What is the freezing point, in °C, of a solution made with 1.31 mol of CHCl₃ in 530.0 g of CCl₄?

Answers

Answer 1

Answer:

-96.6°C

Explanation:

Given:

moles CHCl3 = 1.31 molmass CCl4 = 530.0 g

        mass CCl4 = 530.0 g * (1 kg / 1000)

        mass CCl4 = 0.5300 kg

First calculate the molality (m) of the solute, which is the moles of solute divided by the kg of solvent ([tex]\frac{moles}{kg}[/tex]):

molality CHCl3 = (mass moles CHCl3) / (mass CCl4 in kg)

molality CHCl3 = (1.31 mol) / (0.5300 kg)

molality CHCl3 = 2.471698m

Decrease in freezing point of solvent = (Kf) * (molality of solute)

Decrease in freezing point of CCl4 = (Kf) * (molality CHCl3)

Decrease in freezing point of CCl4 = (29.8°C/m) * (2.47m)

Decrease in freezing point of CCl4 = 73.6566°C

Then use the molality of the solute to calculate the change in freezing point and subtract the change from the original freezing point.

Freezing point of solution =

(freezing point of pure solvent) - (Decrease in freezing point)

Freezing point of solution = (-22.9°C) - (73.66°C)

Freezing point of solution = -96.6°C

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What Is The Freezing Point, In C, Of A Solution Made With 1.31 Mol Of CHCl In 530.0 G Of CCl?
What Is The Freezing Point, In C, Of A Solution Made With 1.31 Mol Of CHCl In 530.0 G Of CCl?

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Is there massless ness in space?Why?

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How many liters of ammonia gas can be formed from 24.5 L of hydrogen gas at STP?

Answers

Answer:  16.3 L of ammonia gas can be formed from 24.5 L of hydrogen gas at STP

Explanation:

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Which of these is a learned behavior of a dog?
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Answers

I think the answer is 2
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Answers

Answer:

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Container 1 particles have lower average kinetic energy and higher temperature than Container 2 particles.

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Answer:

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Answers

Answer:

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Explanation:

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Answer:

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Answer:

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Answer:

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Answer:

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Explanation:

what did you get?

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Explanation:

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Hope this helps! Have a great day!

pls show your work will mark brainliest

Answers

Answer:

0.047 moles

Explanation:

From the question,

Applying

PV = nRT............. Equation 1

Where P = Pressure of CH₄ gas, V = Volume of CH₄ gas, n = number of moles of CH₄ gas, R = Molar gas constant.

make n the subject of the equation

n = PV/RT............. Equation 2

Given: P = 660 torr = (660×0.00131579) atm = 0.868 atm, V = 1300 mL = (1300/1000) L = 1.3 L, T = 20 °C = (20+273) K = 293 K

Constant: R = 0.082 atm.dm³/K.mol

Substitute these values into equation 2

n = (0.868×1.3)/(0.082×293)

n = 1.1284/24.026

n = 0.047 mole

1.
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Answers

Answer Choice: C

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THIS IS A THREE PART QUESTION IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED SO I DONT FAIL.

1. If I add 2.65L of water to 245 grams of sodium acetate, what is the molarity of the NaC2H3O2 solution?
A. 10.8 m
B. 1.13 m
C. 1.08 m
D. 92.4 m

2. If I add 2.65 L of water to 245 grams of sodium acetate, what is the % by mass of NaC2H3O2 in this solution?
A. 8.46%
B.1.12%
C.10.8%
D.9.25%

3. If I add 2.65 L of water to 245 grams of sodium acetate, what is the mol fraction of NaC2H3O2 in this solution?
A. 0.203
B. 0.0846
C. 0.108
D. 0.0199

Answers

Answer:

1. B = 1.13M

2. A. 8.46%

3. D = 0.0199

Explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

Molarity of solution = 2.988 mols/ 2.65 L = 1.13 M

2. Percentage by mass of a substance = mass of substance /mass of solution × 100%

Mass of 2.65 L of water = density × volume

Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

Mass of water = 1000 g/L × 2.65 L = 2650 g

Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g

Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%

3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

Moles of water = mass /molar mass

Mass of water = 2650 g; molar mass of water = 18 g/mol

Moles of water = 2650 g / 18 g/mol = 147.222 moles

Moles of solution = moles of solute + moles of water = 147.222 + 2.988 = 150.21

Moles of NaC₂H₃O₂ =2.988 moles

Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199

WILL MARK BRAINLEST

Which of the following are sources of greenhouse gasses? Select all that apply



a

Bike Riding

b

Cow Farts

c

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d

Burning Coal

e

Car Exhaust

Answers

Answer:

b and c

Explanation:

sorry if i'm wrong

Answer:

b, c, d, e

Explanation:

Clasifica cada uno de los siguientes incisos como sustancia pura (elemento y compuesto) o mezcla (homogénea y heterogénea). Explica brevemente. (a) arroz con leche (b) agua de mar (c) magnesio (d) gasolina

Answers

Answer:

Ver explicacion

Explanation:

Un arroz con leche es una mezcla heterogénea porque está compuesto por partículas sólidas de diferentes tipos y tamaños.

El agua de mar es una mezcla homogénea de sustancias todas en la misma fase.

El magnesio es una sustancia pura y es un elemento.

La gasolina es una mezcla homogénea de sustancias comúnmente utilizadas como combustible.

For the reaction 2H2 + O2 --> 2H2O, how many grams of oxygen are needed to react 3 moles of hydrogen?

Answers

Answer:

48 grams

Explanation:

The chemical equation for the reaction is the following:

2 H₂ + O₂ → 2 H₂O

That means that 2 moles of H₂ react with 1 mol of O₂ to produce 2 moles of H₂O. We convert the moles of oxygen (O₂) by using the molecular weight (MW) as follows:

MW(O₂) = 16 g/mol x 2 = 32 g/mol

mass of O₂ = 1 mol x 32 g/mol = 32 g

So, we have the following stoichiometric ratio: 32 g O₂/2 moles H₂. We have 3 moles of hydrogen (H₂), so we multiply the moles by the stoichiometric ratio to calculate how many grams are needed:

3 moles H₂ x 32 g O₂/2 moles H₂ = 48 g O₂

Therefore, 48 grams of O₂ are needed to react with 3 moles of H₂.

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