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hichkok12 [17]
4 years ago
14

A candy bar has a total mass of 75.0 grams. In a calorimetry experiment, a 1.0-g sample of this candy bar was burned in a calori

meter surrounded by 1000g of water. The temperature of the water in contact with the burning candy bar was measured and found to increase from an initial temperature of 21.2C to a final temperature of 24.3C.
a. Calculate the amount of heat in calories release when the 1.0-g sample burned.
b. Convert the heat in calories to nutritional Calories and then calculate the energy content (fuel value) in Cal/g.
c. Calculate the total caloric content of the candy bar in Calories.
Chemistry
1 answer:
Tcecarenko [31]4 years ago
8 0

<span>A new element, Albanesium, has been discovered.  43.22% of all naturally occurring Albanesium has a mass of 292.0 amu.  46.83% of all Albanesium has a mass of 293.05 amu.  The rest of the Albanesium has a mass of 295.045 amu.  Find the average atomic mass of Albanesium.</span>  <span>Report to 4 significant figures</span>

<span>
</span>

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<h3>Answer:</h3>

5.00 mol O₂

<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>

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

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

3.01 × 10²⁴ atoms O₂

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.01 \cdot 10^{24} \ atoms \ O_2(\frac{1 \ mol \ O_2}{6.022 \cdot 10^{23} \ atoms \ O_2})
  2. Multiply/Divide:                \displaystyle 4.99834 \ mol \ O_2

<u>Step 4: Check</u>

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

4.99834 mol O₂ ≈ 5.00 mol O₂

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From the question, one can work out which states of matter to assign to which species. The trick with organic equations of this nature is to try to balance everything but oxygen first. Make sure you balance oxygen last because it is the easiest to balance.

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