A. Li, N, F belong to same periodSolutionA. Li, N, F belong to.pdf
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A. Li, N, F belong to same period Solution A. Li, N, F belong to same period.
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2H2+O2=2H2O2 moles react with one mole of oxygen to give two moles.pdf
2H2+O2=2H2O
2 moles react with one mole of oxygen to give two moles of water
in this case oxygen is sufficient but hydrogen is less
one mole of oxygen remains unreacted
so 2 moles of water are produced
Solution
2H2+O2=2H2O
2 moles react with one mole of oxygen to give two moles of water
in this case oxygen is sufficient but hydrogen is less
one mole of oxygen remains unreacted
so 2 moles of water are produced.
10 = ( 1+ 65200 ) ^n10 = 1.0115 ^ (n)n = 201.37Solu.pdf
10 = ( 1+ 6/5200 ) ^n
10 = 1.0115 ^ (n)
n = 201.37
Solution
10 = ( 1+ 6/5200 ) ^n
10 = 1.0115 ^ (n)
n = 201.37.
1) Sugar because Payments to Labour has major component in Expenses .pdf
1) Sugar because Payments to Labour has major component in Expenses ;
Solution
1) Sugar because Payments to Labour has major component in Expenses ;.
Yellow fever is caused by the yellow fever virus and is spread by th.pdf
Yellow fever is caused by the yellow fever virus and is spread by the bite of an infected female
mosquito. The mosquitoes feeds during the daytime. Chikungunya is caused by the chikungunya
virus. The virus is spread by the female mosquito.
It might be possible they could become established in the future. But mosquito control and
avoiding mosquito bites will control the endemic spread of the infection. Also vaccinations of
the travelers and immigrants will also avoid the spreading of the disease.
Solution
Yellow fever is caused by the yellow fever virus and is spread by the bite of an infected female
mosquito. The mosquitoes feeds during the daytime. Chikungunya is caused by the chikungunya
virus. The virus is spread by the female mosquito.
It might be possible they could become established in the future. But mosquito control and
avoiding mosquito bites will control the endemic spread of the infection. Also vaccinations of
the travelers and immigrants will also avoid the spreading of the disease..
Values are so important in life. The values kids learn at a young ag.pdf
Values are so important in life. The values kids learn at a young age will be fruitful in their
personal and social life.
Solution
Values are so important in life. The values kids learn at a young age will be fruitful in their
personal and social life..
Time Value of Money (TVM) Value of the money does not remain same i.pdf
Time Value of Money (TVM): Value of the money does not remain same it perishes over a span
of time of it its value go down over a period of time there are various factor one of the major
factor is inflation to compensate inflation and risk interest is given on the lending. So when we
take financial decisions like investment etc. we need to keep in mind the present value of the
inflow of the investment if that money will recovered after certain period of time or if it is series
of the inflow over a certain period of time then present value of all such inflows and also we
need to know what will be the future value of the money after certain period of time or if it is
series of inflow then future value of all such money at end of certain period.
The formula to know the present value of the annuities is PV=P((1-(1+r)^(-n))/r) where P is the
annuity, r is the discount rate and n is the number of period in the same manner formula to know
the future value of the annuity is FV =P *( {(((1+R)^N) - 1) / R}, where ) where P is the annuity,
r is the Interest rate and n is the number of period
Solution
Time Value of Money (TVM): Value of the money does not remain same it perishes over a span
of time of it its value go down over a period of time there are various factor one of the major
factor is inflation to compensate inflation and risk interest is given on the lending. So when we
take financial decisions like investment etc. we need to keep in mind the present value of the
inflow of the investment if that money will recovered after certain period of time or if it is series
of the inflow over a certain period of time then present value of all such inflows and also we
need to know what will be the future value of the money after certain period of time or if it is
series of inflow then future value of all such money at end of certain period.
The formula to know the present value of the annuities is PV=P((1-(1+r)^(-n))/r) where P is the
annuity, r is the discount rate and n is the number of period in the same manner formula to know
the future value of the annuity is FV =P *( {(((1+R)^N) - 1) / R}, where ) where P is the annuity,
r is the Interest rate and n is the number of period.
#include stdio.h #include stdlib.h int main() { int l1.pdf
#include
#include
int main() {
int l1[5]={2,4,6,8,10};
int l2[5]={3,5,7,9,11};
int l3[5]={6,10,14,18,22};
printf(\"%d\ \", print_random(l1));
printf(\"%d\ \", print_random(l2));
printf(\"%d\ \", print_random(l3));
return 0;
}
int print_random(int list){
int n;
n = rand() % sizeof(list) + 1;
return list[n];
}
Solution
#include
#include
int main() {
int l1[5]={2,4,6,8,10};
int l2[5]={3,5,7,9,11};
int l3[5]={6,10,14,18,22};
printf(\"%d\ \", print_random(l1));
printf(\"%d\ \", print_random(l2));
printf(\"%d\ \", print_random(l3));
return 0;
}
int print_random(int list){
int n;
n = rand() % sizeof(list) + 1;
return list[n];
}.
Recommended
2H2+O2=2H2O2 moles react with one mole of oxygen to give two moles.pdf
2H2+O2=2H2O
2 moles react with one mole of oxygen to give two moles of water
in this case oxygen is sufficient but hydrogen is less
one mole of oxygen remains unreacted
so 2 moles of water are produced
Solution
2H2+O2=2H2O
2 moles react with one mole of oxygen to give two moles of water
in this case oxygen is sufficient but hydrogen is less
one mole of oxygen remains unreacted
so 2 moles of water are produced.
10 = ( 1+ 65200 ) ^n10 = 1.0115 ^ (n)n = 201.37Solu.pdf
10 = ( 1+ 6/5200 ) ^n
10 = 1.0115 ^ (n)
n = 201.37
Solution
10 = ( 1+ 6/5200 ) ^n
10 = 1.0115 ^ (n)
n = 201.37.
1) Sugar because Payments to Labour has major component in Expenses .pdf
1) Sugar because Payments to Labour has major component in Expenses ;
Solution
1) Sugar because Payments to Labour has major component in Expenses ;.
Yellow fever is caused by the yellow fever virus and is spread by th.pdf
Yellow fever is caused by the yellow fever virus and is spread by the bite of an infected female
mosquito. The mosquitoes feeds during the daytime. Chikungunya is caused by the chikungunya
virus. The virus is spread by the female mosquito.
It might be possible they could become established in the future. But mosquito control and
avoiding mosquito bites will control the endemic spread of the infection. Also vaccinations of
the travelers and immigrants will also avoid the spreading of the disease.
Solution
Yellow fever is caused by the yellow fever virus and is spread by the bite of an infected female
mosquito. The mosquitoes feeds during the daytime. Chikungunya is caused by the chikungunya
virus. The virus is spread by the female mosquito.
It might be possible they could become established in the future. But mosquito control and
avoiding mosquito bites will control the endemic spread of the infection. Also vaccinations of
the travelers and immigrants will also avoid the spreading of the disease..
Values are so important in life. The values kids learn at a young ag.pdf
Values are so important in life. The values kids learn at a young age will be fruitful in their
personal and social life.
Solution
Values are so important in life. The values kids learn at a young age will be fruitful in their
personal and social life..
Time Value of Money (TVM) Value of the money does not remain same i.pdf
Time Value of Money (TVM): Value of the money does not remain same it perishes over a span
of time of it its value go down over a period of time there are various factor one of the major
factor is inflation to compensate inflation and risk interest is given on the lending. So when we
take financial decisions like investment etc. we need to keep in mind the present value of the
inflow of the investment if that money will recovered after certain period of time or if it is series
of the inflow over a certain period of time then present value of all such inflows and also we
need to know what will be the future value of the money after certain period of time or if it is
series of inflow then future value of all such money at end of certain period.
The formula to know the present value of the annuities is PV=P((1-(1+r)^(-n))/r) where P is the
annuity, r is the discount rate and n is the number of period in the same manner formula to know
the future value of the annuity is FV =P *( {(((1+R)^N) - 1) / R}, where ) where P is the annuity,
r is the Interest rate and n is the number of period
Solution
Time Value of Money (TVM): Value of the money does not remain same it perishes over a span
of time of it its value go down over a period of time there are various factor one of the major
factor is inflation to compensate inflation and risk interest is given on the lending. So when we
take financial decisions like investment etc. we need to keep in mind the present value of the
inflow of the investment if that money will recovered after certain period of time or if it is series
of the inflow over a certain period of time then present value of all such inflows and also we
need to know what will be the future value of the money after certain period of time or if it is
series of inflow then future value of all such money at end of certain period.
The formula to know the present value of the annuities is PV=P((1-(1+r)^(-n))/r) where P is the
annuity, r is the discount rate and n is the number of period in the same manner formula to know
the future value of the annuity is FV =P *( {(((1+R)^N) - 1) / R}, where ) where P is the annuity,
r is the Interest rate and n is the number of period.
#include stdio.h #include stdlib.h int main() { int l1.pdf
#include
#include
int main() {
int l1[5]={2,4,6,8,10};
int l2[5]={3,5,7,9,11};
int l3[5]={6,10,14,18,22};
printf(\"%d\ \", print_random(l1));
printf(\"%d\ \", print_random(l2));
printf(\"%d\ \", print_random(l3));
return 0;
}
int print_random(int list){
int n;
n = rand() % sizeof(list) + 1;
return list[n];
}
Solution
#include
#include
int main() {
int l1[5]={2,4,6,8,10};
int l2[5]={3,5,7,9,11};
int l3[5]={6,10,14,18,22};
printf(\"%d\ \", print_random(l1));
printf(\"%d\ \", print_random(l2));
printf(\"%d\ \", print_random(l3));
return 0;
}
int print_random(int list){
int n;
n = rand() % sizeof(list) + 1;
return list[n];
}.
The shape is angular, because of the two pairs of.pdf
The shape is angular, because of the two pairs of nonbonded electrons in O that
bend the molecule.
Solution
The shape is angular, because of the two pairs of nonbonded electrons in O that
bend the molecule..
Step1 Moles of ions from NaCl = 2x.46 = .92 Step2.pdf
Step1 Moles of ions from NaCl = 2x.46 = .92 Step2 Moles of ions from MgCl2 =3
x.034 =.102 Step3 Moles of ions from MgSO4= 2x.019 =.038 Step4 CaSO4 is sparingly soluble
in water ; moles =.009 Step5 Total concentration of solute ions = 1.069 Moles Step6
Solution
being a dilute one ; molality is almost equal to molarity Step7 Osmotic pressure = 1.069 x .082 x
293 = 25.68 atm.
so that the reaction takes place and forms (=0 )b.pdf
so that the reaction takes place and forms (=0 )bond with carbon in the place of oh
group
Solution
so that the reaction takes place and forms (=0 )bond with carbon in the place of oh
group.
PV = nRT in both cases other one is taken consta.pdf
PV = nRT in both cases other one is taken constant
Solution
PV = nRT in both cases other one is taken constant.
In case of SO2 Dipole - Dipole forces are stron.pdf
In case of SO2 : Dipole - Dipole forces are strongest MgCl2 : Ionic bonding forces
between Mg2+ and Cl-
Solution
In case of SO2 : Dipole - Dipole forces are strongest MgCl2 : Ionic bonding forces
between Mg2+ and Cl-.
D) None of the above. first nitration takes place.pdf
D) None of the above. first nitration takes place when u add hno3 and then
sulphonation in ortho position. which has so3H..so either of nitrated product or sulphonated
product wid SO3H is formed..which is not present in options..
Solution
D) None of the above. first nitration takes place when u add hno3 and then
sulphonation in ortho position. which has so3H..so either of nitrated product or sulphonated
product wid SO3H is formed..which is not present in options...
Decrease in Entropy (S) always occurs when the ph.pdf
Decrease in Entropy (S) always occurs when the phase changes from a less ordered
state (such as a gas), to a more ordered state (such as a solid) Decrease in Enthalpy (H) occurs
when a system loses energy. A gas has a high amount of internal energy (think hundreds of ping
pong balls bouncing around in a container), while a solid has the least internal energy (imagine
those same pingpong balls stacked into a lattice). Solid to gas causes an increase in entropy
(more random) and enthalpy (more energy) Solid to liquid still causes both entropy and enthalpy
to increase. Liquid to gas also causes an increase in both entropy and enthalpy for the same
reasons. The reverses of these reactions: Gas to solid, Gas to liquid, and Liquid to solid all have
a decrease in enthalpy and entropy. Example: Gas to Liquid. As a gas cools, it loses energy.
Eventually it will clump together to form a liquid, which decreases its entropy.
Solution
Decrease in Entropy (S) always occurs when the phase changes from a less ordered
state (such as a gas), to a more ordered state (such as a solid) Decrease in Enthalpy (H) occurs
when a system loses energy. A gas has a high amount of internal energy (think hundreds of ping
pong balls bouncing around in a container), while a solid has the least internal energy (imagine
those same pingpong balls stacked into a lattice). Solid to gas causes an increase in entropy
(more random) and enthalpy (more energy) Solid to liquid still causes both entropy and enthalpy
to increase. Liquid to gas also causes an increase in both entropy and enthalpy for the same
reasons. The reverses of these reactions: Gas to solid, Gas to liquid, and Liquid to solid all have
a decrease in enthalpy and entropy. Example: Gas to Liquid. As a gas cools, it loses energy.
Eventually it will clump together to form a liquid, which decreases its entropy..
The diploid genetic system is more intersting than th monoploid gene.pdf
The diploid genetic system is more intersting than th monoploid genetic system due to followinf
reasons:Completely dominant allelesIncompletely dominant allelesCodominant allelesIn
complete dominance, the effect of one allele in heterozygous genotype completely masks the
other. The allele that masks the other is said to be dominant while the other which has been
masked is called recessive.Incomplete dominance is a form of inheritance in which one allele
for a specific trait is not completely expressed over its paired allele. This results in a third
phenotype in which the physical traits expressed are a combination of both alleles.Allels that are
masked or hidden by dominant alleles are known as recessive alleles. In some situations, both
alleles are expressed equally.Example : The flower\'s on Mendel\'s pea plant.Example: Pink
rosesExample: blood group AB
Solution
The diploid genetic system is more intersting than th monoploid genetic system due to followinf
reasons:Completely dominant allelesIncompletely dominant allelesCodominant allelesIn
complete dominance, the effect of one allele in heterozygous genotype completely masks the
other. The allele that masks the other is said to be dominant while the other which has been
masked is called recessive.Incomplete dominance is a form of inheritance in which one allele
for a specific trait is not completely expressed over its paired allele. This results in a third
phenotype in which the physical traits expressed are a combination of both alleles.Allels that are
masked or hidden by dominant alleles are known as recessive alleles. In some situations, both
alleles are expressed equally.Example : The flower\'s on Mendel\'s pea plant.Example: Pink
rosesExample: blood group AB.
Solution If r t= Multiply both the sides by -1 , we have= -.pdf
Solution
: If r Multiply both the sides by -1 , we have
=> -r >-t
So option B is correct.
Step-1Code the playing card and then sort the cards in a deck.Ste.pdf
Step-1:Code the playing card and then sort the cards in a deck.
Step-2:Code the deck of cards then take a card from the deck.
Step-3:use the deck of cards and then rank distribution of cards.
PROGRAM IN VB.NET
Public Class Form1
02
Dim TheDeck As DeckOfCards
03
Dim Hands(NumberOfHands - 1) As HandOfCards
04
Const NumberOfHands As Integer = 10
05
06
07
Private Sub Form1_Paint(ByVal sender As Object, ByVal e
AsSystem.Windows.Forms.PaintEventArgs) Handles Me.Paint
08
If TheDeck IsNot Nothing Then
09
For i = 0 To NumberOfHands - 1
10
Hands(i).DrawHand(e.Graphics, 25, 25 + i * 65)
11
Next
12
End If
13
14
15
End Sub
16
17
18
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As
System.EventArgs)Handles Button1.Click
19
Console.WriteLine(\"--------\")
20
21
22
TheDeck = New DeckOfCards
23
TheDeck.ShuffleCards()
24
TheDeck.ShuffleCards()
25
\'TheDeck.ShuffleCards()
26
\' TheDeck.ShuffleCards()
27
For i = 0 To NumberOfHands - 1
28
Hands(i) = New HandOfCards
29
For c As Integer = 0 To 4
30
Hands(i).AddCard(TheDeck.TakeCard)
31
Next
32
Hands(i).SortHand()
33
Next
34
Array.Sort(Hands, New HandOfCards.HandComparer)
35
For i = 0 To NumberOfHands - 1
36
Console.WriteLine(\"{0}\" & vbTab & \"[{1}] = {2}\" & vbTab & \" {3}\" & vbTab &
\"[{4}]\" & vbTab & \" ({5})\", i + 1, Hands(i).ToString, Hands(i).SuitValueOfHand,
Hands(i).RankDistrubtion, Hands(i).NameOfHand, Hands(i).ValueOfHand)
37
Next
38
\'Console.WriteLine(\"Deck [{0}]\", TheDeck.DeckState)
39
\'Console.WriteLine(\"Deck [{0}]\", TheDeck.DeckState)
40
\'TheDeck.ShuffleCards()
41
\'Console.WriteLine(\"Deck [{0}]\", TheDeck.DeckState)
42
Me.Refresh()
43
End Sub
44
End Class
Public Class Form1
Solution
Step-1:Code the playing card and then sort the cards in a deck.
Step-2:Code the deck of cards then take a card from the deck.
Step-3:use the deck of cards and then rank distribution of cards.
PROGRAM IN VB.NET
Public Class Form1
02
Dim TheDeck As DeckOfCards
03
Dim Hands(NumberOfHands - 1) As HandOfCards
04
Const NumberOfHands As Integer = 10
05
06
07
Private Sub Form1_Paint(ByVal sender As Object, ByVal e
AsSystem.Windows.Forms.PaintEventArgs) Handles Me.Paint
08
If TheDeck IsNot Nothing Then
09
For i = 0 To NumberOfHands - 1
10
Hands(i).DrawHand(e.Graphics, 25, 25 + i * 65)
11
Next
12
End If
13
14
15
End Sub
16
17
18
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As
System.EventArgs)Handles Button1.Click
19
Console.WriteLine(\"--------\")
20
21
22
TheDeck = New DeckOfCards
23
TheDeck.ShuffleCards()
24
TheDeck.ShuffleCards()
25
\'TheDeck.ShuffleCards()
26
\' TheDeck.ShuffleCards()
27
For i = 0 To NumberOfHands - 1
28
Hands(i) = New HandOfCards
29
For c As Integer = 0 To 4
30
Hands(i).AddCard(TheDeck.TakeCard)
31
Next
32
Hands(i).SortHand()
33
Next
34
Array.Sort(Hands, New HandOfCards.HandComparer)
35
For i = 0 To NumberOfHands - 1
36
Console.WriteLine(\"{0}\" & vbTab & \"[{1}] = {2}\" & vbTab & \" {3}\" & vbTab &
\"[{4}]\" & vbTab & \" ({5})\", i + 1, Hands(i).ToString, Han.
conc . of H+ = conc. of H3O + = 0.250.17 =0.0425.pdf
conc . of H+ = conc. of H3O + = 0.25*0.17 =0.0425M pH + pOH = 14 always and
[H+]*[OH-] is 10^-14 conc . of OH- = 10^-14/0.0425 =2.3529*10^-13 M
Solution
conc . of H+ = conc. of H3O + = 0.25*0.17 =0.0425M pH + pOH = 14 always and
[H+]*[OH-] is 10^-14 conc . of OH- = 10^-14/0.0425 =2.3529*10^-13 M.
Reactivity of the metal Tendency to formcation .pdf
Reactivity of the metal Tendency to formcation
Tendency to loose outer elecrtons
E ooxid
1 / E ored
So the Eo cell = E ocathode - Eoanode is more, then themetal is more reactive to reduce the Cu
2+
Given
I. Zn/Zn 2+ || Cu 2+/Cu 1.072 V
II. Fe/Fe 2+ || Cu 2+/Cu 0.334 V
III. Mg/Mg 2+ || Cu 2+/Cu 1.640 V
IV. Pb/Pb 2+ || Cu 2+/Cu 0.459V
V. Sn/Sn 2+ || Cu 2+/Cu 0.600 V
So the order of reactivity is Mg > Zn > Sn > Pb >Fe
So the Eo cell = E ocathode - Eoanode is more, then themetal is more reactive to reduce the Cu
2+
Given
I. Zn/Zn 2+ || Cu 2+/Cu 1.072 V
II. Fe/Fe 2+ || Cu 2+/Cu 0.334 V
III. Mg/Mg 2+ || Cu 2+/Cu 1.640 V
IV. Pb/Pb 2+ || Cu 2+/Cu 0.459V
V. Sn/Sn 2+ || Cu 2+/Cu 0.600 V
So the order of reactivity is Mg > Zn > Sn > Pb >Fe
Solution
Reactivity of the metal Tendency to formcation
Tendency to loose outer elecrtons
E ooxid
1 / E ored
So the Eo cell = E ocathode - Eoanode is more, then themetal is more reactive to reduce the Cu
2+
Given
I. Zn/Zn 2+ || Cu 2+/Cu 1.072 V
II. Fe/Fe 2+ || Cu 2+/Cu 0.334 V
III. Mg/Mg 2+ || Cu 2+/Cu 1.640 V
IV. Pb/Pb 2+ || Cu 2+/Cu 0.459V
V. Sn/Sn 2+ || Cu 2+/Cu 0.600 V
So the order of reactivity is Mg > Zn > Sn > Pb >Fe
So the Eo cell = E ocathode - Eoanode is more, then themetal is more reactive to reduce the Cu
2+
Given
I. Zn/Zn 2+ || Cu 2+/Cu 1.072 V
II. Fe/Fe 2+ || Cu 2+/Cu 0.334 V
III. Mg/Mg 2+ || Cu 2+/Cu 1.640 V
IV. Pb/Pb 2+ || Cu 2+/Cu 0.459V
V. Sn/Sn 2+ || Cu 2+/Cu 0.600 V
So the order of reactivity is Mg > Zn > Sn > Pb >Fe.
Purple sulfur bacteria use H2S as a source of electrons and protons.pdf
Purple sulfur bacteria use H2S as a source of electrons and protons
H2S = S + 2H(+) + 2e-
Solution
Purple sulfur bacteria use H2S as a source of electrons and protons
H2S = S + 2H(+) + 2e-.
Chemical reaction tkes place Zn + CuSO4 = ZnS04.pdf
Chemical reaction tkes place Zn + CuSO4 = ZnS04 +Cu so the blue coloured
copper sulphate solution turns clear
Solution
Chemical reaction tkes place Zn + CuSO4 = ZnS04 +Cu so the blue coloured
copper sulphate solution turns clear.
C. three bonds and no lone pairs .pdf
C. three bonds and no lone pairs
Solution
C. three bonds and no lone pairs.
c) None, all the anions form a precipitate with s.pdf
c) None, all the anions form a precipitate with some of the cations in the table e.g.
Pb2+ forms an insoluble salt with all the anions listed. d) None, all the anions form a soluble salt
with K+, NH4+ for example.
Solution
c) None, all the anions form a precipitate with some of the cations in the table e.g.
Pb2+ forms an insoluble salt with all the anions listed. d) None, all the anions form a soluble salt
with K+, NH4+ for example..
Option B nonsense-mediated decay, using small RNA molecules to trig.pdf
Option B: nonsense-mediated decay, using small RNA molecules to trigger degradation.
this is a eukaryotic translation regulation and not found in bacteria.
Solution
Option B: nonsense-mediated decay, using small RNA molecules to trigger degradation.
this is a eukaryotic translation regulation and not found in bacteria..
Floating point basicsThe core idea of floating-point representatio.pdf
Floating point basics
The core idea of floating-point representations (as opposed to fixed point representations as used
by, say, ints), is that a number x is written as m*be where m is a mantissa or fractional part, b is a
base, and e is an exponent. On modern computers the base is almost always 2, and for most
floating-point representations the mantissa will be scaled to be between 1 and b. This is done by
adjusting the exponent, e.g.
1 = 1*20
2 = 1*21
0.375 = 1.5*2-2
etc.
Iam writing a program that does some floating addition that uses bit patterns with shifts applied
to the mantissa and such to obtain the sum of the two floating point numbers. Logically and on
paper I can get this to compute the correct sum.
Code:
#include
#include
#include
#include
int isNegative (float f)
{
unsigned int* iptr = (unsigned int*)&f;
return ( ((*iptr) & 0x80000000) ? 1:0);
}
unsigned char getExponent (float f)
{
unsigned int* iptr = (unsigned int*)&f;
return (((*iptr >> 23) & 0xff) - 127);
}
unsigned int getMantissa (float f)
{
unsigned int* iptr = (unsigned int*)&f;
if( *iptr == 0 ) return 0;
return ((*iptr & 0xFFFFFF) | 0x800000 );
}
float sum (float left, float right)
{
unsigned int littleMan;
unsigned int bigMan;
unsigned char littleE;
unsigned char bigE;
unsigned char lexp = getExponent(left);
unsigned char rexp = getExponent(right);
int Dexponent;
if (lexp > rexp)
{
bigE = lexp;
bigMan = getMantissa(left);
littleE = rexp;
littleMan = getMantissa(right);
}
else
{
bigE = rexp;
bigMan = getMantissa(right);
littleE = lexp;
littleMan = getMantissa(left);
}
printf(\"little: %x %x\ \", littleE, littleMan);
printf(\"big: %x %x\ \", bigE, bigMan);
void shift( unsigned int *valToShift, int bitsToShift )
{
// Masks is used to mask out bits to check for a \"sticky\" bit.
static unsigned masks[24] =
{
0, 1, 3, 7, 0xf, 0x1f, 0x3f, 0x7f,
0xff, 0x1ff, 0x3ff, 0x7ff, 0xfff, 0x1fff, 0x3fff, 0x7fff,
0xffff, 0x1ffff, 0x3ffff, 0x7ffff, 0xfffff, 0x1fffff, 0x3fffff, 0x7fffff
};
// HOmasks - masks out the H.O. bit of the value masked by the masks entry.
static unsigned HOmasks[24] =
{
0,
1, 2, 4, 0x8, 0x10, 0x20, 0x40, 0x80,
0x100, 0x200, 0x400, 0x800, 0x1000, 0x2000, 0x4000, 0x8000,
0x10000, 0x20000, 0x40000, 0x80000, 0x100000, 0x200000, 0x400000
};
// shiftedOut- Holds the value that will be shifted out of a mantissa
// during the denormalization operation (used to round a denormalized value).
int shiftedOut;
assert( bitsToShift <= 23 );
// Grabs the bits we\'re going to shift out (so we can determine
// how to round this value after the shift).
shiftedOut = *valToShift & masks[ bitsToShift ];
// Shift the value to the right the specified number of bits:
*valToShift = *valToShift >> bitsToShift;
// If necessary, round the value:
if( shiftedOut > HOmasks[ bitsToShift ] )
{
// If the bits we shifted out are greater than 1/2 the L.O. bit, then
// round the value up by one.
*valToShift = *valToShift + 1;
}
else if( shiftedOut == HOmasks[ bitsToShift ] )
{
// If the bits we shif.
F1 females - All wild typeF1 males – All wild typeF2 females – A.pdf
F1 females - All wild type
F1 males – All wild type
F2 females – All wild type
F2 males – 1/2 Wild type and
1/2 Vermilion
Solution
F1 females - All wild type
F1 males – All wild type
F2 females – All wild type
F2 males – 1/2 Wild type and
1/2 Vermilion.
Azure Interview Questions and Answers PDF By ScholarHat
Azure Interview Questions and Answers PDF By ScholarHat
Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
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The shape is angular, because of the two pairs of.pdf
The shape is angular, because of the two pairs of nonbonded electrons in O that
bend the molecule.
Solution
The shape is angular, because of the two pairs of nonbonded electrons in O that
bend the molecule..
Step1 Moles of ions from NaCl = 2x.46 = .92 Step2.pdf
Step1 Moles of ions from NaCl = 2x.46 = .92 Step2 Moles of ions from MgCl2 =3
x.034 =.102 Step3 Moles of ions from MgSO4= 2x.019 =.038 Step4 CaSO4 is sparingly soluble
in water ; moles =.009 Step5 Total concentration of solute ions = 1.069 Moles Step6
Solution
being a dilute one ; molality is almost equal to molarity Step7 Osmotic pressure = 1.069 x .082 x
293 = 25.68 atm.
so that the reaction takes place and forms (=0 )b.pdf
so that the reaction takes place and forms (=0 )bond with carbon in the place of oh
group
Solution
so that the reaction takes place and forms (=0 )bond with carbon in the place of oh
group.
PV = nRT in both cases other one is taken consta.pdf
PV = nRT in both cases other one is taken constant
Solution
PV = nRT in both cases other one is taken constant.
In case of SO2 Dipole - Dipole forces are stron.pdf
In case of SO2 : Dipole - Dipole forces are strongest MgCl2 : Ionic bonding forces
between Mg2+ and Cl-
Solution
In case of SO2 : Dipole - Dipole forces are strongest MgCl2 : Ionic bonding forces
between Mg2+ and Cl-.
D) None of the above. first nitration takes place.pdf
D) None of the above. first nitration takes place when u add hno3 and then
sulphonation in ortho position. which has so3H..so either of nitrated product or sulphonated
product wid SO3H is formed..which is not present in options..
Solution
D) None of the above. first nitration takes place when u add hno3 and then
sulphonation in ortho position. which has so3H..so either of nitrated product or sulphonated
product wid SO3H is formed..which is not present in options...
Decrease in Entropy (S) always occurs when the ph.pdf
Decrease in Entropy (S) always occurs when the phase changes from a less ordered
state (such as a gas), to a more ordered state (such as a solid) Decrease in Enthalpy (H) occurs
when a system loses energy. A gas has a high amount of internal energy (think hundreds of ping
pong balls bouncing around in a container), while a solid has the least internal energy (imagine
those same pingpong balls stacked into a lattice). Solid to gas causes an increase in entropy
(more random) and enthalpy (more energy) Solid to liquid still causes both entropy and enthalpy
to increase. Liquid to gas also causes an increase in both entropy and enthalpy for the same
reasons. The reverses of these reactions: Gas to solid, Gas to liquid, and Liquid to solid all have
a decrease in enthalpy and entropy. Example: Gas to Liquid. As a gas cools, it loses energy.
Eventually it will clump together to form a liquid, which decreases its entropy.
Solution
Decrease in Entropy (S) always occurs when the phase changes from a less ordered
state (such as a gas), to a more ordered state (such as a solid) Decrease in Enthalpy (H) occurs
when a system loses energy. A gas has a high amount of internal energy (think hundreds of ping
pong balls bouncing around in a container), while a solid has the least internal energy (imagine
those same pingpong balls stacked into a lattice). Solid to gas causes an increase in entropy
(more random) and enthalpy (more energy) Solid to liquid still causes both entropy and enthalpy
to increase. Liquid to gas also causes an increase in both entropy and enthalpy for the same
reasons. The reverses of these reactions: Gas to solid, Gas to liquid, and Liquid to solid all have
a decrease in enthalpy and entropy. Example: Gas to Liquid. As a gas cools, it loses energy.
Eventually it will clump together to form a liquid, which decreases its entropy..
The diploid genetic system is more intersting than th monoploid gene.pdf
The diploid genetic system is more intersting than th monoploid genetic system due to followinf
reasons:Completely dominant allelesIncompletely dominant allelesCodominant allelesIn
complete dominance, the effect of one allele in heterozygous genotype completely masks the
other. The allele that masks the other is said to be dominant while the other which has been
masked is called recessive.Incomplete dominance is a form of inheritance in which one allele
for a specific trait is not completely expressed over its paired allele. This results in a third
phenotype in which the physical traits expressed are a combination of both alleles.Allels that are
masked or hidden by dominant alleles are known as recessive alleles. In some situations, both
alleles are expressed equally.Example : The flower\'s on Mendel\'s pea plant.Example: Pink
rosesExample: blood group AB
Solution
The diploid genetic system is more intersting than th monoploid genetic system due to followinf
reasons:Completely dominant allelesIncompletely dominant allelesCodominant allelesIn
complete dominance, the effect of one allele in heterozygous genotype completely masks the
other. The allele that masks the other is said to be dominant while the other which has been
masked is called recessive.Incomplete dominance is a form of inheritance in which one allele
for a specific trait is not completely expressed over its paired allele. This results in a third
phenotype in which the physical traits expressed are a combination of both alleles.Allels that are
masked or hidden by dominant alleles are known as recessive alleles. In some situations, both
alleles are expressed equally.Example : The flower\'s on Mendel\'s pea plant.Example: Pink
rosesExample: blood group AB.
Solution If r t= Multiply both the sides by -1 , we have= -.pdf
Solution
: If r Multiply both the sides by -1 , we have
=> -r >-t
So option B is correct.
Step-1Code the playing card and then sort the cards in a deck.Ste.pdf
Step-1:Code the playing card and then sort the cards in a deck.
Step-2:Code the deck of cards then take a card from the deck.
Step-3:use the deck of cards and then rank distribution of cards.
PROGRAM IN VB.NET
Public Class Form1
02
Dim TheDeck As DeckOfCards
03
Dim Hands(NumberOfHands - 1) As HandOfCards
04
Const NumberOfHands As Integer = 10
05
06
07
Private Sub Form1_Paint(ByVal sender As Object, ByVal e
AsSystem.Windows.Forms.PaintEventArgs) Handles Me.Paint
08
If TheDeck IsNot Nothing Then
09
For i = 0 To NumberOfHands - 1
10
Hands(i).DrawHand(e.Graphics, 25, 25 + i * 65)
11
Next
12
End If
13
14
15
End Sub
16
17
18
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As
System.EventArgs)Handles Button1.Click
19
Console.WriteLine(\"--------\")
20
21
22
TheDeck = New DeckOfCards
23
TheDeck.ShuffleCards()
24
TheDeck.ShuffleCards()
25
\'TheDeck.ShuffleCards()
26
\' TheDeck.ShuffleCards()
27
For i = 0 To NumberOfHands - 1
28
Hands(i) = New HandOfCards
29
For c As Integer = 0 To 4
30
Hands(i).AddCard(TheDeck.TakeCard)
31
Next
32
Hands(i).SortHand()
33
Next
34
Array.Sort(Hands, New HandOfCards.HandComparer)
35
For i = 0 To NumberOfHands - 1
36
Console.WriteLine(\"{0}\" & vbTab & \"[{1}] = {2}\" & vbTab & \" {3}\" & vbTab &
\"[{4}]\" & vbTab & \" ({5})\", i + 1, Hands(i).ToString, Hands(i).SuitValueOfHand,
Hands(i).RankDistrubtion, Hands(i).NameOfHand, Hands(i).ValueOfHand)
37
Next
38
\'Console.WriteLine(\"Deck [{0}]\", TheDeck.DeckState)
39
\'Console.WriteLine(\"Deck [{0}]\", TheDeck.DeckState)
40
\'TheDeck.ShuffleCards()
41
\'Console.WriteLine(\"Deck [{0}]\", TheDeck.DeckState)
42
Me.Refresh()
43
End Sub
44
End Class
Public Class Form1
Solution
Step-1:Code the playing card and then sort the cards in a deck.
Step-2:Code the deck of cards then take a card from the deck.
Step-3:use the deck of cards and then rank distribution of cards.
PROGRAM IN VB.NET
Public Class Form1
02
Dim TheDeck As DeckOfCards
03
Dim Hands(NumberOfHands - 1) As HandOfCards
04
Const NumberOfHands As Integer = 10
05
06
07
Private Sub Form1_Paint(ByVal sender As Object, ByVal e
AsSystem.Windows.Forms.PaintEventArgs) Handles Me.Paint
08
If TheDeck IsNot Nothing Then
09
For i = 0 To NumberOfHands - 1
10
Hands(i).DrawHand(e.Graphics, 25, 25 + i * 65)
11
Next
12
End If
13
14
15
End Sub
16
17
18
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As
System.EventArgs)Handles Button1.Click
19
Console.WriteLine(\"--------\")
20
21
22
TheDeck = New DeckOfCards
23
TheDeck.ShuffleCards()
24
TheDeck.ShuffleCards()
25
\'TheDeck.ShuffleCards()
26
\' TheDeck.ShuffleCards()
27
For i = 0 To NumberOfHands - 1
28
Hands(i) = New HandOfCards
29
For c As Integer = 0 To 4
30
Hands(i).AddCard(TheDeck.TakeCard)
31
Next
32
Hands(i).SortHand()
33
Next
34
Array.Sort(Hands, New HandOfCards.HandComparer)
35
For i = 0 To NumberOfHands - 1
36
Console.WriteLine(\"{0}\" & vbTab & \"[{1}] = {2}\" & vbTab & \" {3}\" & vbTab &
\"[{4}]\" & vbTab & \" ({5})\", i + 1, Hands(i).ToString, Han.
conc . of H+ = conc. of H3O + = 0.250.17 =0.0425.pdf
conc . of H+ = conc. of H3O + = 0.25*0.17 =0.0425M pH + pOH = 14 always and
[H+]*[OH-] is 10^-14 conc . of OH- = 10^-14/0.0425 =2.3529*10^-13 M
Solution
conc . of H+ = conc. of H3O + = 0.25*0.17 =0.0425M pH + pOH = 14 always and
[H+]*[OH-] is 10^-14 conc . of OH- = 10^-14/0.0425 =2.3529*10^-13 M.
Reactivity of the metal Tendency to formcation .pdf
Reactivity of the metal Tendency to formcation
Tendency to loose outer elecrtons
E ooxid
1 / E ored
So the Eo cell = E ocathode - Eoanode is more, then themetal is more reactive to reduce the Cu
2+
Given
I. Zn/Zn 2+ || Cu 2+/Cu 1.072 V
II. Fe/Fe 2+ || Cu 2+/Cu 0.334 V
III. Mg/Mg 2+ || Cu 2+/Cu 1.640 V
IV. Pb/Pb 2+ || Cu 2+/Cu 0.459V
V. Sn/Sn 2+ || Cu 2+/Cu 0.600 V
So the order of reactivity is Mg > Zn > Sn > Pb >Fe
So the Eo cell = E ocathode - Eoanode is more, then themetal is more reactive to reduce the Cu
2+
Given
I. Zn/Zn 2+ || Cu 2+/Cu 1.072 V
II. Fe/Fe 2+ || Cu 2+/Cu 0.334 V
III. Mg/Mg 2+ || Cu 2+/Cu 1.640 V
IV. Pb/Pb 2+ || Cu 2+/Cu 0.459V
V. Sn/Sn 2+ || Cu 2+/Cu 0.600 V
So the order of reactivity is Mg > Zn > Sn > Pb >Fe
Solution
Reactivity of the metal Tendency to formcation
Tendency to loose outer elecrtons
E ooxid
1 / E ored
So the Eo cell = E ocathode - Eoanode is more, then themetal is more reactive to reduce the Cu
2+
Given
I. Zn/Zn 2+ || Cu 2+/Cu 1.072 V
II. Fe/Fe 2+ || Cu 2+/Cu 0.334 V
III. Mg/Mg 2+ || Cu 2+/Cu 1.640 V
IV. Pb/Pb 2+ || Cu 2+/Cu 0.459V
V. Sn/Sn 2+ || Cu 2+/Cu 0.600 V
So the order of reactivity is Mg > Zn > Sn > Pb >Fe
So the Eo cell = E ocathode - Eoanode is more, then themetal is more reactive to reduce the Cu
2+
Given
I. Zn/Zn 2+ || Cu 2+/Cu 1.072 V
II. Fe/Fe 2+ || Cu 2+/Cu 0.334 V
III. Mg/Mg 2+ || Cu 2+/Cu 1.640 V
IV. Pb/Pb 2+ || Cu 2+/Cu 0.459V
V. Sn/Sn 2+ || Cu 2+/Cu 0.600 V
So the order of reactivity is Mg > Zn > Sn > Pb >Fe.
Purple sulfur bacteria use H2S as a source of electrons and protons.pdf
Purple sulfur bacteria use H2S as a source of electrons and protons
H2S = S + 2H(+) + 2e-
Solution
Purple sulfur bacteria use H2S as a source of electrons and protons
H2S = S + 2H(+) + 2e-.
Chemical reaction tkes place Zn + CuSO4 = ZnS04.pdf
Chemical reaction tkes place Zn + CuSO4 = ZnS04 +Cu so the blue coloured
copper sulphate solution turns clear
Solution
Chemical reaction tkes place Zn + CuSO4 = ZnS04 +Cu so the blue coloured
copper sulphate solution turns clear.
C. three bonds and no lone pairs .pdf
C. three bonds and no lone pairs
Solution
C. three bonds and no lone pairs.
c) None, all the anions form a precipitate with s.pdf
c) None, all the anions form a precipitate with some of the cations in the table e.g.
Pb2+ forms an insoluble salt with all the anions listed. d) None, all the anions form a soluble salt
with K+, NH4+ for example.
Solution
c) None, all the anions form a precipitate with some of the cations in the table e.g.
Pb2+ forms an insoluble salt with all the anions listed. d) None, all the anions form a soluble salt
with K+, NH4+ for example..
Option B nonsense-mediated decay, using small RNA molecules to trig.pdf
Option B: nonsense-mediated decay, using small RNA molecules to trigger degradation.
this is a eukaryotic translation regulation and not found in bacteria.
Solution
Option B: nonsense-mediated decay, using small RNA molecules to trigger degradation.
this is a eukaryotic translation regulation and not found in bacteria..
Floating point basicsThe core idea of floating-point representatio.pdf
Floating point basics
The core idea of floating-point representations (as opposed to fixed point representations as used
by, say, ints), is that a number x is written as m*be where m is a mantissa or fractional part, b is a
base, and e is an exponent. On modern computers the base is almost always 2, and for most
floating-point representations the mantissa will be scaled to be between 1 and b. This is done by
adjusting the exponent, e.g.
1 = 1*20
2 = 1*21
0.375 = 1.5*2-2
etc.
Iam writing a program that does some floating addition that uses bit patterns with shifts applied
to the mantissa and such to obtain the sum of the two floating point numbers. Logically and on
paper I can get this to compute the correct sum.
Code:
#include
#include
#include
#include
int isNegative (float f)
{
unsigned int* iptr = (unsigned int*)&f;
return ( ((*iptr) & 0x80000000) ? 1:0);
}
unsigned char getExponent (float f)
{
unsigned int* iptr = (unsigned int*)&f;
return (((*iptr >> 23) & 0xff) - 127);
}
unsigned int getMantissa (float f)
{
unsigned int* iptr = (unsigned int*)&f;
if( *iptr == 0 ) return 0;
return ((*iptr & 0xFFFFFF) | 0x800000 );
}
float sum (float left, float right)
{
unsigned int littleMan;
unsigned int bigMan;
unsigned char littleE;
unsigned char bigE;
unsigned char lexp = getExponent(left);
unsigned char rexp = getExponent(right);
int Dexponent;
if (lexp > rexp)
{
bigE = lexp;
bigMan = getMantissa(left);
littleE = rexp;
littleMan = getMantissa(right);
}
else
{
bigE = rexp;
bigMan = getMantissa(right);
littleE = lexp;
littleMan = getMantissa(left);
}
printf(\"little: %x %x\ \", littleE, littleMan);
printf(\"big: %x %x\ \", bigE, bigMan);
void shift( unsigned int *valToShift, int bitsToShift )
{
// Masks is used to mask out bits to check for a \"sticky\" bit.
static unsigned masks[24] =
{
0, 1, 3, 7, 0xf, 0x1f, 0x3f, 0x7f,
0xff, 0x1ff, 0x3ff, 0x7ff, 0xfff, 0x1fff, 0x3fff, 0x7fff,
0xffff, 0x1ffff, 0x3ffff, 0x7ffff, 0xfffff, 0x1fffff, 0x3fffff, 0x7fffff
};
// HOmasks - masks out the H.O. bit of the value masked by the masks entry.
static unsigned HOmasks[24] =
{
0,
1, 2, 4, 0x8, 0x10, 0x20, 0x40, 0x80,
0x100, 0x200, 0x400, 0x800, 0x1000, 0x2000, 0x4000, 0x8000,
0x10000, 0x20000, 0x40000, 0x80000, 0x100000, 0x200000, 0x400000
};
// shiftedOut- Holds the value that will be shifted out of a mantissa
// during the denormalization operation (used to round a denormalized value).
int shiftedOut;
assert( bitsToShift <= 23 );
// Grabs the bits we\'re going to shift out (so we can determine
// how to round this value after the shift).
shiftedOut = *valToShift & masks[ bitsToShift ];
// Shift the value to the right the specified number of bits:
*valToShift = *valToShift >> bitsToShift;
// If necessary, round the value:
if( shiftedOut > HOmasks[ bitsToShift ] )
{
// If the bits we shifted out are greater than 1/2 the L.O. bit, then
// round the value up by one.
*valToShift = *valToShift + 1;
}
else if( shiftedOut == HOmasks[ bitsToShift ] )
{
// If the bits we shif.
F1 females - All wild typeF1 males – All wild typeF2 females – A.pdf
F1 females - All wild type
F1 males – All wild type
F2 females – All wild type
F2 males – 1/2 Wild type and
1/2 Vermilion
Solution
F1 females - All wild type
F1 males – All wild type
F2 females – All wild type
F2 males – 1/2 Wild type and
1/2 Vermilion.
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Step1 Moles of ions from NaCl = 2x.46 = .92 Step2.pdf
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so that the reaction takes place and forms (=0 )b.pdf
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In case of SO2 Dipole - Dipole forces are stron.pdf
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Step-1Code the playing card and then sort the cards in a deck.Ste.pdf
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conc . of H+ = conc. of H3O + = 0.250.17 =0.0425.pdf
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Reactivity of the metal Tendency to formcation .pdf
Purple sulfur bacteria use H2S as a source of electrons and protons.pdf
Purple sulfur bacteria use H2S as a source of electrons and protons.pdf
Chemical reaction tkes place Zn + CuSO4 = ZnS04.pdf
Chemical reaction tkes place Zn + CuSO4 = ZnS04.pdf
c) None, all the anions form a precipitate with s.pdf
c) None, all the anions form a precipitate with s.pdf
Option B nonsense-mediated decay, using small RNA molecules to trig.pdf
Option B nonsense-mediated decay, using small RNA molecules to trig.pdf
Floating point basicsThe core idea of floating-point representatio.pdf
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