The green light emitted by a stoplight has a wavelength of 525 nm. What is the frequency of this photon? (c = 3.00 × 10⁸ m/s).
1 answer:
The frequency of this photon is 5.94 x .
Solution:
The speed of light has a value of approximately <u>3.00×108 m/s</u>
The wavelength should have units of meters.
The frequency should have units of Hz or s−1. Hz is hertz or reciprocal seconds.
We know the value of the speed of light and we are given the wavelength. All we have to do is rearrange the equation to solve for the frequency:
<u>v = c/λ</u>
= 3.00× m/s/505×
m
= 5.94Xs-1
= <u>5.94X</u><u> Hz</u>
<u />
Frequency describes the number of waves passing through a particular location at a particular time. A class interval frequency is the number of observations that occur in a particular predefined interval. So, for example, if 20 of her ages 5-9 appear in the survey's data, the interval 5-9 has a frequency of 20. The class interval endpoints are the lowest and highest possible values of the variable.
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Here is the full question
Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH solution after the following additions of titrant (total volume of added base given):
a) 10.00 mL
pH = <u> </u>
b) 20.10 mL
pH = <u> </u>
c) 25.00 mL
pH = <u> </u>
<u />
Answer:
pH = 4.81
pH = 10.40
pH = 12.04
Explanation:
a)
Number of moles of butanoic acid
=
= 0.002000 mol
Number of moles of NaOH added
=
= 0.001000 mol
pKa of butanoic acid = - log Ka
= - log ( 1.54 × 10⁻⁵)
= 4.81
Equation for the reaction is expressed as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
The ICE Table is expressed as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
Initial 0.002000 0.001000 0
Change - 0.001000 - 0.001000 + 0.001000
Equilibrium 0.001000 0 0.001000
Total Volume = (20.00 + 10.00 ) mL
= 30.00 mL = 0.03000 L
Concentration of [CH₃CH₂CH₂COOH] =
= 0.03333 M
Concentration of [CH₃CH₂COO⁻] =
= 0.03333 M
By Henderson- Hasselbalch equation
pH = pKa + log
pH = pKa + log
PH = 4.81 + log
pH = 4.81
Thus; the pH of the resulted buffer solution after 10.00 mL of NaOH was added = 4.81
b )
After the equivalence point, we all know that the pH of the solution will now definitely be determined by the excess H⁺
Number of moles of butanoic acid
=
= 0.002000 mol
Number of moles of NaOH added
=
= 0.002010 mol
Following the previous equation of reaction , The ICE Table for this process is as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
Initial 0.002000 0.002010 0
Change - 0.002000 -0.002000 + 0.002000
Equilibrium 0 0.000010 0.002000
We can see here that the base is present in excess;
NOW, number of moles of base present in excess
= ( 0.002010 - 0.002000) mol
= 0.000010 mol
Total Volume = (20.00 + 20.10 ) mL
= 40.10 mL ×
= 0.04010 L
Concentration of acid [OH⁻] =
=
Using the ionic product of water:
where
=
pH = - log
pH = - log []
pH = 10.40
Thus, the pH of the solution after the equivalence point = 10.40
c)
After the equivalence point, pH of the solution is determined by the excess H⁺.
Number of moles of butanoic acid
=
= 0.002000 mol
Number of moles of NaOH added
=
= 0.002500 mol
From our chemical equation; The ICE Table can be illustrated as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
Initial 0.002000 0.002500 0
Change - 0.002000 -0.002000 +0.002000
Equilibrium 0 0.000500 0.002000
Base is present in excess
Number of moles of base present in excess = [ 0.002500 - 0.002000] mol
= 0.000500 mol
Total Volume = ( 20.00 + 25.00 ) mL
= 45.00 mL
= 45.00 ×
= 0.04500 L
Concentration of acid [OH⁻] =
= 0.01111 M
Using the ionic product of water
=
= M
pH = - log
pH = - log []
pH = 12.04
Thus, the pH of the solution after the equivalence point = 12.04