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likoan [24]
3 years ago
12

Menthol is a crystalline substance with a peppermint taste and odor. When 1.48 g of menthol is dissolved in 25.0 g of cyclohexan

e, the freezing point of the solution is lowered by 7.90 °C. The freezing point and Kf constant for cyclohexane can be found here. Calculate the molar mass of menthol.
Chemistry
1 answer:
pishuonlain [190]3 years ago
4 0

Answer:

Molar mass of menthol is 155.86 g/mol

Explanation:

Let's apply the colligative property of freezing point depression:

ΔT = Kf . m . i

As menthol is an organic compound we assume the i (Van't Hoff factor as non electrolytic, so i = 1)

m = mol /kg (molality)

Kf = Cryoscopic constant, in this case, for cyclohexane

ΔT = T° freezing of pure solvent - T° freezing solution

7.90°C = 20.8°C / m . m . 1

7.90°C / 20.8 m/°C = m

0.380 m = molality

0.380 moles are the moles of solute (menthol) dissolved in 1kg of solvent, but our mass of solvent is 25 g.

Let's convert the g of solvent to kg to work with molality

25 g = 0.025 kg

molal . solvent mass = moles of solute

0.380 mol /kg . 0.025kg = 9.49×10⁻³ moles

Now, we have the moles and the mass of solute. Let's determine the molar mass of menthol (grams / moles)

1.48 g / 9.49×10⁻³ m = 155.86 g/m

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How many chlorine atoms are found in 8.3 moles of chlorine?
erik [133]

Answer:

5*10²⁴ chlorine atoms are found in 8.3 moles of chlorine.

Explanation:

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number represents a quantity without an associated physical dimension, so it is considered a pure number that allows describing a physical characteristic without an explicit dimension or unit of expression. Avogadro's number applies to any substance.

Then you can apply the following rule of three: if 1 mole of the compound contains 6.023 * 10²³ atoms, 8.3 moles of the compound how many atoms does it have?

amount of atoms=\frac{8.3 moles*6.023*10^{23}atoms }{1 mole}

amount of atoms≅ 5*10²⁴ atoms

<u><em>5*10²⁴ chlorine atoms are found in 8.3 moles of chlorine.</em></u>

4 0
2 years ago
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larisa [96]

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8 0
2 years ago
Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f
pishuonlain [190]

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>

<em />

<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>

T₂ = 0.5 T₁. According to Gay-Lussac's law,

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}

P₂ in terms of the ideal gas equation is:

P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}

3 0
3 years ago
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garri49 [273]
Force = mass x gravity

Force = 20 kg x 9.8 m/s²

Force = 196 Newtons

Answer A

hope this helps!

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3 years ago
A 93-L sample of dry air is cooled from 145 oC to -22 oC at a constant pressure of 2.85 atmospheres). What is the final volume?
Aloiza [94]

Answer: 55.84L

Explanation: Please see attachment for explanation.

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