PLEASE HELP ME WITH MY CHEM TEST

Crude oil pumped out of the ground may be accompanied by formation water, a solution that contains high concentrations of nacl a

djverab [1.8K]

solution:

the change in the boiling point is given as,

dTbp =2.30°c

elevation constant for the solvent is given by,

kb=0.512°c/m

$molality=\frac{dTBP}{KB\times m}\\=\frac{2.30}{0.512c/m}$

= 4.49m

8 0

3 years ago

3. Complete the following conversions.

Tasya [4]

A. 0.0105 kg
b. 1570 m
c. 3.5x10^-6
d. 3500000
e. 1000 mL
f. 0.000358 m3
g. 548.6 cm3

3 0

3 years ago

How much eat is absorbed when 50.0g of water changes from 25.0 Celsius to 37.0 Celsius?

Firdavs [7]

27 I think beacuse if you subtract

5 0

3 years ago

Read 2 more answers

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0

2 years ago

Learning Task No. 2: Study the diagram above. Answer the guide questions.

Lyrx [107]

1. "Fn" stands for "Normal Force," "Fg" for "Gravity Force," "Ff" for "Force Friction," and "A" for "Applied Force." 2. As a result of the friction, which causes the ball to stop. 3. The reason the ball rolls is due to the force that is imparted to it.

The definition of force is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. A force that opposes this motion of the hands is activated when they brush against one another. Frictional force is the force resisting motion applied in the direction perpendicular to the direction of motion of the hands.

Learn more about Force here-

brainly.com/question/13191643

#SPJ9

3 0

1 year ago