Answer:
Option D: it's ability to lose electrons
Explanation:
Alkali metals are usually discovered in nature. They have highly reactivity at STP conditions (standard temperature and pressure conditions) and easily lose their outermost electron to form positive ions known that have a charge of +1.
Thus, what can determine the extent of reactivity of an alkali metal, is it's ability to lose electrons
The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as
Mg(IO3)2
MgI2O6
As we know that
Mg has +2
O has -2
So,
(+2) + 2I + 6 (-2)=0
2 + 2I - 12 =0
10+ 2I =0
10 = 2I
I =5
When in doubt pick C I don’t really know the answer but I just pick C
Answer:
CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4
Explanation: