What is the percentage of oxygen in Li(NO2)3
1 answer:
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<h3>Answer:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right<u> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Moles
- Compounds
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<u>Step 1: Define</u>
[RxN - Balanced] 2C + O₂ → 2CO₂
[Given] 0.25 moles O₂
[Solve] moles CO₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol O₂ → 2 mol CO₂
<u>Step 3: Stoichiometry</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
Lewis Structure is drawn in following steps,
1) Calculate Number of Valence Electrons:
# of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
---------
Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.
3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
----------
12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,
1) Calculate Number of Valence Electrons:
# of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
---------
Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.
3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
----------
12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,