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Vinil7 [7]
3 years ago
13

diagram 1 above shows equimolar samples of two gases inside a container fitted with a removable barrier placed so that each gas

occupies the same volume. the barrier is carefully removed as the temperature is held constant. diagram 2 above shows the gases soon after the barrier is removed. which statement describes the changes to the initial pressure of each gas and the final partial pressure of each gas in the mixture and also indicates the final total pressure?
Chemistry
1 answer:
vampirchik [111]3 years ago
8 0

This problem is describing the state two gases have when separated and together as shown on the attached picture. First of all, diagram 1 shows how they are separated in two containers with apparently equal volumes, whereas diagram 2 shows the removal of the barrier so that they get mixed together.

In this case, we can analyze that each gas has its own pressure and due to the removal of the barrier, both pressure and volume undergo a change. Thus, we can infer that the final volume is doubled with respected to the initial one for each gas, causing the pressure of each gas to be halved and the total pressure the half of the added ones, in agreement to the Boyle's law (inversely proportional relationship between pressure and temperature).

Therefore, the correct choice is:

C. The partial pressure of each gas in the mixture is half its initial pressure; the final total pressure is half the sum of the initial pressures of the two gases.

Learn more:

  • brainly.com/question/21184611

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What is the molarity to the nearest hundredth of a solution made by dissolving 1.71 moles of NaOH in enough water to produce an
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When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin
LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine (C_3H_7NO_2) are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is 5.9^oC lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 7.2^oC lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

<u>Answer:</u> The van't Hoff factor for potassium bromide in X is 1.63

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\Delta T_f=i\times K_f\times m

OR

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

  • <u>When alanine is dissolved in mystery liquid X:</u>

\Delta T_f=5.9^oC

i = Vant Hoff factor = 1 (for non-electrolytes)

K_f = freezing point depression constant

m_{solute} = Given mass of solute (alanine) = 177. g

M_{solute} = Molar mass of solute (alanine) = 89 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m

  • <u>When KBr is dissolved in mystery liquid X:</u>

\Delta T_f=7.2^oC

i = Vant Hoff factor = ?

K_f = freezing point depression constant = 2.37^oC/m

m_{solute} = Given mass of solute (KBr) = 177. g

M_{solute} = Molar mass of solute (KBr) = 119 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63

Hence, the van't Hoff factor for potassium bromide in X is 1.63

7 0
3 years ago
Which solution will form a precipitate when mixed with a solution of aqueous na2co3? which solution will form a precipitate when
trapecia [35]

Answer:

CuCl_{2}(aq.) will form a precipitate of insoluble CuCO_{3} when aqueous Na_{2}CO_{3} is added.

Explanation:

According to solubility rule-

all carbonates are insoluble except group IA compounds and NH_{4}^{+}

all salts of sodium are soluble

When Na_{2}CO_{3} is added to given solutions, a double displacement reaction takes place in each solution to form a sodium salt and a carbonate salt.

So, in accordance with solubility rule, addition of Na_{2}CO_{3} into CuCl_{2}(aq.) will result precipitation of insoluble CuCO_{3}

Reaction: Na_{2}CO_{3}(aq.)+CuCl_{2}(aq.)\rightarrow 2NaCl(aq.)+CuCO_{3}(aq.)

4 0
3 years ago
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