An ideal gas sample with 7.20 atm pressure occupies 250ml voulme at 18•c. what volume would it occupy at 1.50atm and 216•c?

The   volume  of  ideal gas  that would   occupy  at 1.50 atm  and 216 c is

2016.49 ml

calculation

  that is P1V1/T1=  P2V2/T2

 P1( pressure 1) =  7.20  atm

V1 ( volume 1) = 250  ml

T1 ( temperature 1) = 18 + 273=291  K

P2 ( pressure 2) = 1.50 atm

T2 ( temperature 2)= 216+273=489 K

V2  (  volume 2) =?

 V2=[(T2 xP1 x V1)/(T1 xP2)]

V2=[ (489 K  x 7.20 atm  x 250 ml)   / ( 291 k  x 1.50 atm)] =  2016.49  Ml

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.