Answer:
a. What makes this an oxidation-reduction reaction? (1 point)
Loss of electron by the aluminum and the gain of electron by the silver
b. Write the half-reactions showing the oxidation and reduction reactions. Identify which is the oxidation reaction and which is the reduction reaction. (3 points)
oxidation half reaction: Al → Al3+ + 3e-
Reduction half reaction: Ag+ + e- → Ag
c. What is oxidized in the reaction? What is reduced? (2 points)
Aluminum is oxidized and silver is reduced
d. In this simple electrochemical cell, what functions as the anode? What is the cathode? (3 points)
In a simple electrochemical cell the electrode where oxidation takes place is the anode. And the electrode where reduction reaction happens is the cathode.
e. Is this a galvanic cell or electrolytic cell? Explain your answer. (2 points)
An electrolytic cell because the reaction converts electrical energy into chemical energy
Not sure about e and f!
Answer:
a push
please thank and brainlest!
Explanation:
Answer: D
Explanation: Bacteria thrives in warm and moist environments.
Elements with three p-electrons....
That would be N, P, As, Sb, and Bi -- elements in group 15
For example, energy diagram showing "empty" orbitals up through the 3p.
.....3p __ __ __
3s __
.....2p __ __ __
2s __
1s __
Energy diagram of phosphorous showing three unpaired electrons in 3p-sublevel
.....3p ↑_ ↑_ ↑_
3s ↑↓
.....2p ↑↓ ↑↓ ↑↓
2s ↑↓
1s ↑↓
According to Hund's rule, the electrons singly occupy the p-orbitals, and all have the same spin.
Answer:
Solution that is 0.100 M CH3COOH (acetic acid)
and 0.100 M NaCH3COO (sodium acetate)
Find pH of buffer solution:
CH3COOH(aq) + H2O ↔ CH3COO-
(aq) + H3O+(aq)
[CH3COOH] [CH3COO-
] [H3O+]
initial 0.100 0.100 ≈0
-x x x
equil 0.100 – x 0.100 + x xFind pH of buffer solution:
CH3COOH(aq) + H2O ↔ CH3COO-
(aq) + H3O+(aq)
Ka = [CH3COO-
][H3O+
]
[CH 3COOH] = (.100 + x)x
(.100 - x) = 1.8 x 10-5
x = 1.80 x 10-5 M
pH = 4.7
Explanation:
