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<h3>Answer:</h3>
495 g K₃N
<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> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
3.77 mol K₃N
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
495.039 g K₃N ≈ 495 g K₃N
Answer:
Explanation:
Let's rewrite the given word equation in its chemical balanced equation representation:
1. Lead(II) nitride is represented by lead, Pb, in an oxidation state of 2+, while nitride is a typical nitrogen anion with a state 3-. As a result, the lowest common multiple between 2 and 3 is 6, meaning 2 lead cations are needed to balance 3 nitrogen anions: .
2. Ammonium sulfate consists of an ammonium cation with a 1+ charge and sulfate anion with a 2- charge, two ammonium cations needed: .
3. Lead(II) sulfate would have one lead cation and one sulfate anion, as they have the same magnitude of charges with opposite signs: .
4. Ammonium nitride would require three amonium cations to balance the nitride anion: .
Let's write the balanced equation: