What is the [OH-],[H3O] and PH for the following
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Answer:
d. Adding 5% HCl solution to the crude reaction mixture will protonate aniline increasing its solubility in the aqueous solution.
Explanation:
On this case, we have to check the <u>structures of each compound</u> (figure 1). For naphthalene we dont have <u>any functional groups</u> therefore, the addition of HCl or NaOH it will not affect naphthalene so <u>we can discard "B" and"C".</u>
When we add HCl solution we will have the production the presence of this <u>hydronium ion will protonate the acid</u>, so we can <u>discard a.</u>
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Finally, for d when we add the <u>hydronium ion will react with aniline</u> (a base) and will produce an <u>ammonium ion</u>. This ammonium ion have a <u>positive charge</u>, therefore the <u>polarity will increase</u> and the molecule would be more soluble on water (figure 2).
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Answer:
The mole fraction of hexane is 0.61
Explanation:
Raoult's law allows calculating the vapor pressure of a substance when it is forming part of an ideal solution, knowing its vapor pressure when it is pure (at the same temperature) and the composition of the ideal solution in terms of mole fraction.
Pi=xi*Pi°
where Pi is the partial pressure of component i in the solution, Pi ° is the vapor pressure of component i pure and xi is the mole fraction of component i in the liquid phase.
In a solution, once the components have reached chemical equilibrium, the total vapor pressure is:
Psolution=P1 + P2=x1*P1° +x2*P2°
In this case:
Psolution= xpentane*Ppentane° +xhexane*Phexane°
Being:
- Psolution= 258 torr
- xpentane + xhexane=1 ⇒ xpentane= 1 - xhexane
- Ppentane°= 425 torr
- xhexane= ?
- Phexane°= 151 torr
and replacing:
258 torr= (1-xhexane)* 425 torr + xhexane* 151 torr
you get:
258 torr= 425 torr - 425 torr*xhexane + xhexane* 151 torr
258 torr - 425 torr= - 425 torr*xhexane + xhexane* 151 torr
-167 torr= -274 torr*xhexane
xhexane= 0.61
<u><em>The mole fraction of hexane is 0.61</em></u>