Use Daltons Law of partial pressures. P(Total).pdf
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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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#include
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struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
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typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
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int compare_node( struct link_node *n1, struct link_node *n2 )
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{
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}
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{
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list=node;
return list;
}
else
{
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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..
The Br was originally neutral, but picks up an ex.pdf
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..
Step1 firstly balance the Phophorous atoms by mul.pdf
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4
Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O------->
4H3PO4
Solution
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4
Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O------->
4H3PO4.
Year 2000 is the period when the productivity growth rate was at min.pdf
Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth
refers to those years where productivity growth rate was in negative. Minimum growth rate
refers to the minimum positive growth rate achieved.
Solution
Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth
refers to those years where productivity growth rate was in negative. Minimum growth rate
refers to the minimum positive growth rate achieved..
You have tab characters in your text file which is making console to.pdf
You have tab characters in your text file which is making console to display following both lines
aligned at middle.
dress
across the porch as the
There are two ways of doing it.
Solution
You have tab characters in your text file which is making console to display following both lines
aligned at middle.
dress
across the porch as the
There are two ways of doing it..
the frequency of the allele I^A in this population-0.22the frequen.pdf
the frequency of the allele I^A in this population-0.22
the frequency of the allele I^B in this population-0.08
Solution
the frequency of the allele I^A in this population-0.22
the frequency of the allele I^B in this population-0.08.
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
The reaction is:
Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g)
Solution
The reaction is:
Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g).
Solution!DOCTYPE html html head style type=textcss.pdf
Solution
IDENameCostOpen SourceDestinationNETBEANSFreeyescross platform IDEINTELLIJ
IDEA$199noit has great moderen UIECLIPSEfreeyesit features huge library of
pluginsJDEVELOPERfreeyesit has fully loades tool set for creating javaDR. JAVAFreeyesit
supports cross platformBLUEJFreeyesit suitable for small software projectsJCREATORfreeyesit
is fast and efficient ideJGRASPfreeyesit is super lightweight ideGREENFOOTfreeyesit supports
designing for 2D games CODEENVY$1(month)nocloud based ide.
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TCP/IP is mostly uaed network protocol in Linux and Mac os.
But appletalk also used in both the os.
Solution
TCP/IP is mostly uaed network protocol in Linux and Mac os.
But appletalk also used in both the os..
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