Acyclopropane-oxygen mixture is used as an anesthetic. if the partial pressure of cyclopropane in the mixture is 334 mmhg and the partial pressure of the oxygen is 1.02 atm, what is the total pressure of the mixture in torr?

Total pressure = 1109.2 torr

Explanation:

Given data:

Partial pressure of cyclopropane = 334 mmHg

Partial pressure pressure of oxygen = 1.02 atm

Total pressure of mixture = ?

Solution:

Formula:

Total pressure = P of C₃H₆ + P of O₂

1 atm = 760 torr = 760 mmHg

1.02 atm × 760 torr / 1atm = 775.2 torr

Total pressure = P of C₃H₆ + P of O₂

Total pressure = 334 torr + 775.2 torr

Total pressure = 1109.2 torr

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied