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3 years ago

What is the thermal energy needed to vaporize 18.02g of water at 100°C? The Heat of Vaporization for water is 40.650 kJ/mol.

Chemistry

1 answer:

podryga [215]3 years ago

5 0

Answer:

40.7 kJ

Explanation:

Applying,

q = c'n.................. Equation 1

Where q = Thermal Heat, c' = Heat of vaporization of water, n = number of mole of water.

But,

n = mass(m)/Molar mass(m')

n = m/m'............... Equation 2

Substitute equation 2 into equation 1

q = c'(m/m')............. Equation 3

Given: c' = 40.650 KJ/mol, m = 18.02 g

Constant: m' = 18 g/mol

Substitute into equation 3

q = 40.650(18.02/18)

q = 40.695 kJ

q ≈ 40.7 kJ

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At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

Answer: The freezing point of solution is -0.974°C

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

8 0

3 years ago

How would Earths revolution be affected if it had a larger orbit??

Sidana [21]

Answer:

When the Earth rotates on its axis, it prevents air currents from going in a straight line to the north and the south from the equator. It results in one of the effects of rotation of the Earth: the Coriolis Effect.

4 0

3 years ago

What method would a geologist use to determine the approximate age of a rock found in a newly discovered formation?

shtirl [24]

Rock paper scissor and that is on Albert Einstein , he made that up because he is so freaking smart so there you go have a wonderful day

8 0

3 years ago

a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel

eimsori [14]

Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

1) What is the partial pressure of SO₂ gas in the larger container?

∵ P = nRT/V.

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

2) What is the partial pressure of N₂ gas in the larger container?

∵ P = nRT/V.

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

3) What is the total pressure in the vessel?

According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0

3 years ago

HURRY HELP 10 POINTS ILL GOVE BRAINLIEST I HAVE TO PASS PICTURE PROVIDED

kramer

Answer:

C₃H₄O₄

Explanation:

In order to get the empirical formula of a compound, we have to follow a series of steps.

Step 1: Divide the percent by mass of each element by its atomic mass.

C: 34.6/12.01 = 2.88

H: 3.9/1.01 = 3.86

O: 61.5/16.00 = 3.84

Step 2: Divide all the numbers by the smallest one, i.e., 2.88

C: 2.88/2.88 = 1

H: 3.86/2.88 ≈ 1.34

O: 3.84/2.88 ≈ 1.33

Step 3: Multiply all the numbers by a number that makes all of them integer

C: 1 × 3 = 3

H: 1.34 × 3 = 4

O: 1.33 × 3 = 4

The empirical formula is C₃H₄O₄.

5 0

3 years ago