressure of 1 atm
Δ G ° = − 69.0 kJ/mol
Step 2: The balanced equation
N2(g) + 3H2(g) ⇆ 2NH3 (g)
Step 3: Calculate Q
we will use the expression: ΔG = ΔG° + RT*ln(Q)
⇒with Q = the reaction coordinate: Q = (PNH3)²/ ((PN2)*(Ph2)³) = 666.67
Step 4: Calculate ΔG
So, ΔG = -69.0 kJ/mol + (0.008314 kJ/mol*K)*(298 K)*ln(666.67) = -52.9 kJ/mol
(R = the gas constant = 8.314 J/mol* K OR 0.008314 kJ/mol*K)