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gayaneshka [121]
3 years ago
6

One electron is removed from both an Na atom and a K atom, producing two ions. Using principles of atomic structure, explain why

the Na ion is much smaller than the K ion. Discuss both ions in your answer.
Chemistry
2 answers:
olganol [36]3 years ago
7 0

Answer:

Explanation:

The Na ion has 2 electron shells while the K ion  has 3 electron shells making it bigger.

Anni [7]3 years ago
3 0

Answer:

In brief, electrons in the K⁺ ion occupy three main electron shells. In comparison, electrons in the Na⁺ ion occupy only two main electron shells.

Explanation:

Consider the electron cloud model for the structure of an atom or an ion. The nucleus was very dense and small. It was highly located at the center of the atom or the ion. Electrons occupy most of the space of the atom or ion.

If there are more than one electrons, they would be located in one or more main energy levels. Generally, for two atoms/ions with a similar number of protons, the one with more filled main electron shells would have a larger radius.

Electrons in a neutral Na atom occupy three main energy shells. The outermost shell contains only one electron. That electron would be removed first, such that in the Na⁺ ion, electrons would occupy only two main shells.

Similarly, electrons in a neutral K atom occupy four main energy shells. The outermost shell contains only one electron. That electron would be removed first, such that in the K⁺ ion, electrons would occupy three main shells.

Since there are more occupied electrons shells in K⁺ than in Na⁺, K⁺ would have a larger ionic radius.

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The half-life of a reaction, t1/2, is the time required for one-half of a reactant to be consumed. It is the time during which t
ycow [4]

<u>Answer:</u> The half life of the reaction is 593.8 seconds

<u>Explanation:</u>

We are given:

Rate constant = 0.0016mol/L.s

The formula for determining the unit of 'k' is:

\text{Unit}=\frac{(Concentration)^{1-n}}{Time}

where, n = order of reaction

The unit of concentration is, M or mole/L

The unit of time is, second or 's'

Evaluating the value of 'n' from above equation:

mol.L^{-1}s^{-1}=\frac{(mol/L)^{1-n}}{s}\\\\n=0

The reaction is zero order reaction.

The equation used to calculate half life for zero order kinetics:

t_{1/2}=\frac{[A_o]}{2k}

where,

k = Rate constant = 0.0016mol/L.s

[A_o] = initial concentration = 1.90 mol/L

Putting values in above equation, we get:

t_{1/2}=\frac{1.90mol/L}{2\times 0.0016mol/L.s}=593.8s

Hence, the half life of the reaction is 593.8 seconds

5 0
3 years ago
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