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stiks02 [169]
4 years ago
15

Using the accompanying chart, calculate the volume of a Densities of Group 14 Elements tin block that has a mass of 95.04 grams

at STP. Your response must include both a numerical setup and the calculated result.
Chemistry
1 answer:
saw5 [17]4 years ago
5 0

<u>Answer:</u> The volume of given mas of tin block at STP is 13.0cm^3

<u>Explanation:</u>

To calculate volume of of the substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Mass of tin = 95.04 g

Density of tin = 7.31g/cm^3

Putting values in above equation, we get:

7.31g/cm^3=\frac{95.04g}{\text{Volume of tin}}\\\\\text{Volume of tin}=\frac{95.04g}{7.31g/cm^3}=13.0cm^3

Hence, the volume of given mas of tin block at STP is 13.0cm^3

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Explanation:

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In a 200 cm3 container, a certain amount of O2 gas is held with a piston at equilibrium with the external pressure being 1 bar.
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Answer:

a) V2 = 200.161 cm³

b)  Wexp = 0.0161 J done by the system

c) ΔU = 1.984 J

d) ΔH = 2.006 J

Explanation:

ΔU = Q + W.....first law

a) at constant temperature (298 K ):

∴ ΔU = 0 ⇒ Q = - W

∴ W = - ∫ P dV

⇒ Q = ∫ PdV

∴ PV = nRT

⇒ P = nRT / V

⇒  Q = ∫ nRT dV/V

⇒ Q = nRT Ln ( V2 / V1 )....assuming n O2 = 1 mol

⇒  Q / nRT = Ln ( V2 / V1 )

⇒ 2 J / (( 1mol) ( 8.314 J/molK * 298 K )) = Ln ( V2 / V1 )

⇒ 8.072 E-4 = Ln ( V2 / V1 )

⇒ 1.001 = V2 / V1

⇒ V2 = 1.001 * 200 cm³

⇒ V2 = 200.161 cm³  ( 2.00161 E-4 m³)

b) Wexp = - ∫ PdV.......work done by the system ( - )

P = 1 bar = 10197.2 Kgf/m²

⇒ W exp = - 10197.2 Kgf/m² * ( 2.00161 E-4 - 2.00 E-4 )m³

⇒ W exp = - 1.642 E-3 Kgf *m ( - 0.0161 J )

c) ΔU = Q + W

⇒ ΔU = 2 J - 0,0161 J

⇒ ΔU = 1.984 J   ( 0.202 Kgf*m )

d) H = U + PV.....ideal gas

⇒ ΔH = Δ ( U + PV )

⇒ ΔH = ΔU + P2V2 - P1V1

⇒ ΔH = 0.202  + (( 10197.2 * 2.00161 e-4 )) - (( 10197.2 * 2.00 E-4 ))

⇒ ΔH = 0.202 + 2.041 - 2.0394

⇒ ΔH = 0,204 Kgf*m ( 2.006 J )

8 0
4 years ago
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