the surface energy of plane of niobium.the heat of atomization is 745 kj/mol

It is wrong.............

mit 3.091 cheat. for shame. so blatant.

you only cheat if you don't do anything at all and then come here and ask for the answer..its an online course so for others outside the US, it is hard to gain 100% from the course because internet is not 100% reliable. so for me, its not cheating..its adapting to the new way of learning. If am in the US especially at MIT attending lectures, or have full access to video lectures downloading them within minutes, i won't come here to get the answer. But i will come here to help others so that when i ask for help, i get one in return...

so if you are not here to help anyone or looking for any help, then mind your own business..i bet you are the top student for the course 3.091

I agree with bonjo..

yes, really it is cheating.

if you disagree, please post your statements to the course management and identify yourselves.

this is an exam question

cheater alert

To find the surface energy of a plane of niobium, we can use the equation:

Surface Energy = ½ * Heat of Atomization

In this case, the given heat of atomization for niobium is 745 kJ/mol. So, we can substitute this value into the equation:

Surface Energy = ½ * 745 kJ/mol

Now, we can calculate the surface energy:

Surface Energy = 372.5 kJ/mol

Therefore, the surface energy of a plane of niobium is 372.5 kJ/mol.