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3 years ago

Un recipiente contiene 224 dm3 de Ozono de masa 4.561 Kg a 51.09 grados celsius. Calcula la presión del Ozono

Physics

1 answer:

kari74 [83]3 years ago

5 0

Answer:

Por lo tanto, la presión del ozono es:

$P=0.011\: atm$

Explanation:

Podemos usar la ecuacion de los gases ideales;

$PV=nRT$ (1)

Tenemos:

El volumen V = 224 dm³ = 224 L

La temperatura T = 51.09 C = 324.09 K

La masa es m = 4.561 kg

Lo necesitamos ahora es calvular n que es el numero de moles;

recordemos que el peso molecular del ozono M = 48 g/mol.

$n=\frac{m}{M}=\frac{4.561}{48}=0.095\: mol$

Finalmente, usando la ecuacion 1 despejamos la presion P

$P=\frac{nRT}{V}$

$P=\frac{0.095*0.082*324.09}{224}$

Por lo tanto, la presion del ozono es:

$P=0.011\: atm$

Espero te haya ayudado!

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Answer:

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Explanation:

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b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

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