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Alik [6]
2 years ago
7

How many liters of a 0.352 M solution of Ca(SO4) would contain 62.1 g of Ca(SO4)?​

Chemistry
1 answer:
konstantin123 [22]2 years ago
4 0

Answer:

1.3 L.

Explanation:

  • Molarity is the no. of moles of solute per 1.0 L of the solution.

<em>M = (no. of moles of CaSO₄)/(Volume of the solution (L))</em>

<em></em>

M = 0.352 M.

no. of moles of CaSO₄ = mass/molar mass = (62.1 g / 136.14 g/mol) = 0.456 mol,

Volume of the solution = ??? L.

∴ (0.352 M) = (0.456 mol)/(Volume of the solution (L))

<em>∴ (Volume of the solution (L) </em>= (0.456 mol)/(0.352 M) = <em>1.296 L ≅ 1.3 L.</em>

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Submit At 25.0 C, a 10.00 L vessel is filled with 5.00 atm of Gas A and 7.89 atm of Gas B. What is the mole fraction of Gas B?
Firlakuza [10]

Answer:

the mole fraction of Gas B is xB= 0.612 (61.2%)

Explanation:

Assuming ideal gas behaviour of A and B, then

pA*V=nA*R*T

pB*V=nB*R*T

where

V= volume = 10 L

T= temperature= 25°C= 298 K

pA and pB= partial pressures of A and B respectively = 5 atm and 7.89 atm

R= ideal gas constant = 0.082 atm*L/(mol*K)

therefore

nA= (pA*V)/(R*T) = 5 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 2.04 mole

nB= (pB*V)/(R*T) = 7.89 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 3.22 mole

therefore the total number of moles is

n = nA +nB= 2.04 mole +  3.22 mole = 5.26 mole

the mole fraction of Gas B is then

xB= nB/n= 3.22 mole/5.26 mole = 0.612

xB= 0.612

Note

another way to obtain it is through Dalton's law

P=pB*xB , P = pA+pB → xB = pB/(pA+pB) = 7.69 atm/( 5 atm + 7.89 atm) = 0.612

5 0
2 years ago
HELP! Ethanol has a density of 0.8 g/cm3. a. What is the mass of 225 cm3 of ethanol? b. What is the volume of 75.0 g of ethanol?
IRINA_888 [86]

Answer:

The correct answers are:

a) 180 g

b) 93.7 cm³

Explanation:

The density of a substance is the mass of the substance per unit of volume. So, it is calculated as follows:

density= mass/volume

From the data provided in the problem:

density = 0.8 g/cm³

a) Given: volume= 225 cm³

mass= density x volume = 0.8 g/cm³ x 225 cm³ = 180 g

b) Given: mass= 75.0 g

volume = mass/density = 75.0 g/(0.8 g/cm³)= 93.75 cm³≅ 93.7 cm³

5 0
1 year ago
Item 5 A solution of methanol, CH3OH, in water is prepared by mixing together 128 g of methanol and 108 g of water. The mole fra
Basile [38]

Answer:

Mole fraction of methanol will be closest to 4.

Explanation:

Given, Mass of methanol = 128 g

Molar mass of methanol = 32.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{128\ g}{32.04\ g/mol}

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Given, Mass of water = 108 g

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The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{108\ g}{18.0153\ g/mol}

Moles\ of\ water= 5.995\ mol

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<u>Mole fraction of methanol will be closest to 4.</u>

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2 years ago
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2 years ago
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