Answer:
0.68 V
Explanation:
For anode;
3Mg(s) ---->3Mg^2+(aq) + 6e
For cathode;
2Al^3+(aq) + 6e -----> 2Al(s)
Overall balanced reaction equation;
3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)
Since
E°anode = -2.356 V
E°cathode = -1.676 V
E°cell=-1.676 -(-2.356)
E°cell= 0.68 V
You would weigh the zinc with a weight, Zinc is a mineral so wether it's a solid or a liquid you would measure it with a zinc.
Answer:
V = 42.6 L
Explanation:
Given data:
Number of moles of Cl₂ = 1.9 mol
Temperature and pressure = standard
Volume occupy = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm × V = 1.9 mol ×0.0821 atm.L /mol.K × 273.15 k
V = 42.6 atm.L / 1 atm
V = 42.6 L
