Answer:
As electro-negativity decreases from Florine to downwards in the group and only Florine is above Chlorine, so Florine should react with sodium chloride solution to produce chlorine.
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Since one atmosphere is 29.9213 inches of mercury, we can simply find this by taking 29.9213 and multiplying that by 1.24. You end up with 37.102412 inches of mercury. I hope this helps you out, and good luck on the rest of your homework.
Option (a) is correct.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In

, Cr has oxidation number 6+ in the L.H.S of the equation, but on R.H.S its oxidation number is 0 i.e. it Cr has gained electrons such that total charge is 0.
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.
Hence, Cr is the oxidizing agent and Al is the reducing agent.
2H₂ + O₂ = 2H₂O
n(H₂)=m(H₂)/M(H₂)
n(H₂)=5g/2.0g/mol=2.5 mol
n(O₂)=m(O₂)/M(O₂)
n(O₂)=40g/32.0g/mol=1.25 mol
H₂ : O₂ = 2 : 1
2.5 : 1.25 = 2 : 1
n(H₂O)=n(H₂)=2n(O₂)=2.5 mol
m(H₂O)=n(H₂O)M(H₂O)
m(H₂O)=2.5mol*18.0g/mol=45.0 g