Answer:
1.mass of Cl in sample = 15.09g
2.volume at STP= 9.53L
3.Temperature at 15L and 875torr= 494.7K
4.pressure at 6.4L and 58 °C= 1.80542atm
Explanation:
(1.) Using the ideal gas law to calculate, PV=nRT
R= 0.08206Latm/mol/K
we will convert P from torr to atm and T from °C to K
P=895torr = 895/760 =1.1776atm
T= 24 °C+273 = 297K
n =[tex]\frac{PV}{RT}[/tex]
n= [tex]\frac{1.1776*8.8}{0.08206*297}[/tex]
n= 0.425 mol
mass= n*molar mass
molar mass of Cl = 35.5g/mol
mass= 0.425 *35.5
mass of Cl in sample = 15.09g
(2.) At stp, T=273K, P=1atm.
Using the combined gas law , [tex]\frac{v1p1}{t1} =\frac{v2p2}{t2}[/tex]
v1 = 8.8L, p1= 1.1776atm, t1= 297K, v2=?, p2= 1atm, t2=273K
[tex]v2= \frac{v1p1t2}{p2t1}[/tex]
[tex]v2= \frac{8.8*1.1776*273}{1*297}[/tex]
v2=9.53L
(3.) t1=? v1=15L p1= 875torr/760 = 1.1513atm,
[tex]t1=\frac{v1p1t2}{v2p2}[/tex]
[tex]t1=\frac{15*1.1513*273}{9.53*1}[/tex]
t1= 494.7K
(4.) p1=? v1=6.4L t1= 58°C+273 =331K
[tex]p1= \frac{v2p2t1}{v1t2}[/tex]
[tex]p1= \frac{9.53*1*331}{6.4*273}[/tex]
p1= 1.80542atm
