Answer:
B. Warm water rises within the pot.?
Explanation
<em>There wasn't enough information given for me to safely determine the correct answer.</em>
Answer:

Explanation:
Hello there!
In this case, according to the given information it will be firstly necessary to set up the chemical equation taking place:

We infer we need to calculate the moles of NH3 by using both of the moles of N2 and H2 at the beginning, in order to identify the limiting reactant:

Thus, since hydrogen yields the fewest moles of ammonia, we conclude that we are just able to yield 4 moles of NH3.
Regards!
Rutherford is the man responsible :)
Answer is: V<span>an't Hoff factor (i) for this solution is 2,26.
</span>Change in freezing point
from pure solvent to solution: ΔT =i · Kf · m.
<span>Kf - molal freezing-point depression constant for water is 1,86°C/m.
</span>m - molality, moles of solute per kilogram of solvent.
n(K₂SO₄) = 16,8 g ÷ 174,25 g/mol
n(K₂SO₄) = 0,096 mol.
m(K₂SO₄) = 0,096 mol/kg.
ΔT = 0,405°C.
i = 0,405 ÷ (1,86 · 0,096)
i = 2,26.
They all have something to do with heat