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iren [92.7K]
3 years ago
8

Which of the following statements about minerals is false?

Chemistry
1 answer:
Eva8 [605]3 years ago
3 0
Satement no. a and d
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Calculate the number of gold atoms in a sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and r
Fudgin [204]

Answer: N = 2.78 × 10^23 atoms

There are N = 2.78 × 10^23 atoms in 70g of Au2cl6

Completed Question:

Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits

Explanation:

Given:

Molar mass of Au2cl6 = 303.33g/mol

Mass of Au2cl6 = 70g

Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol

According to the chemical formula of Au2cl6,

1 mole of Au2cl6 contains 2 moles of Au

Number of moles of Au = 2 × 0.231mol = 0.462mole

There are 6.022 × 10^23 atoms in 1 mole of an element.

Number of Atom of gold in 0.462 mole of gold is:

N = 0.462 mol × 6.022 × 10^23 atoms/mol

N = 2.78 × 10^23 atoms

7 0
3 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
3 years ago
Organic compounds that are synthesized by human cells
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Organic compounds essential to human functioning include carbohydrates, lipids, proteins, and nucleotides. These compounds are said to be organic because they contain both carbon and hydrogen.
4 0
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A jet flies over the ocean. A sound wave travels from the jet to the surface of the ocean. Almost none of the energy from the so
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We need to see the diagram
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3 years ago
How to change 5 % W/V of NaCl to ppm , M ? molar mass = 58.5<br>please clear explain​
11Alexandr11 [23.1K]

Answer:

50000ppm and 0.855M.

Explanation:

ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters

A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.

To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:

<em>mg NaCl:</em>

5g * (1000mg / 1g) = 5000mg

<em>L Solution:</em>

100mL * (1L / 1000mL) = 0.100L

ppm:

5000mg / 0.100L = 50000ppm

To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:

5g * (1mol / 58.5g) = 0.0855moles NaCl

Molarity:

0.0855mol NaCl / 0.100L = 0.855M

7 0
3 years ago
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