Friday
April 18, 2014

Homework Help: chemistry

Posted by Anonymous on Sunday, October 15, 2006 at 9:38pm.

At 9.20 cm cubed air bubble forms in a deep late at a depth where the temperature and pressure are 6 degress Celcius and 4.50 atm, respectively. Assuming that the amount of air in the bubble has not changed, calculate its new volume at STP.


Use the combined gas law for this:

P1V1/T1 = P2V2/T2

change the temps to Kelvins.


With anonymous answering anonymous I'll just agree that p1v1/t1 = p2v2/t2 is the formula to use and remember to change T to Kelvin
C _ 273 = K.

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

Chemistry - a diver exhales a bubble with a volume of 250 mL at a pressure of 2....
Physics - An air bubble of volume 15 cc is formed at a depth of 50 m in a lake. ...
last physics problem. - At 25.0 m below the surface of the sea(density=1025 kg/m...
Physics - A scuba diver takes a tank of air on a deep dive. The tank's volume is...
Chemistry - A snorkeler takes a syringe filled with 16 mL of air from the ...
Chemistry - a helium balloon is partially filled with 28.7 L of gas at 25.0 ...
Physics - A bubble of air, 0.010 m^3 in volume, is formed at the bottom of a ...
chemistry - an air bubble is released at the bottom of a lake where the ...
Chemistry - If a gas has a volume of 400.0 cm cubed at a pressure of 755 mmHg ...
Chemistry - The adult blue whale has a lung capacity of 5.0*10^3L. Calculate the...

Search
Members