Question

Calculate the root mean square speed of hydrogen molecule at STP ????​

Answer 1

Answer:

The number of atoms of hydrogen present in one molecule of hydrogen is 2. So, we will multiply the mass of 1 atom of hydrogen of 2. Now, we will convert this value into kilograms. Hence, the root mean square velocity of hydrogen molecules at STP is 1.845×103ms−1.

Answer 2

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c=

c= M

c= M3PV

c= M3PVP=1 atm=101.3 kPa

c= M3PVP=1 atm=101.3 kPaV=22.4 dm

c= M3PVP=1 atm=101.3 kPaV=22.4 dm 3

c= M3PVP=1 atm=101.3 kPaV=22.4 dm 3 M=2 g mol

M=2 g mol −1

M=2 g mol −1 =0.002 kg mol

M=2 g mol −1 =0.002 kg mol −1

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c=

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.002

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c=

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1 ;T=273 K;

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1 ;T=273 K;M=0.002 kg mol

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1 ;T=273 K;M=0.002 kg mol −1

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1 ;T=273 K;M=0.002 kg mol −1 Substituting the values,

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1 ;T=273 K;M=0.002 kg mol −1 Substituting the values,c=

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1 ;T=273 K;M=0.002 kg mol −1 Substituting the values,c= 0.002

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1 ;T=273 K;M=0.002 kg mol −1 Substituting the values,c= 0.0023×8.314×273

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1 ;T=273 K;M=0.002 kg mol −1 Substituting the values,c= 0.0023×8.314×273 =1845.15 ms

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1 ;T=273 K;M=0.002 kg mol −1 Substituting the values,c= 0.0023×8.314×273 =1845.15 ms −1

M=2 g mol −1 =0.002 kg mol −1 Substituting the values,c= 0.0023×101.3×22.4 =1844.91 ms −1 .Alternative method:c= M3RT R−8.314 kPa dm 3 K −1 mol −1 ;T=273 K;M=0.002 kg mol −1 Substituting the values,c= 0.0023×8.314×273 =1845.15 ms −1 .