Ionization enthalpy, IE, is also called ionization potential is the ability to remove the electron from the neutral gaseous atom. There is a trend observed in the periodic table for the IE value. As we go from left to right in a period, IE vale increases. While moving from top to bottom in a group, IE value decreases.
- The phenomenon of unexpected drop in IE1 values between Groups 2 and 13, in period 2 and period 4 is due to the introduction of d-orbitals in the case of period 4 elements.
- While moving in the period, there is the constant addition of electrons in the nucleus. The shell sie remains constant while electron pull increases from the nucleus, this leads to a reduction in the size of the atom. As the size decreases, it is difficult to remove the electron from the atom, and thus IE value increases in the case of period 2.
- When we study the case of period 4, there is an introduction of d-electrons. As the inner shell electron increases, there is an increase in the shielding effect. This shielding effect tends to decrease the nuclear attraction between the nucleus and outermost electrons. Ultimately this decreases the IE value in the fourth period. Such a phenomenon is absent in the case of group 2 elements.
- If we speak in terms of orbital energy, the IE value decreases while moving from top to bottom in the period. This is due to the fact that, as we go down in the periodic table, the number of shells increases, and the outermost electron is too far from the nuclear attraction, therefore it can be ejected out easily. This marks a decrease in IE value.
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Answer:
<h3>The answer is 0.03 moles</h3>
Explanation:
To find the number of moles given the number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question
N = 2 × 10²² molecules of water
We have
We have the final answer as
<h3>0.03 moles</h3>
Hope this helps you
Answer:
b. 9 g
Explanation:
Solubility Ce2(SO4)3 at 10°C is a little less than 10 g/100 g water.
So, the best answer is b. 9 g.
The answers to your questions are as follows
A) The number of aluminium ions present = 1.62 * 10²³ ions
B) The number of chloride ions present = 9.86 * 10²³ ions
C) The mass of one unit of aluminium chloride = 133.34 grams
<u>Given data :</u>
mass of aluminium chloride = 37.2 g
molar mass of aluminium chloride = 133.34 g/mol
note : I mole of a molecule has 6 * 10²³ molecules
Number of moles = mass / molar mass = 0.27 moles
<h3>Determine the number of aluminum ions and chloride ions present</h3>
A) aluminium ions present
moles of AlCl₃ * 6 * 10²³
= 0.27 * 6 * 10²³ = 1.62 * 10²³ ions
B) Chloride ions present
moles of AlCl₃ * 6 * 10²³ * 3
= 0.27 * 6 * 10²³ * 3
= 4.86 * 10²³ ions
C) The mass of one formula unit of aluminium chloride = 133.34 grams
Hence we can conclude that the answers to your question are as listed above.
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