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finlep [7]
3 years ago
10

In which of the following atoms are valence electrons in the lowest average potential energy states?

Chemistry
1 answer:
Andrej [43]3 years ago
7 0

Answer:

C) F

Explanation:

Valence electrons are the electrons present in the outermost shell of an atom and have the highest energy level of an atom whereas electrons at ground state or lower orbitals have less potential energy.

This is so because <u>orbitals nearby nucleus are strongly bonded with the atomic nucleus and have less energy than an outermost shell.</u>

So, the potential energy states of an atom depend on the number of orbitals. In the given options fluorine with atomic number 9 has less number of orbitals that is 2 orbitals and valence electrons will be present in second orbitals, so fluorine will have the lowest average potential energy states.

Hence, the correct option is "C) F".

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An acid-base indicator is used in what type of reaction?
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Why do different materials have similar properties
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2 years ago
Read 2 more answers
A 2.5 L flask is filled with 0.25 atm SO3, 0.20 atm SO2, and 0.40 atm O2, and allowed to reach equilibrium. Assume at the temper
Anit [1.1K]

Explanation:

Reaction equation for the given chemical reaction is as follows.

      2SO_{3} \rightleftharpoons 2SO_{2} + O_{2}

Equation for reaction quotient is as follows.

         Q = \frac{P^{2}_{SO_{2}} \times P_{O_{2}}}{P^{2}_{SO_{3}}}

             = \frac{(0.20)^{2} \times 0.40}{(0.25)^{2}}

             = 0.256

As, Q > K (= 0.12)

The effect on the partial pressure of SO_{3} as equilibrium is achieved by using Q, is as follows.

  • This means that there are too much products.
  • Equilibrium will shift to the left towards reactants.
  • More SO_{3} is formed.
  • Partial pressure of SO_{3} increases.
4 0
3 years ago
Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1
NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}

6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}

6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

\% = \frac{x}{[HCN]} \times 100

\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100

\% = 3.5 \cdot 10^{-3} \%          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!      

6 0
2 years ago
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