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hjlf
3 years ago
9

which type of chemical bond would be formed between two elements having electron configuration of 1s2 2s2 2p6 3s2 and 1s2 2s2 2p

4​
Chemistry
1 answer:
OLga [1]3 years ago
7 0
The electron configuration
1
s
2
2
s
2
2
p
6
3
s
2
3
p
2
is the element Silicon.
The key to deciphering this is to look at the last bit of information of the electron configuration
3
p
2
.
The '3' informs us that the element is in the 3rd Energy Level or row of the periodic table. The 'p' tells us that the element is found in the p-block which are all of the Groups to the right of the transition metals, columns 13-18. The superscript '2' tells us that the element is found in the 2nd column of the p-block Group 14.
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In an experiment, you measure a solution absorbance of 0.2 with a path length of 1cm. If the molar absorptivity coefficient is 5
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A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.

<h3>What does Beer-Lambert law state?</h3>

The Beer-Lambert law states that for a given material sample, path length and concentration of the sample are directly proportional to the absorbance of the light.

A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, we can calculate the molarity of the solution using the following expression.

A = ε × b × c

c = A / ε × b

c = 0.2 / (59 cm⁻¹ M⁻¹) × 1 cm = 0.003 M

where,

  • A is the absorbance.
  • ε is the path length.
  • b is the molar absorptivity coefficient.
  • c is the molar concentration.

A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.

Learn more about the Beer-Lambert law here: brainly.com/question/12975133

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2 years ago
Please help me with this chemistry
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Answer:

57)a)increase

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7 0
2 years ago
calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?
inessss [21]

<u>Answer:</u>

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}

\Delta E=-1312[0.111-1]KJ mol^{-1}

\Delta E=1166 KJ mol^{-1}

\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}

=193623 \times 10^{-23}  \frac {J}{photon}

\Delta E=1.93623 \times 10^{-18}  \frac {J}{photon}

\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}

h is planck's constant  

c is the speed of light

λ is the wavelength of light  

\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18}  J/photon)}

Wavelength

\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m}  =103 nm (Answer)

<em>Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is </em><u><em>103 nm.</em></u>

7 0
3 years ago
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