Monday 30 November 2015
Friday 27 November 2015
Thursday 26 November 2015
Friday 20 November 2015
Q.No.33:How can we recognise systems that are aromatic ?
Reasoning Question Series:
Q.No.33:How can we recognise systems that are aromatic ?#Ans: By applying the criteria for aromaticity outlined below.
Q.No.33:How can we recognise systems that are aromatic ?#Ans: By applying the criteria for aromaticity outlined below.
Based on the properties of aromatic compounds, there are FOUR criteria about the π system that need to be met inorder for the "special" aromatic stabilisation to be observed:
#Conjugated : there needs to one "p" orbital from each atom in the ring, so each atom must be either sp2 or sp hybridised.
#Cyclic : linear systems are not aromatic, all atoms in the ring must be involved in the π system (i.e. no sp3 atoms)
#Planar : if the ring is planar flat then this means there is good overlap / interaction between the "p" orbitals....not always easy to consider.
#The Huckel Rule..... 4n+2 π electrons in the cyclic conjugated π system (n = 0, 1, 2, 3 etc.) This is equivalent to an odd number of π-electrons pairs).
Malik Xufyan
#Conjugated : there needs to one "p" orbital from each atom in the ring, so each atom must be either sp2 or sp hybridised.
#Cyclic : linear systems are not aromatic, all atoms in the ring must be involved in the π system (i.e. no sp3 atoms)
#Planar : if the ring is planar flat then this means there is good overlap / interaction between the "p" orbitals....not always easy to consider.
#The Huckel Rule..... 4n+2 π electrons in the cyclic conjugated π system (n = 0, 1, 2, 3 etc.) This is equivalent to an odd number of π-electrons pairs).
Malik Xufyan
Q.32: Why SiH4 bond length is less than SiF4 bond length.?
Reasoning Question Series :
Q.34: Why SiH4 bond length is less than SiF4 bond length.?
Ans:
#1. First All we wil Know that Bond length
bond length or bond distance is the average distance between nuclei of two bonded atoms in a molecule.
#2.As we Know That both are Sp3 Hybridized.
but bond length depend upon the forces of attraction between Two Atoms.
as number of share electron increases , Bond length decreases.
#3. Due to electronegaive difference between the Si and Hydrogen,
Si and H have lower electronegative difference : Silicon has 1.9 and Hydrogen is 2.1.
while Si and F have higher Electronegative difference .
3. Thats why Si attracts Four Hydrogen strongly Than Flourine.
then SiH4 Bond Length is less than SiF6 Bond Length.
.if u know another reason .plz comment Below
Regards
Malik Xufyan
Thursday 19 November 2015
Q.No 31: How can electricity be used to make a magnet?
Reasoning Question Series
Q.No 31: How can electricity be used to make a magnet?
Ans:
u can be learn in urdu
Q.No 31: How can electricity be used to make a magnet?
Ans:
u can be learn in urdu
Regards Malik Xufyan
Tuesday 17 November 2015
Sunday 15 November 2015
Saturday 14 November 2015
First year Chemistry - Pairing scheme (2016)
Pairing Scheme First Year Chemistry 1st year (2016)
MCQs===>
2 Question from
CH# 1,3,4,5,6 & 8
One Question from
CH# 2,7,9,10 & 11
Short Questions ====>
Q#2 ===>(8/12)
CH# 1(3)
CH# 2(2)
CH# 7(2)
CH# 8(3)
Q#3 ===>(8/12)
CH# 4(2+2)
CH# 5(4)
CH#6(4)
Q#4 =====>(6/9)
CH# 9(3)
OH# 10(1)
CH#1.1(3)
Long Questions ===>
Q# 5 (1+4)
Q#6 (3+5)
Q#7 (6+7)
Q#8 (8+11)
Q#9 (9+10)
Practicals ===>
Practicals are included in theory
Q# 10 &11
--->physical Experiment(Common Ion
effect, Enthalpy, Crystallization etc.)
---->Chromatography
---> Titration from each type (acid -base,
lodometic. Redox Titrations)
Malik Xufyan
Second Year Chemistry -Pairing Scheme (2016)
F.sc 2nd year chemistry
Pairing scheme
Question 1 MCQs
1 from each chaptr an 2 from chptr 12
Q.2 -Short questions
2 from each chaptr 1, 4,7,10, 13, 15
Q.3- Short questions
2 from each chaptr 2,5,8,11,14,16
Q.4- Short questions
2 from each chaptr 3,6,9 and 3 from chaptr 12
Q.5- Long question 2+8
Q.6- Long question 3+9
Q.7- Long question 4+10
Q.8- Long question 5+11
Q.9- Long question 7+12
PRACTICAL PART
A.
Write the scheme detectiön of basic radical
(Group I, II,III, IV)
B.
Write the scheme for detection of basic radical
( Group V, VI)
C.
a- write the scheme of acid radical.
b. By name test, palm test, ring test e.g.
D.
Detection of functional group/ detection of element.
E.
Preparatiön any one. Iodoform, glucosazone, copperamine complex, aspirine.
3rd chapter exersise short question - 1st year chemistry
(b) What is the physical significance of
van der Waals constant, ‘a’ and ‘b’.
Give their units.
Ans. (b) Physical Significance
of van der Waals constant ‘a’ and ‘b’
(i) Significance of’ a’:
The value of constant ‘a’ is a measure of the intermolecular attractive forces
and greater will be the ease of its liquefaction.
Units of ‘a’ :
The units of ‘a’ are related to the units of pressure, volume and number of
moles.
P=
or
a= =
a= atm dm6 mol-2
In SI
units:
a= ==Nm4 mole -2
(iii)
Significance of ‘b’: The value of
constant ‘b’ us related to the size of the molecule. Larger the size of the
molecule, lager is the value of ‘b’. It is effective volume of the gas
molecules.
Units of ‘b’:
‘b’ is the compressible volume
per mole of gas. So the units of ‘b’ are related to the units of volume and
moles.
V=nb
or b =
b==dm3
mol-1
In SI
units: b===dm3 mol-1
Q. Do you think that 1 mole of H2
and 1 mole NH3 at 0oC and 1 atm pressure will have
equal Avogadro’s number of particles? If not, why?
Justify
that 1 cm3 of H2 and 1 cm3 of CH4 at
STP will have the same number of molecules, when one molecules of CH4 is
8 times heavier than that of hydrogen.
Ans 1 mole of H2 and 1 mole of
NH3 at 0oC and 1 atm pressure will have equal number of
molecules under the same conditions of temperature and pressure. Hence, 1 cm3
of H2 and 1 cm3 of CH4 at STP will have the same number of molecules.
Q15. Explain the following
facts:
(a) The plot of PV versus P is a
straight line at constant temperature and with a fixed number of moles of an
ideal gas.
ans:
At constant temperature and with a fixed number of moles of an ideal gas, when the pressure of the gas is varied, its volume changes, but the product PV remains constant. Thus,
ans:
At constant temperature and with a fixed number of moles of an ideal gas, when the pressure of the gas is varied, its volume changes, but the product PV remains constant. Thus,
P1 V1 = P2 V2 = P3 V3
=
Hence, for any fixed temperature, the product PV when plotted against P.a
straight line parallel to P-axis is obtained. This straight line indicates that
PV remains constant quantity.
(b) The straight line in (a) is parallel to
pressure-axis and goes away from the pressure axis at higher pressure for many
gases.
ans:
ans:
Now, increase the temperature of the same from T1 to
T2 .At
constant temperature T2 and with the same fixed
number of moles
of an ideal gas, when the constant. However, the value of PV
increase with increase in temperature. On plotting graph
between P
on x-axis is obtained. This straight line at T2 will
be away from the
x-axis. This straight line also shows that PV is a constant
quantity.
(c) The van der walls constant ‘b’ of
a gas is four times the molar volume of that gas
Excluded volume, ‘b’ is four times the molar volume fo gas. The excluding with each other as shown in Fig. The spheres are considered to be non-compressible. So the molecules cannot approach each other more closely than the distance, 2r . Therefore, the space indicated by the dotted sphere having radius, 2r will not be available to all other molecules of the gas. In other words, the dotted spherical space is excluded volume per pair of molecules.
Excluded volume, ‘b’ is four times the molar volume fo gas. The excluding with each other as shown in Fig. The spheres are considered to be non-compressible. So the molecules cannot approach each other more closely than the distance, 2r . Therefore, the space indicated by the dotted sphere having radius, 2r will not be available to all other molecules of the gas. In other words, the dotted spherical space is excluded volume per pair of molecules.
Let each molecules be a sphere with
radius =r
Volume
of one molecules (volume of
sphere)= 
The distance of the closest approach of 2
molecules =2 r
The
excluded volume for 2 molecules=
The excluded volume for 1 molecule=
=
=4Vm =b
The excluded volume for ‘n’ molecules=n b
Where Vm is the actual volume of a molecule.
Hence, the excluded volume or co-volume or non-compressible volume is equal to
4 times the actual volume of the molecules of the gas.
(d) Pressure of NH3
gas at given conditions (say 1 atm pressure and room temperature) is less
as calculated by van der Waals equation than that calculated by general gas
equation.
ans:
ans:
The pressure of NH3 calculated by general gas
equation is high
because it is considered as an ideal gas. In an ideal gas,
the
molecules do not exert any force of attraction on one
another. On
the other hand, when the pressure of NH3 is
considered as a real
gas. Actually, NH3 is a real gas. It consists of
polar NH3 molecules
approaches the walls of the container, it experiences an
inward
pull. Clearly, the molecule strikes the wall with a lesser
force than
it would have done it these are no attractive forces. As a
result of
this , the real gas pressure is less than the ideal
pressure.
(e) Water vapors do not behave ideally
at 273 k.
ans:
Water vapors present at 273K do not behave ideally because polar
ans:
Water vapors present at 273K do not behave ideally because polar
water molecules exert force of attraction on one another.
(f) SO2 is comparatively
non-ideal at 273 k but behaves ideally at
327 K.
ans:
At low temperature, the molecules of SO2 possess low kinetic energy. They come close to each other. The e intermolecular attractive forces become very high. So, it behave non-ideally at 273K. At high temperature, the molecules of SO2 have high kinetic energy. The molecules are at larger distances from one other another. The intermolecular attractive forces become very weak. So, it behaves ideally at 327K.
ans:
At low temperature, the molecules of SO2 possess low kinetic energy. They come close to each other. The e intermolecular attractive forces become very high. So, it behave non-ideally at 273K. At high temperature, the molecules of SO2 have high kinetic energy. The molecules are at larger distances from one other another. The intermolecular attractive forces become very weak. So, it behaves ideally at 327K.
Thursday 12 November 2015
1st year chemistry- 4th Chapter exercise short questions answers
1st year chemistry- 4th Chapter exercise short questions answers
Q.5. Explain the following with reasons.
Q.5. Explain the following with reasons.
a) In
the hydrogen bonded structure of HF, which is the stronger bond: the shorter
covalent bond
or
the longer hydrogen bond between different molecules.
ans:
The hydrogen bonded structure of HF is shown in figure. The fluorine has maximum electro negativity.The shorter covalent bond. H-F is shorter than longer hydrogen bond H---F.
ans:
The hydrogen bonded structure of HF is shown in figure. The fluorine has maximum electro negativity.The shorter covalent bond. H-F is shorter than longer hydrogen bond H---F.
b) In
a very cold winter the fish in garden ponds owe their lives to hydrogen
bonding?
ans:
When winter comes, then temperature falls down. The water at 4oC has maximum density. So it goes to the bottom of pond. The surface water freezes into ice. During ice formation the water molecules get regular arrangements due to H-bonds. In these arrangements there are many empty spaces and ice occupies 9% more volume than liquid water. Thus ice has less density and floats on water. Moreover ice is an insulator of heat. So it prevents underneath water from freezing. Therefore fish and other aquatic animals live (survive) under thick blanket of ice.
ans:
When winter comes, then temperature falls down. The water at 4oC has maximum density. So it goes to the bottom of pond. The surface water freezes into ice. During ice formation the water molecules get regular arrangements due to H-bonds. In these arrangements there are many empty spaces and ice occupies 9% more volume than liquid water. Thus ice has less density and floats on water. Moreover ice is an insulator of heat. So it prevents underneath water from freezing. Therefore fish and other aquatic animals live (survive) under thick blanket of ice.
c) Water
and ethanol can mix easily and in all proportions.
ans
Both water and Ethanol (C2H5OH) show H-bonding with each other. It is the reason that they can mix easily in all proportions.
ans
Both water and Ethanol (C2H5OH) show H-bonding with each other. It is the reason that they can mix easily in all proportions.
d) The
origin of the intermolecular forces in water.
Answer:
The
oxygen atom has small size and high electro negativity. There is high electro
negativity difference between oxygen and hydrogen. This high E.N difference
between O-atom and H-atom is the origin of intermolecular forces in water.
Q.14. Explain the following with reasons.
(i) Evaporation causes cooling.
ans : The reason is that during evaporation first of all high energy molecules leave the liquid and low energy molecules are left behind. So temperature of the liquid falls and cooling is produced. To continue the evaporation heat moves from surrounding to the liquid. Thus temperature of surrounding also falls. For example when we put spirit on our hands then spirit evaporates and hand feels cooling.
ans : The reason is that during evaporation first of all high energy molecules leave the liquid and low energy molecules are left behind. So temperature of the liquid falls and cooling is produced. To continue the evaporation heat moves from surrounding to the liquid. Thus temperature of surrounding also falls. For example when we put spirit on our hands then spirit evaporates and hand feels cooling.
(ii) Evaporation takes place at all
temperatures.
ans: Evaporation is a surface process. It takes place at all temperatures. When high energy molecules come at the surface of liquid, they escape out of surface. However by increasing temperature, the average K.E of molecules increases. Hence rate of evaporation increases by increasing the temperature.
ans: Evaporation is a surface process. It takes place at all temperatures. When high energy molecules come at the surface of liquid, they escape out of surface. However by increasing temperature, the average K.E of molecules increases. Hence rate of evaporation increases by increasing the temperature.
(iii) Boiling needs a constant supply of heat.
ans: When we supply heat to a liquid, then K.E of molecules and rate of evaporation increase. At boiling point the K.E of molecules becomes maximum. Thus heat supplied at boiling point is used to break intermolecular forces. So there is no increase in K.E of molecules. Hence temperature remains constant at boiling point. It is the reason that boiling needs a constant supply of heat.
ans: When we supply heat to a liquid, then K.E of molecules and rate of evaporation increase. At boiling point the K.E of molecules becomes maximum. Thus heat supplied at boiling point is used to break intermolecular forces. So there is no increase in K.E of molecules. Hence temperature remains constant at boiling point. It is the reason that boiling needs a constant supply of heat.
(iv) Earthenware vessels keep water cool.
ans: Earthenware vessels have pores in them. Water evaporates from these pores and causes cooling. During evaporation, the escaping molecules get energy from neighbours to overcome intermolecular forces. Thus temperature of remaining water decreases. In the old earthenware’s pores are blocked with dust. So water can not evaporate. Thus they do not keep water cool.
ans: Earthenware vessels have pores in them. Water evaporates from these pores and causes cooling. During evaporation, the escaping molecules get energy from neighbours to overcome intermolecular forces. Thus temperature of remaining water decreases. In the old earthenware’s pores are blocked with dust. So water can not evaporate. Thus they do not keep water cool.
(v) One feels sense of cooling under the
fan after bath.
ans: One feels sense of cooling under the fan after bath. It is due to evaporation process. After a bath some water molecules are present on the body. These molecules get heat from the body and evaporate. Thus body feels cooling.
ans: One feels sense of cooling under the fan after bath. It is due to evaporation process. After a bath some water molecules are present on the body. These molecules get heat from the body and evaporate. Thus body feels cooling.
(vi) Dynamic equilibrium is established
during evaporation of a liquid in a closed vessel at constant temperature.
(vii) The boiling point of water is different
at Murree hills and at Mount Everest.
ans: Atmospheric pressure is different at Murree hills and at Mount Everest. Due to different atmospheric pressure boiling point of water is different at two places. At Murree hills atmospheric pressure is 700 torr and boiling point of water is 98oC. At Mount Everest the atmospheric pressure is further reduced and boiling point of water is 69oC.
ans: Atmospheric pressure is different at Murree hills and at Mount Everest. Due to different atmospheric pressure boiling point of water is different at two places. At Murree hills atmospheric pressure is 700 torr and boiling point of water is 98oC. At Mount Everest the atmospheric pressure is further reduced and boiling point of water is 69oC.
(viii) Vacuum distillation can be used to avoid
decomposition of sensitive liquid.
ans: The distillation which is carried out at low pressure is called Vacuum distillation. It has many advantages.
ans: The distillation which is carried out at low pressure is called Vacuum distillation. It has many advantages.
(i) It decreases the time for distillation.
(ii) It decreases fuel cost for
distillation.
(iii) It prevents decomposition of compounds
For example B.P of glycerine is 290oC. Pressure of 760 torr but at this temperature glycerine decomposes. Hence distillation of glycerine is impossible at 290oC. Therefore its vacuum distillation is done at 120oC at reduced pressure of 50 torr.
For example B.P of glycerine is 290oC. Pressure of 760 torr but at this temperature glycerine decomposes. Hence distillation of glycerine is impossible at 290oC. Therefore its vacuum distillation is done at 120oC at reduced pressure of 50 torr.
(ix) Heat of sublimation of a substance is
greater than that of heat of vaporization.
ans : In sublimation a substance directly changes into vapours. It is two steps process (Solidliquidvapours) on other hand vaporization is single step process. It is the reason heat of sublimation of a substance is greater than that of heat of vaporization.
ans : In sublimation a substance directly changes into vapours. It is two steps process (Solidliquidvapours) on other hand vaporization is single step process. It is the reason heat of sublimation of a substance is greater than that of heat of vaporization.
(x) Heat of sublimation of iodine is very
high as compared to other halogens.
ans: Iodine has biggest atomic size than other halogens (F2, Cl2, Br2). So I2 has high polarizability. Due to high polarizability Iodine has greater London dispersion forces. It is the reason that heat of sublimation of Iodine is very high.
ans: Iodine has biggest atomic size than other halogens (F2, Cl2, Br2). So I2 has high polarizability. Due to high polarizability Iodine has greater London dispersion forces. It is the reason that heat of sublimation of Iodine is very high.
1st year chemistry - 2nd chapter exercise short Question
Q.4.
Why is there a need to crystallize the crude product?
Ans:
When a compound is prepared in laboratory, it may contain impurities. This
impure and un-refined compound is called crude product. It is necessary to
purify the crude product. So there is a need to crystallize the crude product.
Q.5.
A water insoluble organic compound aspirin is prepared by the reaction of
salicylic acid with a mixture of acetic acid and acetic anhydride. How will you
separate the product from the reaction mixture?
Ans:
Aspirin is separated from reaction mixture is added to cold water. The aspirin
forms crystals and other products remain in solution. Finally aspirin is
filtered from water by sintered glass crucible.
Q.6.
A solid organic compound is soluble in water as well as in chloroform. During
its preparation, it remains in aqueous layer. Describe a method to obtain it
from this layer.
Ans: The organic compound can be
extracted by solvent extraction. The aqueous solution of compound is mixed with
carbon tetrachloride (CCl4). The mixture is put into the separating funnel.
Here two layers are formed. The water layer is separated and evaporated to get
the compound.
Q.7. The following figure shows a
developed chromatogram on paper with five spots. (i) Unknown mixture X (ii)
Sample A (iii) Sample B (iv) Sample C (v) Sample D
Find out (i) the composition of
mixture X (ii) which sample is impure & what is its composition.
Ans:
The unknown mixture X contains the components B and C. The sample D is impure.
It contains components A and C. We should know that a pure sample give only one
spot. The samples A and C are pure.
Q.8.
In solvent extraction technique, why repeated extraction using small portions
of solvent are more efficient than using a single extraction but larger volume
of solvent?
Ans:
The solvent extraction technique is based upon distribution coefficient. The
ratio of concentration of a solute dissolved in two immiscible solvents is a
constant. It is called distribution coefficient.
In
repeated extractions, we can get maximum amount of solute from the other
solvent. It is the reason that repeated extractions using small portions of
solvent are more efficient than using a single extraction but larger volume of
solvent.
Wednesday 11 November 2015
Ist year Chemistry : Exercies Short Question Chapter 1 -------- Malik Xufyan
Ist year Chemistry : Exercies Short Question Chapter 1 --------------- Malik Xufyan
Q8: Justify the following statements:(a) 23 g of sodium and 238g of uranium have equal number of atoms in themAns :
(a) 23g of Na =1 mole of Na =6.02x1023 atoms of Na
238g of U =1 mole of U =6.02x1023 atoms of U.
Since equal number of gram atoms(moles) of different elements contain equal number of atoms. Hence, 1 mole (23g ) of sodium and 1 mole (238)g of uranium contain equal number of atoms , i , e ,6.02x1023 atoms.
(b) Mg atom is twice heavier than that of carbon.Ans:
Since the atomic mass of Mg (24) is twice the atomic mass of carbon (12) therefore, Mg atom is twice heavier than that of carbon. Or
Mass of 1 atom of Mg= 24
Mass of 1 atom of C = 12
Since the mass of one atom of Mg is twice the mass of one atom of C, therefore, Mg atom is twice heavier than that of carbon.
(c) 180g of glucose and 342 g of sucrose have the same number of molecules but different number of atoms present in them.Ans:
180 g of glucose = 1 mole of glucose =6.02x1023 molecules of glucose 342 g of sucrose=1mole of sucrose =6.02x1023 molecules of sucrose
Since one mole of different compounds has the same number of molecules.
Therefore 1 mole (180g) of glucose and I mole (342g) of sucrose contain the same number (6.02x1023)of molecules. Because one molecule of glucose, C6H12O6 contains 45 atoms whereas one molecules of glucose, C12 H22 O11 contains 24 atoms. Therefore, 6.02x1023 molecules of glucose contain different atoms as compound to6.02x1023 molecules of sucrose. Hence, 180 g of glucose and 342g og sucrose have the same number of molecules but different number of atoms present in them.
(d) 4.9g of H2 SO4 when completely ionized in water , have equal number of positive and negative charges but the number of positively charged ions are twice the number of negatively charged ions.,Ans: H2 SO4 <------> 2H+ + SO4-2
When one molecules of H2 SO4 completely ionizes in water it produces two H+ ion and one SO ion,. Hydrogen ion carries a unit positive charge whereas SO ion carries a double negative charge. To keep the neutrality, the number of hydrogen are twice than the number of soleplate ions. Similarly the ions produced by complete ionization of 4.8g of H2 SO4 in water will have equal number of positive and negative but the number of positively charged ions are twice the number of negatively charged ions.
(e) One mg of K2 Cr O4 has thrice the number of ions than the number of formula units when ionized in water. Ans:
K2Cr2O7..........> 2k+ + Cr2O7-2
K2 Cr O4 when ionizes in water produces two k+ ions one C O ion. Thus each formula unit of K2 Cr O4produces three ions in solution .Hence one mg of K2 Cr O4 has thrice the number of ion than the number of formula units ionized in water.
(f) Two grams of H2 , 16 g of ch4 and 44g of CO2 occupy separately the volumes of 22.414 dm3 , although the sizes and masses of molecules of three gases are very different from each other.Ans:
2g of H2 =1 mole of H2 =6.02x1023 molecules of H2 at STP =22.414dm3 16g of CH4 =1mole of CH4 =6.02x1023 molecules of CH4 at STP =22.414dm3 144g of CO2 =1mole of CO2 =6.02x1023 molecules of CO2 at STP =22.144dm3
Although H2 , CH4 and CO2 have different masses but they have the same number of moles and molecules . Hence the same number of moles or the same number of molecules of different gases occupy the same volume at STP. Hence 2 g of H2 ,16g of CH4 and 44 g of CO2 occupy the same volume 22.414 dm3 at STP. The masses and the sizes of the molecules do not affect the volumes
Tuesday 10 November 2015
Q.No.30 Why Steel is more Elastic Than Rubber??
Reasoning Question Series
Q.No.30
Why Steel is more Elastic Than Rubber???
Ans:
1. Firstly , we should know about Elasticity and Youngs Modulus
Q.No.30
Why Steel is more Elastic Than Rubber???
Ans:
1. Firstly , we should know about Elasticity and Youngs Modulus
a) Elasticity is the capability of an object to return to its
former shape once a load inducing strain is removed
former shape once a load inducing strain is removed
b) Youngs Modulus = tensile stress /tensile strain
tensile stress = force/ area
tensile strain = change in length / original length
2. So steel has greater youngs modulus than rubber thats why steel can deform easily and restore its original length after stress the steel than rubber.
Comments plz
Malik Xufyan
tensile strain = change in length / original length
2. So steel has greater youngs modulus than rubber thats why steel can deform easily and restore its original length after stress the steel than rubber.
Comments plz
Malik Xufyan
Q.No.29 Why Water is universal solvent ?
Reasoning Question Series
Q.No.29
Why Water is universal solvent ?
Ans:
water does dissolves many compounds than other all liquids thats why it is called excellent and universal solvent.
it is due to the two main properties
1. Polarity nature of water
2. Extensive Nature of Hydrogen Bonding
comments plz
Regards
Malik Xufyan
Q.No.29
Why Water is universal solvent ?
Ans:
water does dissolves many compounds than other all liquids thats why it is called excellent and universal solvent.
it is due to the two main properties
1. Polarity nature of water
2. Extensive Nature of Hydrogen Bonding
comments plz
Regards
Malik Xufyan
Q.No.28 Do the free radicals give reduction and addition reactions ?
Reasoning Question Series
Q.No.28
Do the free radicals give reduction and addition reactions ?
Ans:
Yes ,these are called radical addition reactions
#First Reason:
Addition reactions may occur between radical and non radical or between two radicals.
These reaction can be seen in anti markonikove rule,
in which we discuss
a) initiation
b) propagation
c) fragmentation
Hydrogen bromide to alkene involing radical is example in organic chemistry
self-terminating oxidative radical cyclization is another example of radica addition reactions.
#second reason:
Free radical reactions are redox reactions in which electrons are transferred
Malik Xufyan
Q.No.28
Do the free radicals give reduction and addition reactions ?
Ans:
Yes ,these are called radical addition reactions
#First Reason:
Addition reactions may occur between radical and non radical or between two radicals.
These reaction can be seen in anti markonikove rule,
in which we discuss
a) initiation
b) propagation
c) fragmentation
Hydrogen bromide to alkene involing radical is example in organic chemistry
self-terminating oxidative radical cyclization is another example of radica addition reactions.
#second reason:
Free radical reactions are redox reactions in which electrons are transferred
Malik Xufyan
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