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.
Group 17 is the most readily reduced elements on the periodic table, meaning that they are so close to being a stable elements, only missing 1 electron to complete their valance electron shell. Thus they will essentially react with anything to get that last electron!
Group 1 elements are extremely reactive because they are the most readily oxidized, they are very close to reaching stability by giving up only 1 electron. Thus they will react with almost anything to give up their electron.
Explanation:
It is known that the specific heat capacity of Liver 
is 3.59 kJ 
It is given that :
Initial temperature of Liver = Body temperature = 
= 310 K
Final temperature of Liver = 180 K
Relation between heat energy, mass, and change in temperature is as follows.
Q = 
Now, putting the given values into the above formula as follows.
Q =
Q = 
= 700.05 kJ
Therefore, we can conclude that amount of heat which must be removed from the liver is 700.05 kJ.
Answer:
9 moles
Explanation:
The balanced chemical equation provided in this question is as follows:
2CH₄ + S₈ → 2CS₂ + 4H₂S
In accordance to the above balanced equation, 1 mole of sulphur (S8) produces 4 moles of hydrogen sulfide (H2S).
Therefore, if 2.25mol of S8 is used, 2.25 × 4 = 9 mol
9 moles of H2S is produced.
