11. The least chemically reactive of the hydrocarbon
1 answer:
You might be interested in
Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.
Explanation : Given,
Vapor presume of 1‑Propanol = 20.9 torr
Vapor presume of 2‑Propanol = 45.2 torr
Mole fraction of 1‑Propanol = 0.540
Mole fraction of 2‑Propanol = 1-0.540 = 0.46
First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.
where,
= partial vapor pressure of 1‑Propanol
= vapor pressure of pure substance 1‑Propanol
= mole fraction of 1‑Propanol
and,
where,
= partial vapor pressure of 2‑Propanol
= vapor pressure of pure substance 2‑Propanol
= mole fraction of 2‑Propanol
Thus, total pressure = 11.3 + 20.8 = 32.1 torr
Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.
and,
Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.
The mass of water produced by the reaction of the 23 g of is 13.8 g.
The given chemical reaction;
In the given compound above, we can deduce the following;
- molecular mass of
= 28 + (2 x 16) = 60 g
- molecular mass of
= 2(18) = 36 g
60 g of --------- 36 g of water
23 g of ------------- ? of water
Thus, the mass of water produced by the reaction of the 23 g of is 13.8 g.
- <em>"Your question is not complete, it seems to be missing the following information";</em>
In the reaction of the given compound, , what mass of water (in grams) is produced by the reaction of 23.0 g of SiO2?
Learn more here:brainly.com/question/13644576