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bija089 [108]
3 years ago
11

A periodic table allows you to easily compare the properties of one element ( or a group of element to another element or group

of elements)
Chemistry
2 answers:
Thepotemich [5.8K]3 years ago
7 0
<span>The statement is very true.
The periodic table is a table that classifies elements based on their chemical properties and physical properties into groups (columns) and periods (rows). Elements in the same group have similar chemical properties, for example group I elements (alkali earth metals) have the same valency (electrons in the outermost energy level). The table also shows the physical properties of an element, for example whether it is a metal or a non metal; e.g group 1, 2 and 3 elements are all metals while group 4 are metaloids, group 5,6 and 7 are non metals and group 8 elements are noble gases with an octet configuration. Additionally, the table also reveals the order reactivity depending on the ease of gaining electrons (for non metals) and loosing electrons (for metals).</span>
AfilCa [17]3 years ago
4 0
The statement above is TRUE.
The periodic table refers to the table of chemical elements, in which the elements are arranged in order of their atomic numbers in such a way that, similar elements appear in vertical columns. Elements in the same column usually have similar atomic structures and similar chemical properties. Thus, the periodic table allows one to easily compared the properties of one element to that of another one depending on the placement of the elements in the periodic table.<span />
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2 years ago
Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr
Alecsey [184]

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

E_{n} = -\frac{m_{e}e^{4}}{2(4\pi\epsilon_{0})^2\hbar^{2}}\frac{1}{n^2} = E_{1}\frac{1}{n^{2}}, where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

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Solving for [tex]\lambda[tex] and substituting the required values:

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2 years ago
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