Question

question 6
1 pts
when aqueous solutions of ______ are mixed, a precipitate forms.
nal and kbr
nibr₂ and agno₃
koh and ba(no₃)₂
k₂so₄ and crcl₃
li₂co₃ and csl

question 7
1 pts
the net ionic equation for the reaction between aqueous sulfuric acid and aqueous sodium hydroxide is ______.
h⁺(aq) + oh⁻(aq) → h₂o(l)
so₄²⁻(aq) + 2na⁺(aq) → 2na⁺(aq) + so₄²⁻(aq)
h⁺(aq) + hso₄⁻(aq) + 2na⁺(aq) + 2oh⁻(aq) → 2h₂o(l) + 2na⁺(aq) + so₄²⁻(aq)
2h⁺(aq) + so₄²⁻(aq) + 2na⁺(aq) + 2oh⁻(aq) → 2h₂o(l) + 2na⁺(aq) + so₄²⁻(aq)

Explanation:

Question 6

Step1: Analyze \(NaI\) and \(KBr\)

When \(NaI\) and \(KBr\) are mixed, the possible products are \(NaBr\) and \(KI\). According to solubility rules, all sodium (\(Na^{+}\)) and potassium (\(K^{+}\)) salts are soluble. So, no precipitate forms.

Step2: Analyze \(NiBr_{2}\) and \(AgNO_{3}\)

The reaction is \(NiBr_{2}(aq)+2AgNO_{3}(aq)=2AgBr(s)+Ni(NO_{3})_{2}(aq)\). Silver bromide (\(AgBr\)) is insoluble (as per solubility rules: most bromides are soluble except those of \(Ag^{+}\), \(Pb^{2+}\), \(Hg_{2}^{2 +}\)). So, a precipitate (\(AgBr\)) forms.

Step3: Analyze \(KOH\) and \(Ba(NO_{3})_{2}\)

The possible products are \(KNO_{3}\) and \(Ba(OH)_{2}\). Potassium nitrate (\(KNO_{3}\)) is soluble (all nitrates are soluble). Barium hydroxide (\(Ba(OH)_{2}\)) has some solubility (\(1.67\ g/100\ mL\) at \(20^{\circ}C\)), and no precipitate forms under normal mixing conditions (not a highly insoluble compound like \(AgBr\)).

Step4: Analyze \(K_{2}SO_{4}\) and \(CrCl_{3}\)

The possible products are \(KCl\) and \(Cr_{2}(SO_{4})_{3}\). Potassium chloride (\(KCl\)) is soluble (all potassium salts are soluble). Chromium(III) sulfate (\(Cr_{2}(SO_{4})_{3}\)) is also soluble (most sulfates are soluble except \(BaSO_{4}\), \(PbSO_{4}\), \(CaSO_{4}\) (slightly soluble)). So, no precipitate forms.

Step5: Analyze \(Li_{2}CO_{3}\) and \(CsI\)

The possible products are \(LiI\) and \(Cs_{2}CO_{3}\). Lithium iodide (\(LiI\)) is soluble (most iodides are soluble). Cesium carbonate (\(Cs_{2}CO_{3}\)) is also soluble (all alkali - metal carbonates except \(Li_{2}CO_{3}\) have good solubility). So, no precipitate forms.

Step1: Write the molecular equation

The reaction between sulfuric acid (\(H_{2}SO_{4}\)) and sodium hydroxide (\(NaOH\)) is \(H_{2}SO_{4}(aq)+2NaOH(aq)=Na_{2}SO_{4}(aq)+2H_{2}O(l)\).

Step2: Write the total ionic equation

\(2H^{+}(aq)+SO_{4}^{2 -}(aq)+2Na^{+}(aq)+2OH^{-}(aq)=2Na^{+}(aq)+SO_{4}^{2 -}(aq)+2H_{2}O(l)\)

Step3: Cancel the spectator ions

The spectator ions are \(Na^{+}\) and \(SO_{4}^{2 -}\). After canceling them, we get \(H^{+}(aq)+OH^{-}(aq)=H_{2}O(l)\)

Answer:

\(NiBr_{2}\) and \(AgNO_{3}\)

Question 7