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stepan [7]
3 years ago
15

Which ketone in each pair is more reactive?

Chemistry
1 answer:
KiRa [710]3 years ago
4 0

Answer:

a. 2-heptanone is more reactive than 4-heptanone

b. chloromethyl phenyl ketone is more reactive than bromomethyl phenyl ketone

Explanation:

The reactivity of the carbonyl compound (ketone ) is affected by the steric effect. The steric effect is a hindrance that occurs in the structure or reactivity of a molecule, which is affected by the physical size and the proximity of the adjacent parts of the molecule.

Between 2-heptanone or 4-heptanone, 2-heptanone is more reactive than 4-heptanone. This is because 2-heptanone is less affected by the steric hindrance, unlike the 4-heptanone.

Similarly, the reactivity of the carbonyl compound (ketone) is also affected by the polarity on the carbon compound, which is associated with how electronegative the substituent attached is to the carbonyl compound. From the periodic table, the electronegativity of the Halogen family decreases down the group. Therefore chlorine is more electronegative than bromine.

As such, chloromethyl phenyl ketone is more reactive than bromomethyl phenyl ketone.

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How many molecules are contained in 103.4g of sulfuric acid?
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<h3>Answer:</h3>

1.827 × 10²⁴ molecules H₂S

<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>Chemistry</u>

<u>Compounds</u>

  • Writing Compounds
  • Acids/Bases

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

103.4 g H₂S (Sulfuric Acid)

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

Avogadro's Number

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 103.4 \ g \ H_2S(\frac{1 \ mol \ H_2S}{34.09 \ g \ H_2S})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2S}{1 \ mol \ H_2S})
  2. Multiply:                                                                                                            \displaystyle 1.82656 \cdot 10^{24} \ molecules \ H_2S

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

1.82656 × 10²⁴ molecules H₂S ≈ 1.827 × 10²⁴ molecules H₂S

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3 years ago
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