Hi!Intermolecular bonding refers to forces present between two or .pdf

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Hi! Intermolecular bonding refers to forces present between two or more 1-butanol molecules. You are right, Hydrogen bonding is one of them. You will also get some dipole-dipole interactions and london dispersion forces. This is due to the polar bond between the terminal oxygen and the carbon chain. I hope this helps! Please don\'t forget to rate : ) Solution Hi! Intermolecular bonding refers to forces present between two or more 1-butanol molecules. You are right, Hydrogen bonding is one of them. You will also get some dipole-dipole interactions and london dispersion forces. This is due to the polar bond between the terminal oxygen and the carbon chain. I hope this helps! Please don\'t forget to rate : ).

Hi!
Intermolecular bonding refers to forces present between two or more 1-butanol molecules. You
are right, Hydrogen bonding is one of them. You will also get some dipole-dipole interactions
and london dispersion forces. This is due to the polar bond between the terminal oxygen and the
carbon chain.
I hope this helps! Please don't forget to rate : )
Solution
Hi!
Intermolecular bonding refers to forces present between two or more 1-butanol molecules. You
are right, Hydrogen bonding is one of them. You will also get some dipole-dipole interactions
and london dispersion forces. This is due to the polar bond between the terminal oxygen and the
carbon chain.
I hope this helps! Please don't forget to rate : )

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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. 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.

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#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.

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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.

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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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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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1. Hi! Intermolecular bonding refers to forces present between two or more 1-butanol molecules. You are right, Hydrogen bonding is one of them. You will also get some dipole-dipole interactions and london dispersion forces. This is due to the polar bond between the terminal oxygen and the carbon chain. I hope this helps! Please don't forget to rate : ) Solution Hi! Intermolecular bonding refers to forces present between two or more 1-butanol molecules. You are right, Hydrogen bonding is one of them. You will also get some dipole-dipole interactions and london dispersion forces. This is due to the polar bond between the terminal oxygen and the carbon chain. I hope this helps! Please don't forget to rate : )