Apparent frequency calculation
Source, sound, observer are in same direction
f' = f $ ( $ v-v~^o #_ v-v~^s $ )
if the direction of observer or source changes, then sign changes.
If freq. Appears be double the freq. of source, for a stationary observer, then
v~^s = $v#_2
If for a stationary observer freq. Appears to be half the real freq. Then
v~^s= -v
if the direction of observer or source changes, then sign changes.
If freq. Appears be double the freq. of source, for a stationary observer, then
If for a stationary observer freq. Appears to be half the real freq. Then
v~^s= -v
Simple Harmonic Motion
Velocity of particle executing SHM:
v=ωACosωt = ω√A^2-y^2
Accelaration:
a=-ω^2ASinωt = -ω^2y
Kinetic Energy:
E~^k= $ 1 #_ 2 $ mω^2A^2Cos^2ωt
Potential Energy:
E~^p=$1#_2$mω^2A^2sin^2ωt
Total Energy:
E=2π^2mf^2A^2
Accelaration:
Kinetic Energy:
E~^k= $ 1 #_ 2 $ mω^2A^2Cos^2ωt
Potential Energy:
E~^p=$1#_2$mω^2A^2sin^2ωt
Total Energy:
Conjugate Acid-Base pair
Pair of acid and base that differ by a proton.
Examples:
1. OH^- and O^-^2 # 2. HCO~^3^- and CO~^3^2^-
How to find conjugate base pair:
To form an base, the substance should loose a H+ ion (proton). So, remove one H+ and subtract 1 from the charge.
To form an acid, the substance should recieve a proton. So add an H+ ion.
Examples:
How to find conjugate base pair:
To form an base, the substance should loose a H+ ion (proton). So, remove one H+ and subtract 1 from the charge.
To form an acid, the substance should recieve a proton. So add an H+ ion.
Faraday's laws of Electrolysis
First law:
Amount substance discharged at an electrode during Electrolysis is directly proportional to the quantity of electricity passed.
W α Q
W = ZIt
Z = $ Eq.mass # _ 96,500
Second law:
when same quantity of electricity is passed to solutions of different Electrolytes amount of substances discharged at the electrodes is directly proportional to their equivalet masses.
W~^1 #_ W~^2 $ = $ E~^1 #_ E~^2
Amount substance discharged at an electrode during Electrolysis is directly proportional to the quantity of electricity passed.
W α Q
W = ZIt
Z = $ Eq.mass # _ 96,500
Second law:
when same quantity of electricity is passed to solutions of different Electrolytes amount of substances discharged at the electrodes is directly proportional to their equivalet masses.
Calculation of Number of Pi and Sigma bonds in a Molecule
PI bond is formed by axial overlapping.
SIGMA bond is formed by lateral overlapping
A SIGMA bond alone can exist in a molecule, but PI bond cannot.
How to calculate number of Pi and sigma bonds in molecule.
Every single bond in a molecule is formed by sigma bond.
Every double bond in a molecule is formed by a sigma bond and a Pi bond.
On the basis of this number of sigma and Pi bonds can be calculated.
Eg:CH~^3-CH=CH-CH~^3
Number of Sigma Bonds: 3
Number of π bonds : 1
SIGMA bond is formed by lateral overlapping
A SIGMA bond alone can exist in a molecule, but PI bond cannot.
How to calculate number of Pi and sigma bonds in molecule.
Every single bond in a molecule is formed by sigma bond.
Every double bond in a molecule is formed by a sigma bond and a Pi bond.
On the basis of this number of sigma and Pi bonds can be calculated.
Eg:
Number of Sigma Bonds: 3
Number of π bonds : 1
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