The Br was originally neutral, but picks up an ex.pdf
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The Br was originally neutral, but picks up an extra electron from the N and becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double bonds on both sides of C1, the O, C1, N backbone is linear. Solution The Br was originally neutral, but picks up an extra electron from the N and becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double bonds on both sides of C1, the O, C1, N backbone is linear..
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Explain in detail- why Al2Cl6 is obtained as a dimer but BCl3 as a mon.docx
Explain in detail, why Al2Cl6 is obtained as a dimer but BCl3 as a monomer with special reference to the boning involved in these systems?
Solution
BCl3 and AlCl3 both are electron deficient molecules because there are six electrons present in the valency shell of central atom. There is a strong tendency for both the compounds to gain two electrons and complete their octet. For both the molecules, the central atom has sp2 hybridisation and trigonal planar geometry. The sp2 hybridisation has one s and two p orbitals, and one unhybridised p orbital.
The unhybridised p orbital is perpendicular to the trigonal plane of the molecule. Since it is empty, it can receive two electrons from a donor. Also, Chlorine atoms have three lone pairs.
The deficiency of electrons of Boron in BCl3 is overcome by formation of co-ordinate bond between the lone pair of Chlorine atom and the empty unhybridised p orbital of Boron atom. The bond formed is forming p?–p? and no dimer is formed.
In AlCl3, the electron deficiency of Aluminium atom is overcome through the formation of co-ordinate bond between lone pair of Chlorine atom of another AlCl3 molecule and the empty unhybridised p orbital of Aluminium atom. Hence, dimer is formed.
.
a. BeCl2. The Be has 2 bonds and no lone pairs, so it is linear. b..pdf
a. BeCl2. The Be has 2 bonds and no lone pairs, so it is linear. b. BBr3. The B has 3 bonds and
no lone pairs, so it is trigonal planar. c. HCN. The C is the central atom and has two bonds (a
single to H, triple to N), and no lone pairs, so it is linear. d. The C is the central atom and has
three bonds (single to H, double to O), so it is trigonal planar. e. There are two \"central\" N
atoms. The first has a single bond to H and double to N, and one lone pair, so it is bent. The
second has two bonds (double to each other N), and no lone pairs, so it is linear. H-N=N=N.
The first N is bent, the second is linear (so the 3 N atoms all lie in a line, but not the H).
Solution
a. BeCl2. The Be has 2 bonds and no lone pairs, so it is linear. b. BBr3. The B has 3 bonds and
no lone pairs, so it is trigonal planar. c. HCN. The C is the central atom and has two bonds (a
single to H, triple to N), and no lone pairs, so it is linear. d. The C is the central atom and has
three bonds (single to H, double to O), so it is trigonal planar. e. There are two \"central\" N
atoms. The first has a single bond to H and double to N, and one lone pair, so it is bent. The
second has two bonds (double to each other N), and no lone pairs, so it is linear. H-N=N=N.
The first N is bent, the second is linear (so the 3 N atoms all lie in a line, but not the H)..
Molecular Orbital Theory
Molecular Orbital Theory, Molecular Orbital diagram, Molecular Orbital Theory Assumption, Molecular Orbital Theory for homonuclear diatomic molecules H2,Li2,Be2,B2,C2, N2, O2, F2,Ne2, MOT, Electron density, Bond Order, Diamagnetic, Paramagnetic, Hydrogen molecule, Hydrogen ion, Helium molecule, Lithium molecule, Beryllium molecule, Boron molecule, Carbon molecule, Nitrogen molecule, Oxygen molecule, Oxygen ion, O2- ion Superoxide ion, O2- - ion peroxide ion, Fluorine molecule, Neon molecule,
OK, let me try to talk you through this. The arro.pdf
OK, let me try to talk you through this. The arrow pointing straight up takes a
double bond between the O and the C (all intersections of lines are understood to be carbon
atoms) and forms an O-H bond and leaves a single O-C bond. The curved arrow on the right,
straddling the O takes away the H-O bond and forms a second bond between the O and C. This
separates the H from the O. The curved arrow at the bottom takes away the bond between the
two carbons, separating the molecule. We are left with O=C=O which is carbon dioxide. Carbon
dioxide has two double bonds and is a linear compound. This is the species with the least (we
should say fewest since we can count the atoms) atoms. Then for the first part we have
everything else A central carbon with a single bond to each of two -OH groups and a double
bond to one carbon and that carbon has single bonds to two other carbons. So the central carbon
has for bonds 1 to O, 2 to O, 3 and 4 to C. Then that C has four bonds, 1 and 2 to the central
carbon and then 3 to a carbon and 4 to a carbon.
Solution
OK, let me try to talk you through this. The arrow pointing straight up takes a
double bond between the O and the C (all intersections of lines are understood to be carbon
atoms) and forms an O-H bond and leaves a single O-C bond. The curved arrow on the right,
straddling the O takes away the H-O bond and forms a second bond between the O and C. This
separates the H from the O. The curved arrow at the bottom takes away the bond between the
two carbons, separating the molecule. We are left with O=C=O which is carbon dioxide. Carbon
dioxide has two double bonds and is a linear compound. This is the species with the least (we
should say fewest since we can count the atoms) atoms. Then for the first part we have
everything else A central carbon with a single bond to each of two -OH groups and a double
bond to one carbon and that carbon has single bonds to two other carbons. So the central carbon
has for bonds 1 to O, 2 to O, 3 and 4 to C. Then that C has four bonds, 1 and 2 to the central
carbon and then 3 to a carbon and 4 to a carbon..
Ionic and covalent bonds
This slides demonstrate on the formation of positive and negative ions, followed by formation of ionic bonding and covalent bonding. Chemical bonding have various types of bonding. In the presentation, we're sharing only ionic and covalent bonding only.
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Explain in detail- why Al2Cl6 is obtained as a dimer but BCl3 as a mon.docx
Explain in detail, why Al2Cl6 is obtained as a dimer but BCl3 as a monomer with special reference to the boning involved in these systems?
Solution
BCl3 and AlCl3 both are electron deficient molecules because there are six electrons present in the valency shell of central atom. There is a strong tendency for both the compounds to gain two electrons and complete their octet. For both the molecules, the central atom has sp2 hybridisation and trigonal planar geometry. The sp2 hybridisation has one s and two p orbitals, and one unhybridised p orbital.
The unhybridised p orbital is perpendicular to the trigonal plane of the molecule. Since it is empty, it can receive two electrons from a donor. Also, Chlorine atoms have three lone pairs.
The deficiency of electrons of Boron in BCl3 is overcome by formation of co-ordinate bond between the lone pair of Chlorine atom and the empty unhybridised p orbital of Boron atom. The bond formed is forming p?–p? and no dimer is formed.
In AlCl3, the electron deficiency of Aluminium atom is overcome through the formation of co-ordinate bond between lone pair of Chlorine atom of another AlCl3 molecule and the empty unhybridised p orbital of Aluminium atom. Hence, dimer is formed.
.
a. BeCl2. The Be has 2 bonds and no lone pairs, so it is linear. b..pdf
a. BeCl2. The Be has 2 bonds and no lone pairs, so it is linear. b. BBr3. The B has 3 bonds and
no lone pairs, so it is trigonal planar. c. HCN. The C is the central atom and has two bonds (a
single to H, triple to N), and no lone pairs, so it is linear. d. The C is the central atom and has
three bonds (single to H, double to O), so it is trigonal planar. e. There are two \"central\" N
atoms. The first has a single bond to H and double to N, and one lone pair, so it is bent. The
second has two bonds (double to each other N), and no lone pairs, so it is linear. H-N=N=N.
The first N is bent, the second is linear (so the 3 N atoms all lie in a line, but not the H).
Solution
a. BeCl2. The Be has 2 bonds and no lone pairs, so it is linear. b. BBr3. The B has 3 bonds and
no lone pairs, so it is trigonal planar. c. HCN. The C is the central atom and has two bonds (a
single to H, triple to N), and no lone pairs, so it is linear. d. The C is the central atom and has
three bonds (single to H, double to O), so it is trigonal planar. e. There are two \"central\" N
atoms. The first has a single bond to H and double to N, and one lone pair, so it is bent. The
second has two bonds (double to each other N), and no lone pairs, so it is linear. H-N=N=N.
The first N is bent, the second is linear (so the 3 N atoms all lie in a line, but not the H)..
Molecular Orbital Theory
Molecular Orbital Theory, Molecular Orbital diagram, Molecular Orbital Theory Assumption, Molecular Orbital Theory for homonuclear diatomic molecules H2,Li2,Be2,B2,C2, N2, O2, F2,Ne2, MOT, Electron density, Bond Order, Diamagnetic, Paramagnetic, Hydrogen molecule, Hydrogen ion, Helium molecule, Lithium molecule, Beryllium molecule, Boron molecule, Carbon molecule, Nitrogen molecule, Oxygen molecule, Oxygen ion, O2- ion Superoxide ion, O2- - ion peroxide ion, Fluorine molecule, Neon molecule,
OK, let me try to talk you through this. The arro.pdf
OK, let me try to talk you through this. The arrow pointing straight up takes a
double bond between the O and the C (all intersections of lines are understood to be carbon
atoms) and forms an O-H bond and leaves a single O-C bond. The curved arrow on the right,
straddling the O takes away the H-O bond and forms a second bond between the O and C. This
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two carbons, separating the molecule. We are left with O=C=O which is carbon dioxide. Carbon
dioxide has two double bonds and is a linear compound. This is the species with the least (we
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bond to one carbon and that carbon has single bonds to two other carbons. So the central carbon
has for bonds 1 to O, 2 to O, 3 and 4 to C. Then that C has four bonds, 1 and 2 to the central
carbon and then 3 to a carbon and 4 to a carbon.
Solution
OK, let me try to talk you through this. The arrow pointing straight up takes a
double bond between the O and the C (all intersections of lines are understood to be carbon
atoms) and forms an O-H bond and leaves a single O-C bond. The curved arrow on the right,
straddling the O takes away the H-O bond and forms a second bond between the O and C. This
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two carbons, separating the molecule. We are left with O=C=O which is carbon dioxide. Carbon
dioxide has two double bonds and is a linear compound. This is the species with the least (we
should say fewest since we can count the atoms) atoms. Then for the first part we have
everything else A central carbon with a single bond to each of two -OH groups and a double
bond to one carbon and that carbon has single bonds to two other carbons. So the central carbon
has for bonds 1 to O, 2 to O, 3 and 4 to C. Then that C has four bonds, 1 and 2 to the central
carbon and then 3 to a carbon and 4 to a carbon..
Ionic and covalent bonds
This slides demonstrate on the formation of positive and negative ions, followed by formation of ionic bonding and covalent bonding. Chemical bonding have various types of bonding. In the presentation, we're sharing only ionic and covalent bonding only.
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to minimise the length, term with \'m\' should be reduced so first to absolute values
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17-The Y-Chromosome DNA testing helps in the examination of the male specific portion of the
biological evidence. This is useful where small amount of male DNA is found amidst a large
amount of female DNA , as in the sexual assault cases.It can also help in the missing person
investigation as it extends the range of potential reference samples. Sine the Y chromosome is
passed on from the fathers to their sone (unchanged, except in the case of some mutations),
therefore all the males of the same paternal lineage will possess common Y chromosome
haplotype.
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genes found on the short arm of the Ychromosome are vital to male sex determination,
spermatogenesis andother male related functions. The Y-STRs are polymorphic among the
nonrelated males.
Sometimes when sepoarating the mal ecomponent and the female component becomes difficult,
then the YSTRs help in the determination. Since The Y-STRs are not present in the females,
therefore in a case where the male and female are involved, it becomes clear that as the Y-STRs
are present the male component is easily detected as only this part will be amplified in the PCR.
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masking effect the YSTR will help to identify all the males who have contributed to the
evidence.
ADVANTAGES-
1. extremely discriminating.
2. Extremely sensitive.e. low amounts of DNA can be used for determination.
3. Not onlu body fluids, but alsp touched materials or handled surfaces, where skinn cells ahev
been left can be used for DNA profiling.
4. Multiple testing systems ar available, which increases the likelyhood of successfully
generating DNA profiles.
5. Apart from standart=d STR enem mt DNA or YSTRs are used for profiling.
6. With the presence of the current technologies, from amidst mixed DNA samples, the DNA of
a particular individual can be profiled.
LIMITATIONS
1. Random match probability cannot be excluded.
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3.The important question while dealing with low DNA quantity is that whether the DNA that
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4.Searching evidence is challenging.
5.Sometimes finding the evidence is altogether impossibl.
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share the same profile.
7. DNA mixtures are very complex.
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PULSE CHASE EXPERIME.
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=node;
return list;
}
else
{
link_node *prev=list;
while(temp!=NULL&&compare_node(node,temp)>=0)
{
prev=temp;
temp=temp->next;
}
prev->next=node;
node->next=temp;
return list;
}
}
//method definition for displaying the list
void display_list( struct link_node *head )
{
link_node *temp;
while(head)
{
printf(\"%s\ \",head->node_str);
temp=head;
head=head->next;
free(temp);
}
}
Sample output:
Enter the string:
Geneva
America
Solution
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=no.
The pH and pOH of a solution are defined as pH .pdf
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74
Solution
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74.
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, 8, 6, 3, 4, 5] import sys global
count def display_board( state ): #print \"-------------\" print \" %i %i %i \" %
(state[0], state[3], state[6]) #print \"-------------\" print \" %i %i %i \" % (state[1],
state[4], state[7]) #print \"-------------\" print \" %i %i %i \" % (state[2], state[5],
state[8]) #print \"-------------\" def move_up( state ): new_state = state[:] index =
new_state.index( 0 ) if index not in [0, 3, 6]: temp = new_state[index - 1]
new_state[index - 1] = new_state[index] new_state[index] = temp return
new_state else: return None def move_down( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [2, 5, 8]: temp = new_state[index + 1]
new_state[index + 1] = new_state[index] new_state[index] = temp
return new_state else: return None def move_left( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [0, 1, 2]: temp = new_state[index
- 3] new_state[index - 3] = new_state[index] new_state[index] = temp
return new_state else: return None def move_right( state ): new_state =
state[:] index = new_state.index( 0 ) if index not in [6, 7, 8]: temp =
new_state[index + 3] new_state[index + 3] = new_state[index]
new_state[index] = temp return new_state else: return None def
create_node( state, parent, operator, depth, cost ): return Node( state, parent, operator,
depth, cost ) def expand_node( node, nodes ): expanded_nodes = []
expanded_nodes.append( create_node( move_up( node.state ), node, \"u\", node.depth + 1, 0 ) )
expanded_nodes.append( create_node( move_down( node.state ), node, \"d\", node.depth + 1,
0 ) ) expanded_nodes.append( create_node( move_left( node.state ), node, \"l\", node.depth
+ 1, 0 ) ) expanded_nodes.append( create_node( move_right( node.state), node, \"r\",
node.depth + 1, 0 ) ) # Filter the list and remove the nodes that are impossible (move
function returned None) expanded_nodes = [node for node in expanded_nodes if node.state
!= None] #list comprehension! expanded_nodes = [node for node in expanded_nodes if
node not in nodes] #list comprehension! #expanded_nodes = [node for node in
expanded_nodes if node.state!= ((node.parent).parent).state] #list comprehension!
expanded_nodes1 = [] for node in expanded_nodes: temp=node
temp=temp.parent temp=temp.parent if temp==None:
expanded_nodes1.extend(expanded_nodes) break if node.state
!=temp.state: expanded_nodes1.append(node) return
expanded_nodes1 def bfs( start, goal ): nodes = [] count=0 # Create the queue
with the root node in it. nodes.append( create_node( start, None, None, 0, 0 ) ) while
True: # We\'ve run out of states, no solution. if len( nodes ) == 0:
return None # take the node from the front of the queue node = nodes.pop(0)
count=count+1 #display_board(node.state) # Append the move we made to
moves # if this node is the goal, return the moves it took to get here. if node.state
== goal: #display_board(node.state) moves = []
temp = node print cou.
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7 steps are required as 9 is the 7th element and the search is linear
Solution
7 steps are required as 9 is the 7th element and the search is linear.
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(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
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(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
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Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
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Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
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Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm
Solution
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm.
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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.
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
More Related Content
More from ANJALIENTERPRISES1
H2SO4 may act as 1) an acidexample 2NaOH + H2SO4 - Na2SO4 +.pdf
to minimise the length, term with \'m\' should be reduced so first to absolute values
must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4
Solution
to minimise the length, term with \'m\' should be reduced so first to absolute values
must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4.
Computer graphics are pictures and movies produced use computers fre.pdf
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken=
(10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years
Solution
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken=
(10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years.
Circle.javaimport java.text.DecimalFormat;public class Circle {.pdf
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate
Solution
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate.
C is correct. Only the host that the unicast message is addressed to.pdf
mol L-1 s-1 is the answer
Solution
mol L-1 s-1 is the answer.
According to the seriousness level, the following organs areThymu.pdf
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14
Solution
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14.
Among 40 subjects randomly choose 20 subjects and assign themSol.pdf
First substance = Ionic solid Second substance = Covalent solid
Solution
First substance = Ionic solid Second substance = Covalent solid.
AdvantagesThe main objective of business combination is to elimina.pdf
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 -
CH2 - CH2 - CH2=O
Solution
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 -
CH2 - CH2 - CH2=O.
a. Germ cell speciication in Drosophila is primarily a maternlly con.pdf
coordination no is the no of ligands in the coordination sphere and in this case it is 6
Solution
coordination no is the no of ligands in the coordination sphere and in this case it is 6.
A is correct. The packets to be filtered would be heading into the r.pdf
At the beginning of the 11th year, there would be 1,024,000 rabbits.
Solution
At the beginning of the 11th year, there would be 1,024,000 rabbits..
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
1). A) ipsilateral
Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs).
So, the nociceptive information can be found on the same side of the body.
Solution
1). A) ipsilateral
Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs).
So, the nociceptive information can be found on the same side of the body..
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
17-The Y-Chromosome DNA testing helps in the examination of the male specific portion of the
biological evidence. This is useful where small amount of male DNA is found amidst a large
amount of female DNA , as in the sexual assault cases.It can also help in the missing person
investigation as it extends the range of potential reference samples. Sine the Y chromosome is
passed on from the fathers to their sone (unchanged, except in the case of some mutations),
therefore all the males of the same paternal lineage will possess common Y chromosome
haplotype.
A set of Short tandem repeats (STR) are found on the male specific Y chromosome.The coding
genes found on the short arm of the Ychromosome are vital to male sex determination,
spermatogenesis andother male related functions. The Y-STRs are polymorphic among the
nonrelated males.
Sometimes when sepoarating the mal ecomponent and the female component becomes difficult,
then the YSTRs help in the determination. Since The Y-STRs are not present in the females,
therefore in a case where the male and female are involved, it becomes clear that as the Y-STRs
are present the male component is easily detected as only this part will be amplified in the PCR.
The YSTRs are also helpful when a number of males are involved. As regular STRs cause
masking effect the YSTR will help to identify all the males who have contributed to the
evidence.
ADVANTAGES-
1. extremely discriminating.
2. Extremely sensitive.e. low amounts of DNA can be used for determination.
3. Not onlu body fluids, but alsp touched materials or handled surfaces, where skinn cells ahev
been left can be used for DNA profiling.
4. Multiple testing systems ar available, which increases the likelyhood of successfully
generating DNA profiles.
5. Apart from standart=d STR enem mt DNA or YSTRs are used for profiling.
6. With the presence of the current technologies, from amidst mixed DNA samples, the DNA of
a particular individual can be profiled.
LIMITATIONS
1. Random match probability cannot be excluded.
2. Sometimes even when crime is not committed, we do get ,low levels of DNA on surfaces.
3.The important question while dealing with low DNA quantity is that whether the DNA that
was deposited was deposited during the crime or at some other time.
4.Searching evidence is challenging.
5.Sometimes finding the evidence is altogether impossibl.
6. In case of mtDNA /T STRs , as they are passed on from generation to generation either from
fathers to sons or mothers to daughters and son, therefore sometimes unrelated persons are also
share the same profile.
7. DNA mixtures are very complex.
8.Interpretation is required in many cases.
16- from Cyanobacterial to humans, many terristrial organisms have acquired the circadian
rhythms to adapt to sunlight to increase their survival rate.Circadian Rhythm is a biological clock
that displays an endogenous, entrainable oscillation of 24 hours.Thesse oscillations are driven by
circadian clocks.
PULSE CHASE EXPERIME.
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=node;
return list;
}
else
{
link_node *prev=list;
while(temp!=NULL&&compare_node(node,temp)>=0)
{
prev=temp;
temp=temp->next;
}
prev->next=node;
node->next=temp;
return list;
}
}
//method definition for displaying the list
void display_list( struct link_node *head )
{
link_node *temp;
while(head)
{
printf(\"%s\ \",head->node_str);
temp=head;
head=head->next;
free(temp);
}
}
Sample output:
Enter the string:
Geneva
America
Solution
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=no.
The pH and pOH of a solution are defined as pH .pdf
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74
Solution
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74.
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, 8, 6, 3, 4, 5] import sys global
count def display_board( state ): #print \"-------------\" print \" %i %i %i \" %
(state[0], state[3], state[6]) #print \"-------------\" print \" %i %i %i \" % (state[1],
state[4], state[7]) #print \"-------------\" print \" %i %i %i \" % (state[2], state[5],
state[8]) #print \"-------------\" def move_up( state ): new_state = state[:] index =
new_state.index( 0 ) if index not in [0, 3, 6]: temp = new_state[index - 1]
new_state[index - 1] = new_state[index] new_state[index] = temp return
new_state else: return None def move_down( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [2, 5, 8]: temp = new_state[index + 1]
new_state[index + 1] = new_state[index] new_state[index] = temp
return new_state else: return None def move_left( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [0, 1, 2]: temp = new_state[index
- 3] new_state[index - 3] = new_state[index] new_state[index] = temp
return new_state else: return None def move_right( state ): new_state =
state[:] index = new_state.index( 0 ) if index not in [6, 7, 8]: temp =
new_state[index + 3] new_state[index + 3] = new_state[index]
new_state[index] = temp return new_state else: return None def
create_node( state, parent, operator, depth, cost ): return Node( state, parent, operator,
depth, cost ) def expand_node( node, nodes ): expanded_nodes = []
expanded_nodes.append( create_node( move_up( node.state ), node, \"u\", node.depth + 1, 0 ) )
expanded_nodes.append( create_node( move_down( node.state ), node, \"d\", node.depth + 1,
0 ) ) expanded_nodes.append( create_node( move_left( node.state ), node, \"l\", node.depth
+ 1, 0 ) ) expanded_nodes.append( create_node( move_right( node.state), node, \"r\",
node.depth + 1, 0 ) ) # Filter the list and remove the nodes that are impossible (move
function returned None) expanded_nodes = [node for node in expanded_nodes if node.state
!= None] #list comprehension! expanded_nodes = [node for node in expanded_nodes if
node not in nodes] #list comprehension! #expanded_nodes = [node for node in
expanded_nodes if node.state!= ((node.parent).parent).state] #list comprehension!
expanded_nodes1 = [] for node in expanded_nodes: temp=node
temp=temp.parent temp=temp.parent if temp==None:
expanded_nodes1.extend(expanded_nodes) break if node.state
!=temp.state: expanded_nodes1.append(node) return
expanded_nodes1 def bfs( start, goal ): nodes = [] count=0 # Create the queue
with the root node in it. nodes.append( create_node( start, None, None, 0, 0 ) ) while
True: # We\'ve run out of states, no solution. if len( nodes ) == 0:
return None # take the node from the front of the queue node = nodes.pop(0)
count=count+1 #display_board(node.state) # Append the move we made to
moves # if this node is the goal, return the moves it took to get here. if node.state
== goal: #display_board(node.state) moves = []
temp = node print cou.
7 steps are required as 9 is the 7th element and the search is line.pdf
7 steps are required as 9 is the 7th element and the search is linear
Solution
7 steps are required as 9 is the 7th element and the search is linear.
(D) the number of moles of hydroxide ion added and the number of mol.pdf
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
initially present
Solution
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
initially present.
Look for changes in oxidation numbers. These occ.pdf
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
Thus, reactions (c) and (e) are REDOX reactions.
Solution
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
Thus, reactions (c) and (e) are REDOX reactions..
Use Daltons Law of partial pressures. P(Total).pdf
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm
Solution
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm.
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H2SO4 may act as 1) an acidexample 2NaOH + H2SO4 - Na2SO4 +.pdf
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C is correct. Only the host that the unicast message is addressed to.pdf
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17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
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#includestdio.h#includestring.h#includestdlib.h#define M.pdf
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The pH and pOH of a solution are defined as pH .pdf
The pH and pOH of a solution are defined as pH .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
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7 steps are required as 9 is the 7th element and the search is line.pdf
7 steps are required as 9 is the 7th element and the search is line.pdf
(D) the number of moles of hydroxide ion added and the number of mol.pdf
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Look for changes in oxidation numbers. These occ.pdf
Look for changes in oxidation numbers. These occ.pdf
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Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
A Strategic Approach: GenAI in Education
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Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
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Introduction to AI for Nonprofits with Tapp Network
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Presentation required for the master in Education.
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How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
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Embracing GenAI - A Strategic Imperative
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
The approach at University of Liverpool.pptx
How libraries can support authors with open access requirements for UKRI funded books
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Unit 2- Research Aptitude (UGC NET Paper I).pdf
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Honest Reviews of Tim Han LMA Course Program.pptx
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Instructions for Submissions thorugh G- Classroom.pptx
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
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The Br was originally neutral, but picks up an ex.pdf
- 1. The Br was originally neutral, but picks up an extra electron from the N and becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double bonds on both sides of C1, the O, C1, N backbone is linear. Solution The Br was originally neutral, but picks up an extra electron from the N and becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double bonds on both sides of C1, the O, C1, N backbone is linear.