The freezing point depression is a colligative property which means that it is proportional to the number of particles dissolved.
The number of particles dissolved depends on the dissociation constant of the solutes, when theyt are ionic substances.
If you have equal concentrations of two solutions on of which is of a ionic compound and the other not, then the ionic soluton will contain more particles (ions) and so its freezing point will decrease more (will be lower at end).
In this way you can compare the freezing points of solutions of KCl, Ch3OH, Ba(OH)2, and CH3COOH, which have the same concentration.
As I explained the solution that produces more ions will exhibit the greates depression of the freezing point, leading to the lowest freezing point.
In this case, Ba(OH)2 will produce 3 iones, while KCl will produce 2, CH3OH will not dissociate into ions, and CH3COOH will have a low dissociation constant.
Answer: Then, you can predict that Ba(OH)2 solution has the lowest freezing point.
A. Fireworks exploding is the answer
The answer to this question would be: 3.125%
Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
Answer:
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Explanation:
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Answer:
The formula of the compound is:
N2H2
Explanation:
Data obtained from the question:
Nitrogen (N) = 93.28%
Hydrogen (H) = 6.72%
Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:
N = 93.28%
H = 6.72%
Divide by their molar mass
N = 93.28 /14 = 6.663
H = 6.72 /1 = 6.7
Divide by the smallest
N = 6.663 / 6.663 = 1
H = 6.72 /6.663 = 1
Therefore, the empirical formula is NH.
Now, we can obtain the formula of the compound as follow:
The formula of a compound is simply a multiple of the empirical formula.
[NH]n = 30.04
[14 + 1]n = 30.04
15n = 30.04
Divide both side by 15
n = 30.04/15
n = 2
Therefore, the formula of the compound is:
[NH]n => [NH]2 => N2H2
