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Leokris [45]
3 years ago
13

A certain substance X has a normal freezing point of 5.6 °C and a molal freezing point depression constant Kf-7.78 °C-kg·mol-1.

A solution is prepared by dissolving some urea ((NH2)2CO) in 550. g of Χ. This solution freezes at-0.9 °C. Calculate the mass of urea that was dissolved. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Harrizon [31]3 years ago
7 0

Answer:

27.60 g urea

Explanation:

The <em>freezing-point depression</em> is expressed by the formula:

  • ΔT= Kf * m

In this case,

  • ΔT = 5.6 - (-0.9) = 6.5 °C
  • Kf = 7.78 °C kg·mol⁻¹

m is the molality of the urea solution in X (mol urea/kg of X)

First we<u> calculate the molality</u>:

  • 6.5 °C = 7.78 °C kg·mol⁻¹ * m
  • m = 0.84 m

Now we<u> calculate the moles of ure</u>a that were dissolved:

550 g X ⇒ 550 / 1000 = 0.550 kg X

  • 0.84 m = mol Urea / 0.550 kg X
  • mol Urea = 0.46 mol

Finally we <u>calculate the mass of urea</u>, using its molecular weight:

  • 0.46 mol * 60.06 g/mol = 27.60 g urea

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What is the percent composition each element in c6h8o6
QveST [7]

Answer:

Percentage of oxygen = 30%

Percentage of carbon = 30%

Percentage of hydrogen = 40%

Explanation:

Formula:

Percentage of element = given amount / total amount × 100

Given compound:

C₆H₈O₆

Number of atoms of carbon = 6

Number of atoms of hydrogen = 8

Number of atoms of oxygen = 6

Total number of atoms = 20

Percentage of carbon = 6/20 × 100

Percentage of carbon = 30%

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3 years ago
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4 years ago
A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following
marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

\frac{V_1}{n_1} =\frac{V_2}{n_2}

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}

<em>V₂ = 4,00L</em>

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}

<em>V₂ = 16,00L</em>

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}

<em>V₂ = 12,31L</em>

I hope it helps!

6 0
4 years ago
URGENT PLZ HELP
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I don’t want to be wrong but I’m gonna say 2p
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