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Ainat [17]
3 years ago
12

A certain substance X has a normal boiling point of 117.8°C and a molal boiling point elevation constant =Kb·0.63°C·kgmol−1. A s

olution is prepared by dissolving some urea NH22CO in 900.g of X. This solution boils at 118.3°C. Calculate the mass of urea that was dissolved. Round your answer to 1 significant digit.
Chemistry
1 answer:
OlgaM077 [116]3 years ago
5 0

Answer:

42.8 g was the mass of dissolved urea.

Explanation:

Boiling point elevation to solve this:

ΔT = Kb . m

where ΔT → Boiling T° of solution - Boiling T° of pure solvent

Let's replace the data given:

118.3°C - 117.8°C = 0.63°C/m . m

0.5°C / 0.63 m/°C = m → 0.793 mol/kg

Mass of solvent → 900 g → g to kg → 900 g . 1kg/1000 = 0.9 kg

Molality . kg = moles → 0.793 mol/kg . 0.9 kg = 0.714 mol

These are the moles of urea we used. Let's determine the mass dissolved.

0.714 mol . 60 g /1mol = 42.8 g

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How many moles are in 187.54 grams of magnesium chlorate?
Svetlanka [38]

Hey there!

Magnesium chlorate: Mg(ClO₃)₂

Find molar mass.

Mg: 1 x 24.305 = 24.305

Cl: 2 x 35.453 = 70.906

O: 6 x 16 = 96

------------------------------------

                      191.211 g/mol

We have 187.54 grams.

187.54 ÷ 191.211 = 0.9808

There are 0.9808 moles in 187.54 grams of magnesium chlorate.

Hope this helps!

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Using the graph below please draw a reaction potential energy diagram for a reaction with the
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2 years ago
15. Which sample of argon gas has the same number of atoms as a 100.-milliliter sample of helium gas at 1.0 atm and 300. K? A) 5
OLga [1]

The sample of argon gas that has the same number of atoms as a 100 milliliter sample of helium gas at 1.0 atm and 300 is 100. mL at 1.0 atm and 300. K

The correct option is D.

<h3>What is the number of moles of gases in the given samples?</h3>

The number of moles of gases in each of the given samples of gas is found below using the ideal gas equation.

The ideal gas equation is: PV/RT = n

where;

  • P is pressure
  • V is volume
  • n is number of moles of gas
  • T is temperature of gas
  • R is molar gas constant = 0.082 atm.L/mol/K

Moles of gas in the given helium gas sample:

P = 1.0 atm, V = 100 mL or 0.1 L, T = 300 K

n =  1 * 0.1 / 0.082 * 300

n = 0.00406 moles

For the argon gas sample:

A. n =  1 * 0.05 / 0.082 * 300

n = 0.00203 moles

B. n =  0.5 * 0.05 / 0.082 * 300

n = 0.00102 moles

C. n =  0.5 * 0.1 / 0.082 * 300

n = 0.00203 moles

D. n =  1 * 0.1 / 0.082 * 300

n = 0.00406 moles

Learn more about ideal gas equation at: brainly.com/question/24236411

#SPJ1

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