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scoray [572]
3 years ago
12

How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?

Chemistry
1 answer:
zlopas [31]3 years ago
4 0
<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.60 × 10⁻⁵ g Au (Gold)

<u>Step 2: Identify Conversions</u>

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})
  2. Multiply:                            \displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

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Answer:

m H2(g) = 2.241 g H2(g)

Explanation:

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limit reagent:

∴ Mw Al = 26.982 g/mol

∴ Mw H2SO4 = 98.0785 g/mol

⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al

⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4

⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2

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3 years ago
The sum of all chemical reactions that occur in a cell is called:.
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A gas is collected in a 34.3L container at a temperature of 31.5°C. Later, the container has a volume of 29.2L, a temperature of
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Answer:

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Step-by-step explanation:

To solve this problem, we can use the <em>Combined Gas Laws</em>:

p₁V₁/T₁ = p₂V₂/T₂             Multiply each side by T₁

   p₁V₁ = p₂V₂ × T₁/T₂      Divide each side by V₁

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(a) Convert temperatures to <em>kelvins </em>

T₁ = (31.5 + 273.15) K = 304.65 K

T₂ = (21.0 + 273.15) K = 294.15 K

(b) Calculate the <em>pressure </em>

p₁ = 122.2 kPa × (29.2/34.3) × (304.65/294.15)  

   = 122.2 kPa × 0.8542 × 1.0357

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