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stellarik [79]
3 years ago
12

A solution is prepared by condensing 4.00 L of a gas,

Chemistry
1 answer:
vredina [299]3 years ago
5 0

Answer:

-2.3 ºC

Explanation:

Kf (benzene) = 5.12 ° C kg mol – 1

1st - We calculate the moles of condensed gas using the ideal gas equation:

n = PV / (RT)

P = 748/760 = 0.984 atm

T = 270 + 273.15 = 543.15 K

V = 4 L

R = 0.082 atm.L / mol.K

n = (0.984atm * 4L) / (0.082atm.L / K.mol * 543.15K) = 0.088 mol

Then, you calculate the molality of the solution:

m = n / kg solvent

m = 0.088 mol / 0.058 kg = 1.52mol / kg

Then you calculate the decrease in freezing point (DT)

DT = m * Kf

DT = 1.52 * 5.12 = 7.8 ° C

Knowing that the freezing point of pure benzene is 5.5 ºC, we calculate the freezing point of the solution:

T = 5.5 - 7.8 = -2.3 ºC

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PLEASE PLEASE HELP!
valentinak56 [21]

Answer: The number of grams of H_2 in 1620 mL is 1.44 g

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = 1620 ml = 1.62 L  (1L=1000ml)

n = number of moles = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =273K

n=\frac{PV}{RT}

n=\frac{1atm\times 16.2L}{0.0821Latm/K mol\times 273K}=0.72moles

Mass of hydrogen =moles\times {\text {Molar mass}}=0.72mol\times 2g/mol=1.44g

The number of grams of H_2 in 1620 mL is 1.44 g

8 0
3 years ago
A compound is found to contain 73.23% xenon name 26.77% oxygen by mass. What is the empirical formula for this compound ?
Luba_88 [7]

The empirical formula is XeO₃.

<u>Explanation:</u>

Assume 100 g of the compound is present. This changes the percents to grams:

Given mass in g:

Xenon = 73.23 g

Oxygen = 26.77 g

We have to convert it to moles.

Xe = 73.23/   131.293 = 0.56 moles

O = 26.77/ 16 = 1.67 moles

Divide by the lowest value, seeking the smallest whole-number ratio:

Xe = 0.56/ 0.56 = 1

O = 1.67/ 0.56 = 2.9 ≈3

So the empirical formula is XeO₃.

6 0
3 years ago
When substances combine to form mixtures they keep their properties.<br> O<br> True<br> O<br> False
Gnesinka [82]

Answer:

True

Explanation:

8 0
3 years ago
Read 2 more answers
Write ionic formulas for the following compounds:
lakkis [162]

You may find the chemical formulas for the ionic compounds below.

Explanation:

a. Calcium Bromide - CaBr₂

b. Potassium Phosphide  - K₃P

c. Iron (III) Oxide  - Fe₂O₃

d. Iron (II) Oxide  - FeO

e. Magnesium Hydroxide  - Mg(OH)₂

f. Nickel (III) Sulfate​ - Ni₂(SO₄)₃

Learn more about:

 chemical formulas of ionic compounds

brainly.com/question/11448164

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brainly.com/question/7460355

#learnwithBrainly

6 0
3 years ago
A 5.21 mass % aqueous solution of urea (CO(NH2)2) has a density of 1.15 g/mL. Calculate the molarity of the solution. Give your
Alex777 [14]

Answer:

Molarity is 0.99 M

Explanation:

5.21% by mass, is a sort of concentration which shows the mass of solute in 100 g of solution.

Molarity is a sort of concentration that indicates the moles of solute in 1 L of solution (mol/L)

Let's find out the volume of solution by density.

Solution density = Solution mass / Solution volume

1.15 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.15 g/mL → 86.9 mL

We must have the volume of solution in L, so let's convert it.

86.9 mL / 1000 = 0.0869 L

Now, we have to determine the moles of solute (urea)

5.21 g . 1 mol / 60 g = 0.0868 moles

Mol/L = Molarity → 0.0868 moles / 0.0869L  = 0.99 M

4 0
3 years ago
Read 2 more answers
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