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alexandr402 [8]
3 years ago
14

Consider the following ionic compounds: CdCO3, Na2S, PbSO4, (NH4)3PO4, and Hg2Cl2. Which compounds will be soluble when added to

water?
a. Na2S and (NH4)3PO4
b. Hg2Cl2
c. Hg2Cl2, Na2S, and (NH4)3PO4
d. CdCO3, PbSO4, and Hg2Cl2
Chemistry
1 answer:
Grace [21]3 years ago
8 0

Answer:

a. Na₂S and (NH₄)₃PO₄ .

Explanation:

To know which salts are soluble we need to remember some <em>solubility rules</em>. And apply these as follows:

* All compounds of group 1 (Li⁺, Na⁺, K⁺, etc.) are soluble, Na₂S is soluble.

* All ammonium salts are soluble,

<h3>(NH₄)₃PO₄ is soluble.</h3>

* All Cl⁻, I⁻, Br⁻ are soluble except when are with Ag⁺, Hg₂²⁺, Pb²⁺, Hg₂Cl₂ is <em>insoluble</em>

* Carbonates are insoluble except those with group 1 or ammonium, CdCO₃ is <em>insoluble</em>

* All sulfates are soluble except those with Ca²⁺, Sr²⁺, Ba²⁺, Pb²⁺, Ag⁺. PbSO₄ is <em>insoluble. </em>

Right solution is:

<h3>a. Na₂S and (NH₄)₃PO₄ </h3>

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fiasKO [112]
<h3>Answer:</h3>

1.2 × 10⁻⁸ mol Pb

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
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<u>Stoichiometry</u>

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<u>Step 1: Define</u>

[Given] 7.2 × 10¹⁵ atoms Pb

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 7.2 \cdot 10^{15} \ atoms \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ atoms \ Pb})
  2. [DA] Multiply/Divide [Cancel out units]:                                                            \displaystyle 1.19562 \cdot 10^{-8} \ mol \ Pb

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb

6 0
2 years ago
Hello how to draw the electron dot structure of H2S and F2?​
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1. Attachment

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5 0
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