Answer:
![2Mg^0 + O_2^0\rightarrow2Mg^{2+}O^{2-}](https://tex.z-dn.net/?f=2Mg%5E0%20%2B%20O_2%5E0%5Crightarrow2Mg%5E%7B2%2B%7DO%5E%7B2-%7D)
Explanation:
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In this case, according to the rules for the oxidation states in chemical reactions, it is possible to realize that lone elements have 0 and since magnesium is in group 2A, it forms the cation Mg⁺² as it loses electrons and oxygen is in group 6A so it forms the anion O⁻²; therefore resulting oxidation numbers are:
![2Mg^0 + O_2^0\rightarrow2Mg^{2+}O^{2-}](https://tex.z-dn.net/?f=2Mg%5E0%20%2B%20O_2%5E0%5Crightarrow2Mg%5E%7B2%2B%7DO%5E%7B2-%7D)
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Answer:
2 KClO3 (s) = 2 KCl (s) + 3 O2 (g)
2.5 g x g
Explanation:
x g O2 = 2.5 g KClO3 x (1 mol KClO3) x (3 mol O2) x (32 g O2) = 0.98 g O2
(122.5 g KClO3) (2 mol KClO3) (1 mol O2)
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
2.5 g x g
x g KCl = 2.5 g KClO3 x (1 mol KClO3) x (2 mol KClO3) x (74.5 g KCl) = 1.52 g KCl
(122.5 g KClO3) (2 mol KClO3) (1 mol KCl)
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
x mol 10 mol
x mol KClO3 = 10 mol O2 x (2 mol KClO3) = 6.7 mol KClO3
(3 mol O2)
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As a sidenote, you posted this in Chemistry, when it actually belongs in another topic. Please be sure to post questions only where they belong. Thanks! :)