Answer:
6.1×10^8
Explanation:
The reaction is;
Sn^2+(aq) + Cd(s) -----> Sn(s) + Cd^2+(aq)
E°cell = E°cathode - E°anode
E°cathode= -0.14 V
E°anode= -0.40 V
E°cell = -0.14-(-0.40)
E°cell= -0.14+0.40
E°cell= 0.26 V
But
E°cell= 0.0592/n log K
E°cell= 0.0592/2 log K
0.26= 0.0296log K
log K = 0.26/0.0296
log K= 8.7838
K= Antilog (8.7838)
K= 6.1×10^8
Answer:
They Are all O's/oberavtions, because inference is using facts and reasoning, which is not the case here.
Explanation:
Answer:
6,500 gm of steam require= 3,510 kilo calories (approx)
Explanation:
Every 1 gram of water at 100° C absorb 540 calories
So,
Total water = 6.5 kg = 6,500 gram
So,
6,500 gm of steam require = 6,500 x 540
6,500 gm of steam require= 3,510 kilo calories (approx)
Answer:
0.581 L or 581 mL
Explanation:
As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)
Write down the amounts you are given.
V1 = 0.152 L (I was taught to always convert milliliters to liters)
P1 = 717 mmHg
T1 = 315 K
V2 = ?
P2 = 463 mmHg
T2 = 777 K
The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.
(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)
0.346 = (463*V2) / (777)
0.346*777 = (463*V2) / (777)*777
268.842 = 463*V2
268.842/463 = (463*V2)/463
V2 = 0.581
Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.
Answer:
Enthalpy of vaporization = 30.8 kj/mol
Explanation:
Given data:
Mass of benzene = 95.0 g
Heat evolved = 37.5 KJ
Enthalpy of vaporization = ?
Solution:
Molar mass of benzene = 78 g/mol
Number of moles = mass/ molar mass
Number of moles = 95 g/ 78 g/mol
Number of moles = 1.218 mol
Enthalpy of vaporization = 37.5 KJ/1.218 mol
Enthalpy of vaporization = 30.8 kj/mol
