A gas with volume of 700.0 mL at STP (0.0 0C, 1.00 atm) is compressed to a volume of 200.0 mL and the temperature is increased to 30.0 0C. What is the new pressure in kPa?

393.1 KPa

Explanation:

Now we have to use the general gas equation;

P1V1/T1 = P2V2/T2

P1= initial pressure of the gas= 1.00atm

V1 = initial volume= 700.0 ml

T1= initial temperature= 0.0°C +273= 273K

V2= final volume= 200.0 ml

T2= final temperature =30.0°C +273= 303 K

P2= final pressure =

Therefore;

P2= P1V1T2/ V2T1

P2= 1.00 × 700.0 × 303/ 200.0 × 273

P2= 3.88 atm

But

1 atm = 101.325 kilopascals

Therefore 3.88 atm = 3.88 × 101.325 kilopascals = 393.1 KPa

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.