for each of the following formula, name it appropriately. (spelling does matter)
1) hclo₃ (aq)
2) p₂o₅
3) feso₄
4) sio₂
5) cobr₂
6) bbr₃
7) caso₄
8) hbr (aq)
9) cr₂(co₃)₃
10) ag₃p
11) pbs
12) ch₄
13) nabr
14) h₂s (aq)
15) al₂s₃
16) hc₂h₃o₂

Question

Accepted Answer

### 1) $\boldsymbol{\ce{HClO_3(aq)}}$
# Brief Explanations:
$\ce{HClO_3}$ is an oxyacid. The anion is $\ce{ClO_3^-}$, named chlorate. So the acid is chloric acid (since “ate” anions form “ic” acids).
# Answer:
Chloric acid

### 2) $\boldsymbol{\ce{P_2O_5}}$
# Brief Explanations:
This is a covalent compound (non - metal + non - metal). The prefix for 2 is “di - ” for P (phosphorus) and for 5 is “penta - ” for O (oxygen, named as oxide). So it's diphosphorus pentoxide.
# Answer:
Diphosphorus pentoxide

### 3) $\boldsymbol{\ce{FeSO_4}}$
# Brief Explanations:
$\ce{FeSO_4}$ has $\ce{Fe^{2+}}$ (since $\ce{SO_4^{2 - }}$ has a 2 - charge, and to balance, Fe is +2) and $\ce{SO_4^{2 - }}$ (sulfate ion). So it's iron(II) sulfate (or ferrous sulfate).
# Answer:
Iron(II) sulfate (or Ferrous sulfate)

### 4) $\boldsymbol{\ce{SiO_2}}$
# Brief Explanations:
Covalent compound. Si is silicon, O is oxide. Prefix for 1 (Si) is “mono - ” (but not used for the first element), prefix for 2 (O) is “di - ”. So silicon dioxide.
# Answer:
Silicon dioxide

### 5) $\boldsymbol{\ce{CoBr_2}}$
# Brief Explanations:
$\ce{CoBr_2}$ has $\ce{Co^{2+}}$ (since $\ce{Br^-}$ has a 1 - charge, two $\ce{Br^-}$ give a 2 - charge, so Co is +2) and $\ce{Br^-}$ (bromide ion). So cobalt(II) bromide.
# Answer:
Cobalt(II) bromide

### 6) $\boldsymbol{\ce{BBr_3}}$
# Brief Explanations:
Covalent compound. B is boron, Br is bromide. Prefix for 3 (Br) is “tri - ”. So boron tribromide.
# Answer:
Boron tribromide

### 7) $\boldsymbol{\ce{CaSO_4}}$
# Brief Explanations:
$\ce{CaSO_4}$ has $\ce{Ca^{2+}}$ (calcium ion) and $\ce{SO_4^{2 - }}$ (sulfate ion). It's an ionic compound, so calcium sulfate.
# Answer:
Calcium sulfate

### 8) $\boldsymbol{\ce{HBr(aq)}}$
# Brief Explanations:
$\ce{HBr}$ in aqueous solution is a binary acid (H + non - metal). The non - metal is Br (bromine), so the acid is hydrobromic acid.
# Answer:
Hydrobromic acid

### 9) $\boldsymbol{\ce{Cr_2(CO_3)_3}}$
# Brief Explanations:
$\ce{CO_3^{2 - }}$ is carbonate (2 - charge). Let the charge of Cr be $x$. Then $2x+3	imes(- 2)=0$, so $2x = 6$, $x = + 3$. So $\ce{Cr^{3+}}$ (chromium(III)) and $\ce{CO_3^{2 - }}$ (carbonate). So chromium(III) carbonate.
# Answer:
Chromium(III) carbonate

### 10) $\boldsymbol{\ce{Ag_3P}}$
# Brief Explanations:
$\ce{Ag_3P}$ has $\ce{Ag^+}$ (silver ion, charge +1) and $\ce{P^{3 - }}$ (phosphide ion). Three $\ce{Ag^+}$ balance one $\ce{P^{3 - }}$. So silver phosphide.
# Answer:
Silver phosphide

### 11) $\boldsymbol{\ce{PbS}}$
# Brief Explanations:
$\ce{PbS}$ has $\ce{Pb^{2+}}$ (since $\ce{S^{2 - }}$ has a 2 - charge) and $\ce{S^{2 - }}$ (sulfide ion). So lead(II) sulfide (or plumbous sulfide).
# Answer:
Lead(II) sulfide (or Plumbous sulfide)

### 12) $\boldsymbol{\ce{CH_4}}$
# Brief Explanations:
$\ce{CH_4}$ is a covalent compound, and it's the simplest hydrocarbon. Its common name is methane.
# Answer:
Methane

### 13) $\boldsymbol{\ce{NaBr}}$
# Brief Explanations:
$\ce{NaBr}$ has $\ce{Na^+}$ (sodium ion) and $\ce{Br^-}$ (bromide ion). Ionic compound, so sodium bromide.
# Answer:
Sodium bromide

### 14) $\boldsymbol{\ce{H_2S(aq)}}$
# Brief Explanations:
$\ce{H_2S}$ in aqueous solution is a binary acid. The non - metal is S (sulfur), so the acid is hydrosulfuric acid.
# Answer:
Hydrosulfuric acid

### 15) $\boldsymbol{\ce{Al_2S_3}}$
# Brief Explanations:
$\ce{Al_2S_3}$ has $\ce{Al^{3+}}$ (aluminum ion, charge +3) and $\ce{S^{2 - }}$ (sulfide ion). Ionic compound, so aluminum sulfide.
# Answer:
Aluminum sulfide

### 16) $\boldsymbol{\ce{HC_2H_3O_2}}$
# Brief Explanations:
$\ce{HC_2H_3O_2}$ is acetic acid (its common name) or ethanoic acid (IUPAC name). The formula is also written as $\ce{CH_3COOH}$, and it's a weak acid.
# Answer:
Acetic acid (or Ethanoic acid)

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QUESTION IMAGE

Question

Explanation:

1) $\boldsymbol{\ce{HClO_3(aq)}}$

Answer:

2) $\boldsymbol{\ce{P_2O_5}}$