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GalinKa [24]
2 years ago
6

Glycerol (C3H8O3, 92.1 g/mol) is a nonvolatile nonelectrolyte substance. Consider that you have an aqueous solution that contain

s 34.4 % glycerol by mass. If the vapor pressure of pure water is 23.8 torr at 25oC, what is the vapor pressure of the solution at 25oC?
Chemistry
1 answer:
pychu [463]2 years ago
6 0

Answer:

The vapor pressure of the solution at 25°C is 26.01 Torr

Explanation:

This is a usual excersise of colligative properties. In this case we apply the vapor pressure lowering formula:

ΔP = Xst . P°

Where ΔP is the diferrence between  Pressure of solution - Pressure of pure solvent.

And Xst  the molar fraction.

P° is Pressure of pure solvent.

So the formula will be:

Pressure Solution - P° = Xst . P°

Pressure Solution - 23.8 Tor = Xst . 23.8 Torr

Xst : Mole fraction ( Moles of solute or solvent / Total moles)

34.4 % m/m means that in 100 g of solution I have 34.4 g of solute

If I have 34.4 g of solute and the mass of 100 g in solution, I can know the mass of solvent, and finally the moles.

100 g solution - 34.4 g solute = 65.6 g (mass of solvent)

Molar mass of water : 18 g/m

Moles of water: Mass of water / Molar mass

65.6 g / 18g/m = 3.64 moles

Moles of glycerol : Mass glycerol / Molar mass glycerol

34.4 g / 92.1 g/m = 0.373 moles

Total moles: moles of glycerol + moles of water

0.373 m + 3.64 m = 4.01 m

So Xst = 0.373 m / 4.01 m → 0.093

Pressure Solution - 23.8 Tor = 0.093 . 23.8 Torr

Xst HAVE NO UNITS

Pressure Solution = (0.093 . 23.8 Torr ) + 23.8 Tor

Pressure Solution = 26.01 Torr

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A rod, X has a positive charge of 8. An otherwise identical rod, Y has a negative charge of 4. The rods are touched together, an
gtnhenbr [62]

Answer:

1.  electrons

2. From "Y" to "X"

Explanation:

1. Electrons move between the rod since the electrons are the only charge carriers which are free to move.

2. The particles move from from "Y" to "X"  since the electrons are the only charge carriers which are free to move. The positive charge on rod x is due to a deficit of electrons while the negative charge on rod Y is due to the excess of electrons. When the rods come together, the electrons move from "Y" to "X" since the electrons are the only charge carriers which are free to move.

4 0
1 year ago
Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

5 0
2 years ago
A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net
Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

4 0
2 years ago
Write the reduction reaction of glucose to form sorbitol. List and explain the side effects caused by too much sorbitol consumpt
Lemur [1.5K]

Answer:

The product of reduction of glucose is sorbitol

The side effects caused by too much sorbitol consumption include: Diarrhea, Nausea, stomach discomfort

Explanation:

Please find attached the reaction of glucose with NADPH to produce sorbitol

5 0
2 years ago
Read 2 more answers
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
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


4 0
2 years ago
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