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Chapter 5: Problem 92

Write the formula of each of the following ionic substances. a. sodium dihydrogen phosphate b. lithium perchlorate c. copper(II) hydrogen carbonate d. potassium acetate e. barium peroxide f. cesium sulfite

Short Answer

Expert verified
a. \(NaH_2PO_4\) b. \(LiClO_4\) c. \(Cu(HCO_3)_2\) d. \(KCH_3COO\) e. \(BaO_2\) f. \(Cs_2SO_3\)

Step by step solution

01

a. sodium dihydrogen phosphate

Since sodium (Na) has a +1 charge and dihydrogen phosphate (H2PO4) has a -1 charge, one sodium ion can effectively cancel the charge of one dihydrogen phosphate ion. Therefore, the formula for sodium dihydrogen phosphate is simply NaH2PO4.
02

b. lithium perchlorate

Similar to the previous case, lithium (Li) has a +1 charge and perchlorate (ClO4) has a -1 charge. So, the formula for lithium perchlorate is LiClO4.
03

c. copper(II) hydrogen carbonate

Copper(II) (Cu) has a +2 charge, and hydrogen carbonate (HCO3) has a -1 charge. To balance these charges, we need two hydrogen carbonate ions for each copper ion. Thus, the formula for copper(II) hydrogen carbonate is Cu(HCO3)2.
04

d. potassium acetate

Both potassium (K) and acetate (CH3COO) ions have charges of +1 and -1, respectively. So, one of each ion will cancel out each other's charges, giving us the formula KCH3COO for potassium acetate.
05

e. barium peroxide

Barium (Ba) has a +2 charge, and peroxide (O2) has a -2 charge. These charges directly cancel each other out, so the formula for barium peroxide is BaO2.
06

f. cesium sulfite

Cesium (Cs) has a +1 charge, while sulfite (SO3) has a -2 charge. To balance their charges, we need two cesium ions for each sulfite ion, resulting in the formula Cs2SO3 for cesium sulfite.

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Most popular questions from this chapter

Under what physical state at room temperature do each of the halogen elements exist?

Name each of the following compounds, which contain polyatomic ions. a. \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) b. \(\mathrm{Ba}(\mathrm{OH})_{2}\) c. \(\mathrm{K}_{2} \mathrm{O}_{2}\) d. \(\mathrm{Al}\left(\mathrm{HSO}_{4}\right)_{3}\) e. \(AgCN\) f. \(\mathrm{CaHPO}_{4}\)

For each of the negative ions listed in column \(1,\) use the periodic table to find in column 2 the total number of electrons the ion contains. A given answer may be used more than once. Column 1 \(\left[\begin{array}{l}1\end{array}\right] \mathrm{Se}^{2-}\) \(\left[\begin{array}{l}2\end{array}\right] S^{2-}\) \(\left[\begin{array}{l}3\end{array}\right] \mathrm{P}^{3-}\) [4] \(\mathrm{O}^{2-}\) \(\left[\begin{array}{lll}5 & \mathrm{N}^{3-}\end{array}\right.\) \(\left[\begin{array}{lll}6\end{array}\right] \mathrm{I}^{-}\) \(\left[\begin{array}{l}7\end{array}\right] \mathbf{F}^{-}\) [8] \(\mathrm{Cl}^{-}\) \(\left[\begin{array}{l}9\end{array}\right] \mathbf{B} \mathbf{r}^{-}\) [10] \(\mathrm{At}^{-}\) Column 2 [a] 18 \([b] 35\) \([\mathrm{cl}] \mathrm{s} 2\) \(\left[\begin{array}{ll}1 & 34\end{array}\right.\) [e] 36 \([\mathrm{f}] 54\) \([g] 10\) \([\mathrm{h}] 9\) \([\mathrm{i}] 53\) \([j] 86\)

What is an oxyanion? How is a series of oxyanions named? (Give an example.)

Write the formula for each of the following compounds that contain polyatomic ions. Be sure to enclose the polyatomic ion in parentheses if more than one such ion is needed. a. barium sulfite b. calcium dihydrogen phosphate c. ammonium perchlorate d. sodium permanganate e. iron(III) sulfate f. cobalt(II) carbonate g. nickelous hydroxide h. zinc chromate
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