Chemistry, 12.12.2019 04:31 damiangomzzz903

Naming compounds test review practice name the following ionic compounds: nh4cl fe(no3)3 tibr3 cu3p snse2 gaas pb(so4)2 be(hco3)2 mn2(so3)3 al(cn)3

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Answer from: jthollis1348

Answer :

The name of NH_4Cl is ammonium chloride.

The name of Fe(NO_3)_3 is iron(III) nitrate.

The name of TiBr_3 is titanium(III) bromide.

The name of Cu_3P is copper(I) phosphide.

The name of SnSe_2 is tin(IV) selenide.

The name of GaAs is gallium arsenide.

The name of Pb(SO_4)_2 is lead(IV) sulfate.

The name of Be(HCO_3)_2 is beryllium hydrogen carbonate.

The name of Mn_2(SO_3)_3 is manganese(III) sulfite.

The name of Al(CN)_3 is aluminium cyanide.

Explanation :

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

Ionic compound are usually formed when a metal reacts with a non-metal.

The nomenclature of ionic compounds is given by:

1. Positive is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

3. In case of transition metals, the oxidation state are written in roman numerals in bracket in-front of positive ions.

(a) NH_4Cl

It is an ionic compound because it is formed by the combination of one polyatomic cation (ammonium ion) and anion (chloride ion). The charge on ammonium ion is (+1) and the the charge on chloride ion is (-1).

Thus, the name of NH_4Cl is ammonium chloride.

(b) Fe(NO_3)_3

It is an ionic compound because it is formed by the combination of one polyatomic anion (nitrate ion) and cation (iron ion). The charge on iron ion is (+3) and the the charge on nitrate ion is (-1).

Thus, the name of Fe(NO_3)_3 is iron(III) nitrate.

It is an ionic compound because it is formed by the combination of titanium cation and bromide anion. The charge on titanium is (+3) and the the charge on bromide ion is (-1).

Thus, the name of TiBr_3 is titanium(III) bromide.

(d) Cu_3P

It is an ionic compound because it is formed by the combination of copper cation and phosphorous anion. The charge on copper is (+1) and the the charge on phosphorous ion is (-3).

Thus, the name of Cu_3P is copper(I) phosphide.

(e) SnSe_2

It is an ionic compound because it is formed by the combination of tin cation and selenium anion. The charge on tin is (+4) and the the charge on selenium ion is (-2).

Thus, the name of SnSe_2 is tin(IV) selenide.

(f) GaAs

It is an ionic compound because it is formed by the combination of gallium cation and arsenic anion. The charge on gallium is (+3) and the the charge on arsenic ion is (-3).

Thus, the name of GaAs is gallium arsenide.

(g) Pb(SO_4)_2

It is an ionic compound because it is formed by the combination of one polyatomic anion (sulfate ion) and cation (lead ion). The charge on lead ion is (+4) and the the charge on sulfate ion is (-2).

Thus, the name of Pb(SO_4)_2 is lead(IV) sulfate.

(h) Be(HCO_3)_2

It is an ionic compound because it is formed by the combination of one polyatomic anion (hydrogen carbonate ion) and cation (beryllium ion). The charge on beryllium ion is (+2) and the the charge on hydrogen carbonate ion is (-1).

Thus, the name of Be(HCO_3)_2 is beryllium hydrogen carbonate.

(i) Mn_2(SO_3)_3

It is an ionic compound because it is formed by the combination of one polyatomic anion (sulfite ion) and cation (manganese ion). The charge on manganese ion is (+3) and the the charge on sulfite ion is (-2).

Thus, the name of Mn_2(SO_3)_3 is manganese(III) sulfite.

(j) Al(CN)_3

It is an ionic compound because it is formed by the combination of one polyatomic anion (cyanide ion) and cation (aluminium ion). The charge on aluminium ion is (+3) and the the charge on cyanide ion is (-1).

Thus, the name of Al(CN)_3 is aluminium cyanide.

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Answer from: Quest

In any atom of any element, you will find protons and neutrons in the nucleus of any atom

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Naming compounds test review practice name the following ionic compounds: nh4cl fe(no3)3 tibr3 c...

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