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Mariana [72]
3 years ago
13

Magnesium sulfate heptahydrate is heated until all the water is driven off. The sample loses 11.80 grams upon heating. What was

the mass of the original sample?
Chemistry
1 answer:
iren [92.7K]3 years ago
6 0
Hello!

The initial mass of Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is 23,08 g

The chemical reaction for the dehydrating of Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is the following:

MgSO₄·7H₂O(s) + Δ → MgSO₄(s) + 7H₂O(g)

We know that the sample loses 11,80 g upon heating. That mass is the mass of Water that is released as vapor. Knowing that piece of information, we can apply the following conversion factor to go from the mass of water to the moles of water and back to the mass of the original compound (mi).

mi=11,80gH_2O* \frac{1 mol H_2O}{18gH_2O}* \frac{1molMgSO_4.7H_2O}{7molH_2O}* \frac{246,47 g MgSO_4.7H_2O}{1 mol MgSO_4.7H_2O}  \\  \\ mi=23,08gMgSO_4.7H_2O

Have a nice day!

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A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V

The mole fraction of nitrogen in the mixture is:

X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

Learn more: brainly.com/question/2060778

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3 years ago
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Formula units in 450 g of Na_{2} So_{4} is 1.93 × 10²⁴ formula units.

<u>Explanation:</u>

First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,

$\frac{450 g}{142.04 g/mol} =  3.2 moles

Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.

3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.

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