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Chapter 4: Problem 46

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\operatorname{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Mg} \mathrm{I}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\operatorname{AlBr}_{3}(a q)\)

Short Answer

Expert verified
a. A precipitate of Cu(Hg2)(SO4) will form b. No precipitate will form c. A precipitate of MgCO3 will form d. No precipitate will form

Step by step solution

01

Identify ions present

In the first solution, we have Hg2 + ions and NO3 - ions. In the second solution, we have Cu2 + ions and SO4 2- ions.
02

Formulate possible double replacement reactions

Possible combinations are Hg2 + CuSO4 -> Cu(Hg2)(SO4), and NO3 - + CuSO4 -> Cu(NO3)2.
03

Use solubility rules to determine if precipitates will form

Using the solubility rules, we find that Cu(NO3)2 is soluble in water and no precipitate will form. However, Cu(Hg2)(SO4) is insoluble in water, meaning that a precipitate will form. b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\)
04

Identify ions present

In the first solution, we have Ni2+ ions and NO3- ions. In the second solution, we have Ca2+ ions and Cl- ions.
05

Formulate possible double replacement reactions

Possible combinations are Ni(NO3)2 + CaCl2 -> Ca(NO3)2 + NiCl2.
06

Use solubility rules to determine if precipitates will form

Using the solubility rules, we find that both Ca(NO3)2 and NiCl2 are soluble in water, meaning no precipitate will form. c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Mg} \mathrm{I}_{2}(a q)\)
07

Identify ions present

In the first solution, we have K+ ions and CO3 2- ions. In the second solution, we have Mg2+ ions and I- ions.
08

Formulate possible double replacement reactions

Possible combinations are K2CO3 + MgI2 -> MgCO3 + 2KI.
09

Use solubility rules to determine if precipitates will form

Using the solubility rules, we find that MgCO3 is insoluble in water, meaning a precipitate will form. KI is soluble in water, so no precipitate will form for that compound. d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\operatorname{AlBr}_{3}(a q)\)
10

Identify ions present

In the first solution, we have Na+ ions and CrO4 2- ions. In the second solution, we have Al3+ ions and Br- ions.
11

Formulate possible double replacement reactions

Possible combinations are Na2CrO4 + AlBr3 -> Al(CrO4)3 + 3NaBr.
12

Use solubility rules to determine if precipitates will form

Using the solubility rules, we find that NaBr and Al(CrO4)3 are soluble in water, and no precipitate will form. #Summary# a. A precipitate of Cu(Hg2)(SO4) will form b. No precipitate will form c. A precipitate of MgCO3 will form d. No precipitate will form

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Most popular questions from this chapter

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. \(0.100\) mole of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(100.0 \mathrm{~mL}\) of solution b. \(2.5\) moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in \(1.25 \mathrm{~L}\) of solution c. \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(500.0 \mathrm{~mL}\) of solution d. \(1.00 \mathrm{~g} \mathrm{~K}_{2} \mathrm{PO}_{4}\) in \(250.0 \mathrm{~mL}\) of solution

Douglasite is a mineral with the formula \(2 \mathrm{KCl} \cdot \mathrm{FeCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\). Calculate the mass percent of douglasite in a \(455.0-\mathrm{mg}\) sample if it took \(37.20 \mathrm{~mL}\) of a \(0.1000-M \mathrm{AgNO}_{3}\) solution to precipitate all the \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\). Assume the douglasite is the only source of chloride ion.

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. \(0.0200\) mole of sodium phosphate in \(10.0 \mathrm{~mL}\) of solution b. \(0.300\) mole of barium nitrate in \(600.0 \mathrm{~mL}\) of solution c. \(1.00 \mathrm{~g}\) of potassium chloride in \(0.500 \mathrm{~L}\) of solution d. \(132 \mathrm{~g}\) of ammonium sulfate in \(1.50 \mathrm{~L}\) of solution

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{FeSO}_{4}(a q)+\mathrm{KCl}(a q)\) b. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) c. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) d. \(\mathrm{K}_{2} \mathrm{~S}(a q)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

Chlorisondamine chloride \(\left(\mathrm{C}_{4} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{~N}_{2}\right)\) is a drug used in the treatment of hypertension. A \(1.28-\mathrm{g}\) sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, \(0.104 \mathrm{~g}\) silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.
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