endothernic by 26.4 kJ/mol. The solution process is spontaneous because A) osmot

Question

endothernic by 26.4 kJ/mol. The solution process is spontaneous because A) osmotic properties predict this behavior B) the vapor pressure of the water decreases upon addition of the solute C of the increase in disorder u D) of the increase in enthalpy upon dissolution of this strong electrolyte E) of the decrease in enthalpy upon addition of the solute upon dissolution of this strong electrolyte 8) 8) 8) Which of the following substances is more likely to dissolve in water? A) CHa(CH2)8CH2OH B) CHlCH2)9CH 9) 9) Which of the following aqueous solutions will have the highest boilling point? A) 0.25 m sucrose B)0.10 m Na2S04 0.20 m glucose D) 0.10 m SrS04 E) 0.10 m NaC 10) Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 10) - 2N02 2NO + O2 In a particular experiment at 300°C,[NO21 drops from 0.0100 to 0.00650 M in 100& The rate of disappearance of NO2 for this period is M/s. A) 0.35 B)1.8×10-3 93.5 x 10-5 D) 3.5 * 10-3 E)70+10-3 11) Consider the following reaction: 3A2B The average rate of appearance of B is given by AlByst. Comparing the rate of appearance of B and the rate of disappearance of A, we get [B1At =--m(_ AlAt) A)-23 C) -3/2 D) +1 E) 3/2 4-2

Explanation / Answer

You have multiple questions in a single post, and due to guidelines, I cannot answer multiple questions in one single post. I will answer the first 4 questions, and the rest of them post them in a new question thread per separate (3 or 4 questions at a time).

Question 1.

As the solubility is 70g/100 mL or simply 0.7 g/mL, and you have 75 g/100 mL means that there are a remaining of 5 g undissolved, and when a solution has more solute that it can solve at a determined temperature, it means that the solution is supersaturated.

Question 2.

Mole fraction (Xa) = na / na+nb. Let a be urea and b the water. let's calculate moles for both of them.

na = 16 / 60 = 0.2667 moles

nb = 39 / 18 = 2.1667 moles

Xa = .02667 / 2.1667+0.2667 = 0.1096 or simply 0.11.

Question 3.

m of solution = 16+´39 = 55 g

V of solution = 55 g / 1.3 g/mL = 42.30 mL (0.0423 L)

M = moles / V

M = 0.2667 / 0.0423 = 6.3 M

Question 4.

using the expression:

Tf° - Tf = Kf * mass solute / MM solute * kg solvent solve for Tf. Tf° is the freezing point of ethanol. MM for glycerin is 92 g/mol

-114.6 - Tf = 2 * 50 / 92 * 0.2

-114.6 - Tf = 5.43

Tf = -120.03 °C

For the rest of your questions, like I said before, post them in a new question thread per separate so you can be answered faster and better.

Hope this helps

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