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1 year ago

how did the boiling point of plain water compare to that of water with salt? Compared to water with sugar?

Chemistry

1 answer:

elixir [45]1 year ago

7 0

The boiling point of plain water is less than the boiling point of both salt and sugar water.

<h3>What is boiling point?</h3>

Boiling point can be defined as the point when the pressure exerted by the surroundings upon a liquid is equal to the pressure exerted by the vapour of the liquid.

The boiling point of plain water is 100°C which increases upon addition of solute substances such as salt and sugar.

But salts are usually made up of ionic bonds while sugar are made up of covalent bonds. This means that more energy would be required to boil salt solution due to its ionic bonds.

Therefore, the boiling point of salt water is highest following sugar water before plain water which is the lowest.

Learn more boiling point here:

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Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined

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The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

$Concentration=\frac{number of moles of solute}{Volume}$

You know the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

$Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}$

The concentrations are:

[PCl₃]= $\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}$

[Cl₂]= $\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}$

[PCl₅]= $\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}$

Replacing:

$Q=\frac{0.08*0.16}{0.04}$

Solving:

Q= 0.32

<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>

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