Nobles gases, since they all have 8 e- on their last layer of electrons.
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!
0.4 g/ml.............................
The stoichiometry of the reaction gives the molar ratio in which the reactants react with each other and the ratio in which products are formed.
The coefficients of the reactants in the reaction follow the stoichiometry
the balanced chemical equation for the reaction is as follows;
2C₃H₆(g) + 9O₂(g) ---> 6CO₂(g) + 6H₂O(l)