Group 1A(1), the alkali metals, includes lithium, sodium, and potassium. Group 7A(17) the halogens, includes chlorine, bromine, and iodine. hope this helps:)
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.
Answer:
A simple example of decomposition reaction is hydrolysis of water where a water molecule is broken down into hydrogen and oxygen gas.
Answer:
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Don't front, no needExplanation:
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Answer:
W = -10.3 kJ
Explanation:
During combustion, the system performs work and releases heat. Therefore, the change in internal energy is negative, and the change in enthalpy, which is equal to heat at constant pressure, is also negative. Work is then calculated by rearranging the equation for the change in internal energy:
w=ΔE−qp=−5084.3 kJ−(−5074.0 kJ)
The release of heat is much greater than the work performed by the system on its surroundings. The potential energy stored in the bonds of octane explains why considerably large amounts of energy can be lost by the system during combustion.
