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natulia [17]
3 years ago
9

A sample of neon gas occupies a volume of 752ml at 25k. What volume will the gas occupy at 80k if the pressure remains constant

Chemistry
1 answer:
Anettt [7]3 years ago
7 0

Answer:

The gas occupy 2406.4 mL at 80 K.

Explanation:

Given data:

Initial volume of gas = 752 mL

Initial temperature = 25 K

Final temperature = 80 K

Final volume = ?

Solution:

The given problem is solved by using charle's law.

V₁/T₁ = V₂/T₂

V₂ =  V₁. T₂ /T₁

V₂ = 752 mL × 80 k / 25 K

V₂ = 60160 mL. k/25 K

V₂ =  2406.4 mL

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marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

5 0
3 years ago
Identify the following reaction as oxidation and reduction F2(g)+2e-/2F-(aq)​
sertanlavr [38]

Answer:

Reduction

Explanation:

The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

In given reaction fluorine gas gain two electron and form fluoride ions.

F₂(g) + 2e⁻    →   2F⁻(aq)

The given reaction is reduction because oxidation state is decreased from zero to -1.

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3 years ago
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