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3 years ago

Equal quantities of different liquids are placed in closed manometers at 20°

Chemistry

1 answer:

horrorfan [7]3 years ago

7 0

NOTE: Above question is incomplete:
Complete question is attached as an attached (image),along with the answer.
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Vapor pressure is defined as the pressure exerted by a vapor which is in equilibrium with liquid, at a given temperature, in a closed system. If vapor pressure is high, it suggest that escaping tendency of liquid is high. This will result in higher pressure level in manometer.

In present case liquid A (i.e option F) has highest level in manometer, suggesting that it has highest vapor pressure.

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aleksandr82 [10.1K]

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2 years ago

How will adding NaCl affect the freezing point of a solution?

lord [1]

Answer is: adding NaCl will lower the freezing point of a solution.

A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water).

The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.

Equation describing the change in freezing point:

ΔT = Kf · b · i.

ΔT - temperature change from pure solvent to solution.

Kf - the molal freezing point depression constant.

b - molality (moles of solute per kilogram of solvent).

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3 years ago

Read 2 more answers

What is the molarity of the potassium hydroxide if 25.25 mL of KOH is required to neutralize 0.500 g of oxalic acid, H2C2O4? H2C

Greeley [361]

Answer:

0.444 mol/L

Explanation:

First step is to find the number of moles of oxalic acid.

n(oxalic acid) = $\frac{0.5g}{90.03 g/mol} = 5.5537*10^{-3} mol\\$

Now use the molar ratio to find how many moles of NaOH would be required to neutralize $5.5537*10^{-3} mol\\$ of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

1 : 2 (we get this from the balanced equation)

$5.5537*10^{-3} mol\\$ : x

x = 0.0111 mol

Now to calculate what concentration of KOH that would be in 25 mL of water:

$c = \frac{number of moles}{volume} = \frac{0.0111}{0.025} = 0.444 mol/L$

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4 years ago

When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t

Harlamova29_29 [7]

Answer: Molarity of $MgCl_2$ in the original sample was 1.96M

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{\text{no of moles}}{\text{Volume in L}}$

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{186g}{95g/mol}=1.96$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1.96}{1.00L}$

$Molarity=1.96mol/L$

Thus molarity of $MgCl_2$ in the original sample was 1.96M

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3 years ago

The density of gold is 19.32 g/mL. What is the volume, in mL, of a nugget of gold with a mass of 55.07 g?

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Explanation:

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2 years ago