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PSYCHO15rus [73]
3 years ago
12

A scuba diver 25.0m below the sea surface inhales 75.0mL of compressed air in a scuba tank at a pressure of 3.50atm and a temper

ature of 5.0C. What is the pressure of the air in the lungs if the gas expands to 215.0mL at the body temperature of 37.2C?
Chemistry
1 answer:
julsineya [31]3 years ago
4 0

Answer:

Explanation:

Use the combined gas law

P(V)/T=P(V)/T

P is pressure

V is volume

T is temperature

3.5 atm(75 mL)/(5 C)=P(215 mL)=(37.2 C)

yu can change celcius to kelvin by adding 273.

And then just solve the equation

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Second Option: </span><span>2-methylbutane: I have drawn its structure below, compare it with your structure.

Third Option: </span>4-methylbutene:<span> I have drawn the structure for this name. But this name is also against IUPAC rules, and its correct name is 1-Pentene.

Fourth Option: </span><span>4,3-methylbutyne: Again incorrect name, 4,3 are two positions, but here only one substituent is given (methyl). I have drawn structure of 3,4-Dimethylbutene, this name is incorrect, and the correct name for this compound is 3-Methyl-1-pentene.

Fifth Option: </span><span>4,4-methylbutane: Again incorrect name, 4,4 means at 4 position two substituents, but in name only methyl is given. Anyhow lets make it two and draw a structure, for 4,4-Dimethylbutane. Ooops!! This name is also incorrect, and the correct name for this compound is 2-Methylpentane.

Result:
           Among all options only, option B (2-Methylbutane) is correctly named, and this will be the correct name for the structure given to you. </span>

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