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svet-max [94.6K]
3 years ago
13

Several billion years ago, which gas was the least prevalent in the atmosphere?

Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
7 0
The least prevalent gas in the atmosphere was oxygen several billion years ago. This is because it was only around 2.4-3 billion years ago when the first photosynthetic bacteria evolved, meaning they produce food from inorganic compounds such as carbon dioxide and water to produce glucose and oxygen. Oxygen is then released in the atmosphere and this is called <em>The Great Oxygenation Event</em>. 
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3 years ago
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How many moles of Al would be produced from 20 moles of Al2O3?<br> 2Al2O3<br> -&gt;<br> 4A1 + 302
Evgesh-ka [11]
<h3>Answer:</h3>

\displaystyle 40 \ mol \ Al

<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>Stoichiometry</u>

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

<u>Step 1: Define</u>

[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂

[Given] 20 mol Al₂O₃

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

[RxN] 2 mol Al₂O₃ → 4 mol Al

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                     \displaystyle 20 \ mol \ Al_2O_3(\frac{4 \ mol \ Al}{2 \ mol \ Al_2O_3})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 40 \ mol \ Al

<u>Step 4:Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

Since our final answer already has 1 sig fig, there is no need to round.

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2 years ago
a closed flask of air (0.250 L) contains 5.00 "puffs" of particles. The pressure probe on the flask reads 93 kPa. A student uses
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Answer: New pressure inside the flask would be 148.8 kPa.

Explanation: The combined gas law equation is given by:

PV=nRT

As the flask is a closed flask, so the volume remains constant. Temperature is constant also.

So, the relation between pressure and number of moles becomes

P=n\\or\\\frac{P}{n}=constant

\frac{P_1}{n_1}=\frac{P_2}{n_2}

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P_1=93kPa\\n_1=5\text{ puffs}

  • Final conditions: When additional 3 puffs of air is added

P_2=?kPa\\n_2=8\text{ puffs}

Putting the values, in above equation, we get

\frac{93}{5}=\frac{P_2}{8}\\P_2=148.8kPa

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