Odpowiedź:
II II
a) MgO
mMgO = 24u + 16u = 40u
mMg = 24u
mO= 16u
mMg 24u
---------- *100% = -------- * 100% = 6*10% = 60%
mMgO 40u
Mg= 60% 100%-60%= 40% O= 40%
III II
b) Al203
mAl203= 27u*2 + 16u*3 = 54u * 32u = 86u
mAl = 54u
mO= 32u
mAl 54u
-------------*100% = -------- * 100% = 62.79%
mAl203 86u
Al = 62.79% 100%- 62.79% = 37.21%
O= 37.21%
