Question 169519
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It follows Ideal Gas Law --------> {{{PV=nRT}}}, Working Eqn
where ---> {{{system(P=mmHg,V=350ml=0.350L,highlight(n=moles),R=GasConstant=0.0821,T=31+273.15=304.15K)}}}
Solving for {{{n}}}:
{{{n=m/highlight(MW)}}}
where, {{{MW=(NumberofAtoms)(AtomicWeight)}}}
In {{{CO[2]}}}, there are "2" atoms of Oxygen and "1" atom of Carbon.
Look on Periodic Table, we see {{{Oxygen}}} has Atomic Weight --->{{{16}}}. Therefore, its Molecular Weight is expressed as {{{MW[O]=2*16=32}}}
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For Carbon, look on Periodic Table and the Atomic Weight is ---->{{{12}}}.
It's Molecular Weight is expressed as {{{MW[C]=1*12=12}}}
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Therefore, {{{MW=MW[O]+MW[C]=32+12=44}}}
Subst.: {{{n=mass/MW=39cross(g)/(44cross(g)/mol)}}}
{{{n=0.886mol}}}
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Going Back our Working Eqn,
{{{P(0.350)=(0.886)(0.0821)(304.15)}}}
{{{P*cross(0.350)/cross(0.350)=cross(22.124)63.211581/cross(0.350)}}}
{{{P=63.212atm}}} ----------> mmHg?
{{{63.212cross(atm)(1mmHg/.001315789cross(atm))}}}
{{{highlight(48044mmHg=Pressure)}}}, Answer
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Thank you,
Jojo