QuestionJuly 10, 2025

Question 6 (Mandatory)(1 point) Which of the following two solutions will result in the formation of a precipitate when they are mixed? A) Cal_(2)(aq)+AgNO_(3)(aq) B) Sr(C_(2)H_(3)O_(2))_(2)(aq)+NaNO_(3)(aq) C) (NH_(4))_(2)CO_(3)(aq)+KCl(aq) D) BaBr_(2)(aq)+NaOH(aq)

Question 6 (Mandatory)(1 point) Which of the following two solutions will result in the formation of a precipitate when they are mixed? A) Cal_(2)(aq)+AgNO_(3)(aq) B) Sr(C_(2)H_(3)O_(2))_(2)(aq)+NaNO_(3)(aq) C) (NH_(4))_(2)CO_(3)(aq)+KCl(aq) D) BaBr_(2)(aq)+NaOH(aq)
Question 6 (Mandatory)(1 point)
Which of the following two solutions will result in the formation of a precipitate when they are
mixed?
A) Cal_(2)(aq)+AgNO_(3)(aq)
B) Sr(C_(2)H_(3)O_(2))_(2)(aq)+NaNO_(3)(aq)
C) (NH_(4))_(2)CO_(3)(aq)+KCl(aq)
D) BaBr_(2)(aq)+NaOH(aq)

Solution
4.7(264 votes)

Answer

None of the solutions will result in the formation of a precipitate when mixed. Explanation 1. Identify possible precipitates Use solubility rules to determine if any combination forms an insoluble compound. 2. Analyze each option A) Ca^{2+} and Ag^+ do not form insoluble compounds with NO_3^- or I^-. No precipitate. B) Sr^{2+} and Na^+ do not form insoluble compounds with C_2H_3O_2^- or NO_3^-. No precipitate. C) NH_4^+ and K^+ do not form insoluble compounds with CO_3^{2-} or Cl^-. No precipitate. D) Ba^{2+} can form Ba(OH)_2, which is slightly soluble, but BaBr_2 and NaOH do not form a precipitate. No precipitate.

Explanation

1. Identify possible precipitates<br /> Use solubility rules to determine if any combination forms an insoluble compound.<br />2. Analyze each option<br /> A) $Ca^{2+}$ and $Ag^+$ do not form insoluble compounds with $NO_3^-$ or $I^-$. No precipitate.<br /> B) $Sr^{2+}$ and $Na^+$ do not form insoluble compounds with $C_2H_3O_2^-$ or $NO_3^-$. No precipitate.<br /> C) $NH_4^+$ and $K^+$ do not form insoluble compounds with $CO_3^{2-}$ or $Cl^-$. No precipitate.<br /> D) $Ba^{2+}$ can form $Ba(OH)_2$, which is slightly soluble, but $BaBr_2$ and $NaOH$ do not form a precipitate. No precipitate.
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