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maria [59]
3 years ago
14

t the R group in the reagent over the arrow be isopropyl. (i.e. The reagent is LiN[CH(CH3)2]2.) Now, fill in the blanks. This re

action will follow ______ rule and will form the _________ substituted alkene. a. Zaitzev's; more b. Zaitzev's; less c. Hofmann's; less d. Hofmann's; more
Chemistry
2 answers:
vladimir2022 [97]3 years ago
6 0

Answer:. A

Explanation:

T the R group in the reagent over the arrow be isopropyl. (i.e. The reagent is LiN[CH(CH3)2]2.) Now, fill in the blanks. This reaction will follow Zaitzev'srule and will form the more substituted alkene.

Tanzania [10]3 years ago
3 0

Answer:c. Hofmann's; less

Explanation:

The Hofmann elimination process, named after its discoverer, the German chemist August Wilhelm Von Hofmann rule states that the major alkene product is the <u>least substituted </u>and least stable product when it comes to asymmetrical amines.  The Hofmann elimination can be illustrated in t e formation of “Hoffman” products in elimination reactions  using a less common, LDA LiN[CH(CH3)2]2  Lithium Di-isopropyl Amide (LDA) usually with base  potassium t-butoxide  which will give rise to a less substituted alkene.

This reaction will follow Hofmann's_____ rule and will form the ___less______ substituted alkene.

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2C3H7OH + 9O2 --&gt; 6CO2 + 8H2O
Ne4ueva [31]
<h3>Answer:</h3>

733 g CO₂

<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>

[RxN - Balanced]   2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

[Given]   5.55 mol C₃H₇OH

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

[RxN] 2 mol C₃H₇OH → 6 CO₂

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                    \displaystyle 5.55 \ mol \ C_3H_7OH(\frac{6 \ mol \ CO_2}{2 \ mol \ C_3H_7OH})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})
  2. Multiply/Divide:                                                                                               \displaystyle 732.767 \ g \ CO_2

<u>Step 4: Check</u>

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

732.767 g CO₂ ≈ 733 g CO₂

8 0
3 years ago
What numbers are shown in the symbol for a radioactive nuclide
Bingel [31]
The atomic # and the mass #. 
6 0
2 years ago
Read 2 more answers
A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new
Artyom0805 [142]

Answer: There are now 2.07 moles of gas in the flask.

Explanation:

PV=nRT

P= Pressure of the gas = 697 mmHg = 0.92 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = ?

n = number of moles = 1.9

T = Temperature of the gas = 21°C=(21+273)K= 294 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

V=\frac{nRT}{P}=\frac{1.9\times 0.0821 \times 294}{0.92}=49.8L

When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

PV=nRT

P= Pressure of the gas = 775 mmHg = 1.02 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

n=\frac{PV}{RT}=\frac{1.02\times 49.8}{0.0821\times 299}=2.07moles

Thus the now the container contains 2.07 moles.

6 0
2 years ago
During the processes of erosion and deposition, sediments that are the ________ in size will be carried the greatest distances b
Ahat [919]

The full sentences are given below:

1. During the process of erosion and deposition, sediments that are the SMALLEST in size will be carried the greatest distance before being deposited.

Erosion and deposition are the methods by which sand and rock particles are moved from one place to another. The erosion can be caused by water or wind. Water and wind have the capacity to transport particle from one location and deposit them in another location. How far the erosion is able to move the particles depend on the weight of the particles. It is easier for erosion to carry small particles over a long distance than for it to carry large particles over the same distance.

2. Most METAMORPHIC rocks form under conditions found a few kilometer under the earth surface.

Metamorphic rocks generally are formed from existing rocks. The existing rocks are usually subjected to heat and pressure, which cause radical changes in the chemical and physical properties of the rock. Metamorphic rocks can be formed underneath the earth surface if they are subjected to high temperature and pressure by the rock layers above them.

3 0
3 years ago
Read 2 more answers
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
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