Răspuns :
Răspuns:
configuratii electronice
Na 1s2 2s2 2p6 3s1 5e tips 6e tip p total 11e
K 1s2 2s2 2p6 3s2 2p6 4s1 7e tip s 12e tip p total 19e
aliaj alcatuit din a moli Na si b moli K
(5a+7b)/(6a+12b)=3/4---> 2a=8b----> a=4b
total e: 11a+ 19b= 151,7544x10²³/6,022x10²³---> 11a+ 19b=25,2
rezolvand sistemul de ecuatii, rezulta b=9,4mol K a= 1,6molNa
a.
2mol aliaj.....0,4molK.....1,6molNa
100mol............x.............y
x=20% y= 80%
b
m= 0,4x39gK + 1,6molx23gNa= 52,4g
Explicație: