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jonny [76]
3 years ago
5

Write balanced equations and solubility product expressions for the following compounds

Chemistry
1 answer:
lyudmila [28]3 years ago
8 0

<u>Explanation:</u>

Solubility product is defined as the product of the concentrations or solubilities of the ions each raised to the power their stoichiometric coefficients present in the solution. It is expressed as K_{sp}.

(a): CuBr

The chemical equation for the dissociation of CuBr into its ions follows:

CuBr\rightleftharpoons Cu^++Br^-

The expression of K_{sp} for CuBr follows:

K_{sp}=[Cu^+][Br^-]

(b): ZnC_2O_4

The chemical equation for the dissociation of ZnC_2O_4 into its ions follows:

ZnC_2O_4\rightleftharpoons Zn^{2+}+C_2O_4^{2-}

The expression of K_{sp} for ZnC_2O_4 follows:

K_{sp}=[Zn^{2+}][C_2O_4^{2-}]

(c): Ag_2CrO_4

The chemical equation for the dissociation of Ag_2CrO_4 into its ions follows:

Ag_2CrO_4\rightleftharpoons 2Ag^{+}+CrO_4^{2-}

The expression of K_{sp} for Ag_2CrO_4 follows:

K_{sp}=[Ag^{+}]^2[CrO_4^{2-}]

(d): ZnC_2O_4

The chemical equation for the dissociation of Hg_2Cl_2 into its ions follows:

Hg_2Cl_2\rightleftharpoons 2Hg^{+}+2Cl^{-}

The expression of K_{sp} for Hg_2Cl_2 follows:

K_{sp}=[Hg^{+}]^2[Cl^{-}]^2

(e): AlCl_3

The chemical equation for the dissociation of AlCl_3 into its ions follows:

AlCl_3\rightleftharpoons Al^{3+}+3Cl^{-}

The expression of K_{sp} for AlCl_3 follows:

K_{sp}=[Al^{3+}][Cl^{-}]^3

(f): ZnC_2O_4

The chemical equation for the dissociation of Mn_3(PO_4)_2 into its ions follows:

Mn_3(PO_4)_2\rightleftharpoons 3Mn^{2+}+2PO_4^{3-}

The expression of K_{sp} for Mn_3(PO_4)_2 follows:

K_{sp}=[Mn^{2+}]^3[PO_4^{3-}]^2

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The only true statement from the given options is that "an orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy." Inner orbitals which are also known to contain core electrons feels the bulk of the nuclear pull on them compared to the outermost orbitals containing the valence electrons.

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Answer:

5.702 mol K₂SO₄

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Compounds
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 993.6 g K₂SO₄

[Solve] moles K₂SO₄

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of K: 39.10 g/mol

[PT] Molar Mass of S: 32.07 g/mol

[PT] Molar mass of O: 16.00 g/mol

Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 993.6 \ g \ K_2SO_4(\frac{1 \ mol \ K_2SO_4}{174.27 \ g \ K_2SO_4})
  2. [DA] Divide [Cancel out units]:                                                                         \displaystyle 5.7015 \ mol \ K_2SO_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

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