O2 is present in one litre flask at a pressure of 7.6 10−10
mm of Hg.
Calculate no. of O2 molecule
Answers
Explanation:
Using ideal gas equation,
Using ideal gas equation,n=
Using ideal gas equation,n= RT
Using ideal gas equation,n= RTPV
Using ideal gas equation,n= RTPV
Using ideal gas equation,n= RTPV ....(i)
Using ideal gas equation,n= RTPV ....(i)P=
Using ideal gas equation,n= RTPV ....(i)P= 760
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atm
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n=
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×273
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 =
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 = 0.0821×273
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 = 0.0821×27310
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 = 0.0821×27310 −12
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 = 0.0821×27310 −12
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 = 0.0821×27310 −12
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 = 0.0821×27310 −12 ×6.023×10
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 = 0.0821×27310 −12 ×6.023×10 23
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 = 0.0821×27310 −12 ×6.023×10 23
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 = 0.0821×27310 −12 ×6.023×10 23 =2.68×10
Using ideal gas equation,n= RTPV ....(i)P= 7607.6×10 −10 =10 −12 atmV=1 L,T=273 K,R=0.0821 L atm K −1 mol −1 From eq. (i) n= 0.0821×27310 −12 ×1 Number of molecules =n×6.023×10 23 = 0.0821×27310 −12 ×6.023×10 23 =2.68×10 10
Answer:
2.66 x 10 10 is the answer
Explanation:
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