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
The answer is A.Hope this helps
Answer:
SrSO4
Explanation:
According to solubility rules, we know that the sulphates of the elements of group two are insoluble in water. The solubility rules describe what chemical species are soluble in water and what species are not soluble in water.
Generally, all chlorides are soluble in water with exception of chlorides such as silver chloride. The chlorides of group one elements are usually highly soluble in water.
Since SrSO4 is a sulphate of a group two element (strontium) it will be the insoluble solid product of the double displacement reaction described in the question.
Answer:
Explanation:
Then, multiply the number of moles of Na by the conversion factor 6.02214179×1023 atoms Na/ 1 mol Na, with 6.02214179×1023 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na.Aug 15, 2020