Answer : The mole fraction and partial pressure of
and
gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.
Explanation : Given,
Moles of
= 1.79 mole
Moles of
= 1.20 mole
Moles of
= 3.71 mole
Now we have to calculate the mole fraction of
and
gases.


and,


and,


Thus, the mole fraction of
and
gases are, 0.267, 0.179 and 0.554 respectively.
Now we have to calculate the partial pressure of
and
gases.
According to the Raoult's law,

where,
= partial pressure of gas
= total pressure of gas = 5.78 atm
= mole fraction of gas


and,


and,


Thus, the partial pressure of
and
gases are, 1.54, 1.03 and 3.20 atm respectively.
If you are talking about just pure regular water, the answer is false. BUT, some salts dissolved IN WATER, can act as electrolytes. But regular water, no.
Sorry if I'm wrong but I think that it is B.
The major classes are monosaccharides, oligosaccharides and polysaccharides.
The types are sugar, starch, and fibersugar.
Hope I helped!!
The best and most correct answer among the choices provided by your question is the second choice.
<span>Based on the reaction given above, as the temperature decreases, the formation of the reactants will increase.</span>
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