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GarryVolchara [31]
4 years ago
13

Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points.

Chemistry
1 answer:
OverLord2011 [107]4 years ago
7 0
The freezing point depression is a colligative property, which means that it depends on the number of particles of solute disolved in the solution.

When you have solutes that are ionic compounds they dissociate in water into ions, then the compound that dissociates more ions will produce more particles and will decrease the freezing point the most.

Given theses aqueous solutions Na2 CO3, Co Cl3, and Li NO3 you can predict the order of the freezing points.

First, write the dissociation equations>

Na2CO3 -> 2Na(+) + CO3 (2-)  These are 3 ions: two of Na(+) and one of CO3(2-)

The number inside parenthesis are number of charge not number of molecules.

Co Cl3 -> Co(3+) + 3 Cl (1-) Those are 4 ions: one of Co (+) and three of Cl (-)

Li NO3 -> Li (+) + NO3 (-) those are two ions: one of Li (+) and one of NO3(-)

Then the ionic compound that dissociates into more ions give the solution with lower freezing point, and these is the rank from higher to lower freezing point:

Li NO3 > Na2 CO3 > Co Cl3.
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A 52-gram sample of water that has an initial temperature of 10.0 °C absorbs 4,130 joules. If the specific heat of water is 4.18
FinnZ [79.3K]

Answer:

The final temperature of the water is 28.98 degree Celsius.

Explanation:

It is given that,

Mass of sample of water, m = 52 grams

Initial temperature, T_i=10^{\circ}C

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We need to find the final temperature of the water. The heat absorbed is given by the formula as follows :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\T_f=\dfrac{Q}{mc}+T_i\\\\T_f=\dfrac{4130}{52\times 4.184 }+10\\\\T_f=28.98^{\circ} C

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You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,
telo118 [61]

Answer:

\frac{[F^{-}]}{[HF]} is larger

Explanation:

pK_{a}=-logK_{a} , where K_{a} is the acid dissociation constant.

For a monoprotic acid e.g. HA, K_{a}=\frac{[H^{+}][A^{-}]}{[HA]} and \frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}

So, clearly, higher the K_{a} value , lower will the the pK_{a}

In this mixture, at equilibrium, [H^{+}] will be constant.

K_{a} of HF is grater than K_{a} of HCN

Hence, (\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})

So, \frac{[F^{-}]}{[HF]} is larger

8 0
4 years ago
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