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2 years ago

True or False: You can make a concentrated solution more dilute by adding solvent.

2 answers:

6 0

That statement is true because to make a diluted solution, you have to add more solvent than the solute in a concentrated solution. This is extra information, but to make a diluted solution more concentrated, you have to add more solute since a concentrated solution has an excess amount of solute as stated before.

mamaluj [8]2 years ago

5 0

You can make a concentrated solution more dilute by adding solvent.
I think that statement is true

hope this helps

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3 years ago

Give the four symbols for physical states of reactants and products, and tell what each means.

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2 years ago

2.Which term best describes the processes that change rocks and mountains on Earth’s surface over time?

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2 years ago

One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc

Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

$\text{The new partial pressure for }N_2 \ gas}$

$P_1V_1=P_2V_2$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar$

$\text{The new partial pressure for }Ar \ gas}$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar$

$Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.225 \ Bar$

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

$2.1 \ bar = 2.07 \ atm \\ \\3.4 \ bar = 3.36 \ atm$

For moles N₂:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}$

$n = 0.08297 \ mol \ N_2$

For moles of Ar:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}$

$n = 0.2172 \ mol \ Ar$

$\mathtt{total \ moles = moles \ of \ N_2 + moles \ of \ Ar}$

$=0.08297 mol + 0.2037 mol \\ = 0.2867 mol gases$

Finally;

The final pressure of the mixture is:

$PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}$

P = 2.217 atm

P ≅ 2.24 bar

7 0

3 years ago

A 16.9 gram sample of KCl is used to mak a 250 mL KCl solution. What is the molarity of the KCl solution?

Genrish500 [490]

Answer:

0.906 gm/l

Explanation:

We know that molarity of the a solution is given by,

$M= \frac{w}{m} \times \frac{1000}{v}$ , where 'M' is the molarity, 'w' is the weight of the sample, 'm' is the molar mass, and 'v' is the volume of the solution.

Molarity of a solution tells us the concentration of the solute in the solvent.

Molar mass of KCl is = 74.55

Putting the values we get,

$M = \frac{16.9}{74.55} \times \frac{1000}{250}=0.906 gm/l$

So the molarity of KCl solution is 0.906 gm/l.

5 0

2 years ago

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