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MrMuchimi
3 years ago
12

Four balloons, each with a mass of 10.0 g, are inflated to a volume of 20.0 L, each with a different gas: helium, neon, carbon m

onoxide, or nitrogen monoxide. If the temperature is 25.0°C and the atmospheric pressure is1.00 atm, what is the density of each filled balloon? (Note: The density of the filled balloon refers to both the contents of the balloon and the balloon itself.)
Chemistry
2 answers:
RideAnS [48]3 years ago
7 0

Answer:

T=25^o C+273 K + 25= 298 K(0^oC=273 K)

Pressure ,P = 1 atm

Volume, V = 20.0 L

R = 0.0821 atm L/mol K

Balloon 1

PV=n_{He}RT=\frac{\text{mass of He}}{\text{molar mass of He}}\times RT

\text{mass of He}=\frac{PV\times \text{molar mass of He}}{RT}=\frac{1atm \times 20.0 L\times 4 g/mol}{0.0821 atm L/mol K\times 298 K}=3.26 grams

Density of balloon 1 = \rho _1=\frac{\text{mass of balloon+mass of He}}{volume}=\frac{10.0 g+3.26 g}{20 L}=0.663 g/L

Balloon 2

PV=n_{Ne}RT=\frac{\text{mass of Ne}}{\text{molar mass of Ne}}\times RT

\text{mass of Ne}=\frac{PV\times \text{molar mass of Ne}}{RT}=\frac{1atm \times 20.0 L\times 20.18 g/mol}{0.0821 atm L/mol K\times 298 K}=16.49 grams

Density of balloon 2= \rho _2=\frac{\text{mass of balloon+mass of Ne}}{volume}=\frac{10.0 g+16.49 g}{20 L}=1.32 g/L

Balloon 3

PV=n_{CO}RT=\frac{\text{mass of CO}}{\text{molar mass of CO}}\times RT

\text{mass of CO}=\frac{PV\times \text{molar mass of CO}}{RT}=\frac{1atm \times 20.0 L\times 28 g/mol}{0.0821 atm L/mol K\times 298 K}=22.88 grams

Density of balloon 3 = \rho _3=\frac{\text{mass of balloon+mass of CO}}{volume}=\frac{10.0 g+22.88 g}{20 L}=1.64 g/L

Balloon 4

PV=n_{NO}RT=\frac{\text{mass of NO}}{\text{molar mass of NO}}\times RT

\text{mass of NO}=\frac{PV\times \text{molar mass of NO}}{RT}=\frac{1atm \times 20.0 L\times 30 g/mol}{0.0821 atm L/mol K\times 298 K}=24.52 grams

Density of balloon 4= \rho _4=\frac{\text{mass of balloon+mass of NO}}{volume}=\frac{10.0 g+24.52 g}{20 L}=1.72 g/L

weeeeeb [17]3 years ago
6 0
On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ <span>0,082 L</span>·<span>atm/K</span>·<span>mol </span>· 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.
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