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neonofarm [45]
3 years ago
13

A mixture of argon and krypton gases, in a 5.38 L flask at 67 °C, contains 6.18 grams of argon and 9.66 grams of krypton. The pa

rtial pressure of krypton in the flask is________atm and the total pressure in the flask is______atm?
Chemistry
1 answer:
kramer3 years ago
5 0

Answer:

The partial pressure of krypton in the flask is 0.59 atm and the total pressure in the flask is 1.39 atm

Explanation:

This must be solved with the Ideal Gas Law equation.

First of all we need the moles or Ar and Kr in the mixture

Moles = Mass / Molar mass

Molar mass Ar 39.95g/m

Moles Ar = 6.18 g/39.95 g/m → 0.154 moles

Molar mass Kr 83.8 g/m

Moles Kr = 9.66 g/ 83.8g/m → 0.115 moles

Total moles in the mixture: 0.154 moles + 0.115 moles = 0.269moles

Now, we have the total moles, we can calculate the total pressure.

P . V = n . R . T

(T° in K = T° in C + 273)

P. 5.38L = 0.269mol . 0.082 L.atm/mol.K . 340K

P = (0.269mol . 0.082 L.atm/mol.K . 340K) / 5.38 L

P = 1.39 atm

Now we have the total pressure, we can apply molar fraction so we can know the partial pressure of Kr.

Kr pressure / Total Pressure = Kr moles / Total moles

Kr pressure / 1.39 atm = 0.115 moles / 0.269 moles

Kr pressure = (0.115 moles / 0.269 moles) / 1.39atm

Kr pressure = 0.59 atm

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The extent of ionization of a weak electrolyte is increased by adding to the solution a strong electrolyte that has an ion in co
rjkz [21]

Answer:

The statement is FALSE.

Explanation:

It is known as the ion effect common to the displacement of an ionic equilibrium when the concentration of one of the ions that are involved in said equilibrium changes, due to the presence in the dissolution of a salt that is dissolved in it.

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AB (s) ⇔ A⁺(aq) + B⁻ (aq)

The equilibrium constant of the reaction is:

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Given the reaction of the dissociation reaction of a stong electrolyte:

CB (s) ⇒ C⁺(aq) + B⁻ (aq)

If the electrolyte CB is added to the medium in which electrolyte AB is found, the medium will have a common ion B⁻:

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CB (s) ⇒ C⁺(aq) + B⁻ (aq)

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Hence, the extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.

4 0
3 years ago
The molecular weight of KMnO4 is 158 g/mol. If you wanted to make 500 mL of a 0.1M stock solution of KMnO4, how much solid KMnO4
labwork [276]

Solid KMnO₄ needed = 7.9 g

<h3>Further explanation</h3>

Given

MW KMnO₄ = 158 g/mol

500 mL(0.5 L) of a 0.1M stock solution of KMnO₄

Required

solid KMnO₄

Solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}

Input the value :

n = M x V

n = 0.1 M x 0.5 L

n = 0.05 mol

Mass KMnO₄ :

= mol x MW

= 0.05 x 158 g/mol

= 7.9 g

5 0
3 years ago
What is the mass, in grams, of 0.0490 mol of iron(III) phosphate
horrorfan [7]

Answer:

m = 7.39 g.

Explanation:

Hello!

In this case, since the molar mass of iron (III) phosphate is 150.82 g/mol based on its molecular formula (FePO₄), we can compute the mass in grams of 0.0490 moles of this compound by setting up the following dimensional analysis:

m=0.0490mol*\frac{150.82g}{1mol} \\\\m=7.39 g

Best regards!

7 0
3 years ago
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