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Lelu [443]
3 years ago
5

How many constitutionally isomeric monochlorination products are possible from the following?a. 2-Methylpentane b. 3-Methylpenta

ne c. 2,2-Dimethylbutane d. 2,3-Dimethylbutane
Chemistry
1 answer:
Verdich [7]3 years ago
3 0

Answer:

a. 5

b. 4

c. 3

d. 2

Explanation:

How many constitutionally isomeric monochlorination products are possible from the following?

a. 2-Methylpentane

In 2-methylpentane, there are five possible monochlorinated substitutional products in the reaction of chlorination of 2-methylpentane in the presence of sunlight.

b. In 3-methylpentane there are four possible monochlorinated substitutional products in the reaction of chlorination of 3-methylpentane in the presence of sunlight.

c. In 2,2-Dimethylbutane there are three possible monochlorinated substitutional products in the reaction of chlorination of 2,2-Dimethylbutane in the presence of sunlight.

d. In 2,3-Dimethylbutane there are two possible monochlorinated substitutional products in the reaction of chlorination of 2,3-Dimethylbutane  in the presence of sunlight.

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Given: 36.7 grams of CaF2 is added to 300 mL water. Find molarity?
BigorU [14]
<h3>Answer:</h3>

2 M

<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>Unit 0</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

<u>Aqueous Solutions</u>

  • Molarity = moles of solute / liters of solution
<h3>Explanation:</h3>

<u>Step 1: Define</u>

36.7 g CaF₂

300 mL H₂O

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

Molar Mass of Ca - 40.08 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of CaF₂ - 40.08 + 2(19.00) = 78.08 g/mol

1000 mL = 1 L

<u>Step 3: Convert</u>

<em>Solute</em>

  1. Set up:                               \displaystyle 36.7 \ g \ CaF_2(\frac{1 \ mol \ CaF_2}{78.08 \ g \ CaF_2})
  2. Multiply:                             \displaystyle 0.470031 \ mol \ CaF_2

<em>Solution</em>

  1. Set up:                              \displaystyle 300 \ mL \ H_2O(\frac{1 \ L \ H_2O}{1000 \ mL \ H_2O})
  2. Multiply:                            \displaystyle 0.3 \ L \ H_2O

<u>Step 4: Find Molarity</u>

  1. Substitute [M]:                    \displaystyle x \ M = \frac{0.470031 \ mol \ CaF_2}{.3 \ L \ H_2O}
  2. Divide:                                \displaystyle x = 1.56677 \ M

<u>Step 5: Check</u>

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

1.56677 M ≈ 2 M

8 0
3 years ago
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