Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
Answer:
The outside temperature is -45.8°C
Explanation:
When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).
V1 / T1 = V2 / T2
2.95L/298K = 2.25L / T2
(2.95L/298K ) . T2 = 2.25L
T2 = 2.25L . 298K / 2.95L
T2 = 227.2K
T°K - 273 = T°C
227.2K - 273 = -45.8°C
Answer:
CH4 + 4Cl2 --> CCl4 + 4HCl
Explanation: Message me if you need a thorough explanation
Answer:
1.88 × 10²² Molecules of CO
Explanation:
At STP for an ideal gas,
Volume = Mole × 22.4 L/mol
Or,
Mole = Volume / 22.4 L/mol
Mole = 0.7 L / 22.4 L/mol
Mole = 0.03125 moles
Now,
No. of Molecules = Moles × 6.022 × 10²³ Molecules/mol
No. of Molecules = 0.03125 × 6.022 × 10²³ Molecules/mol
No. of Molecules = 1.88 × 10²² Molecules of CO
