Chemistry, 24.09.2019 23:30 Jerry4891

For ir radiation with û = 1,130 cm 1, v=__thz

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Answer from: imknutson962

The frequency of the radiation is 33.9 THz

Explanation:

We are given:

Wave number of the radiation, $\bar{\nu}=1130cm^{-1}$

Wave number is defined as the number of wavelengths per unit length.

Mathematically,

$\bar{\nu}=\frac{1}{\lambda}$

where,

$\bar{\nu}$ = wave number = $1130cm^{-1}$

$\lambda$ = wavelength of the radiation = ?

Putting values in above equation, we get:

$1130cm^{-1}=\farc{1}{\lambda}\\\\\lambda=\frac{1}{1130cm^{-1}}=8.850\times 10^{-4}cm$

Converting this into meters, we use the conversion factor:

1 m = 100 cm

So, $8.850\times 10^{-4}cm=8.850\times 10^{-4}\times 10^{-2}=8.850\times 10^{-6}m$

The relation between frequency and wavelength is given as:

$\nu=\frac{c}{\lambda}$

where,

c = the speed of light = $3\times 10^8m/s$

$\nu$ = frequency of the radiation = ?

Putting values in above equation, we get:

$\nu=\frac{3\times 10^8m/s}{8.850\times 10^{-4}m}$

$\nu=0.339\times 10^{14}Hz$

Converting this into tera Hertz, we use the conversion factor:

$1THz=1\times 10^{12}Hz$

So, $0.339\times 10^{14}Hz\times \frac{1THz}{1\times 10^{12}Hz}=33.9THz$

Hence, the frequency of the radiation is 33.9 THz

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

Answer from: Quest

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.

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