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3 years ago

What is the freezing point of an aqueous solution that boils at 103.7 degrees Celsius?

1 answer:

7 0

For this, we use equations from the colligative properties of solutions specifically boiling point elevation and freezing point depression. The equations for these are expressed as:

ΔTb = kb m

where k is a boiling point elevation constant and m is the concentration in terms of molality

ΔTf = kf m

where k is a freezing point elevation constant and m is the concentration in terms of molality

We use both expression to solve for the freezing point. For this case, concentration is the same. The equation will then be:

ΔTf = kf ( ΔTb / kb )
0-Tf = 1.86 (103.7 - 100 / 0.512 )
Tf = -13.4°C

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3 years ago

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3 years ago

You have a 2.0 L balloon filled with air at a temperature of 273 K.

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You have a 2.0 L balloon filled with air at a temperature of 273 K.

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3 years ago

Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2

quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

The formula used for isothermally irreversible expansion is :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

The expression used for work done in reversible isothermal expansion will be,

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $25^oC=273+25=298K$

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(2.5atm)\times (10-1)L$

$w=-22.5L.atm=-22.5\times 101.3J=-2279.25J$

(2) A free isothermal expansion from 1 L to 100 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})$

$w=-11409.6J$

(3) A reversible isothermal expansion from 0.5 L to 4 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})$

$w=-5151.97J$

(4) A reversible isothermal expansion from 0.5 L to 40 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})$

$w=-10856.8J$

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(0.5atm)\times (100-1)L$

$w=-49.5L.atm=-49.5\times 101.3J=-5014.35J$

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

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3 years ago

Which pair of elements would you expect to exhibit the greatest similarity in their physical and chemical properties? o,s si, p?

aliya0001 [1]

I believe the appropriate pair is O and S, Periodic table represents arrangements of elements ordered by their atomic number,physical and chemical properties. Both sulfur and oxygen are in the same group of the periodic table (group 6), they have a valency of 2, This means they have a greater similarity in chemical and physical properties.

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3 years ago