For Ar :
1 mol ------------ 22.4 L ( at STP )
7.6 mol ---------- x L
x = 7.6 * 22.4
x = 170.24 L
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For C2H3:
1 mol ------------ 22.4 ( at STP)
0.44 mol --------- y L
y = 0.44 * 22.4
y = 9.856 L
hope this helps !.
You have to calculate the oxidation estates of the atoms in each compound.
I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.
In K2Cr2O7:
- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.
- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.
That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.
In Cr2O3:
- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6
- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.
So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.
Answer: Cr has a change in oxidation number of - 3.
Answer:
Ground state
the state with the smallest amount of energy.
I believe it might be A, hope this helps!
Answer:
The whole molecule is polar because Sulfur has lone pairs but Carbon doesn't. Lone pairs count more toward polarity, shifting dipole toward S.
Explanation:
Even though carbon and sulfur have identical values of electronegativities, the molecule, 
is polar because of the presence of the lone pairs on the sulfur atom.
The C-S bond is not polar because the both the atoms have electronegatiivty. <u>But S has lone pairs which can attract the bond pairs of the bond between the S and H and thus acquires slightly negative charge and H acquires slightly positive charge.</u>
