. how many lone pairs of electrons are present in the lewis structure of calcium sulfide?
1 answer:
Answer:
<u>Four</u>
Explanation:
Calcium Sulfide is an ionic compound made up of Ca²⁺ and S²⁻.
Ca²⁺ is formed as,
Ca → Ca²⁺ + 2 e⁻
These two electrons are accepte by Sulfur as,
S + 2 e⁻ → S²⁻
So, before accepting 2 electrons S was having six valence electrons, after accepting two electrons from Ca it has 8 electrons which are present in four pairs as shown below,
<u>Four</u>
Explanation:
Calcium Sulfide is an ionic compound made up of Ca²⁺ and S²⁻.
Ca²⁺ is formed as,
Ca → Ca²⁺ + 2 e⁻
These two electrons are accepte by Sulfur as,
S + 2 e⁻ → S²⁻
So, before accepting 2 electrons S was having six valence electrons, after accepting two electrons from Ca it has 8 electrons which are present in four pairs as shown below,
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Answer:
Explanation:
First, find the mass of empirical formula, CH. 12.01 g/mol is for carbon, and 1.008 g/mol is for hydrogen. 12.01+1.008=13.018 G/mol CH. Divide 78.110 G/mol by 13.018 g/mol. You get approximately 6. Multiply that by the subscript of each element. 6(CH)=
<h3>Answer:</h3>
9.6724 g MgO
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Step 1: Define</u>
[RxN - Balanced] 2Mg + O₂ → 2MgO
[Given] 5.8332 g Mg
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Mg = 2 mol MgO
Molar Mass of Mg - 24.31 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of MgO - 24.31 + 16.00 = 40.31 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 5 sig figs.</em>
9.67241 g MgO ≈ 9.6724 g MgO