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3 years ago

How does atomic radius change from top to bottom in a group in the periodic table?

2 answers:

4 0

The atomic radius increases as you would go down a particular group on the periodic table of elements. This is because along with a greater number of protons, there would also be electrons as well, and thus the need of electron shells surrounding the atom would also be required, to compensate for the more electrons, as according to the bohr model, each shell contains 8 electrons in its electron shell. Thus the distance from the nucleus to the outermost shell increases, the atomic radius.

vfiekz [6]3 years ago

4 0

Answer: Atomic radii increases from top to bottom in a group.

Explanation:

Atomic radius of an atom is defined as the total distance from the nucleus to the outermost orbital of electron.

As moving from top to bottom, there is an addition of shell around the nucleus and the outermost orbital gets far away from the nucleus and hence, the distance between the nucleus and outermost orbital increases.

This in turn increases the atomic radii of the element from moving top to bottom in a group.

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Is a free electron the same as an electron

Sliva [168]

I don’t know about this ‍♀️

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3 years ago

Read 2 more answers

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

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3 years ago

Which development was a direct result of the

gladu [14]

"(2) increase in suburbanization" was a direct result of the <span>baby boom that followed World War II, but of course there were other factors that resulted as well. </span>

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3 years ago

Answer all parts of the following questions on Particle Stoichiometry.

Ugo [173]

<span>1. Translate, predict the products, and balance the equation above.

Li + Cu(NO3)2 = Li(NO3)2 + Cu

2. How many particles of lithium are needed to produce 125 g of copper?

125 g Cu ( 1 mol / 63.55 g ) (1 mol Li / 1 mol Cu ) ( 6.022 x 10^23 particles / 1 mol ) = 1.18x10^24 Li particles

3. How many grams of lithium nitrate are produced from 4.83E24 particles of copper (II) nitrate?

</span>4.83E24 particles of copper (II) nitrate ( 1 mol / 6.022x10^23 particles ) (1 mol Li(NO3)2 / 1 mol Cu(NO3)2 ) ( 130.95 g / 1 mol ) = 1043.77 grams Li(NO3)2

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3 years ago

Electromagnetic radiation with a wavelength of 575 nm appears as yellow light to the human eye. The energy of one photon of this

Alecsey [184]

Answer:

9.12 * 10^20 photons

Explanation:

Given that;

E=n⋅h⋅ν

Where;

E= energy of the electromagnetic radiation

n = number of photons

h = Plank's constant

ν = frequency of electromagnetic radiation

Hence;

n = E/hν

n = 3.46 × 10 -19/6.6 * 10^-34 * 575 * 10^-9

n = 3.46 × 10 -19/3795 * 10^-43

n= 9.12 * 10^20 photons

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2 years ago