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Vikentia [17]
3 years ago
6

Complete and balance the molecular equation for the reaction of aqueous ammonium bromide, NH4Br , and aqueous lead(II) acetate,

Pb(C2H3O2)2 . Include physical states. molecular equation: NH_{4}Br(aq) + Pb(C_{2}H_{3}O_{2})_{2}(aq) -> NH4Br(aq)+Pb(C2H3O2)2(aq)⟶
Chemistry
2 answers:
sergeinik [125]3 years ago
8 0

Answer:alll u. Gotta do is ask the teacher for the answer hope this help

Explanation:

Bacon chacon

frutty [35]3 years ago
3 0

Answer : The complete balanced molecular equation will be,

2NH_4Br(aq)+Pb(C_2H_3O_2)_2(aq)\rightarrow 2NH_4C_2H_3O_2(aq)+PbBr_2(s)

Explanation :

Balanced chemical equation : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The complete balanced molecular equation will be,

2NH_4Br(aq)+Pb(C_2H_3O_2)_2(aq)\rightarrow 2NH_4C_2H_3O_2(aq)+PbBr_2(s)

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8 0
3 years ago
The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
maw [93]
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

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I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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