Question

When the evacuated tube of volume 400 is filled with a gas at 300k and 101kpa the mass of tube is 0.65g what could be the identity of gas?

Answer 1

Answer:

Explanation:

P volume = 0.112 L

volume = 0.112 Lpressure = 3.75 atm

volume = 0.112 Lpressure = 3.75 atmtemp = 273K

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRT

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn=

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×273

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 moles

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measured

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23 =112.5×10

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23 =112.5×10 20

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23 =112.5×10 20

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23 =112.5×10 20 Hence, the correct option is C

volume = 0.112 Lpressure = 3.75 atmtemp = 273Kgas law is PV=nRTn= 0.0821×2733.75×0.112 n=0.01874 molesA mole has 6.022×10 23 atoms or molecules of the pure substance being measuredno. of molecules =0.01874×6.022×10 23 =112.5×10 20