At an ocean depth of 10.0m, a diver's lung capacity is 2.40L. The air temperature is 32.0°C and the pressure is 101.30 kPa. What is the volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa?

The volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa is 1.66 L

Explanation:

The combined gas equation is,

where,

= initial pressure of gas = 101.30 kPa

= final pressure of gas = 141.20 kPa

= initial volume of gas = 2.40 L

= final volume of gas = ?

= initial temperature of gas =

= final temperature of gas =

Now put all the given values in the above equation, we get:

The volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa is 1.66 L

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2