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3 years ago

How would you describe the size of atoms and molecules in your own words ?

Chemistry

2 answers:

topjm [15]3 years ago

8 0

Answer:

tomic size is the distance from the nucleus to the valence shell where the valence electrons are located. Atomic size is difficult to measure because it has no definite boundary. The electrons surrounding the nucleus exist in an electron cloud.

Explanation:

Mazyrski [523]3 years ago

8 0

Answer:

Atomic size is the distance from the nucleus to the valence shell where the valence electrons are located. The separation that occurs because electrons have the same charge. The number of protons in the nucleus. The core electrons in an atom interfere with the attraction of the nucleus for the outermost electrons. and Molecular size is a measure of the area a molecule occupies in three-dimensional space. The amount of space any mass takes up in three-dimensional space is known specifically as its volume.

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3 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$

= $9.669 \times 10^{-10}$ m

or, = $9.7 A^{o}$

= 1 nm (approx)

Also, it is known that $\lambda_{D}$ = $\sqrt \frac{1}{n^{o}}$

Hence, we can conclude that addition of 0.1 $mol/dm^{3}$ of KCl in 0.1 $mol/dm^{3}$ of NaBr " $\lambda_{D}$ " will decrease but not significantly.

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3 years ago

Determine the formula unit for the compound formed when each pairs of ions interact.

maksim [4K]

The formula unit of compounds formed depends on the valency of the combining ions.

The formula unit is the smallest unit of an ionic substance that gives us a picture of the ratio in which ions are combined in the ionic compound.

Usually, the formula unit is written based on the valency of the ions.

For the ions listed;

Li+ and 02- forms Li2O

Mg2+ and S2- forms MgS

A13+ and CI- forms AlCl3

Na+ and N3- forms Na3N

Learn more: brainly.com/question/9743981

3 0

2 years ago