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shepuryov [24]
4 years ago
6

A gas mixture contains 1.25 g n2 and 0.85 g o2 in a 1.55 l c ontainer at 18 °c. calculate the mole fraction and partial pressure

of each component in the gas mixture.
Chemistry
1 answer:
KatRina [158]4 years ago
8 0

Answer:- mole fraction of nitrogen is 0.626 and mole fraction of oxygen is 0.374.

partial pressure of nitrogen is 0.687 atm and partial pressure of oxygen is 0.410 atm.

Solution:- Moles of nitrogen = 1.25 g x (1mol/28g) = 0.0446 mol

moles of oxygen = 0.85 g x (1mol/32g) = 0.0266 mol

Total moles of gases in the container = 0.0446 + 0.0266 = 0.0712

mole fraction of a gas = moles of gas/total moles of the gases

so, mole fraction of nitrogen = 0.0446/0.0712 = 0.626

mole fraction of oxygen = 0.0266/0.0712 = 0.374

Volume of the container is 1.55 L and the temperature is 18 degree C that is 18 + 273 = 291 K

From ideal gas law equation, PV = nRT

P = nRT/V

Partial pressure of nitrogen = (0.0446 x 0.0821 x 291)/1.55 = 0.687 atm

and the partial pressure of oxygen = (0.0266 x 0.0821 x 291)/1.55 = 0.410 atm

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Kruka [31]

Answer:

Strong acids are assumed 100% dissociated in water- True

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The weaker the acid, the stronger the conjugate base- true

Explanation:

An acid is regarded as a strong acid if it attains 100% or complete dissociation in water.

The pOH decreases as a solution becomes more basic (as OH^- concentration increases).

Ka refers to the dissociation of an acid HA into H3O^+ and A^-.

The greater the base dissociation constant, the greater the base strength.

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7 0
3 years ago
You have two compounds that you have spotted on the TLC plate. One compound is more polar than the other. You ran the TLC plate
goldenfox [79]

Answer:

we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf

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A manufacturing plan has been found guilty of polluting the nearby river. this is pollution
solniwko [45]
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___Al+NaOH=>__Na3AlO3+H2
MaRussiya [10]

Explanation:

<em><u>2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2</u></em>

<em><u>...</u></em>

8 0
3 years ago
If 45 mL of water are added to 250 mL of a 0.75 M K2SO4 solution, what will the molarity of the diluted solution be?
krok68 [10]

Answer:

\large\boxed{\large\boxed{0.64M}}

Explanation:

When you form a <em>diluted solution</em> from a mother (concentrated) solution, the moles of solute are determined by the mother solution.

The main equation is:

Molarity=\dfrac{\text{moles of solute}}{\text{volume of the solution in liters}}

Then, since the moles of solute is the same for both the mother solution and the diluted solution:

          \text{Molarity mother solution }\times\text{ volume mother solution}=\\\\=\text{Molarity diluted solution }\times\text{ volume diluted solution}

Substitute and solve for the molarity of the diluted solution:

           250mL\times 0.75M=(45mL+250mL)\times M\\\\\\M=\dfrac{250mL\times 0.75M}{295mL}=0.64M

8 0
3 years ago
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