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lisabon 2012 [21]
3 years ago
8

Last lesson we calculated that propane has an energy of combustion of 2220 kilojoules per mole while hydrogen gas (rocket fuel)

has an energy of combustion of 286 kilojoules per mole. However, hydrogen gas has a much higher energy of combustion per gram than propane. Why is this? ​

Chemistry
1 answer:
3241004551 [841]3 years ago
5 0

Answer:

1 or a

Explanation:

The number of grams in each mole of hydrogen gas is smaller than the number of grams in each mole of propane gas.

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<h3>Answer:</h3>

28 mol CaF

<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>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.7 × 10²⁵ molecules CaF

[Solve] moles CaF

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                          \displaystyle 1.7 \cdot 10^{25} \ molecules \ CaF(\frac{1 \ mol \ CaF}{6.022 \cdot 10^{23} \ molecules \ CaF})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 28.2298 \ moles \ CaF

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

28.2298 mol CaF ≈ 28 mol CaF

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