Answer:
E) Two of the above statements are true.
Explanation:
The options are:
A) Before the solution is titrated with HCl it is pink and when the color changes from pink to colorless, the moles of H*(aq) equals the moles of OH"(aq) used in the hydrolysis of the neutralized aspirin. <em>TRUE. </em>Before the solution is titrated, there is an excess of NaOH (Basic solution, phenolphtalein is pink). Then, at equivalence point, after the addition of HCl, the pH is acidic and phenolphtalein is colorless.
B) Before the solution is titrated with HCl it is colorless and when the color changes from colorless to pink, the moles of H*(aq) equals the excess moles of OH(aq) added. <em>FALSE. </em>As was explained, before the titration, the solution is pink.
C) 25.0 mL of 0.100 M NaOH was added to the sample to hydrolyze the neutralized aspirin in the solution. The titration with HCl allows us to determine the moles of excess OH(aq) added. Once we determine the moles of excess OH(aq), we can determine moles of OH"(aq) used in the hydrolysis of the neutralized aspirin, which is equal to the moles of aspirin in the recrystallized aspirin. <em>TRUE. </em>Aspirin requires an excess of base (NaOH) for a complete dissolution (Hydrolysis). Then, we add H+ as HCl to know the excess moles of OH-. As we know the added moles of OH-, we can find the moles of OH that reacted = Moles of aspirin.
D) We can determine the moles of aspirin in the recrystallized aspirin by titrating with the 0.100 M NaOH to the neutralization point. The purpose of the hydrolysis of the neutralized aspirin and the back-titration with the 0.100 M HCl is to confirm the moles of aspirin in the recrystallized aspirin. <em>FALSE. </em>NaOH can be added directly unyil neutralization point because, initially, aspirin can't be dissolved completely
E) Two of the above statements are true. <em>TRUE</em>
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Right option is:
<h3>E) Two of the above statements are true.</h3>
Answer:
composting scraps
recycling is the action or process of converting waste into reusable material.
Answer:
Pb(s) + 2Ag+(aq) → 2Ag(s) + Pb2+(aq)
Explanation:
Step 1: The unbalanced equation
Pb(s) + AgNO3aq) → Ag(s) + Pb(NO3)2(aq)
Step 2: Balancing the equation
Pb(s) + AgNO3aq) → Ag(s) + Pb(NO3)2(aq)
On the right side we have 2x NO3, on the left side we have 1x NO3.
To balance the amount of NO3 on both sides, we have to multiply AgNO3 (on the left side) by 2.
Pb(s) + 2AgNO3aq) → Ag(s) + Pb(NO3)2(aq)
On the left side we have 2x Ag, on the right side we have 1x Ag.
To balance the amount of Ag on both sides, we have to multiply Ag (on the right side) by 2. Now the equation is balanced.
Pb(s) + 2AgNO3(aq) → 2Ag(s) + Pb(NO3)2(aq)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will look like this
Pb(s) + 2Ag+(aq) → 2Ag(s) + Pb2+
Answer:
let number of carbon is an
Explanation:
hence 12 and + 2n is equal to 126
14 and is equal to 126
n is equal to 9
