σ1s² σ*1s² σ2s² σ*2s²(π2p²x=π2p²y) σ2pz
Mo^3+ = [Kr] 4d^3
N2+ and N2- I just did it on mastering chem and it worked I'm pretty sure its because when you count the valence electrons in N2+ and N2- you get 9 and 11 respectively because these are odd there has to be an unpaired electron in each
The reaction is :- 3Li + 1/2 N2 -> Li3N Therefore 0.483 M of lithium nitride are produced
0 in N2
0 in N2
Sipsipin mo titi ko
Mo^3+ = [Kr] 4d^3
putang inah nyong bumasa n2..haha bobo binasa kasi . . . . . . . . . . . . . . . . . . sangala binasa tlaga bobo mo gagu...
[Kr] 5s1 4d5Mo is an exception to the usual rules of electron configuration. You would think that the configuration would be[Kr] 5s2 4d4But Mo is more stable with and extra electron in it's d orbital, rather than it's s orbital.
n2-1 and n2-4 are trivial cases because of n2-m2=(n-m)(n+m). So the only prime of the form n2-1 is 3 and of the form n2-4 is 5.
N2+ and N2- I just did it on mastering chem and it worked I'm pretty sure its because when you count the valence electrons in N2+ and N2- you get 9 and 11 respectively because these are odd there has to be an unpaired electron in each
If it is an element, it could be either chromium (Cr) or molybdenum (Mo). You need to know more to know which. The electronic configuration for Cr is:[Ar]3d54s1and for Mo it is[Kr]4d55s1So both elements have 5 d-electrons and 1 s-electron. I'm not sure if that what you mean however by "s1d5"...
n2 + n2 = 2 n2
1 mole N2 = 28.0134g 1 mole N2 = 6.022 x 1023 molecules N2 28.0134g N2 = 6.022 x 1023 molecules N2 (4.00 x 1023 molecules N2) x (28.0134g/6.022 x 1023 molecules) = 18.6g N2
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
0 in N2
0 in N2