A 25.0 mL bubble forms at the ocean depths where the pressure is 10.0 atm and the temperature is 5.00 C .What is the volume of that bubble at the ocean surface when the pressure is 760.0 mm Hg and the temperature is 250 C ? help asap!

470.3 mL

Explanation:

Given that:

Initial volume of bubble V1 = 25.0mL

Initial Temperature of bubble T1 = 5.00°C

Convert Celsius to Kelvin

(5.00°C + 273 = 278K)

Pressure P1 = 10.0 atm

Convert pressure in atmosphere to mmHg

(Recall that 1 atm = 760 mmHg

So, 10.0 atm = 10 x 760 = 7600 mmHg)

New Volume of bubble V2 = ?

New temperature of bubble T2 = 250°C

(250°C + 273 = 523K)

New pressure of bubble P2 =760 mmHg

Since, pressure, volume and temperature are given, apply the combined gas gas equation

(P1V1)/T1 = (P2V2)/T2

(7600 mmHg x 25.0mL)/278K = (760 mmHg x V2)/523K

190000 mmHg•mL/278K= 760 mmHg•V2/523K

Cross multiply

190000 mmHg•mL x 523K = 760 mmHg•V2 x 278K

99370000 mmHg•mL•K = 211280mmHg•V2•K

V2 = (99370000 mmHg•mL•K / 211280mmHg•K

V2 = 470.3 mL

Thus, the volume of that bubble at the ocean surface is 470.3 mL

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.