<h3>
Answer:</h3>
1.69 g Mg₃N₂
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] Mg + N₂ → Mg₃N₂
[RxN - Balanced] 3Mg + N₂ → Mg₃N₂
[Given] 1.22 g Mg
[Solve] grams Mg₃N₂
<u>Step 2: Identify Conversions</u>
[RxN] 3 mol Mg → Mg₃N₂
[PT] Molar Mass of Mg - 24.31 g/mol
[PT] Molar Mass of N - 14.01 g/mol
Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol
<u>Step 3: Stoich</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂
Octet means presence of a total of 8 electrons in its valence shell while in case of duplet only 2 electrons are present in valence shell.
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Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>