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dedylja [7]
3 years ago
10

How many particles of CuCr2O7 are present in a 64.5 gram sample?

Chemistry
1 answer:
kow [346]3 years ago
8 0
<h3>Answer:</h3>

1.39 × 10²³ particles CuCr₂O₇

<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
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

[Given] 64.5 g CuCr₂O₇

[Solve] particles CuCr₂O₇

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

Avogadro's Number

[PT] Molar Mass of Cu - 63.55 g/mol

[PT] Molar Mass of Cr - 52.00 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of CuCr₂O₇ - 63.55 + 2(52.00) + 7(16.00) = 279.55 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 64.5 \ g \ CuCr_2O_7(\frac{1 \ mol \ CuCr_2O_7}{279.55 \ g \ CuCr_2O_7})(\frac{6.022 \cdot 10^{23} \ particles \ CuCr_2O_7}{1 \ mol \ CuCr_2O_7})
  2. [DA] Divide/Multiply [Cancel out units]:                                                         \displaystyle 1.38944 \cdot 10^{23} \ particles \ CuCr_2O_7

<u>Step 4: Check</u>

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

1.38944 × 10²³ particles CuCr₂O₇ ≈ 1.39 × 10²³ particles CuCr₂O₇

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Answer:

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Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

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<u> Step 2:</u> Calculate heat absorbed by the calorimeter

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q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

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⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25  = 8.2 °C

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<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

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The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

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