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3 years ago

What can a drop of liquid mercury be described as?

Chemistry

1 answer:

Fiesta28 [93]3 years ago

3 0

Answer:

a pure substance and an element

Explanation:

A drop of liquid mercury be described as a pure substance and an element.

Mercury is one of the known elements in the periodic table. The drop of liquid mercury is pure since it has constant and definite composition as well as distinct chemical properties.

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Which substance gets broken down in a homogeneous mixture? O colloid solution solute Osolvent

astra-53 [7]

Answer:

Solute

Explanation:

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3 years ago

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Arisa [49]

Answer:

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2 years ago

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

When should you<br> use a fume hood?

AfilCa [17]

Answer:

if you are working with hazardous materials.

Explanation:

A properly operating and correctly used fume hood can reduce or eliminate exposure to volatile liquids, dusts, and mists. It is advisable to use a laboratory hood when working with all hazardous substances.

4 0

3 years ago

a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill

baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

Also,

Moles = mass (m) / Molar mass (M)

Density (d) = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRT$

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>

5 0

3 years ago