Answer:option D= 82.6°C
Explanation:
Step 1: determine the molality
Molality = moles of solute/ kg of solvent.
(Recall: number of moles= mass / molar mass).
Therefore, 375/256.5 g/mol = 1.4620 mole.
Substituting into the molality equation we have,
1.4620/ 0.1250kg = 11.696.
∆T(b) = k(b) × molality
∆T(b) = 82.6°C
the answer is A......
it is supported by practical evidence and examples. this is the answer because he tried and tested many different ways to see what would happen so he is happy with the conclusion that what he tested is what he gets.
Answer:
-145.2kJ
Explanation:
Enthalpy is an extensive property as its value depends on the amount of substance present in the system.
If the enthalpy for one mole of methanol = -726 kJ/mol;
The Enthalpy for 0.2 mol is given as;
Enthalpy = 0.200 * 726
Enthalpy = -145.2kJ
It would take -145.2kJ for 0.200 mol of methanol to undego the combustion reaction.
The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
It would be solid
hope this helps
