Answer: <span>A reaction progress curve has three peaks and two valleys between the peaks. This curve describes a reaction mechanism that involves
<u>three elementary reactions</u>.
Explanation: I have drawn the progress curve with three peaks and two valleys. In fact the peaks shows higher energy and valleys show lower energies. So, Let suppose we react
A and
B. This reaction between A and B results in the formation of
C. In this reaction the energies of A and B are less, and during the progress of reaction they cross a transition state of higher energy and forms product C with lower energy which is present at lower valley. This was first reaction. Other two reactions will be followed by conversion of C to
D and conversion of D into
E.</span>
Answer:
Period
Explanation:
Periodic trend along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction.
The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.The electron affinity also increases along period because of greater positive charge.
Answer:
- <u>Yes, it is 14. g of compound X in 100 ml of solution.</u>
Explanation:
The relevant fact here is:
- the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.
That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.
Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.
With that, the solubility can be calculated from the followiing proportion:
- 84. g solute / 600 ml solution = y / 100 ml solution
⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.
The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.
<u>The answer is 14. g of solute per 100 ml of solution.</u>
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