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Genrish500 [490]
2 years ago
10

Please Help Im Struggling And Failing I Need All The Help I Can Get!!!!

Chemistry
2 answers:
Margaret [11]2 years ago
7 0
2 floors because they are double the work.
luda_lava [24]2 years ago
5 0
2 floors because you would have to do it twice
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10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of
Dimas [21]

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and 7.2m^3 respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

P_1 = initial pressure of gas = 10 kPa

P_2 = final pressure of gas = 15 kPa

V_1 = initial volume of gas = 10m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 50^oC=273+50=323K

T_2 = final temperature of gas = 75^oC=273+75=348K

Now put all the given values in the above equation, we get:

\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}

V_2=7.2m^3

The new volume of carbon dioxide gas is 7.2m^3

Now we have to calculate the new density of carbon dioxide gas.

PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}

Formula for new density will be:

\rho_2=\frac{P_2M}{RT_2}

where,

P_2 = new pressure of gas = 15 kPa

T_2 = new temperature of gas = 75^oC=273+75=348K

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

\rho = new density

Now put all the given values in the above equation, we get:

\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}

\rho_2=0.2281g/L

The new density of carbon dioxide gas is 0.2281 g/L

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Acids which contain only hydrogen and a single other element are referred to as.
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\huge{\green}\fcolorbox{blue}{cyan}{\bf{\underline{\red{\color{red}Answer}}}}

  • A binary acid is an acid that consists of hydrogen and one other element.
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Write the isotope notation for an element with 42 protons and 96 neutrons.
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What is the molality of a solution of water and KCl if the freezing point of the solution is –3°C? (Kf = 1.86°C/m; molar mass of
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Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. Calculations are as follows:

<span> ΔT(freezing point)  = (Kf)mi
3  = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg</span>

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