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Olenka [21]
3 years ago
12

Rank the following atoms in order of decreasing first ionization energies (i.e., highest to lowest): Li, Be, Ba, F.

Chemistry
1 answer:
Minchanka [31]3 years ago
5 0

Answer:

The order is:

F >Be >Li >Ba

Explanation:

Electrons are held in atoms by their  attraction to the nucleus which means that to remove an electron from the atom energy is needed.

The ionization energy is the minimum energy necessary to remove an electron from an atom in the gas phase and ground state, the electron removed being the outermost, that is, the furthest from the nucleus. The further away the electron is from the nucleus, the easier it is to remove it, that is, the less energy is needed.

By increasing the atomic number of the elements of the same group, the nuclear attraction on the outermost electron decreases, since the atomic radius increases. Then the ionization energy decreases. In other words, in a group it decreases from top to bottom because the size of the atom increases and it is easier to remove an external electron.

By increasing the atomic number of the elements of the same period, the nuclear attraction on the outermost electron increases, since the atomic radius decreases. Therefore, in a period, as the atomic number increases, the ionization energy increases. In summary, in a period it increases from left to right as the effective nuclear charge increases and it increases thanks to the decrease in the size of the atom.

Taking these considerations into account, the order is:

<u><em>F >Be >Li >Ba</em></u>

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A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are
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Answer: Resulting solution will not be neutral because the moles of OH^-ions is greater. The remaining concentration of [OH^-]ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

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[Ca(OH)_2] =0.0100 M

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Step 1: Calculating the total moles of H^+ ion from both the acids

moles of H^+ in HCl

HCl\rightarrow {H^+}+Cl^-

if 1 L of HClsolution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 \times0.1 moles= 0.005 moles    (1L=1000mL)

moles of H^+ in HCl = 0.005 moles

Similarliy

moles of H^+ in HNO_3

HNO_3\rightarrow H^++NO_3^-}

If 1L of HNO_3 solution= 0.200 moles

Then 0.1L of HNO_3 solution= 0.1 \times 0.200 moles= 0.02 moles

moles of H^+ in HNO_3 =0.02 moles

so, Total moles of H^+ ions  = 0.005+0.02= 0.025 moles     .....(1)

Step 2: Calculating the total moles of [OH^-] ion from both the bases

Moles of OH^-\text{ in }Ca(OH)_2

Ca(OH)_2\rightarrow Ca^2{+}+2OH^-

1 L of Ca(OH)_2= 0.0100 moles

Then in 0.5 L Ca(OH)_2 solution = 0.5 \times0.0100 moles = 0.005 moles

Ca(OH)_2 produces two moles of OH^- ions

moles of OH^- = 0.005 \times 2= 0.01 moles

Moles of OH^- in RbOH

RbOH\rightarrow Rb^++OH^-

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2 \times 0.100 moles = 0.02 moles

Moles of OH^- = 0.02 moles

so,Total moles of OH^- ions = 0.01 + 0.02=0.030 moles      ....(2)

Step 3: Comparing the moles of both H^+\text{ and }OH^- ions

One mole of H^+ ions will combine with one mole of OH^- ions, so

Total moles of H^+ ions  = 0.005+0.02= 0.025 moles....(1)

Total moles of OH^- ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of H^+ ions = total moles of OH^- ions

0.025 moles H^+ will neutralize the 0.025 moles of OH^-

Moles of OH^- ions is in excess        (from 1 and 2)

The remaining moles of OH^- will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of OH^- ions = \frac{\text{Moles remaining}}{\text{Total volume}}

[OH^-]=\frac{0.005}{0.85}=0.0058M

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