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Which of the following solutions would have the lowest freezing point? 1.0 nacl 3.0 nacl 2.0 nacl or 5.0 nacl

Scilla [17]

The solution that will have the lowest freezing point is 5.0 SODIUM CHLORIDE.
Adding solute to solvents usually result in the depression of the freezing point. The higher the quantity of the solute that is added, the lower the freezing point of the solution.

5 0

4 years ago

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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

Zina [86]

<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:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

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

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

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = 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.

4 0

3 years ago

2. A chemical analysis of a sample provides the following elemental data:

Vadim26 [7]

Answer:

C3 H6 O2

Explanation:

first divide their mass by their respective molar mass, we get:

30.4 moles of C

61.2 moles of H

20.25 moles of O

now divide everyone by the smallest one of them then we get

C= 1.5

H= 3

O= 1

since our answer of C is not near to any whole number so we will multiply all of them by 2

so,

C3 H6 O2 is our answer

3 0

2 years ago

One mole of an ideal gas<br><br> occupies a volume of 22.4<br><br> liters at

-Dominant- [34]

Answer : One mole of an ideal gas occupies a volume of 22.4 liters at STP.

Explanation :

As we know that 1 mole of substance occupies 22.4 liter volume of gas at STP conditions.

STP stands for standard temperature and pressure condition.

At STP, pressure is 1 atm and temperature is 273 K.

By using STP conditions, we get the volume of 22.47 liter.

Hnece, the one mole of an ideal gas occupies a volume of 22.4 liters at STP.

3 0

4 years ago

You place 12.0 milliliters of water in a graduated cylinder. Then you add 15.0 grams of metal to the water and the new water lev

Marrrta [24]

Answer:

<h3>The answer is 7.85 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

volume = final volume of water - initial volume of water

volume = 13.91 - 12 = 1.91 mL

We have

$density = \frac{15}{1.91} \\ = 7.853403141...$

We have the final answer as

Hope this helps you

8 0

3 years ago

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