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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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Remember pH=-log(H ions). So it would be pH=-log(10^-7).
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Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH
uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

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7 0
3 years ago
For the following reaction, 33.7 grams of bromine are allowed to react with 13.0 grams of chlorine gas.
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Explanation:

The reaction is given as;

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From the equation;

1 mol of Br2 reacts with 1 mol of Cl2

Converting the masses given to moles, using the formular;

Number of moles = Mass / Molar mass

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Number of moles = 33.7 g / 159.808 g/mol = 0.21088 mol

Cl2;

Number of moles = 13.0 g / 70.906 g/mol = 0.18334 mol

From the values;

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1 mol of Cl2 produces 2 mol of Bromine Monochloride

0.18334 mol of Cl2 would produce x

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<em />

<em>What is the FORMULA for the limiting reagent?</em>

<em />

The limiting reagent is Cl2 as it determines the amount of product formed. The moment the reaction uses up Cl2, the reaction stops.

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The number of moles left is;

0.02754 mol of Br2

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Answer:

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Hypotonic solution will make the red blood cell grow in size if in contact.
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