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Nina [5.8K]
3 years ago
7

Br NaOCH2CH3 + CH3CH-OH + NaBr CH3 CH3 a. Identify the mechanism of the reaction. b. Suggest steps for the mechanism of this rea

ction. Use curved arrows to show the electron motions. c. How would the rate be affected if the concentration of sodium ethoxide, NaOCH CH3 is increased? Justify your answer.​

Chemistry
1 answer:
Ulleksa [173]3 years ago
3 0

Answer:

a) The mechanism of the reaction is the Elimination Bimolecular or E2.

b) Steps for the mechanism of this reaction is given as follows,

c) Reaction rate = K[Organic compound][NaoCH_{2} CH_{3}].  

Explanation:

a) The mechanism of the reaction is the Elimination Bimolecular or E2.

c) This is an E2 reaction, so it depends on the concentration of both substrate and reactant. If we increase the concentration of NaoCH_{2} CH_{3}, the reaction rate will be increased.

Reaction rate = K[Organic compound][NaoCH_{2} CH_{3}].  

b) Steps for the mechanism of this reaction is given as follows,

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The CRC Handbook, a large reference book of chemical and physical data, lists two isotopes of indium (). The atomic mass of 4.28
ivanzaharov [21]

<u>Answer:</u> The mass of second isotope of indium is 114.904 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

  .....(1)

Let the mass of isotope 2 of indium be 'x'

  • <u>For isotope 1:</u>

Mass of isotope 1 = 112.904 amu

Percentage abundance of isotope 1 = 4.28 %

Fractional abundance of isotope 1 = 0.0428

  • <u>For isotope 2:</u>

Mass of isotope 2 = x amu

Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %

Fractional abundance of isotope 2 = 0.9572

Average atomic mass of indium = 114.818 amu

Putting values in equation 1, we get:

114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu

Hence, the mass of second isotope of indium is 114.904 amu

4 0
3 years ago
Determine the number of grams of carbon 3.14x 1023 atoms of carbon
tiny-mole [99]
<h3>Answer:</h3>

6.26 g C

<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
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

[Given] 3.14 × 10²³ atoms C

[Solve] grams C

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

Avogadro's Number

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

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 3.14 \cdot 10^{23} \ atoms \ C(\frac{1 \ mol \ C}{6.022 \cdot 10^{23} \ atoms \ C})(\frac{12.01 \ g \ C}{1 \ mol \ C})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 6.26227 \ g \ C

<u>Step 4: Check</u>

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

6.26227 g C ≈ 6.26 g C

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Answer:

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Explanation:

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Explanation:

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