3.32 Balance the following equations and then write the net ionic equation. (a)(NH 4 ) 2 CO 3 (aq)+ Cu(NO 3 ) 2 (aq)( CuCO 3 (s)+ NH 4 NO 3 (aq) (b) Pb(OH) 2 (s)+ HCl(aq) ( PbCl 2 (s) + H 2 O (l) (c) BaCO 3 (s) + HCl(aq)( BaCl 2 (aq) + H 2 O (l) + CO 2 (g) (d) CH 3 CO 2 H (aq)+ Ni(OH) 2 (s)( Ni(CH 3 CO 2 ) 2 (aq) + H 2 O(l)
3.32 Balance the following equations and then write the net ionic equation. (a)(NH 4 ) 2 CO 3 (aq)+ Cu(NO 3 ) 2 (aq)( CuCO 3 (s)+ NH 4 NO 3 (aq) (b) Pb(OH) 2 (s)+ HCl(aq) ( PbCl 2 (s) + H 2 O (l) (c) BaCO 3 (s) + HCl(aq)( BaCl 2 (aq) + H 2 O (l) + CO 2 (g) (d) CH 3 CO 2 H (aq)+ Ni(OH) 2 (s)( Ni(CH 3 CO 2 ) 2 (aq) + H 2 O(l)
To determine the balanced chemical equation and net ionic equation.
Explanation of Solution
Given Information:
The unbalanced chemical equation is (NH4)2CO3 + Cu(NO3)2→CuCO3 + NH4NO3
The Balanced chemical equation is-
(NH4)2CO3 + Cu(NO3)2 = CuCO3 + 2NH4NO3
In the Balanced chemical equation N, H, C, O, & Cu atoms are same on both the side. Copper ion is the spectator ions here so the net ionic equation will be as follows −
2NH4++2NO3**#x2212;→2NH4NO3
(b)
Expert Solution
Interpretation Introduction
To Determine:
To determine the balanced chemical equation and net ionic equation.
Explanation of Solution
Given Information:
The unbalanced chemical equation is Pb(OH)2 + HCl → PbCl2 +H2O
The Balanced chemical equation is-
Pb(OH)2 + 2HCl = PbCl2 +2H2O
In the Balanced chemical equation Pb, H, Cl, & O atoms are same on both the side. There is no spectator ions here so the net ionic equation will be as follows −
2OH-+2H++Pb2++2Cl**#x2212;→PbCl2+2H2O
(c)
Expert Solution
Interpretation Introduction
To Determine:
To determine the balanced chemical equation and net ionic equation.
Explanation of Solution
Given Information:
The unbalanced chemical equation is BaCO3+ HCl→ BaCl2 + H2O + CO2
The Balanced chemical equation is-
BaCO3+ 2HCl= BaCl2 + H2O + CO2
In the Balanced chemical equation Ba, H, C, O, & Cl atoms are same on both the side. There are no spectator ions here so the net ionic equation will be as follows −
Ba2++CO32**#x2212;+2H++2Cl−→BaCl2+H2O+CO2
(d)
Expert Solution
Interpretation Introduction
To Determine:
To determine the balanced chemical equation and net ionic equation.
Explanation of Solution
Given Information:
The unbalanced chemical equation is CH3CO2H+ Ni(OH)2→Ni(CH3CO2)2+ H2O
The Balanced chemical equation is-
2CH3CO2H+ Ni(OH)2=Ni(CH3CO2)2+ 2H2O
In the Balanced chemical equation H, C, O, & Ni atoms remain the same on both sides. Copper ion is the spectator ion here so the net ionic equation will be as follows −
2CH3COO**#x2212;+2H++Ni2++2OH−→Ni(CH3CO2)2+ 2H2O
Conclusion
The balanced chemical equations are as follows-
(NH4)2CO3 + Cu(NO3)2 = CuCO3 + 2NH4NO3
Net ionic equation 2NH4++2NO3−→2NH4NO3
Pb(OH)2 + 2HCl = PbCl2 +2H2O
Net Ionic equation 2OH-+2H++Pb2++2Cl−→PbCl2+2H2O
BaCO3+ 2HCl= BaCl2 + H2O + CO2
Net Ionic equation Ba2++CO32−+2H++2Cl−→BaCl2+H2O+CO2
2CH3CO2H+ Ni(OH)2=Ni(CH3CO2)2+ 2H2O
Net Ionic equation 2CH3COO−+2H++Ni2++2OH−→Ni(CH3CO2)2+ 2H2O
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In addition to the separation techniques used in this lab (magnetism, evaporation, and filtering), there are other commonly used separation techniques. Some of these techniques are:Distillation – this process is used to separate components that have significantly different boiling points. The solution is heated and the lower boiling point substance is vaporized first. The vapor can be collected and condensed and the component recovered as a pure liquid. If the temperature of the mixture is then raised, the next higher boiling component will come off and be collected. Eventually only non-volatile components will be left in the original solution.Centrifugation – a centrifuge will separate mixtures based on their mass. The mixture is placed in a centrifuge tube which is then spun at a high speed. Heavier components will settle at the bottom of the tube while lighter components will be at the top. This is the technique used to separate red blood cells from blood plasma.Sieving – this is…
Briefly describe a eutectic system.
13.53 Draw all stereoisomers formed when each compound is treated with HBr in the presence of peroxides.
a.
b.
C.
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell