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2 years ago

A solution that contains less than the amount of solute that would be dissolved at equilibrium is considered to be

Chemistry

1 answer:

kirill115 [55]2 years ago

5 0

Answer:

dilute solution

A solution containing less solute than the equilibrium amount is called a dilute solution. The solvent has a limited capacity to dissolve a solute.

Explanation:

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A gas has a volume of 490. mL at a temperature of -35.0 degrees C. What volume would the gas occupy at 42.0 degrees Celsius? Ple

miskamm [114]

Answer:

648.5 mL

Explanation:

Here we will assume that the pressure of the gas is constant, since it is not given or specified.

Therefore, we can use Charle's law, which states that:

"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically:

$\frac{V}{T}=const.$

where

V is the volume of the gas

T is its absolute temperature

The equation can be rewritten as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where in this problem we have:

$V_1=490 mL$ is the initial volume of the gas

$T_1=-35.0^{\circ} + 273 = 238 K$ is the initial temperature

$T_2=42.0^{\circ}+273=315 K$ is the final temperature

Solving for V2, we find the final volume of the gas:

$V_2=\frac{V_1 T_2}{T_1}=\frac{(490)(315)}{238}=648.5 mL$

8 0

3 years ago

_________ rock's form when hot, molten rock cools and becomes a solid

zavuch27 [327]

The answer to fill in the blank is (igneous)

I hope I helped.

3 0

3 years ago

Please answer fast, shouldn't take long to do ^-^ (will give brainliest)

Anna [14]

Answer:

Explanation:

5 0

2 years ago

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

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2 years ago

Match the following aqueous solutions with the appropriate letter.

Alla [95]

Answer:

Explanation:

When a salt is dissolved , it increases the boiling point . Increase in boiling point depends upon number of ions . So it is a colligative property .

.19 m AgNO₃ . Each molecule will ionize into two ions . So effective molar concentration is 0.19 x 2 = .38 m

0.17 m CrSO4.Each molecule will ionize into two ions . So effective molar concentration is 0.17 x 2 = .34 m

0.13 m Mn(NO₃)₂. Each molecule will ionize into three ions . So effective molar concentration is 0.13 x 3 = .39 m

0.31 m Sucrose(nonelectrolyte). Molecules will not ionize . So effective molar concentration is 0.31 x 1 = .31 m

Higher the molar concentration , greater the depression in boiling point .

So lowest boiling point is 0.13 m Mn(NO₃)₂.

second highest boiling point is 0.19 m AgNO3.

Third lowest boiling point is 0.17 m CrSO4

Highest boiling point or lowest depression 0.31 m Sucrose.

a . 4

b . 1

c . 2

d . 3

6 0

2 years ago