Nomadic people received income from where
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Answer:
Here's what I get
Explanation:
You may have done a Williamson synthesis of guaifenesin by reacting guaiacol with 3-chloropropane-1,2-diol.
A. Mechanism
Step 1
NaOH converts guaiacol into a phenoxide ion.
Step 2
The phenoxide acts as the nucleophile in an SN2 reaction to displace the Cl from the alkyl halide.
B. Improve the yield
You probably carried out the reaction in ethanol solution — a polar protic solvent.
You might try doing the reaction in a polar aprotic solvent— perhaps DMSO.
A polar aprotic solvent does not hydrogen bond to nucleophiles, so they become stronger.
C. Another method of ether synthesis —dehydration of alcohols
Sulfuric acid catalyzes the conversion of primary alcohols to ethers.
This is also a nucleophilic displacement reaction.
Protonation of the OH converts it into a better leaving group.
Attack by a second molecule of alcohol forms the protonated ether.
A molecule of water then removes the proton.
Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the
amount amount of solute that will be deposited is 1,927.413 grams.
<h3>How can the amount of solute deposited be found?</h3>
The volume of water 1.33 dm³ of water 70 °C.
The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C = 2.25 moles
At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles
Therefore;
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;
1.33 dm³ contains 2.25 moles.
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;
1.33 dm³ contains 0.53 moles
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles
The number of moles that precipitate out = The amount of solute deposited
Which gives;
Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles
The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g
The molar mass of Pb(NO₃)₂ = 331 g/mol
The amount of solute deposited = Number of moles × Molar mass
Which gives;
The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>
Learn more about saturated solutions here: