The question is incomplete, the complete question is
Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V
Answer:
only IV and V
Explanation:
If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.
On the other hand, solve has a positive reduction potential indicating a tendency towards reduction.
This implies that vanadium must be oxidized and silver reduced and not the not her way ground? Hence the answer above.
Answer:
596K
Explanation:
Using Charles law equation;
V1/T1 = V2/T2
Where;
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question,
V1 = 3.00 L
V2 = double of V1 = 2 × 3.00 = 6.00 L
T1 = 25°C = 25 + 273 = 298K
T2 = ?
Using V1/T1 = V2/T2
3/298 = 6/T2
Cross multiply
298 × 6 = 3 × T2
1788 = 3T2
T2 = 1788 ÷ 3
T2 = 596K
