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MrMuchimi
3 years ago
9

15. Discuss the various factors which affect the rate of evaporation. Latent heat

Chemistry
1 answer:
Georgia [21]3 years ago
8 0

Answer:

The correct answer is liquid B.

Explanation:

Latent heat of vaporization also known as the latent heat of evaporation. This latent heat transforms the particles of liquid into a gas without affecting its temperature. For example, the latent heat of evaporation for water is 40.8 kJ per mole, that is, 40.8 kJ per mole of heat is needed to transform water into vapor at 373 K.  

It is known that latent heat of evaporation of a liquid is directly proportional to the cooling effect it generates, that is, more the latent heat of evaporation more will be its cooling effect. Thus, it is clear that liquid B will show the more cooling effect as the latent heat of evaporation of liquid B is more in comparison to liquid A. Thus, more heat will be captivated by liquid B and will generate more cooling effect in comparison to liquid A.  

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A liquid has an empirical formula CCl2, and a boiling point of 1 21 oC. When vapourised, the gaseous compound has a density of 4
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Based on the data given, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

<h3>How can molar mass of a gas be obtained from density, temperature and pressure?</h3>

The molar mass of a gas can be obtained from density, temperature and pressure using the formula below:

  • molar mass = density × molar gas constant × temperature/pressure

Molar gas constant, R = R = 0.082 L.atm/mol/K.

Temperature = 150 °C = 423 K

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density = 4.93 g/L

molar mass of gas = 4.93 × 0.082 × 423/1.033

molar mass of gas = 165.5 g/mol

Then, molecular weight of the gas = 165.5 amu

Therefore, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

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1 year ago
Write the empirical formula of at least four binary iconic compounds that could be formed from the following ions:
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3 years ago
By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask
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Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = V_1=11.00 mL

Concentration of stock solution = M_1=0.823 M

Volume of stock solution after dilution = V_2=50.00 mL

Concentration of stock solution after dilution = M_2=?

M_1V_1=M_2V_2 ( dilution )

M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L

0.823 Molar is the molarity of the solution.

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