Answer:
Explanation:
<u>1) Equilibrium equation (given):</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
<u>2) Write the concentration changes when some concentration, A, of CH₂Cl₂ (g) sample is introduced into an evacuated (empty) vessel:</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
A - x x x
<u>3) Replace x with the known (found) equilibrium concentraion of CCl₄ (g) of 0.348 M</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
A - 0.3485 0.348 0.348
<u>4) Write the equilibrium constant equation, replace the known values and solve for the unknown (A):</u>
- Kc = [ CH₄ (g) ] [ CCl₄ (g) ] / [ CH₂Cl₂ (g) ]²
- A² = 56.0 / 0.348² = 462.
The smallest functional and structural unit of an organism, usually microscopic and consisting of cytoplasm and a nucleus in a membrane.
<h2>
Answer:</h2>
390 g KNO₃
<h2>
General Formulas and Concepts:</h2><h3><u>Chemistry</u></h3>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3><u>Math</u></h3>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h2>
Explanation:</h2>
<u>Step 1: Define</u>
2.3 × 10²⁴ formula units KNO₃
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g.mol
Molar Mass of KNO₃ - 39.10 + 14.01 + 3(16.00) = 101.11 g/mol
<u>Step 3: Convert</u>
<u />
= 386.172 g KNO₃
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules and round.</em>
386.172 g KNO₃ ≈ 390 g KNO₃
Answer:
carbon, hydrogen, and oxygen atoms
Explanation:
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Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
