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Vinil7 [7]
3 years ago
15

El volumen del aire en los pulmones de una persona es de 615 ml aproximadamente, a una presión de 760 mm Hg. La inhalación ocurr

e cuando la presión de los pulmones desciende a 752 mm Hg. ¿A qué volumen se expanden los pulmones?
Chemistry
1 answer:
Ann [662]3 years ago
6 0

The volume of air in a person's lungs is about 615 ml, at a pressure of 760 mm Hg. Inhalation occurs when the pressure in the lungs drops to 752 mm Hg. To what volume do the lungs expand?

Answer:

621.5mL

Explanation:

Given parameters:

Initial volume  = 615mL

Initial pressure  = 760mmHg

Final pressure  = 752mmHg

Unknown;

Final volume  = ?

Solution:

To solve this problem, we simply apply Boyle's law.

"the volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant".

    Mathematically;

              P₁V₁  = P₂V₂

P and V are the pressure

1 and 2 are the initial and final states

 Now insert the parameters and start;

               760 x 615  = 752 x V₂

                   V₂ = 621.5mL

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What happens to the partial pressure of oxygen in a sample of air if the temperature is increased? It increases. It decreases. I
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The partial pressure of oxygen in a sample of air increases if the temperature is increased.

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<u>Explanation: </u>

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If the temperature increases, the partial pressure and the pressure of the mixture of gas also tend to increase. As it can be seen that at higher altitudes, the low temperature leads to the decrease in oxygen's partial pressure in the air.

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3 years ago
PLEASE PLEASE PLEASE ANSWER THIS CORRECTLY ASAP 50 POINTS
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4 years ago
The carbon cycle is closely linked to which process?
ser-zykov [4K]

Answer:

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Explanation:

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4 0
3 years ago
Se ha añadido un evaporador para una alimentación de 11500 kg/dia de zumo de pomelo de forma que evapore 3000 kg/dia de agua por
mamaluj [8]

Answer:

37 %.

Explanation:

¡Hola!

En este caso, para el problema descrito, conocemos la corriente de entrada y la de salida del agua, por lo que podemos obtener el flujo de la corriente que contiene el zumo a la salida una vez el agua fue evaporada:

F_{sol}=11500kg/dia-3000kg/dia=8500 kg/dia

Luego, por medio de un balance de zumo de limón en el evaporador en el cual la cantidad que entra es igual a la que sale con sus respectivas concentraciones:

x_z^{entra}*11500kg/dia=x_z^{sale}*8500kg/dia

Como la concentración del zumo a la salida es del 50 % (0.50), la de entrada es:

x_z^{entra}=\frac{x_z^{sale}*8500kg/dia}{11500kg/dia} =\frac{0.50*8500kg/dia}{11500 kg/dia}\\ \\x_z^{entra}=0.37

Que es igual al 37%.

¡Saludos!

6 0
4 years ago
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