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podryga [215]
3 years ago
5

1. In the investigation of an unknown alcohol, there was a positive Jones test and a negative Lucas test. What deductions may be

made as to the nature of the alcohol? State reasons for your deductions. 2. Draw the structures of the products formed from each of the knowns in the Lucas test. If no product is formed, indicate that with the statement "no reaction" in place of products. 3. Repeat question 2 with the Jones test.

Chemistry
1 answer:
pogonyaev3 years ago
7 0

Answer:

1. When observing a positive test for the jones reagent and negative for the Lucas test, it indicates that it is in the presence of a primary alcohol.

Jones reagent behaves like a strong oxidant, where it transforms the primary alcohols into carboxylic acids and the secondary alcohols into ketones. Tertiary alcohols do not react.

With the Lucas test, tertiary alcohols react immediately producing turbidity, while secondary alcohols do so in five minutes. Primary alcohols do not react significantly with Lucas reagent at room temperature.

2. No reaction (See the attached drawing)

3. (see the attached drawing)

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Entropy means that _________.A. the quantity of usable energy declines with each transformationB. energy can be neither created
son4ous [18]

Answer:

answer A

Explanation:

A) the quantity of usable energy declines with each transformation → True . Since the entropy increases , the amount of energy that can not be converted to useful energy increases and since the total amount of energy is conserved, the quantity of useful energy decreases.

B) energy can be neither created nor destroyed → False in the context of entropy , since the energy is conserved regardless of the changes in entropy (First law → conservation of energy vs second law → increase of entropy)

C)  life should be impossible → False . Since the second law states that the entropy of the <u>universe </u>increases with time . Then the system (life) can experience a decrease in entropy  at the expense of a larger increase in entropy of the surroundings ( so the net increase is positive)

D) it is not possible to observe an increase in molecular organisation → False . Same as C. A system  can experience a decrease in entropy at the expense of a larger increase in entropy of the surroundings ( so the net increase is positive)

7 0
3 years ago
Tris has a molecular weight of 121 g/mol. How many grams of Tris would you need to make 100 mL of a 100 mM solution of Tris
Zepler [3.9K]

Answer:

1.21 g of Tris

Explanation:

Our solution if made of a solute named Tris

Molecular weight of Tris is 121 g/mol

[Tris] = 100 mM

This is the concentration of solution:

(100 mmoles of Tris in 1 mL of solution) . 1000

Notice that mM = M . 1000  We convert from mM to M

100 mM . 1 M / 1000 mM = 0.1 M

M = molarity (moles of solute in 1 L of solution, or mmoles of solute in 1 mL of solution). Let's determine the mmoles of Tris

0.1 M = mmoles of Tris / 100 mL

mmoles of Tris = 100 mL . 0.1 M → 10 mmoles

We convert mmoles to moles → 10 mmol . 1mol / 1000mmoles = 0.010 mol

And now we determine the mass of solute, by molecular weight

0.010 mol . 121 g /mol = 1.21 g

8 0
3 years ago
Organic Chemistry, 7e by L. G. Wade, Jr. Reactions of Alkenes Christine Hermann Radford University Radford VA Copyright © 2010 P
RSB [31]

Answer: Christine Herman & L.G Wade Jr., "2010". Organic Chemistry: Reaction of Alkane, 7e, Pearson Education, Radford University, Radford, VA.

Explanation:

This is an edited book. The Harvard reference style was used in the following order:

Authors name

Year of publication

Title

Edition

Publisher

Place of publication.

Note that the title of book should be italicized with capitalization of first word.

7 0
3 years ago
A sample of oxygen has a volume of 7.84 mL at a pressure of 71.8 mmHg and a
Andru [333]

here is an attached photo with a detailed explanation, good luck!

4 0
3 years ago
How many liters of ammonia (NH3), at 3.2 atm and 23C, must be used to produce of 2.65 grams of calcium hydride (CaH2). 6 Ca(s)
7nadin3 [17]

Answer:

The answer to your question is    V = 0.32 L

Explanation:

Data

Volume of NH₃ = ?

P = 3.2 atm

T = 23°C

mass of CaH₂ = 2.65 g

Balanced chemical reaction

               6Ca  +  2NH₃   ⇒   3CaH₂  +  Ca₃N₂

Process

1.- Convert the mass of CaH₂ to moles

-Calculate the molar mass of CaH₂

 CaH₂ = 40 + 2 = 42 g

                             42 g ------------------ 1 mol

                              2.65 g --------------  x

                              x = (2.65 x 1)/42

                              x = 0.063 moles

2.- Calculate the moles of NH₃

                     2 moles of NH₃ --------------- 3 moles of CaH₂

                      x                        --------------- 0.063 moles

                                x = (0.063 x 2) / 3

                                x = 0.042 moles of NH₃

3.- Convert the °C to °K

Temperature = 23°C + 273

                      = 296°K

4.- Calculate the volume of NH₃

-Use the ideal gas law

              PV = nRT

-Solve for V

                V = nRT / P

-Substitution

                V = (0.042)(0.082)(296) / 3.2

-Simplification

               V = 1.019 / 3.2

-Result

               V = 0.32 L

7 0
2 years ago
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