Calcium sulfateCalcium hydroxide and sulfuric acid are strong base.pdf
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Calcium sulfate Calcium hydroxide and sulfuric acid are strong base and acid respectively. So the conjugate acid and base are both weak. Hence the salt is neutral KNO2 KOH is a strong base but HNO2 is a weak acid So the conjugate base of HNO2 will be strong and the salt will be basic Fe(ClO4)3 Fe(OH)3 is a weak base while HClO4 is a strong acid So the conjugate acid of Fe(OH)3 will be strong and the salt will be acidic NaI NaOH is a strong base while HI is weak acid. So the conjugate base will be strong and the salt will be basic. Solution Calcium sulfate Calcium hydroxide and sulfuric acid are strong base and acid respectively. So the conjugate acid and base are both weak. Hence the salt is neutral KNO2 KOH is a strong base but HNO2 is a weak acid So the conjugate base of HNO2 will be strong and the salt will be basic Fe(ClO4)3 Fe(OH)3 is a weak base while HClO4 is a strong acid So the conjugate acid of Fe(OH)3 will be strong and the salt will be acidic NaI NaOH is a strong base while HI is weak acid. So the conjugate base will be strong and the salt will be basic..
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properties of acids base and the law given by scientists
All acid base concept for First Year B.pharm
Generally the formula for an acid is a hydrogen atom bonded with eit.pdf
Generally the formula for an acid is a hydrogen atom bonded with either another atom that is an
anion, such as Cl- (chlorine anion) or another molecule that is an anion, such as SO42- (sulfate
anion). What makes an acid an acid is that the hydrogen cation can fall off the molecule, known
as protonation, leaving behind a relatively stable anion. This anion is known as the acid\'s
conjugate base. A strong acid, such as HCl, has a weak conjugate base. A weak acid, such as HF,
would have a strong conjugate base. A strong acid is dictated by how much it protonates in
water, that is, how much the H+ cation dissociates from the anion in water. In HCl\'s case, the
acid protonates almost completely, making it a strong acid. The acidity of an acid is measured by
the well known pH scale. As you go upwards in difficulty of chemistry, molecules with multiple
hydrogens can have different acidities because when different hydrogens leave, they can form a
more or less stable anion behind.
I hope this helps!
Solution
Generally the formula for an acid is a hydrogen atom bonded with either another atom that is an
anion, such as Cl- (chlorine anion) or another molecule that is an anion, such as SO42- (sulfate
anion). What makes an acid an acid is that the hydrogen cation can fall off the molecule, known
as protonation, leaving behind a relatively stable anion. This anion is known as the acid\'s
conjugate base. A strong acid, such as HCl, has a weak conjugate base. A weak acid, such as HF,
would have a strong conjugate base. A strong acid is dictated by how much it protonates in
water, that is, how much the H+ cation dissociates from the anion in water. In HCl\'s case, the
acid protonates almost completely, making it a strong acid. The acidity of an acid is measured by
the well known pH scale. As you go upwards in difficulty of chemistry, molecules with multiple
hydrogens can have different acidities because when different hydrogens leave, they can form a
more or less stable anion behind.
I hope this helps!.
Recommended
properties of acids base and the law given by scientists
All acid base concept for First Year B.pharm
Generally the formula for an acid is a hydrogen atom bonded with eit.pdf
Generally the formula for an acid is a hydrogen atom bonded with either another atom that is an
anion, such as Cl- (chlorine anion) or another molecule that is an anion, such as SO42- (sulfate
anion). What makes an acid an acid is that the hydrogen cation can fall off the molecule, known
as protonation, leaving behind a relatively stable anion. This anion is known as the acid\'s
conjugate base. A strong acid, such as HCl, has a weak conjugate base. A weak acid, such as HF,
would have a strong conjugate base. A strong acid is dictated by how much it protonates in
water, that is, how much the H+ cation dissociates from the anion in water. In HCl\'s case, the
acid protonates almost completely, making it a strong acid. The acidity of an acid is measured by
the well known pH scale. As you go upwards in difficulty of chemistry, molecules with multiple
hydrogens can have different acidities because when different hydrogens leave, they can form a
more or less stable anion behind.
I hope this helps!
Solution
Generally the formula for an acid is a hydrogen atom bonded with either another atom that is an
anion, such as Cl- (chlorine anion) or another molecule that is an anion, such as SO42- (sulfate
anion). What makes an acid an acid is that the hydrogen cation can fall off the molecule, known
as protonation, leaving behind a relatively stable anion. This anion is known as the acid\'s
conjugate base. A strong acid, such as HCl, has a weak conjugate base. A weak acid, such as HF,
would have a strong conjugate base. A strong acid is dictated by how much it protonates in
water, that is, how much the H+ cation dissociates from the anion in water. In HCl\'s case, the
acid protonates almost completely, making it a strong acid. The acidity of an acid is measured by
the well known pH scale. As you go upwards in difficulty of chemistry, molecules with multiple
hydrogens can have different acidities because when different hydrogens leave, they can form a
more or less stable anion behind.
I hope this helps!.
TOPIC 8 : Acids and Bases
UNDERSTANDING 8.1 Theories of Acids and Bases
UNDERSTANDING 8.2 Properties of Acids and Bases
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.
More Related Content
Similar to Calcium sulfateCalcium hydroxide and sulfuric acid are strong base.pdf
TOPIC 8 : Acids and Bases
UNDERSTANDING 8.1 Theories of Acids and Bases
UNDERSTANDING 8.2 Properties of Acids and Bases
Similar to Calcium sulfateCalcium hydroxide and sulfuric acid are strong base.pdf (6)
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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