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blagie [28]
3 years ago
8

The successive ionization energies of a certain element are I1 = 589.5 kJ/mol, I2 =1145 kJ/mol, I3= 4900 kJ/mol, I4 = 6500 kJ/mo

l, and I5 = 8100 kJ/mol. This pattern of ionization energies suggests that the unknown element is:_________
a. K
b. Si
c. As
d. Ca
e. S
Chemistry
2 answers:
stepladder [879]3 years ago
8 0

Answer:

=D

Explanation:

Calcium has 2 Valence electrons which are easy to remove. From the ionization energy values, I1 = 589.5 kJ/mol, I2 =1145 kJ/mol, I3= 4900 kJ/mol, I4 = 6500 kJ/mol, and I5 = 8100 kJ/mol, the energy difference between the 2nd and 3rd electron is the highest because the energy required to remove the 3rd electron is geeatest which implies that the 3rd electron must be a completed octet-orbital electron which is more difficult to remove than other electrons

Gnom [1K]3 years ago
7 0

Answer:

As

Explanation:

For any element to exhibit the pattern of ionization energy shown in the question, it must possess five electrons in its outermost shell. These five electrons are not lost at once. They are lost progressively until the valence shell becomes empty. The ionization energy increases steadily as more electrons are lost from the valence shell.

The only pentavalent element among the options in arsenic, hence the answer.

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Answer :

<h2>Multicellular organisms</h2>

Explanation :

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4 0
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All of the following contribute to the large, negative, free-energy change upon hydrolysis of "high-energy" compounds except: a.
klasskru [66]

Answer:

C) low activation energy of forward reaction

Explanation:

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and 
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8 0
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Classify CH3CH2NHCH3 as a strong base or a weak base
harina [27]

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5 0
2 years ago
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In a hexagonal-close-pack (hcp) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes is 1:1:2. What i
Alex777 [14]

Explanation:

In a hexagonal-close-pack (HCP) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes =  1 : 1 : 2

let the :

Number of lattice point = 1x.

Number of octahedral points = 1x

Number of tetrahedral  points = 2x

If anions occupy the HCP lattice points and cations occupy half of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B =  \frac{2x}{2}=x

The formula of the compound will be = A_{1x}B_{1x}=AB

If anions occupy the HCP lattice points and cations occupy all of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B =  2x

The formula of the compound will be = A_{1x}B_{2x}=AB_2

6 0
3 years ago
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