A marshmallow has an initial volume of 0.027 L at standard pressure (1.0 atm). If the marshmallow is placed in a vacuum chamber and the final volume is 0.205L, what is the pressure inside the chamber?

A marshmallow has an initial volume of 0.027 L at standard pressure (1.0 atm). If the marshmallow is placed in a vacuum chamber and the final volume is 0.205L, what is the pressure inside the chamber?

The pressure inside the chamber is 0.132 atm

Explanation:

Step 1: Data given

The initial volume = 0.027 L

The initial pressure = 1.0 atm

The volume changes to 0.205 L

Step 2: Calculate the pressure

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.0 atm

⇒with V1 = the initial volume = 0.027 L

⇒with P2 = the changed pressure = TO BE DETERMINED

⇒with V2 = the changed volume = 0.205 L

1.0 atm *0.027 L = P2*0.205 L

P2 = 0.027/0.205

P2 = 0.132 atm

The pressure inside the chamber is 0.132 atm

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)

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally