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Write the formulas for the following compounds:(a) rubidium nitrite, (b) potassium sulfide, (c) sodium hydrogen sulfide, (d) magnesium phosphate, (e) calcium hydrogen phosphate, (f) potassium dihydrogen phosphate, (g) iodine heptafluoride, (h) ammonium sulfate, (i) silver perchlorate, (j) boron trichloride.

(a) $\mathrm{RbNO} 2$(b) $\mathrm{K} 2 \mathrm{S}$ (c) NaHS(d) $\operatorname{Mg} 3(\mathrm{PO} 4) 2 (\mathrm{e}) \mathrm{CaHPO4}$(f) $\mathrm{KH} 2 \mathrm{PO} 4$(g) IF7(h) $(\mathrm{NH} 4) 2 \mathrm{SO} 4$(i) $\mathrm{AgClO} 4$(j) $\mathrm{BCl} 3$

Chemistry 101

Chapter 2

Atoms, Molecules, and Ions

Atoms, Molecules and Ions

Brown University

University of Toronto

Lectures

02:03

Write the formulas for the…

00:57

Write the chemical formula…

05:42

Write the formula for each…

03:08

The goal of the problem is to determine the chemical formulas of the following compounds. Let's first start with rubidium nitrate. Rubidium is R B plus and nitrate is n 02 muchness. The next step would be to cross the charges. What I mean by this is that the charge of the rubidium would end up being the sub script for the nitrite. And the charge of the nitrate, which is the minus, would be the sub script for the rubidium. So when we read it, we would rate it as R B N 02 Since both of them have a one charge plus and minus justice. Potassium sulphide. Potassium is K plus, while sulfide isn't s tu minus. So when you crossed the charges, the two would be the sub script for K. And as would have a sub script of one leading to K two s. Next we have sodium. Hydrogen sulfide sodium has a plus one charge. Hydrogen has a plus one charge as well, and sulphide has a to minus charge. Now, since there's three ions, we have to balance the charges. The N A and H have a post post charge and the s has a to minus charge. Therefore the charges will be balanced and you just need to write the ions as it's so we write any H s and that would be our eye on a compound. Next, we have magnesium phosphate. Magnesium has a plus to charge, and phosphate is P O for three minus. So when we crossed the charges, MG would get the three minus charge. Who would get MJ three? NPR four would get the plus to charge. So are full compound would be mg Threepio for two. Next we had calcium. Hydrogen phosphate calcium has a plus to charge. Hydrogen had the plus charge and philosophy has a three minus church similar to a previous problem. You just need to add up the charges and balanced. Um So since C A and H both have plus and plus, they add up to three plus, which would bounce out with the three minus. So you write the ions, as is C. A. H P 04 would be your ionic compound, but the next one potassium has a plus one charge, so K plus di hydrogen means we have to hydra agents, so we would rate it as an age to and phosphate is P o for three minus so similarly be after balance out the charges. So potassium has a positive charge and H two would be considered a two plus charge. Since there's two H molecules and each one is plus one and the P 04 has it three minus so one plus two would end up balancing out with the three minus and we can write as is K H two. P 04 would be the compound next for iodine helped a fluoride. We have to non metals, so the naming is slightly different. You have to pay attention to the prefixes. Since I don't doesn't have any prefixes, you start it with I. And in front of florid there's helped, which means seven. So we would have to put F seven for the sub script in the final company B I F seven. Thanks. We have ammonium self. It ammonium is an H four plus and sulphate is s 04 to minus. When you cross the charges, ammonium would get the tu minus. So NH four tu minus and the self it would get a plus. So S +04 and that would be your compound. Next we have silver poor claret. Silver is a G plus and per chlorate is C l +04 minus. When you put it together, you would get a G c l +04 and that would be your compound last. We have boron trick lord, which is also not Ionic, and you have to look at the prefix is you start with boron B and since it's try chloride, you have to have three chloride ions and you would end up with B L C l three.

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