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UNO [17]
3 years ago
10

The approximate water temperature at a depth of 125 meters would be closest to: (1) 15°C (2) 8°C (3) 13°C (4) 3°C

Biology
1 answer:
photoshop1234 [79]3 years ago
7 0

Answer:

(2) 8 ºC

Explanation:

The range of the temperatures from the uppermost layers of the ocean - around 100 to 150 m -, which in this case is 125 m, is not always constant as they fluctuate due to the atmospheric heating and cooling, and also due to seasonal variations.

However, if we base our answer on the table below, we would determinate that the approximate water temperature at a depth of 125 meters is 8 ºC (2). This is because, if we observe, the temperature at 100 m is 12 ºC, while at 150 m is 5 ºC. Therefore, the temperature at 125 m should be higher than 5 ºC and lower than 12 ºC, that is 8 ºC.

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Gas exchange between the air in the lungs and the blood takes place in the alveoli.

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Growth patterns have been found in which of the following animals? Select all that apply.
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12.36Un balón lleno contiene 12 kg de oxígeno, O2, bajo una presión ma-nométrica de 40 atm. Determine la masa de oxígeno que se
const2013 [10]

Answer:

Quedan 7.317 kilogramos de oxígeno en el balón y se extraen 4.683 kilogramos de oxígeno en el proceso.

Explanation:

Asúmase que el balón es rígido. Supongamos que el oxígeno se comporta como un gas ideal, entonces la ecuación de estado es la siguiente:

P\cdot V = \frac{m\cdot R_{u}\cdot T}{M} (1)

Donde:

P - Presión, medida en atmósferas.

V - Volumen, medido en litros.

m - Masa, medida en kilogramos.

M - Masa molar, medida en kilogramos por kilomol.

T - Temperatura, medida en Kelvin.

R_{u} - Constante de los gases ideales, medida en atmósfera-litros por kilomol-Kelvin.

El aire experimenta un proceso isocórico e isotérmico, entonces obtenemos la siguiente relación:

\frac{P_{1}}{m_{1}} = \frac{P_{2}}{m_{2}} (2)

Nótese que los subíndices 1 y 2 representan los estados inicial y final de proceso.

Si sabemos que P_{1} = 41\,atm, m_{1} = 12\,kg y P_{2} = 25\,atm, entonces la masa final del aire es:

m_{2} = \frac{P_{2}}{P_{1}} \cdot m_{1}

m_{2} = \left(\frac{25\,atm}{41\,atm}\right)\cdot (12\,kg)

m_{2} = 7.317\,kg

La masa que se ha extraído del balón es:

\Delta m = m_{1}-m_{2}

\Delta m = 12\,kg-7.317\,kg

\Delta m = 4.683\,kg

Quedan 7.317 kilogramos de oxígeno en el balón y se extraen 4.683 kilogramos de oxígeno en el proceso.

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