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3 years ago

The pressure on a 200-milliliter sample of CO2(g) at

Chemistry

1 answer:

motikmotik3 years ago

3 0

Answer:

A) 100 mL

Explanation:

At constant temperature and number of moles, Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 200 mL

V₂ = ?

P₁ = 60 kPa

P₂ = 120 kPa

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${60\ kPa}\times {200\ mL}={120\ kPa}\times {V_2}$

${V_2}=\frac{{60}\times {200}}{120}\ mL$

${V_2}=100\ mL$

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marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. 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.

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

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

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

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