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How many moles of nitrogen gas will occupy a volume of 347 mL at 6680 torr and 27 degrees celsius?
Wiki User
∙ 12y ago
Assuming that Nitrogen will be an ideal gas at this temperature we may use the ideal gas law:
PV=nRT
To use this law we have to have a value of R that agrees with the units that we use.
1 Torr = 1/760 atmosphere so 6680 tor = 8.789 ATM
27' C = 300.15'K
347ml = .347L
so
R= 8.314472 J/'Kmol
(8.789)(.347)/(8.314472)(300.15) = n =0.00122 moles
and of course a real man has to use scientific notation. 1.22x10-3 moles of nitrogen
Wiki User
∙ 12y ago
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