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raketka [301]
3 years ago
7

Consider the conversion of 2-naphthol and 1-bromobutane into an ether via the williamson ether synthesis. list the procedural st

eps from start to finish.
Chemistry
1 answer:
Lemur [1.5K]3 years ago
6 0

Answer:

Following are the solution to this question:

Explanation:

Please find the complete question in the attachment.

Start of Laboratory

Dissolve 2-naphthol in the round bottom flask with ethanol.

Add pellets of sodium hydroxide and hot chips. Attach a condenser.

Heat for 20 minutes under reflux, until the put a burden dissolves.

After an additional hour, add 1-Bromobutane and reflux.

Pour the contents into a beaker with ice from a round bottom flask.

On a Bachner funnel, absorb the supernatant by vacuum filtration.

Utilizing cold water to rinse the material and dry that on the filter.

Ending of the Lab

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The volume of a gas is 450 mL when its pressure is 1.00 atm. If the temperature of the gas does not change, what is the
NikAS [45]

Answer: 0.225 atm

Explanation:

For this problem, we have to use Boyle's Law.

Boyle's Law: P₁V₁=P₂V₂

Since we are asked to find P₂, let's manipulate the equation.

P₂=(P₁V₁)/V₂

P_{2} =\frac{(1.00atm)(0.45L)}{2.00L}

With this equation, the liters cancel out and we will be left with atm.

P₂=0.225 atm

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3 years ago
The pH of a neutral solution is below 7 above 7 7 7.5
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The pH scale ranges from 0 to 14 and goes from acidic to alkaline, where 7 - the center of the scale - is the most neutral.

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3 years ago
1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.
solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2) The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = S=3.1\times 10^-2M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

                            S     2S

The solubility product of the lead(II) chloride = K_{sp}

K_{sp}=[Pb^{2+}][Cl^-]^2

K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}

The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =1\timed 0.000010 M=0.000010 M

Solubility of aluminium hydroxide in aluminum nitrate solution = S

Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)

                            S     3S

The solubility product of the aluminium nitrate = K_{sp}=1.0\times 10^{-33}

K_{sp}=[Al^{3+}][OH^-]^3

1.0\times 10^{-33}=(0.000010+S)\times (3S)^3

S=1.6\times 10^{-10} M

The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3.

Molarity=\frac{Moles}{Volume (L)}

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = \frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol

Volume of the solution = 0.250 L

[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

[Cl^-]=[NaCl]=0.00024 M

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

n=0.12 M]\times 0.250 L=0.030 mol

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of lead(II) chloride = K_{sp}=1.2\times 10^{-4}

Ionic product of the lead chloride in solution :

Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}

Q_i ( no precipitation)

The given statement is false.

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Which is a symbol that represents SI units for temperature.
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Answer:

A.'C

Explanation:

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Answer: An oxygen atom in heavy water has an extra neutron. A hydrogen atom in heavy water has an extra proton.

Explanation:

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