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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? |
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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 |
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