A) ionic compounds generally formed between the c.pdf
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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.
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Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and
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This is a basic concept because there is a pattern to be followed in each type of reaction.
More types of chemical reactions will be given on my next set of presentation entitled, "Everything You Want to Know About Chemical Reactions."
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More from deepua8
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 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 fact that a molecule vibrates does not in its.pdf
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.
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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The compounds with low oxidation states (O.S.) be.pdf
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A ethers ethers contain R-O-R linkage not carbony.pdf
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The fact that a molecule vibrates does not in its.pdf
The fact that a molecule vibrates does not in its.pdf
The compounds of interest are Na2S and H2SO4. Th.pdf
The compounds of interest are Na2S and H2SO4. Th.pdf
sucrose has a formula of C12H22O11while ammonia h.pdf
sucrose has a formula of C12H22O11while ammonia h.pdf
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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A) ionic compounds generally formed between the c.pdf
- 1. 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