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Ad libitum [116K]
3 years ago
11

How many atoms are in a 158.69 g sample of nickel?

Chemistry
1 answer:
Alchen [17]3 years ago
8 0
<h3>Answer:</h3>

1.6283 × 10²⁴ atoms Ni

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Use Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

158.69 g Ni

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

Avogadro's Number

Molar Mass of Ni - 58.69 g/mol

<u>Step 3: Convert</u>

  1. Set up:                      \displaystyle 158.69 \ g \ Ni(\frac{1 \ mol \ Ni}{58.69 \ g \ Ni})(\frac{6.022 \cdot 10^{23} \ atoms \ Ni}{1 \ mol \ Ni})
  2. Multiply/Divide:        \displaystyle 1.62827 \cdot 10^{24} \ atoms \ Ni

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

1.62827 × 10²⁴ atoms Ni ≈ 1.6283 × 10²⁴ atoms Ni

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Where : N = amount of radioisotope after time "t"

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Plugging value of T_\frac{1}{2} in above formula (equation 2) :

28.8 yrs = \frac{ln 2}{ k }

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Dividing both side by 28.8 yrs

\frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs}

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k = 0.0241 yrs⁻¹

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Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹

Plugging these value in equation (1) as :

ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t

ln (0.0971 ) = -0.0241 yrs ^-^1 * t

(ln 0.0971 = - 2.33 )

Dividing both side by - 0.0241 yrs⁻¹

\frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1}

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

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