Answer:
Explanation:
1. the 1/2 reaction that occurs at the cathode
3Cl2(g) +6e^- -------------> 6Cl^- (aq)
2 the 1/2 reaction that occurs at the anode
2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-
2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-
E0 = -0.59v
3Cl2(g) +6e^- -------------> 6Cl^- (aq)
E0 = 1.39v
3Cl2 (g) + 2MnO2 (s) + 8OH^(−) (aq)---------> 6Cl^(−) (aq) + 2MnO4^(−) (aq) + 4H2O (l)
E0cell = 0.80v
Answer:
3750 cm.
Explanation:
You multiply the three side measurements to find the volume.
25cm·10cm·15cm
375cm·10cm
3750 cm.
<em><u>Hope this helps!</u></em>
Answer:
Mass = 1.33 g
Explanation:
Given data:
Mass of argon required = ?
Volume of bulb = 0.745 L
Temperature and pressure = standard
Solution:
We will calculate the number of moles of argon first.
Formula:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
By putting values,
1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K
0.745 atm. L = n × 22.43 atm.L/mol
n = 0.745 atm. L / 22.43 atm.L/mol
n = 0.0332 mol
Mass of argon:
Mass = number of moles × molar mass
Mass = 0.0332 mol × 39.95 g/mol
Mass = 1.33 g
