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alexira [117]
3 years ago
9

A gas system contains 2.00 moles of O2 and CO2 gas, has an initial temperature of 25.0 oC and is under 1.00 atm of pressure. If

the pressure remains constant and the temperature is raised to 45.0 oC, then what will the new volume (assuming a closed system) be?
Chemistry
1 answer:
Ainat [17]3 years ago
8 0

Answer:

V_2=52.2L

Explanation:

Hello there!

In this case, by bearing to to mind the given conditions, it is firstly possible to determine the initial volume of the closed system via the ideal gas equation:

PV=nRT\\\\V=\frac{2.00mol*0.08206\frac{atm*L}{mol*K}*298.15K}{1.00 atm} \\\\V=48.9L

Which is V1 in the Charles' law:

\frac{T_2}{V_2} =\frac{T_1}{V_1}

And of course, T1 is 298.15 (25+273.15). Therefore, by solving for V2 as the final volume, we obtain:

V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{48.9L*(45+273.15)K}{(25+273.15)K}  \\\\V_2=52.2L

Best regards!

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The sp of pbbr2 is 6. 60×10−6. what is the molar solubility of pbbr2 in pure water?
Inessa05 [86]

Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.

Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.

Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.

Let the molar solubility be S upon dissociation.

PbBr₂ or Lead Bromide dissociates in pure water as follows:

                          PbBr₂ ----------> Pb⁺² + Br⁻

                                                     S      2S

Ksp = [Pb⁺²] [ Br⁻]

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Hence, the Molar solubility S is 0.0118M.

Learn more about Molar solubility here, brainly.com/question/16243859

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2 years ago
At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil
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Answer:

The volume of air at where the pressure and temperature are  52 kPa, -5.0 ºC is 3.64 m^3.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 104 kPa

P_2 = final pressure of gas = 52 kPa

V_1 = initial volume of gas = 2.0m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 21.1^oC=273+21.1=294.1K

T_2 = final temperature of gas = -5.0^oC=273+(-5.0)=268 K

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\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}

V_2=3.64 m^3

The volume of air at where the pressure and temperature are  52 kPa, -5.0 ºC is 3.64 m^3.

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