# Solving question

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You have to do each question every step. You have to show your work step by step. Don't miss any part. I don't have to time do it. Thats why i put here. This homework is effecting me %65 of my grade.

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TutorElice
School: New York University

Hi, here is a copy of the assignment that your requested me to assist you with, in case of any clarification please fell free to contact me through the appropriate channel.

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1

CHEMISTRY HOMEWORK

Express 450nm
a) In angstrom;
1 nm= 10 angstrom
450nm=?
Therefore, 450nmx10 angstrom/1nm
= 4500 angstrom
b) As frequency in, Hz
We can get Frequency by dividing the speed of light by wavelength
Whereby the speed of light=3x 108 ms-1 in vacuum,
Therefore 1m=1x109 nm
?=450nm
450nmx 1m/109 =4.5x10-7 m
Frequency=3x108 ms-1 /4.5x10-7m
=6.667x1014 HZ
c) As a wavenumber
Wavenumber is equal to one divided by wavelength
Therefore, wavenumber=1/4.5x10-7 m
=2,222,222.222m-1
D) As energy in kilojoules per mole
Energy is equal to Planck constant multiply by the frequency
Hence energy=6.63 × 10-34 J x 6.667x1014 HZ
=4.42x10-19 J
But one mole is equal to the 6.022×1023 particles (Gibbs free energy)

2

CHEMISTRY HOMEWORK

3

Therefore 4.42x10-22 J/6.022x1023 moles
=7.34x10-46 KJ/mole
E) In what region of electromagnetic spectrum does 450nm falls?
It falls visible which has a range of between 400nm and 700nm
8. If 10 mL of solution A was mixed with 10 mL of solution B which wavelength should be
used to measure the absorbance in solution B? Justify your answer? The relationship between
absorbance and concentration is determine using the beer lambert law whereby extinction
coefficient=absorbance/ (length x molar concentration)
Since the extinction coefficient is 5x106 L (mol.cm) at 442nm implying that the wavelength
that can be used to measure the absorbance of solution B is 442nm.
b) The analyst found, out that when it was measured in appropriate wavelength solution A
had absorbance of 0.2if solution A was cobalt nitrate, Co (NO 3 )2 , determine its concentration
in mg L-1
M(Co (NO 3 )2 )=18.9 gmol-1 1 mM=10-3 M
Absorbance= extinction coefficient x concentration x length
0.2=18.9gmol-1M-1 x10-3 x concentration
Concentration= (0.2)/(18.9x 10-3 )gmol-1
=10.58gmol-1
c) In another mixture pink compound in solution A and green compound in solution B each
have concentrati...

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Anonymous
Good stuff. Would use again.

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