Answer:
1. 0.97 V
2. 
Explanation:
In this case, we can start with the <u>half-reactions</u>:


With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

I hope it helps!
C
The elements are Lithium and Beryllium— just look at the mass number on the periodic table
They are called Periods. Lesser known as series
Answer:
a. 0.182
b. 1.009
c. 1.819
Explanation:
Henderson-Hasselbach equation is:
pH = pKa + log [salt / acid]
Let's replace the formula by the given values.
a. 3 = 3.74 + log [salt / acid]
3 - 3.74 = log [salt / acid]
-0.74 = log [salt / acid]
10⁻⁰'⁷⁴ = 0.182
b. 3.744 = 3.74 + log [salt / acid]
3.744 - 3.74 = log [salt / acid]
0.004 = log [salt / acid]
10⁰'⁰⁰⁴ = 1.009
c. 4 = 3.74 + log [salt / acid]
4 - 3.74 = log [salt / acid]
0.26 = log [salt / acid]
10⁰'²⁶ = 1.819
Elephants are really cool