Im taking my chemistry final right

a snorkeler takes a breath of air so that his lungs are 1,159.49 ml in volume when the pressure is 764.96 mm of hg. if he dives to a depth where the pressure is 1.16 atm, what volume, in ml, will his lungs have? assume the temperature of the water is constant.

When the diver dives to a depth where the pressure is 1.16 atm, he has 1006.1 mL of air in his lungs.

Explanation:

Step 1: Data given

Volume in his lungs = 1159.49 mL

The pressure = 764.96 mmHg = 764.96 / 760 = 1.00652646 atm

When he dives the pressure changes to 1.16 atm.

Step 2: Calculate final volule

(P1*V1)/T1 = (P2*V2)/T2

 ⇒ Since the temperature will be constant we can simplify as:

P1*V1 = P2*V2

⇒ with P1 = The initial pressure 1.00652646 atm

 ⇒ with V1 = The initial volume = 1159.49 mL

 ⇒ with P2 = The pressure after diving = 1.16 atm

 ⇒ with V2 = The volume of air in his lungs after diving

V2 = (P1*V1)/P2

V2 =(1.00652646 * 1159.49)/1.16

V2 = 1006.1 mL

When the diver dives to a depth where the pressure is 1.16 atm, he has 1006.1 mL of air in his lungs.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

air

in a tire you fill a tire with air

then a couple days later you check the air pressure in the tire

hope this mark me brainliest