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I need help with this question please. I thought of using the formula to find specific latent heat but I'm not sure. Can you please help or advise me?
Substance “X” has a molar mass of 0.98 kg mol-1 and each molecule in the liquid state has 6 nearest neighbours. Given that the binding energy, ε0, of a pair of molecules is 2×10-20 J, calculate the latent heat of vaporisation (in kJ kg-1) for this substance.
Thankyou
Substance “X” has a molar mass of 0.98 kg mol-1 and each molecule in the liquid state has 6 nearest neighbours. Given that the binding energy, ε0, of a pair of molecules is 2×10-20 J, calculate the latent heat of vaporisation (in kJ kg-1) for this substance.
Thankyou
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(Original post by ahmed2017)
I need help with this question please. I thought of using the formula to find specific latent heat but I'm not sure. Can you please help or advise me?
Substance “X” has a molar mass of 0.98 kg mol-1 and each molecule in the liquid state has 6 nearest neighbours. Given that the binding energy, ε0, of a pair of molecules is 2×10-20 J, calculate the latent heat of vaporisation (in kJ kg-1) for this substance.
Thankyou
I need help with this question please. I thought of using the formula to find specific latent heat but I'm not sure. Can you please help or advise me?
Substance “X” has a molar mass of 0.98 kg mol-1 and each molecule in the liquid state has 6 nearest neighbours. Given that the binding energy, ε0, of a pair of molecules is 2×10-20 J, calculate the latent heat of vaporisation (in kJ kg-1) for this substance.
Thankyou
Since 0.98 kg is 1 mol of X, this means
NA molecules of X are found in 0.98 kg.
Then use the information given to work out the latent heat of vaporisation
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(Original post by BobbJo)
You can find the number of molecules in 1 kg of X.
Since 0.98 kg is 1 mol of X, this means
NA molecules of X are found in 0.98 kg.
Then use the information given to work out the latent heat of vaporisation
You can find the number of molecules in 1 kg of X.
Since 0.98 kg is 1 mol of X, this means
NA molecules of X are found in 0.98 kg.
Then use the information given to work out the latent heat of vaporisation
Last edited by ahmed2017; 2 years ago
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(Original post by ahmed2017)
to find the number of molecules, do I multiply mole by avogadros constant. And then find the number of molecules for 1 kg?
to find the number of molecules, do I multiply mole by avogadros constant. And then find the number of molecules for 1 kg?
you can also do 1 mol -> 0.98 kg
therefore 0.98 kg -> 6.02 x 10^23
hence 1kg ...
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[QUOTE=BobbJo;80888220]Since molar mass is 0.98 kg mol^-1, you can find number of moles in 1 kg. Then multiply by Avogadro's constant and you have number of molecules
you can also do 1 mol -> 0.98 kg
therefore 0.98 kg -> 6.02 x 10^23
hence 1kg Y
Yes, that's what I did. Thank you so much
you can also do 1 mol -> 0.98 kg
therefore 0.98 kg -> 6.02 x 10^23
hence 1kg Y
(Original post by BobbJo)
Since molar mass is 0.98 kg mol^-1, you can find number of moles in 1 kg. Then multiply by Avogadro's constant and you have number of molecules
you can also do 1 mol -> 0.98 kg
therefore 0.98 kg -> 6.02 x 10^23
hence 1kg ...
Since molar mass is 0.98 kg mol^-1, you can find number of moles in 1 kg. Then multiply by Avogadro's constant and you have number of molecules
you can also do 1 mol -> 0.98 kg
therefore 0.98 kg -> 6.02 x 10^23
hence 1kg ...
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