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vazorg [7]
3 years ago
13

Match.

Chemistry
1 answer:
Akimi4 [234]3 years ago
8 0

Answer:

Molality, Solvent, Solute, Mole fraction, Molarity.

Explanation:

The expression of concentration that provides the moles of solute per kilograms of solvent is Molality.  This in the only expression referred to the solvent.

A solution is made up of 0.15 grams of sodium chloride in 1 liter of water. For this solution, the Solvent is water.  When water is present, it is usually considered the solvent.

A solution is made up of 0.15 grams of sodium chloride in 1 liter of water. For this solution, the Solute is sodium chloride.  There can be 1 or more solutes in a solution.

If you place 5 moles of sodium chloride and 4 moles of sucrose into 11 moles of water, the Mole fraction of sodium chloride would be 0.25.  The mole fraction is equal to the moles of a substance divided by the total number of moles.

A way to express concentration that provides the moles of solute per liter of solution is Molarity.

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Answer:

0

Explanation:

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An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr
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<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa

Rate of flow of ideal gas , n = 4 kmol/hr = \frac{4\times 1000mol}{3600s}=1.11mol/s    (Conversion factors used:  1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW     (Conversion factor:  1 kW = 1000 W)

We know that:

\Delta U=0   (For isothermal process)

So, by applying first law of thermodynamics:

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Now, calculating the work done for isothermal process, we use the equation:

\Delta W=nRT\ln (\frac{P_1}{P_2})

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P_2 = final pressure = 50 kPa

Putting values in above equation, we get:

\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

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