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3 years ago

PLs heLp Me

Chemistry

2 answers:

olchik [2.2K]3 years ago

7 0

Answer:

Opaque → a.no light travels through the material; all light is reflected or absorbed

Translucent → b.some light travels through the material and the image cannot be seen clearly

Transparent → c.all light travels through the material and the image can be seen clearly

Explanation:

Hope this helps :)

emmainna [20.7K]3 years ago

6 0

A. is opaque b. is translucent c. is transparent

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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

3 years ago

Write a paragraph

steposvetlana [31]

I think It’s 55 but that’s just me

6 0

3 years ago

Write Lewis structures for the following molecules: (a) ICl, (b) PH3, (c) P4 (each P is bonded to three other P atoms), (d) H2S,

spin [16.1K]

Answer : The Lewis-dot structure for the following molecules are shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

Now we have to determine the Lewis-dot structure for the following molecules.

(a) The given molecule is, $ICl$

As we know that iodine and chlorine have '7' valence electrons.

Therefore, the total number of valence electrons in $ICl$ = 7 + 7 = 14

According to Lewis-dot structure, there are 2 number of bonding electrons and 12 number of non-bonding electrons.

(b) The given molecule is, $PH_3$

As we know that phosphorous has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $PH_3$ = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

(c) The given molecule is, $P_4$

As we know that phosphorous has '5' valence electrons.

Therefore, the total number of valence electrons in $P_4$ = 4(5) = 20

According to Lewis-dot structure, there are 6 number of bonding electrons and 14 number of non-bonding electrons.

(d) The given molecule is, $H_2S$

As we know that sulfur has '6' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $H_2S$ = 6 + 2(1) = 8

According to Lewis-dot structure, there are 4 number of bonding electrons and 4 number of non-bonding electrons.

(e) The given molecule is, $N_2H_4$

As we know that nitrogen has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $N_2H_4$ = 2(5) + 4(1) = 14

According to Lewis-dot structure, there are 10 number of bonding electrons and 4 number of non-bonding electrons.

(f) The given molecule is, $HClO_3$

As we know that chlorine has '7' valence electrons, oxygen has '6' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in $HClO_3$ = 1 + 7 + 3(6) = 26

According to Lewis-dot structure, there are 12 number of bonding electrons and 14 number of non-bonding electrons.

(g) The given molecule is, $COBr_2$

As we know that bromine has '7' valence electrons, oxygen has '6' valence electrons and carbon has '4' valence electrons.

Therefore, the total number of valence electrons in $COBr_2$ = 4 + 6 + 2(7) = 24

According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.

3 0

3 years ago

BRAINLIEST

mars1129 [50]

I believe that it most likely would be C,

6 0

3 years ago

The red balls represent _____ and have a positive charge.

AysviL [449]

The answer is protons. Neutrons have no charge and electrons have a negative charge so the positive charge must be protons.

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3 years ago