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WARRIOR [948]
9 months ago
8

this is the chemical formula for methyl tert-butyl ether (the clean-fuel gasoline additive mtbe):a chemical engineer has determi

ned by measurements that there are moles of carbon in a sample of methyl tert-butyl ether. how many moles of oxygen are in the sample?be sure your answer has the correct number of significant digits
Chemistry
1 answer:
Alex17521 [72]9 months ago
6 0

The Number of moles required for MTBE is 2 as per molarity based chemistry theories.

  1. A flammable liquid with a distinct, unpleasant smell is called methyl tert-butyl ether (MTBE).
  2. Since the 1980s, it has been added to unleaded gasoline as a fuel additive to promote more effective burning.
  3. It is created by mixing chemicals like isobutylene and methanol. Gallstones can be removed with MTBE as well.
  4. In this type of treatment, surgically inserted special tubes are used to deliver MTBE directly to the gall bladders of the patients.
  5. Methyl tert-butyl ether is a colorless liquid with a distinct anesthetic-like smell. Vapors are narcotic and heavier than air. 131 °F boiling point. 18 °F is the flash point.
  6. It is miscible in water and less dense than water. Boosts the octane of gasoline.
  7. Therefore it has 2 moles of oxygen.

To study about isobutylene -

<u>brainly.com/question/8409160</u>

#SPJ4

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What is the mass of 1.58 moles of CH4
HACTEHA [7]
<h3>Answer:</h3>

25.4 g CH₄

<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

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

1.58 mol CH₄

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

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 1.58 \ mol \ CH_4(\frac{16.05 \ g \ CH_4}{1 \ mol \ CH_4})
  2. Multiply/Divide:                 \displaystyle 25.359 \ g \ CH_4

<u>Step 4: Check</u>

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

25.359 g CH₄ ≈ 25.4 g CH₄

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Answer: Yes

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With more water, the molecules of the substance have more water molecules to form bonds with, thus they are dissolved even faster at that same particular temperature.

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