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Rudiy27
3 years ago
9

The study of the structures of deep-sea vents and of the life around them represents an intersection of which two fields of scie

nce? A. geology and meteorology B. geology and oceanography C. astronomy and meteorology D. astronomy and oceanography
Chemistry
1 answer:
Bond [772]3 years ago
6 0

Answer: Option (B)

Explanation:

Geology is the study of all the earth dynamic processes, about the various types of rocks, its physical and chemical properties, about the ancient lives, and covers many more topics.

Whereas, Oceanography refers to the study of the marine ecosystem, its biological properties as well as its physical and chemical properties.

This two fields of science namely geology and oceanography is required to study about the deep sea vent structures  as well as the life forms that exists around them.

Deep sea vents are the cracks and openings found underwater from which the hot water comes out in association with numerous gases.

Thus, the correct answer is option (B).

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Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

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V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

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