Explain why smell is a physical property?

VMariaS [17]

Answer:

2. a. All three solutes are nonelectrolytes.

3. a. the solution of solute X

Explanation:

The freezing point depression (ΔTf) and boiling point elevation (ΔTb) are colligative properties: they depend on the numbers of particles.

The formula for ΔTf is

ΔTf = iKf·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For sucrose,

Sucrose (s) ⟶ sucrose (aq)

1 mole sucrose ⟶ 1 mol particles i = 1

For NaCl

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol 1 mol + 1 mol i = 2

For Ca(NO₃)₂

Ca(NO₃)₂(s) ⟶ Ca²⁺(aq) + 2NO₃⁻(aq)

1 mol 1 mol + 2 mol i = 3

===============

2. Freezing points

For a nonelectrolyte, i = 1.

Kf = 1.86 °C·kg⁻¹mol⁻¹

b = 1 mol/kg Calculate ΔTf

ΔTf = 1 × 1.86 × 1

ΔTf = 1.86 °C

Tf = Tf⁰ - ΔTf =0.00 °C – 1.86 °C = -1.86 °C

All the other solutions have lower freezing points, so the solutes must be electrolytes.

===============

3. Boiling points

The formula for ΔTb is

ΔTb = iKb·m

The solution with the highest boiling point will have the highest value of i.

In other words, the solution with the highest boiling point will be the one with the lowest freezing point.

That’s the solution of solute X.

5 0

3 years ago

Please answer quickly!!

FromTheMoon [43]

Answer:

i do not no but i do no .nb dncvejx euvcir

5 0

2 years ago

0.268/6.234 significant figures

den301095 [7]

The answer is 0.0430 following the rules of dividing and multiplying significant figures.

7 0

3 years ago

Calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid.

olga2289 [7]

The question is incomplete,the complete question :

Calculate the molality of a 10.0% (by mass) aqueous solution of hydrochloric acid:

a) 0.274 m

b) 2.74 m

c) 3.05 m

d) 4.33 m

e) the density of the solution is needed to solve the problem

Answer:

The molality of a 10.0% (by mass) aqueous solution of hydrochloric acid is 3.05 mol/kg.

Explanation:

10.0% (by mass) aqueous solution of hydrochloric acid.

10 grams of HCl is present in 100 g of solution.

Mass of HCl = 10 g

Mass of solution = 100 g

Mass of solution = Mass of solute + Mass of water

Mass of water = 100 g - 10 g = 90 g

Moles of HCl = $\frac{10 g}{36.5 g/mol}=0.2740 mol$

Mass of water in kilograms = 0.090 kg

Molality = $\frac{0.2740 mol}{0.090 kg}=3.05 mol/kg$

The molality of a 10.0% (by mass) aqueous solution of hydrochloric acid is 3.05 mol/kg.

6 0

3 years ago

Xas shown in table 15.2, kp for the equilibrium n21g2 + 3 h21g2 δ 2 nh31g2 is 4.51 * 10-5 at 450 °c. for each of the mixtures li

Agata [3.3K]

According to the balanced equation of this reaction:
N2(g) + 3H2(g) ↔ 2NH3(g)
and when we have Kp = 4.51 x 10^-5 so, in the Kp equation we will substitute by the value of the P for each gas to compare the value with Kp = 4.51x10^-5

a) when we have 98 atm NH3, 45 atm N2, 55 atm H2 by substitution in Kp equation:
Kp= [p(NH3)]^2 / [p(N2)]*[p(H2)]^3 = [98]^2 / [45]*[55]^3
= 1.28x10^-3
So here the value is higher than the value of the given Kp.
so the reaction will go leftwards toward the reactants ( to reduce the value of Kp) to reach the equilibrium.
b) When 57 atm NH3, 143 atm N2, No H2 so like a) by substitution:
Kp = [57]^2 / [143] = 22.7
So the reaction will go leftwards toward the reactants to reduce the value of Kp to reach equilibrium.
c) when 13 atm NH3, 27 atm N2, 82 H2
Kp = [13]^2 / [27]*[82]^3 = 1.135 x 10^-5 So this value is lower than the Kp which is given.
so, the reaction will go towards the right toward the products to increase the value of Kp to reach the equilibrium.

8 0

3 years ago