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Ronch [10]
3 years ago
5

A helium balloon has a volume of 36 L. The pressure of the helium gas in the balloon is 132 kPa and its temperature is 27 degree

s Celsius. How many moles of helium are in the balloon?
Chemistry
1 answer:
Verdich [7]3 years ago
3 0

Answer:

n =1.905mol

Explanation:

Hello!

In this case, by considering helium gas as an ideal one, we can use the following equation:

PV=nRT

Whereas P should go in atmospheres and T in Kelvins; thus we proceed as follows:

n=\frac{PV}{RT}=\frac{132kPa*\frac{1atm}{101.325kPa}*36L}{0.08206\frac{atm*L}{mol*K}(27+273)K} =1.905mol

Best regards!

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Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

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A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }

So; in terms of vanadium, the Concentration of iron is:

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V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}

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SO; replacing V_c with a^3 ; \rho_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} } ; A_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} } and

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a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big)  }

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