Answer:
From the ideal gas equation,
PV = nRT
(1)(20) = n*(0.082)(273+25)
Therefore
n = 20/298*0.082 = 0.818 moles
So,
No.of molecules of CO2 = 0.818 * 6.022*10^23
= 4.926*10^23 molecules of CO2 will be present.
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Answers & Comments
PV = nRT
(1)(20) = n*(0.082)(273+25)
Therefore
n = 20/298*0.082 = 0.818 moles
So,
No.of molecules of CO2 = 0.818 * 6.022*10^23
= 4.926*10^23 molecules of CO2 will be present.
Answer:
From the ideal gas equation,
PV = nRT
(1)(20) = n*(0.082)(273+25)
Therefore
n = 20/298*0.082 = 0.818 moles
So,
No.of molecules of CO2 = 0.818 * 6.022*10^23
= 4.926*10^23 molecules of CO2 will be present.