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Chapter 4: Problem 52

Mssign oxidation numbers to each element in (a) \(\mathrm{HIO}_{3}\) (b) \(\mathrm{NaMnO}_{4}\) (c) \(\mathrm{SnO}_{2}\) (d) NOF (e) \(\mathrm{NaO}_{2}\)

Short Answer

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Answer: The oxidation numbers of the elements in the given compounds are as follows: - HIO3: H (+1), I (+5), and O (-2) - NaMnO4: Na (+1), Mn (+7), and O (-2) - SnO2: Sn (+4) and O (-2) - NOF: N (+3), O (-2), and F (-1) - NaO2: Na (+1) and O (-1/2)

Step by step solution

01

HIO3: Assigning Oxidation Numbers for Hydrogen, Iodine, and Oxygen

1. Hydrogen generally has an oxidation number of +1. 2. Oxygen generally has an oxidation number of -2. 3. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Iodine. \(+1*x -2*3+y=0\), x being the amount of H-atoms and y being the amount of I-atoms. In this compound: x=1 and y=1 Hence, the oxidation number of Iodine is +5. So, the oxidation numbers of the elements are: H (+1), I (+5), and O (-2).
02

NaMnO4: Assigning Oxidation Numbers for Sodium, Manganese, and Oxygen

1. Sodium generally has an oxidation number of +1. 2. Oxygen generally has an oxidation number of -2. 3. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Manganese. \(+1-2*4+y=0\), y being the oxidation number for Mn. Hence, the oxidation number of Manganese is +7. So, the oxidation numbers of the elements are: Na (+1), Mn (+7), and O (-2).
03

SnO2: Assigning Oxidation Numbers for Tin and Oxygen

1. Oxygen generally has an oxidation number of -2. 2. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Tin. \(y-2*2=0\), y being the oxidation number for Sn. Hence, the oxidation number of Tin is +4. So, the oxidation numbers of the elements are: Sn (+4) and O (-2).
04

NOF: Assigning Oxidation Numbers for Nitrogen, Oxygen, and Fluorine

1. Oxygen generally has an oxidation number of -2. 2. Fluorine generally has an oxidation number of -1. 3. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Nitrogen. \(y-1-2=0\), y being the oxidation number for N. Hence, the oxidation number of Nitrogen is +3. So, the oxidation numbers of the elements are: N (+3), O (-2), and F (-1).
05

NaO2: Assigning Oxidation Numbers for Sodium and Oxygen

1. Sodium generally has an oxidation number of +1. 2. Knowing that the overall charge of the compound is 0, we will determine the oxidation number of Oxygen. \(+1+y*2=0\), y being oxidation number for O. Hence, the oxidation number of Oxygen is -1/2 (in this case, oxygen is in a peroxide form). So, the oxidation numbers of the elements are: Na (+1) and O (-1/2).

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Most popular questions from this chapter

Wlassify each of the following half-equations as oxidation or reduction and balance. (a) (acidic) \(\quad \mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)\) (b) (basic) \(\quad \mathrm{CrO}_{4}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (c) (basic) \(\quad \mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)\) (d) (acidic) \(\quad \mathrm{ClO}_{2}^{-}(a q) \longrightarrow \mathrm{ClO}^{-}(a q)\)

The percentage of sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), in a powder for stomach upsets is found by titrating with \(0.275 M\) hydrochloric acid. If \(15.5 \mathrm{~mL}\) of hydrochloric acid is required to react with \(0.500 \mathrm{~g}\) of the sample, what is the percentage of sodium hydrogen carbonate in the sample? The balanced equation for the reaction that takes place is $$ \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$

Write net ionic equations for the formation of (a) a precipitate when solutions of magnesium nitrate and potassium hydroxide are mixed. (b) two different precipitates when solutions of silver(I) sulfate and barium chloride are mixed.

For an acid-base reaction, what is the reacting species, that is, the ion or molecule that appears in the chemical equation, in the following acids? (a) perchloric acid (b) hydriodic acid (c) nitrous acid (d) nitric acid (e) lactic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right)\)

Aluminum ions react with carbonate ions to form an insoluble compound, aluminum carbonate. (a) Write the net ionic equation for this reaction. (b) What is the molarity of a solution of aluminum chloride if \(30.0 \mathrm{~mL}\) is required to react with \(35.5 \mathrm{~mL}\) of \(0.137 \mathrm{M}\) sodium carbonate? (c) How many grams of aluminum carbonate are formed in (b)?
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