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Ghella [55]
3 years ago
13

Among the elements of the main group, the general trend in the first ionization energy moving across a period is not followed be

tween group 2 and group 13. Which best explains these exceptions?
The ionization energy decreases because the full s orbital shields the electron entering the p orbital.
The ionization energy increases because the full s orbital shields the electron entering the p orbital.
The ionization energy decreases because the stability of the half-full p subshell is increased.
The ionization energy increases because the stability of the half-full p subshell is increased.
Chemistry
2 answers:
Maru [420]3 years ago
8 0

Group 2 and group 13 do not follow these rules because in their case, s orbitals block electrons from going into the p orbital. So A is the answer.

Radda [10]3 years ago
5 0
<span>The Ionization energy decreases because the full s orbital shields the electron entering the p orbital. This is known as "shielding", When each new electron experiences attraction from the nucleus and repulsion forces from the S orbitals, the net force on outer shell electrons is significantly smaller. Therefore, ionization energy decreases during these groups.</span>
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For the following reaction, 4.07 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 10.4 grams of
torisob [31]

Answer:

Theoretical yield = 13.7 g

% yield =76 %

Explanation:

For Al_2O_3

Mass of Al_2O_3  = 4.07 g

Molar mass of Al_2O_3  = 101.96 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{4.07\ g}{101.96\ g/mol}

Moles\ of\ Al_2O_3= 0.0399\ mol

According to the reaction:

Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O

1 mole of Al_2O_3  on reaction produces 1 mole of Al_2(SO_4)_3

So,  

0.0399 mole of Al_2O_3  on reaction produces 0.0399 mole of Al_2(SO_4)_3

Moles of Al_2(SO_4)_3  obtained = 0.0399 mole

Molar mass of Al_2(SO_4)_3 = 342.2 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0399= \frac{Mass}{342.2\ g/mol}

Mass= 13.7\ g

<u>Theoretical yield = 13.7 g</u>

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100

Given , Values from the question:-

Theoretical yield = 13.7 g

Experimental yield = 10.4 g

Applying the values in the above expression as:-

\%\ yield =\frac{10.4}{13.7}\times 100

<u>% yield =76 %</u>

6 0
3 years ago
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