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.

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b

answer seismograph;