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Ede4ka [16]
3 years ago
13

3. Identify the reagents you would use to convert 1-bromopentane into each of the following compounds: (a) Pentanoic acid (b) He

xanoic acid (c) Pentanoyl chloride (d) Hexanamide (e) Pentanamide (f) Ethyl hexanoate
Chemistry
1 answer:
zhuklara [117]3 years ago
3 0

Answer:

Explanation:

a )

CH₃CH₂CH₂CH₂CH₂Br + KOH   ⇒ CH₃CH₂CH₂CH₂CH₂OH

CH₃CH₂CH₂CH₂CH₂OH  + acidic potassium dichromate ⇒  CH₃CH₂CH₂CH₂COOH

b )

CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN  Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH .

c )

CH₃CH₂CH₂CH₂CH₂Br + KOH   ⇒ CH₃CH₂CH₂CH₂CH₂OH

CH₃CH₂CH₂CH₂CH₂OH  + acidic potassium dichromate ⇒  CH₃CH₂CH₂CH₂COOH + SOCl₂ ( thionyl  chloride ) ⇒ CH₃CH₂CH₂CH₂COCl

d )

CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN  Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + PCC ( NH₃ ) ⇒ CH₃CH₂CH₂CH₂CH₂CONH₂

e )

CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN  Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + C₂H₅OH ( Ethyl alcohol + H⁺ )⇒  

CH₃CH₂CH₂CH₂CH₂COOC₂H₅ ( ethyl hexanoate )

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KCl and KBr are both ionic solids. A mixture of KCl and KBr has a mass of 3.595 g. When this mixture is heated in the presence o
monitta

Answer:

The percentage (by mass) of KBr in the original mixture was 33.1%.

Explanation:

The mixture of KCl and KBr has a mass of 3.595g, thus the sum of the moles of KCl (<em>x</em>) multiplied by it molar mass (74.5g/mol) and the moles of KBr (<em>y</em>) multiplied by it molar mass (119g/mol) is the total mass of the mixture:

x.74.5g/mol + y.119g/mol = 3.595g

Also, after the conversion of KBr into KCl, the total mass of 3.129 g is only from KCl moles, hence

\frac{3.129g}{74.5g/mol} = 0.042 moles

But the 0.042 moles came from the originals KCl and KBr moles, thus

x + y = 0.042moles

Now it is possible to propose a system of equations:

x.74.5g/mol + y.119g/mol = 3.595g

x + y = 0.042moles

Solving the system of equations,

x=0.032moles\\y=0.010 moles

0.010 moles of KBr multiplied it molar mass is

0.010molesx119g/mol = 1.19g

Therefore, the percentage (by mass) of KBr in the original mixture was:

\frac{1.19g}{3.595g}x100% = 33.1%%

4 0
3 years ago
Which substance cannot be decomposed by a chemical change?
coldgirl [10]

Copper substance cannot be decomposed by a chemical change.

<h3 />
  • When heated, the copper to carbonate breaks down into copper to oxide. The copper carbonate, which is dark in colour, releases carbon dioxide as well.Because they are the simplest chemically, elements cannot be broken down by chemical processes.

  • Elements are those pure compounds that cannot be broken down by reactions, heating, electrolysis, or other common chemical processes. Examples of elements are oxygen, gold, and silver. Its makeup stays the same, though.  One instance of a physical change is melting.  A physical change is when a sample of matter experiences a change in some of its qualities but not in its identity. Water turns into water vapour when it is heated.

Learn more about copper here:

brainly.com/question/493292

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6 0
2 years ago
Consider the following combustion reaction:
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Answer:

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3 years ago
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How would you prepare 500 mL of 0.360 M solution of CaCl2 from<br> solid CaCl2?
LenKa [72]

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

0.500 L \times \frac{0.360mol}{L} = 0.180 mol

Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

0.180 mol \times \frac{110.98g}{mol} = 20.0 g

To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

You can learn more about solutions here: brainly.com/question/2412491

4 0
2 years ago
Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem
belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

     Fe                        4.38 g

    H₂O                      2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

Hence, the value of equilibrium constant for given equation is 1.0\times 10^{-4}

6 0
4 years ago
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