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Zanzabum
3 years ago
7

Express the concentration of a 0.0390 M aqueous solution of fluoride, F − , in mass percentage and in parts per million (ppm). A

ssume the density of the solution is 1.00 g/mL.
Chemistry
1 answer:
Vadim26 [7]3 years ago
5 0

Answer:

Mass percentage → 0.074 %

[F⁻] = 741 ppm

Explanation:

Aqueous solution of flouride → [F⁻] = 0.0390 M

It means that in 1L of solution, we have 0.0390 moles of F⁻

We need the mass of solution and the mass of 0.0390 moles of F⁻

Mass of solution can be determined by density:

1g/mL = Mass of solution / 1000 mL

Note: 1L = 1000mL

Mass of solution: 1000 g

Moles of F⁻ → 0.0390 moles . 19g /1 mol = 0.741 g

Mass percentage → (Mass of solute / Mass of solution) . 100

(0.741 g / 1000 g) . 100 = 0.074 %

Ppm = mass of solute . 10⁶ / mass of solution (mg/kg)

0.741 g . 1000 mg/1g = 741 mg

1000 g . 1 kg/1000 g = 1kg

741 mg/1kg = 741 ppm

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Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

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where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0
3 years ago
During decomposition of water,10g of oxygen gas was formed.How much hydrogen gas was formed
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40 g of CaCO3 is how many moles of CaCO3?<br> 10 moles<br> 0.4 moles<br> 40 moles<br> 100 moles
AleksandrR [38]

Answer:

0.4 moles

Explanation:

To convert between moles and grams you need the molar mass of the compound. The molar mass of of CaCO3 is 100.09g/mol. You use that as the unit converter.

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This rounds to 0.4 moles CaCO3

8 0
3 years ago
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