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NNADVOKAT [17]
3 years ago
11

The mass of a single molybdenum atom is 1.59×10-22 grams. How many molybdenum atoms would there be in 40.6 milligrams of molybde

num?
Chemistry
1 answer:
iogann1982 [59]3 years ago
8 0

Answer:

2.55×10²⁰ atoms of Mo are contained in 40.6 mg

Explanation:

1 atom of Mo has a mass of 1.59×10⁻²² g

So let's think a rule of three for this question.

Firstly, we convert the mass in mg to g

40.6 mg / 1000 = 0.0406 g

In the mass of 1.59×10⁻²² g, there is 1 atom of Mo

In the mass of 0.0406 g there are (0.0406  / 1.59×10⁻²²) = 2.55×10²⁰ atoms

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Otrada [13]

Answer:

3626.76dm³

Explanation:

Given parameters:

Number of moles of Nitrogen in tank = 17moles

Temperature of the gas = 34°C

Pressure on the gas = 12000Pa

Unkown:

Volume of the tank, V =?

Converting the parameters to workable units:

We take the temperature from °C to Kelvin

K = 273 +  °C  = 273 + 34 = 307k

Taking the pressure in Pa to atm:

101325Pa = 1atm

12000Pa = 0.118atm

Solution:

To solve this problem, we employ the use of the ideal gas equation. The ideal gas law combines three gas laws which are the Boyle's law, Charles's law and the Avogadro's law.

            It is expressed as PV = nRT

The unknown is the Volume and we make it the subject of the formula

             V = \frac{nRT}{P}

Where R is called the gas constant and it is given as 0.082atmdm³mol⁻¹K⁻¹

Therefore  V = \frac{17 x 0.082 x 307 }{0.118} = 3626.76dm³

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What is the total pressure of air in lungs of an individual with oxygen at 100. mmHg, nitrogen at 573 mmHg, carbon dioxide at 0.
ryzh [129]

Answer: The total pressure of air in lungs of an individual is 760.28 mm Hg

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_A+p_B+p_C...

Given : p_{total} =total pressure of gases = ?

p_{O_2} = partial pressure of oxygen = 100 mm Hg

p_{N_2} = partial pressure of nitrogen = 573 mm Hg

p_{CO_2} = partial pressure of Carbon dioxide = 0.053 atm = 40.28 mm Hg(1 atm = 760 mmHg)

p_{H_2O} = partial pressure of water vapor = 47 torr = 47 mm Hg  (1torr=1 mm Hg)

putting in the values we get:

p_{total}=(100+573+40.28+47)mmHg

p_{total}=760.28mmHg

Thus the total pressure of air in lungs of an individual is 760.28 mm Hg

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Answer:

50,662.5 kPa

Explanation:

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