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GarryVolchara [31]
3 years ago
13

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

Chemistry
1 answer:
OverLord2011 [107]3 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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1. <em>Balancing nuclear equations </em>

Answer:

_{18}^{37}\text{Ar} + _{-1}^{0}\text{e} \rightarrow _{17}^{37}\text{Cl}

Explanation:

The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts</em> (the mass numbers) and the <em>subscripts</em> (the nuclear charges) <em>must balance</em>.

Mass numbers: 37 + 0 = 37; balanced.

Charges: 18 + 1 = 17; balanced

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C is <em>wron</em>g. Mass numbers not balanced. 254 + 4 ≠ 258 + 2(1).

D is <em>wron</em>g. Mass numbers not balanced. 14 + 4 ≠ 17 + 2.

===============

2. <em>Amount remaining </em>

Answer:

D. 5.25 g

Explanation:

The half-life of Th-234 (24 da) is the time it takes for half the Th to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:  

  No. of                  Fraction       Amount  

<u>half-lives</u>   <u>t/(da</u>)  <u>remaining</u>  <u>remaining/g</u>  

      1             24            ½                21.0  

      2            48            ¼                10.5  

      3             72           ⅛                 5.25  

      4             96          ¹/₁₆                 2.62  

We see that 72 da is three half-lives, and the amount of Th-234 remaining is 5.25 g.  

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3.<em> Calculating the half-life </em>

Answer:

a. 2.6 min

Explanation:

The fraction of the original mass remaining is 1.0 g/4.0 g ≈ ¼.

We saw from the previous table that it takes two half-lives to decay to ¼ of the original amount.

2 half-lives = 5.2 min       Divide both sides by 2

  1 half-life = 5.2 min/2 = 2.6 min

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Hello there!

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