Answer:
87.3 calories of heat is required.
Explanation:
Heat = mcΔT
m= mass, c = specific heat of silver, T = temperature
H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)
H = 57.8 * 0.057 * 26.5
H = 87.3069 cal.
The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.
Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18