After the completion of a gas forming reaction, the column of water remaining in the collection flask was measured to me 55mm high, the vapor pressure of water is 0.0313atm at 25oc, and the atmospheric pressure that day was measured as 0.950 atm, what is the partial pressure of the gas produced by the reaction? 1mmhg=13.595mm h2o and 1atm=760mmhg (see sample calculation found in prelab slides.)

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.

you didnt put any of the following

(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