The fact that a molecule vibrates does not in its.pdf
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The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipoles exactly cancel. However, when CO2 undergoes a bending vibration, its dipole moment changes from zero to some non-zero value. This vibration produces a change in dipole moment and is IR active. These vibrational modes are responsible for the \"greenhouse\" effect in which heat radiated from the earth is absorbed (trapped) by CO2 molecules in the atmosphere. The arrows indicate the directions of motion. Vibrations labeled A and B represent the stretching of the chemical bonds, one in a symmetric (A) fashion, in which both C=O bonds lengthen and contract together (in-phase), and the other in an asymmetric (B) fashion, in which one bond shortens while the other lengthens. The asymmetric stretch (B) is infrared active because there is a change in the molecular dipole moment during this vibration. Solution The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipo.
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CCCCCCCCHHHHHHHHH.docx
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octane - unbranched (straight-chain)
4-ethyl-2-methyloctane - branched
ethylcyclohexane - cyclic
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cyclic hydrocarbons.
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Organic chemistry is the chemistry of the
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compounds are called inorganic and are formed
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is almost unique in its ability to form long
chains with other carbon atoms. [Carbon’s
neighbor in the periodic table, silicon, can do
this but rarely does.] These chains with one
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joined to a third, etc., can be branched, ie,
chains of carbon atoms can be attached to
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Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and
therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular
weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =-
114.1/(2*1.14225) =40 kJ
Solution
Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and
therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular
weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =-
114.1/(2*1.14225) =40 kJ.
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It is basically slides of corrosion andr
Corrosion is a natural process that deteriorates materials, commonly metals, due to chemical or electrochemical reactions with their environment. It's a significant concern across various industries, including infrastructure, manufacturing, and transportation. The effects of corrosion can range from minor aesthetic damage to catastrophic structural failure, leading to enormous economic costs and safety hazards.
Several factors influence corrosion, including environmental conditions such as moisture, temperature, pH levels, and the presence of corrosive agents like oxygen, sulfur compounds, and salts. Additionally, the material's composition and microstructure play crucial roles in its susceptibility to corrosion.
To mitigate corrosion and prolong the lifespan of materials, various protection methods are employed:
Barrier Protection: This involves applying coatings or barriers to physically isolate the material from its environment. Common barrier materials include paints, polymer coatings, and enamels. These coatings create a protective layer that prevents corrosive agents from reaching the underlying material.
Cathodic Protection: This method involves making the metal to be protected the cathode of an electrochemical cell, thus reducing its corrosion rate. Cathodic protection can be achieved through sacrificial anodes, where a more reactive metal (such as zinc or magnesium) is connected to the metal to be protected, sacrificing itself to protect the base metal.
Anodic Protection: Conversely, anodic protection works by polarizing the metal to be protected to make it the anode in an electrochemical cell. This method is suitable for metals that exhibit passivity, such as stainless steel. By maintaining the metal in its passive state, its corrosion rate is significantly reduced.
Inhibitors: Corrosion inhibitors are chemicals that are added to the environment surrounding the metal to reduce its corrosion rate. Inhibitors work by adsorbing onto the metal surface, forming a protective layer that blocks corrosive agents from reaching the metal. Common inhibitors include organic compounds, chromates, and phosphates.
Alloying: Alloying involves mixing the base metal with other elements to improve its corrosion resistance. For example, stainless steel contains chromium, which forms a passive oxide layer on the surface, protecting the underlying metal from corrosion.
Design Modification: Sometimes, corrosion can be mitigated through design modifications that minimize exposure to corrosive environments or improve drainage to prevent the accumulation of moisture.
Each protection method has its advantages and limitations, and the choice of method depends on factors such as the material, the environment, cost considerations, and the required durability. In many cases, a combination of protection methods may be employed to provide optimal corrosion resistance.
Overall, effective corrosion protection is essential for maintaining the integrity and longevity of
specrtoscopy.pptx
Dipole moment is represented by the equation μ=q*l, where q is the charge and l is the distance between the charges. A polar bond forms when one atom in a molecule is more electronegative than the other, causing the electrons to be pulled slightly closer to that atom. This gives one end of the bond a partial negative charge and the other end a partial positive charge, creating a separation of charge called a dipole moment. Rotational spectroscopy measures the energies of molecular rotational transitions using electromagnetic radiation. It can only be used on molecules with a dipole moment, as the field exerts a torque. Infrared spectroscopy involves using IR radiation to identify chemicals based on the frequencies at which they absorb infrared light
UV-Vis molecular absorption spectroscopy- BSc-Lect 5.pdf
1. UV-Vis spectroscopy detects electronic transitions in molecules when photons are absorbed, promoting electrons to higher energy states. Transitions involve π or n electrons and occur in the 200-700nm region.
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Unit 1 (Part A).pptx
Addition reactions involve two or more molecules combining to form a larger molecule. There are two main types of addition reactions for compounds with multiple bonds: electrophilic addition and nucleophilic addition. Electrophilic addition reactions are characteristic of alkenes, in which the alkene acts as a nucleophile that reacts with an electrophile. When hydrogen halides are added to unsymmetrical alkenes, the halogen attaches to the carbon with fewer hydrogen atoms according to Markovnikov's rule.
ORGANIC REACTIONS AND THEIR MECHANISMS
Information about different organic reactions and their mechanism of action. It is basically a part of medicinal chemistry.
3.2 molecular fluorescence and phosphorescence spectroscopy
This document discusses molecular fluorescence and phosphorescence spectroscopy. It begins with an introduction to the principles and terms, explaining that fluorescence occurs when emission takes place within 10-8 seconds of absorption, while phosphorescence occurs after more than 10-8 seconds. The document then covers electronic transitions, factors that affect fluorescence and phosphorescence like temperature, pH, and solvent, and instrumentation including components of fluorimeters.
Electrochemistry lecture
The document discusses various topics in electrochemistry including: ionic motion in electrolytic conduction; electrolytes and electrolysis; electrolytic cells; conductance; specific conductance; equivalent conductance; molar conductance; variation of molar conductance with dilution; ionic mobility; Faraday's laws of electrolysis; and Ohm's law as applied to electrolytic conductors. It also describes migration of ions and the factors that influence ionic mobility such as size, charge, hydration, and temperature.
Sem I A) Electronic displacements by Dr Pramod R Padole
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CCCCCCCCHHHHHHHHH.docx
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C
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octane - unbranched (straight-chain)
4-ethyl-2-methyloctane - branched
ethylcyclohexane - cyclic
Figure 1 - Unbranched, branched, and
cyclic hydrocarbons.
Experiment #3 - Hydrocarbons
Introduction
Organic chemistry is the chemistry of the
compounds of carbon. Currently over twenty
million compounds have been reported in the
chemical literature; about 90% of them are
organic, ie they contain carbon. The remaining
compounds are called inorganic and are formed
from the other elements, of which there are
about 100. That carbon so dominates
compound formation is a result of the fact that it
is almost unique in its ability to form long
chains with other carbon atoms. [Carbon’s
neighbor in the periodic table, silicon, can do
this but rarely does.] These chains with one
carbon joined to a second and the second
joined to a third, etc., can be branched, ie,
chains of carbon atoms can be attached to
carbons in the original chain. It is also possible
for one carbon in a chain to become bonded to
another carbon in that chain, resulting in a
closed ring of atoms. We call these compounds
cyclic.
Since there are so many organic
compounds it is fortunate that we can organize
them into various groups that have some
similarity to each other. For example, one large
group of organic compounds is known as the
hydrocarbons because members of this group
contain only carbon and hydrogen and no other
elements. Figure 1 shows examples of branched, unbranched and cyclic hydrocarbons.
It is possible to subdivide the hydrocarbon group of compounds based on the
bonding between the carbons. If all the carbon-carbon bonds are single, the compound
is an alkane. If at least one of the carbon-carbon bonds in the compound is a double
bond, and the remaining carbon-carbon bonds are single, the compound is an alkene. If
at least one of the carbon-carbon bonds in the compound is a triple bond, and the
remaining carbon-carbon bonds are single, the compound is an alkyne. If the compound
contains a six carbon ring that has alternating double and single bonds around the ring
(three double and three single), we say that ring, and the compound, is aromatic. An
aromatic ring looks like an alkene with three double bonds because that’s the way we
draw it using Lewis structures. However, the actual bonding in such a ring is
considerably different from that in alkenes and, consequently, many of the chemical
properties are different also. Therefore, we place these compounds in a separate family.
By the way, the term aromatic as used here has nothing to do with fragrance.
Experiment #3 Hydrocarbons Page 2
Hydrocarbons may be saturated or unsaturated. A saturated hydrocarbon is one
that is maxed out in terms of the number of hydrogens that can be present given the
nu ...
Chemistry zimsec chapter 20 lattice energy
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2) Born-Haber cycles allow calculation of lattice energy by considering standard enthalpy changes in the step-wise formation of an ionic solid from its elements.
3) Ion polarization occurs when cation charge density and anion size lead to distortion of the anion electron cloud, influencing thermal stability.
Emission spectrum of hydrogen
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2. Bohr postulated that electrons orbit in stable, quantized energy levels and emit or absorb photons of specific frequencies when transitioning between levels.
3. Bohr's model accounted for the Rydberg formula and emission spectrum of hydrogen and was later extended to ions of other elements.
1.b IR spec.ppt.........................
Infrared spectroscopy measures the absorption of infrared radiation by molecules to determine their structure. Infrared radiation excites molecular vibrations in covalent bonds. For a vibration to be infrared active, it must induce a change in the dipole moment of the molecule.
Infrared spectroscopy instruments consist of an infrared radiation source, a sample holder, a detector, and a readout system. Common infrared radiation sources include incandescent lamps, Nernst glowers, and globars. Monochromators like prism and grating monochromators are used to select specific wavelengths from the radiation source.
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Internal energy ok1294990369
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Types of bonds
This document discusses different types of molecular bonds including ionic and covalent bonds. It explains that ionic bonds form when ions interact via electrostatic attraction, as seen in NaCl where Na donates an electron to Cl. Covalent bonds form when electrons are shared between atoms, as in H2, O2, and CO where the bonding cannot be explained by ion separation. The document uses scientific models and concepts like ionization energy, electron affinity, and Coulomb force to describe bond formation and dissociation.
Unit 1 State of matter.pptx
Forces of attraction determine the different states of matter. Intramolecular forces like ionic bonds and covalent bonds occur within molecules. Intermolecular forces like van der Waals forces, hydrogen bonding, and London dispersion forces occur between molecules. These intermolecular forces are responsible for solids, liquids, and gases having distinct properties based on the strength of attraction between molecules. In solids, strong intermolecular forces create a rigid structure. In liquids, weaker intermolecular forces allow molecules to flow freely past one another. In gases, intermolecular forces are very weak, causing molecules to move randomly in all directions.
Wave mechanics
The document discusses wave-particle duality and Louis de Broglie's hypothesis that all matter has both wave-like and particle-like properties. It summarizes key experiments that supported this idea, including Davisson and Germer's 1927 experiment in which electron beams were diffracted by crystal lattices, demonstrating their wave-like behavior. The document also explains how de Broglie's hypothesis resolved issues with early atomic models by introducing the concept of electron standing waves within atoms.
Vibrational Spectrroscopy
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IR spectroscopy . P.K.Mani, BCKV, West Bengal, India
This document provides an introduction to infrared (IR) spectroscopy, including:
1. IR spectra originate from the vibrational and rotational motions of molecules, which can absorb IR radiation if there is a change in dipole moment.
2. Molecules absorb specific frequencies that correspond to their natural vibrational frequencies. Stretching and bending vibrations within different functional groups absorb in characteristic regions of the IR spectrum.
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A2 Chemistry Unit 5
My notes for A2 Chemistry Unit 5, typed by me and compiled from various sources.
I cannot trace back where everything came from but again shall any intellectual property rights be violated, please comment /contact me and I will try my best to rectify them as soon as possible.
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Balancing ANY chemical equation is done exactly t.pdf
Balancing ANY chemical equation is done exactly the same. Easiest way is to start
with a primary element such as Carbon, then just work your way back and forth between
products and reactants until everything is balanced. Be sure to check your equation once you\'re
done to make sure everything is balanced completely. Types of Chemical reactions: 1.)
Combustion: A combustion reaction is when oxygen combines with another compound to form
water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An
example of this kind of reaction is the burning of propane: C3H8 + 5O2 --> 3CO2 + 4H2O
*Hint: Combustion reactions always have O2 in the reactants and CO2 and H2O somewhere in
the products. 2.) Synthesis: A synthesis reaction is when two or more simple compounds
combine to form a more complicated one. These reactions come in the general form of: A + B --
-> AB One example of a synthesis reaction is the combination of iron and sulfur to form iron
(II) sulfide: 8 Fe + S8 ---> 8 FeS 3.) Decomposition: A decomposition reaction is the opposite
of a synthesis reaction - a complex molecule breaks down to make simpler ones. These reactions
come in the general form: AB ---> A + B One example of a decomposition reaction is the
electrolysis of water to make oxygen and hydrogen gas: 2 H2O ---> 2 H2 + O2 4.) Single
displacement: This is when one element trades places with another element in a compound.
These reactions come in the general form of: A + BC ---> AC + B One example of a single
displacement reaction is when magnesium replaces hydrogen in water to make magnesium
hydroxide and hydrogen gas: Mg + 2 H2O ---> Mg(OH)2 + H2 5) Double displacement: This is
when the anions and cations of two different molecules switch places, forming two entirely
different compounds. These reactions are in the general form: AB + CD ---> AD + CB One
example of a double displacement reaction is the reaction of lead (II) nitrate with potassium
iodide to form lead (II) iodide and potassium nitrate: Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3 6)
Acid-base: This is a special kind of double displacement reaction that takes place when an acid
and base react with each other. The H+ ion in the acid reacts with the OH- ion in the base,
causing the formation of water. Generally, the product of this reaction is some ionic salt and
water: HA + BOH ---> H2O + BA One example of an acid-base reaction is the reaction of
hydrobromic acid (HBr) with sodium hydroxide: HBr + NaOH ---> NaBr + H2O *Hint: An acid
plus a base always produces a salt and H2O as a product.
Solution
Balancing ANY chemical equation is done exactly the same. Easiest way is to start
with a primary element such as Carbon, then just work your way back and forth between
products and reactants until everything is balanced. Be sure to check your equation once you\'re
done to make sure everything is balanced completely. Types of Chemical reactions: 1.)
Combustion: A combustion reaction is when oxygen combines with an.
36.Kovacs reagent is used in Indole test. Kovacs reagent is 4 (p)-.pdf
Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and
therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular
weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =-
114.1/(2*1.14225) =40 kJ
Solution
Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and
therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular
weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =-
114.1/(2*1.14225) =40 kJ.
10.Real number1.Baseb. Consists of a set and rule for combining2.Bin.pdf
pH = pKa + log ( [ HEPES] / [HEPES protonated] 7.8 = 7.47 + log ( [HEPES] /
[HEPES protonated]) 7.8-7.47 = log ( [HEPES] / [HEPES protonated]) [HEPES] / [HEPES
protonated] = 10 ^ 0.33 = 2.138 [HEPES] = 2.138 * [HEPES protonated] Given [HEPES] =
0.025M [HEPES protonated] = 0.025/2.138 =0.0117M For preparation of solution amount of
HEPES = 0.0117+0.025 = 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required
0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required has to be used
Solution
pH = pKa + log ( [ HEPES] / [HEPES protonated] 7.8 = 7.47 + log ( [HEPES] /
[HEPES protonated]) 7.8-7.47 = log ( [HEPES] / [HEPES protonated]) [HEPES] / [HEPES
protonated] = 10 ^ 0.33 = 2.138 [HEPES] = 2.138 * [HEPES protonated] Given [HEPES] =
0.025M [HEPES protonated] = 0.025/2.138 =0.0117M For preparation of solution amount of
HEPES = 0.0117+0.025 = 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required
0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required has to be used.
1. Yeasts grow by budding. The cell buds and separates into 2 cells..pdf
O-N=O is a resonance structure. In a resonance structure, the atoms themselves do
not move, but the bonds move. In O-O, the elements cannot change to a different element. In
N=C, another atom cannot be added. And in O=N-F, the atoms cannot be moved around.
Solution
O-N=O is a resonance structure. In a resonance structure, the atoms themselves do
not move, but the bonds move. In O-O, the elements cannot change to a different element. In
N=C, another atom cannot be added. And in O=N-F, the atoms cannot be moved around..
1. 252.125Solution1. 252.125.pdf
Most carbon compounds are poor conductors of electricity as we have seen . From
the data on the boiling and melting points of the above compounds, we can conclude that the
forces of attraction between these molecules are not very strong. Since these compounds are
largely nonconductors of electricity, we can conclude that the bonding in these compounds does
not give rise to any ions.
Solution
Most carbon compounds are poor conductors of electricity as we have seen . From
the data on the boiling and melting points of the above compounds, we can conclude that the
forces of attraction between these molecules are not very strong. Since these compounds are
largely nonconductors of electricity, we can conclude that the bonding in these compounds does
not give rise to any ions..
(B) 0.815Solution(B) 0.815.pdf
It\'s due to the reverse of the formation reaction of CdS CdCl2 + H2S(g) <--> CdS
+ 2HCl(aq) if we make the HCl strong enough we can force the equilibrium the other way, and
CdCl2 is soluble.
Solution
It\'s due to the reverse of the formation reaction of CdS CdCl2 + H2S(g) <--> CdS
+ 2HCl(aq) if we make the HCl strong enough we can force the equilibrium the other way, and
CdCl2 is soluble..
Suppose AFnSolution Suppose AFn.pdf
Has Bond in the product that are weaker than the reactants
Solution
Has Bond in the product that are weaker than the reactants.
C code on linked list #include stdio.h #include stdlib.h.pdf
The C code defines functions to create and manipulate a linked list. It takes 5 user-input values, inserts them into an empty linked list, prints the list, takes another input value and inserts it, then prints the final list.
Definition of Log-Normal DistributionA statistical distr.pdf
e =mc^2. m is the defected mass, or mass that isn\'t there and is instead used as
binding energy. For Mg, you have 25 nucleons: 12 protons and 13 neutrons. So your expected
mass would be 1.007825x12 + 1.008665x13 = 25.206545. The defected mass is the supposed
mass minus the actual mass. so its 25.206545-24.985839= .220706 amu. Then you have to
convert this into kilograms. so, divide 220706 amu by 1 kg = 6.022 x 10^26 and you get:
3.664996*10^-28
Solution
e =mc^2. m is the defected mass, or mass that isn\'t there and is instead used as
binding energy. For Mg, you have 25 nucleons: 12 protons and 13 neutrons. So your expected
mass would be 1.007825x12 + 1.008665x13 = 25.206545. The defected mass is the supposed
mass minus the actual mass. so its 25.206545-24.985839= .220706 amu. Then you have to
convert this into kilograms. so, divide 220706 amu by 1 kg = 6.022 x 10^26 and you get:
3.664996*10^-28.
Step1 NaOH (aq) ----- Na(+)(aq) + Cl(-)(aq) Ste.pdf
calculate the mol of NaOH so mol= concentration * volume n=C x V n= 0.175 x
0.0206 = 0.0036 then you get the ratio of acid to base from the equation above which is 1:1 so
the number of mol of base is equal to the number of mol of acid n(acid)=0.0036 the
concentration of that which is C=n/V C= 0.0036/0.025 C=0.144M
Solution
calculate the mol of NaOH so mol= concentration * volume n=C x V n= 0.175 x
0.0206 = 0.0036 then you get the ratio of acid to base from the equation above which is 1:1 so
the number of mol of base is equal to the number of mol of acid n(acid)=0.0036 the
concentration of that which is C=n/V C= 0.0036/0.025 C=0.144M.
The compounds with low oxidation states (O.S.) be.pdf
The oxidation states of p-block main group elements like groups 13, 14, 15 and 16 become more stable at lower oxidation states further down the column. For group 14 elements C, Si, and Ge, the +4 oxidation state is most stable while the +2 state requires kinetic stabilization. For Sn, both +2 and +4 states are important but +4 is more common, and for Pb, the +2 state is most stable due to the inert pair effect caused by relativistic contraction of the s orbital. For transition metals, higher oxidation states also become less favorable down the column as ionization energies increase.
A) ionic compounds generally formed between the c.pdf
A) ionic compounds generally formed between the combination of 1st and 2nd
group elements with 7th group elements . ex: NaCl, MgF, CaCl2 and so on B) Molecular
compounds means vanderwall forces acting btw the molecule these may be crystals or liguids .ex
Iodine(I2) is molecular crystal ,Quartz, phosphorous(P4), sulphur(S8) C) hydrocarbons in
general contain carbon and hydrogen in some cases hetero atoms may present in the compound .
ex CH4 , C2H6, C2H2, C6H5N02, C6H12O6 (glucose) etc D) in general 5th, 6th, 7th groups
oxides , oxy acids behaves as acid . ex: N20, N205, F20,HNO3, HClO4, HCl, HF, H2SO4,
H2S208 etc along with these metal with lower oxidation state can also act as acidic compound
ex: MnO MnO2 MnO3 in the above three oxides oxidation states of metals in orderly +2,+4,+6.
so first compound can act as an acid last one is a base
Solution
A) ionic compounds generally formed between the combination of 1st and 2nd
group elements with 7th group elements . ex: NaCl, MgF, CaCl2 and so on B) Molecular
compounds means vanderwall forces acting btw the molecule these may be crystals or liguids .ex
Iodine(I2) is molecular crystal ,Quartz, phosphorous(P4), sulphur(S8) C) hydrocarbons in
general contain carbon and hydrogen in some cases hetero atoms may present in the compound .
ex CH4 , C2H6, C2H2, C6H5N02, C6H12O6 (glucose) etc D) in general 5th, 6th, 7th groups
oxides , oxy acids behaves as acid . ex: N20, N205, F20,HNO3, HClO4, HCl, HF, H2SO4,
H2S208 etc along with these metal with lower oxidation state can also act as acidic compound
ex: MnO MnO2 MnO3 in the above three oxides oxidation states of metals in orderly +2,+4,+6.
so first compound can act as an acid last one is a base.
A ethers ethers contain R-O-R linkage not carbony.pdf
A ethers ethers contain R-O-R linkage not carbonyl
Solution
A ethers ethers contain R-O-R linkage not carbonyl.
The compounds of interest are Na2S and H2SO4. Th.pdf
The compounds of interest are Na2S and H2SO4. The molecular equation then is:
Na2S + H2SO4 ---> Na2SO4 + H2S. Just exchange one anion for the other. Now, we need to
know a little bit about the properties of these compounds. Almost all sodium salts are soluble, so
both Na2S and Na2SO4 are fully dissociated: 2 Na+ + S2- + 2 H+ + SO42- ---> 2 Na+ + SO42-
+ H2S. Crossing out terms that appear on both sides, we obtain the net ionic equation: 2 H+ +
S2- ----> H2S. Conveniently, H2S is a gas, and escapes from solution. Thus, it selectively
removes H+ and S2- ions from the equilibrium, driving the reaction to completion (favoring
Na2SO4)
Solution
The compounds of interest are Na2S and H2SO4. The molecular equation then is:
Na2S + H2SO4 ---> Na2SO4 + H2S. Just exchange one anion for the other. Now, we need to
know a little bit about the properties of these compounds. Almost all sodium salts are soluble, so
both Na2S and Na2SO4 are fully dissociated: 2 Na+ + S2- + 2 H+ + SO42- ---> 2 Na+ + SO42-
+ H2S. Crossing out terms that appear on both sides, we obtain the net ionic equation: 2 H+ +
S2- ----> H2S. Conveniently, H2S is a gas, and escapes from solution. Thus, it selectively
removes H+ and S2- ions from the equilibrium, driving the reaction to completion (favoring
Na2SO4).
sucrose has a formula of C12H22O11while ammonia h.pdf
sucrose has a formula of C12H22O11while ammonia has a formula of NH3. NH3 is
polar and has anelectronegative N which can conduct charge.
Solution
sucrose has a formula of C12H22O11while ammonia has a formula of NH3. NH3 is
polar and has anelectronegative N which can conduct charge..
reduction strength Fe Pb As .pdf
The document states that iron has a greater reduction strength than lead, which has a greater reduction strength than arsenic. The reduction strengths from highest to lowest are: Fe > Pb > As.
sp3d since it has 1s 3p and 1d orbitals available.pdf
sp3d since it has 1s 3p and 1d orbitals available for hybridization
Solution
sp3d since it has 1s 3p and 1d orbitals available for hybridization.
No standard potential data given. .pdf
This document does not provide any standard potential data. It simply states that no standard potential data is given in two brief sentences without further context or details.
x2 + 4xSolutionx2 + 4x.pdf
This short document provides the equation x2 + 4x without showing any work or solution steps. It appears to present a quadratic equation but does not give any context or solve the equation.
moles of Al = 1227 Since chlorine is in excess ,.pdf
moles of Al = 12/27 Since chlorine is in excess , Moles of Alcl3 = 12/27 * 1 = 4/9
Solution
moles of Al = 12/27 Since chlorine is in excess , Moles of Alcl3 = 12/27 * 1 = 4/9.
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36.Kovacs reagent is used in Indole test. Kovacs reagent is 4 (p)-.pdf
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10.Real number1.Baseb. Consists of a set and rule for combining2.Bin.pdf
1. Yeasts grow by budding. The cell buds and separates into 2 cells..pdf
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C code on linked list #include stdio.h #include stdlib.h.pdf
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sp3d since it has 1s 3p and 1d orbitals available.pdf
sp3d since it has 1s 3p and 1d orbitals available.pdf
moles of Al = 1227 Since chlorine is in excess ,.pdf
moles of Al = 1227 Since chlorine is in excess ,.pdf
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LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
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- 1. The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipoles exactly cancel. However, when CO2 undergoes a bending vibration, its dipole moment changes from zero to some non-zero value. This vibration produces a change in dipole moment and is IR active. These vibrational modes are responsible for the "greenhouse" effect in which heat radiated from the earth is absorbed (trapped) by CO2 molecules in the atmosphere. The arrows indicate the directions of motion. Vibrations labeled A and B represent the stretching of the chemical bonds, one in a symmetric (A) fashion, in which both C=O bonds lengthen and contract together (in-phase), and the other in an asymmetric (B) fashion, in which one bond shortens while the other lengthens. The asymmetric stretch (B) is infrared active because there is a change in the molecular dipole moment during this vibration. Solution The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipoles exactly cancel. However, when CO2 undergoes a bending vibration, its dipole moment changes from
- 2. zero to some non-zero value. This vibration produces a change in dipole moment and is IR active. These vibrational modes are responsible for the "greenhouse" effect in which heat radiated from the earth is absorbed (trapped) by CO2 molecules in the atmosphere. The arrows indicate the directions of motion. Vibrations labeled A and B represent the stretching of the chemical bonds, one in a symmetric (A) fashion, in which both C=O bonds lengthen and contract together (in-phase), and the other in an asymmetric (B) fashion, in which one bond shortens while the other lengthens. The asymmetric stretch (B) is infrared active because there is a change in the molecular dipole moment during this vibration.