Explanation:
FeCl3 + <u>3</u> KSCN ➡ <u>3</u> KCl + Fe(SCN)3
Hope it helps
The answer to this question would be: 3.125%
Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
Answer:
a. -0.63 V
b. No
Explanation:
Step 1: Given data
- Standard reduction potential of the anode (E°red): -1.33 V
- Minimum standard cell potential (E°cell): 0.70 V
Step 2: Calculate the required standard reduction potential of the cathode
The galvanic cell must provide at least 0.70V of electrical power, that is:
E°cell > 0.70 V [1]
We can calculate the standard reduction potential of the cathode (E°cat) using the following expression.
E°cell = E°cat - E°an [2]
If we combine [1] and [2], we get,
E°cat - E°an > 0.70 V
E°cat > 0.70 V + E°an
E°cat > 0.70 V + (-1.33 V)
E°cat > -0.63 V
The minimum E°cat is -0.63 V and there is no maximum E°cat.
