Oxygen gas O2 was placed into a closed container at 273.15K. What is the change in temperature that is required in order to increase the pressure from 2.50atm to 7.50atm and decrease the volume from 3.50L to 1.50L?

351.19K

Explanation: This is a case were the combined gas equation can be applied.

Combined gas equation is given as; P1V1/T1  = P2V2/T2,

where P1 = Initial pressure = 2.50 atm

          P2=  Final pressure = 7.50 atm

          V1 = Initial volume = 3.50L

          V2 = Final Volume =  1.50L

          T1 =  Initial Temperature = 273.15K

          T2 =  Final Temperature =  ?

∴  2.50atm  x  3.50L  /  273.15K   =  7.50atm  x  1.50L /  T2

   Making T2 the subject of  Formular; we then have;

   T2 =  P2 V2 T1 / P1 V1   =  7.50atm  x  1.50L x 273.15K /  2.50atm  x  3.50L

        =   3072.94 / 8.75 = 351.19K

The change in temperature (T2) is 351.19K.

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.