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faltersainse [42]
3 years ago
7

The first ionization energies (kl/mol) of hydrogen (H), nitrogen (N), fluroine (F), and oxygen (O) are 1,312, 1,402, 1,742, and

1,314, respectively. From which electrically neutral atom does it take the least energy to remove the first electron?
• Hydrogen
• Fluorine
• Nitrogen
• Oxygen
Chemistry
1 answer:
Oksanka [162]3 years ago
5 0
Fluorine because of the spin pair repulsion fron the p orbitals
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Answer:

K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}

Explanation:

Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:

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  • in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
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Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:

K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}

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