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Give the oxidation numbers of the underlined atoms in the following molecules and ions: (a) $\mathrm{Mg}_{3} \mathrm{N}_{2}$(b) $\operatorname{Cs} \underline{\mathrm{O}}_{2},(\mathrm{c}) \mathrm{Ca} \underline{\mathrm{C}}_{2},(\mathrm{d}) \underline{\mathrm{C}} \mathrm{O}_{3}^{2-},(\mathrm{e}) \underline{\mathrm{C}}_{2} \mathrm{O}_{4}^{2-},(\mathrm{f}) \mathrm{Zn} \underline{\mathrm{O}}_{2}^{2-}$(g) $\mathrm{NaBH}_{4},$ (h) $\underline{\mathrm{W}} \mathrm{O}_{4}^{2-}$

(a) $\mathrm{Mg}_{3} \mathrm{N}_{2} \rightarrow-3$(b) $\mathrm{Cs} \mathrm{O}_{2} \rightarrow-\frac{1}{2}$(c) $\mathrm{Ca}_{\underline{\mathrm{C}}_{2}} \rightarrow-1$$(d) \underline{C O}_{3}^{2-} \rightarrow+4$(e) $\underline{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow+3$(f) $\mathrm{ZnO}_{2}^{2-} \rightarrow-2$$(\mathrm{g}) \mathrm{NaBH}_{4} \rightarrow+3$(h) $\underline{W} \mathrm{O}_{4}^{2-} \rightarrow+6$

Chemistry 102

Chapter 4

Reactions in Aqueous Solutions

Solutions

Aqueous Equilibria

Rice University

Drexel University

University of Maryland - University College

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Lectures

03:58

In chemistry, a solution is a homogeneous mixture composed of two or more substances. The term "solution" is also used to refer to the resultant mixture. The solution is usually a fluid. The particles of a solute are dispersed or dissolved in the solvent. The resulting solution is also called the solvent. The solvent is the continuous phase.

00:41

In chemistry, an ion is an atom or molecule that has a non-zero net electric charge. The name was coined by John Dalton for ions in 1808, and later expanded to include molecules in 1834.

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Give oxidation numbers for…

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Give the oxidation numbers…

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Okay, question 50: question 50 requires to calculate to find the oxidation number of the elements is under line and we have to show the first 1 keep in mind. The total ocidation number of the add atoms in 1 compar equals 0. If that's compound neutral. So here we see its neutral and we don't know that require us to find the ocidationof number of nitrogen. So we put us here- and we know magnesium in blood to uptossed oxidizer mansoul 2. So from this information we can calculate we have this aside number of magnesium and and plus oxidation number of nitrogen. This should be equal 0 and from this equation we know equal 3. Therefore, nitrogen minus 3 continue to be sumideright and they required to call the oxygen. We know casium broke 1, so the oxidation should be plus 1 and it is for this 1. The same is the neutral compounds we have plus 1 plus 2 times is equal 0 and shown this equation. We know s equals minus 1 concussy, you minus 1 hat again for c s. We calculate the compare for calci capoul. You have to try to writ aloud the name of the molecule to to have the formula okay, so they require for calcium and carbon, and here we see we know the group, calcium and group 2, then to do here as here, and we have the 2 o 2 is equals. 0 is equal, minus 1 point. So from this problem we know calcium in the in the comparith. The metal is get. The ocidation ocidain number is negative: minus 1. Here they got to the wag carbonatesame. But this i because the iron, multiple iron and the o c ratio of the multiple iron is equal. The total oscillation number of the lama in the multiple iron equal, the jack of the iron safe right here, so we can do that is x, plus 3 times 2 minus equal to minus. Here, like i the parenthesis here, so you can mangtynow, we solve the problem as a co plus 4 and carbon here plus 4. We see completive on carbon in the oxidation in the oxygen carbon plus 4 or carbon in the comportment and hydrogen carbon gas. Minus oxidation number now not a we have asia c, 2042 minus so see here so 2. We know already so at 2 s plus 4 times minus 2 equals 2 point so solve the problem. We know as a colt 3 carbon this case plus 3 and net 5 a minute now the same we colitooxygen and then we know our way we blood to so. From this we know blood 2 time for 2 times 2 minus equal minus here, so the i, but i see it's a reacher that asks should be equal minus 1. Now, as i calisto, we got 2 non 0, so oxygen in the sky 2 point and next we go to see the borohyrite g, sodium borenhyriokay and sodium 1 plus 1 hydrogen in the metal with metal in the comte hundreds and we'll get minus 1. So s will be here as 1 plus x, plus 4 power. Minus 1 equals 0 from the equation. We can get s equal, minus 3, plus 3, so bar on here 3 and the last 1, the twits time 4 time minus 2 equals 2 point so that here i go quiccosan here, a.

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