Lewis acid is the electron pair acceptorLewis base is the electron.pdf
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Lewis acid is the electron pair acceptor Lewis base is the electron pair donar a) OH- + Al(OH)3 -----> Al(OH)4- Al(OH)3 is the lewis acid OH- is the lewis base b) SO3 + H2O --------> H2SO4 H2O is the lewis acid SO3 is the lewis base H2O is the lewis acid SO3 is the lewis base c) Co3+ + 6NH3 -----> Co(NH3)63+ Co 3+ is the lewis acid NH3 is the lewis base Co 3+ is the lewis acid NH3 is the lewis base Solution Lewis acid is the electron pair acceptor Lewis base is the electron pair donar a) OH- + Al(OH)3 -----> Al(OH)4- Al(OH)3 is the lewis acid OH- is the lewis base b) SO3 + H2O --------> H2SO4 H2O is the lewis acid SO3 is the lewis base H2O is the lewis acid SO3 is the lewis base c) Co3+ + 6NH3 -----> Co(NH3)63+ Co 3+ is the lewis acid NH3 is the lewis base Co 3+ is the lewis acid NH3 is the lewis base.
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This is a fun problem. How often do you get to think about a compoun.pdf
This is a fun problem. How often do you get to think about a compound that is more soluble in
strong base and strong acid? Pretty nifty!
a. In very acidic solution La Chateliers shows that reaction will shift to products
Al(OH)3(s) + 3H+ ----> Al3+ + 3H2O EQN 1
or if you start with acidic solution and then add ions
Al3+ + OH- + H+ ----> Al3+ + H2O
In very basic solution there are so many OH- ions that probability is high that four will be near
each AL3+
Al 3+ +4OH- -----> Al(OH)4-
b. S has two terms, the first is large when [H+] is high the second one is large when [H+] is low.
For the first term using chemical EQN 1
[Al-] [H2O]3
Ksp = ----------------
[H+]3
To get [Al-] all alone in the numerator we multiply both sides by con of H+ cubed and divide by
water cubed gets [Al-], so
[Al-] = [H+]^3 * Ksp/Kwater^3
For the second term
K is given as[Al(OH-)4]/[OH-]
If this is multiplied by [H+][OH-] , which is Kwater, and divided by [H+] the result is
[Al(OH-)4][H+][OH-]/[OH-][H+] = [Al(OH-)]
adding the first and seond terms gives
S = [Al3+] + [Al(OH-)]
Solution
This is a fun problem. How often do you get to think about a compound that is more soluble in
strong base and strong acid? Pretty nifty!
a. In very acidic solution La Chateliers shows that reaction will shift to products
Al(OH)3(s) + 3H+ ----> Al3+ + 3H2O EQN 1
or if you start with acidic solution and then add ions
Al3+ + OH- + H+ ----> Al3+ + H2O
In very basic solution there are so many OH- ions that probability is high that four will be near
each AL3+
Al 3+ +4OH- -----> Al(OH)4-
b. S has two terms, the first is large when [H+] is high the second one is large when [H+] is low.
For the first term using chemical EQN 1
[Al-] [H2O]3
Ksp = ----------------
[H+]3
To get [Al-] all alone in the numerator we multiply both sides by con of H+ cubed and divide by
water cubed gets [Al-], so
[Al-] = [H+]^3 * Ksp/Kwater^3
For the second term
K is given as[Al(OH-)4]/[OH-]
If this is multiplied by [H+][OH-] , which is Kwater, and divided by [H+] the result is
[Al(OH-)4][H+][OH-]/[OH-][H+] = [Al(OH-)]
adding the first and seond terms gives
S = [Al3+] + [Al(OH-)].
there are several theory on defining acid and base, Lewis acids and .pdf
there are several theory on defining acid and base, Lewis acids and bases is the broadest. The
species donate electron pair is lewis base and the species accept electron pair is lewis acid. so for
the second one. the oxygen in OH- has 3 pair of electrons. It donate one pair to the Al which has
empty p-orbits thus Al accept one pair of electons. So OH- is lewis base, Al(OH)3 is lewis acid.
For the first one, it would be easier when you think their product in water solution. SO2 will give
H2SO3 and CaO will give Ca(OH)2, H2SO3 is clearly a brownsted acid(also lewis acid) and
Ca(OH)2 is a brownsted base(also lewis base).
Solution
there are several theory on defining acid and base, Lewis acids and bases is the broadest. The
species donate electron pair is lewis base and the species accept electron pair is lewis acid. so for
the second one. the oxygen in OH- has 3 pair of electrons. It donate one pair to the Al which has
empty p-orbits thus Al accept one pair of electons. So OH- is lewis base, Al(OH)3 is lewis acid.
For the first one, it would be easier when you think their product in water solution. SO2 will give
H2SO3 and CaO will give Ca(OH)2, H2SO3 is clearly a brownsted acid(also lewis acid) and
Ca(OH)2 is a brownsted base(also lewis base)..
Recommended
This is a fun problem. How often do you get to think about a compoun.pdf
This is a fun problem. How often do you get to think about a compound that is more soluble in
strong base and strong acid? Pretty nifty!
a. In very acidic solution La Chateliers shows that reaction will shift to products
Al(OH)3(s) + 3H+ ----> Al3+ + 3H2O EQN 1
or if you start with acidic solution and then add ions
Al3+ + OH- + H+ ----> Al3+ + H2O
In very basic solution there are so many OH- ions that probability is high that four will be near
each AL3+
Al 3+ +4OH- -----> Al(OH)4-
b. S has two terms, the first is large when [H+] is high the second one is large when [H+] is low.
For the first term using chemical EQN 1
[Al-] [H2O]3
Ksp = ----------------
[H+]3
To get [Al-] all alone in the numerator we multiply both sides by con of H+ cubed and divide by
water cubed gets [Al-], so
[Al-] = [H+]^3 * Ksp/Kwater^3
For the second term
K is given as[Al(OH-)4]/[OH-]
If this is multiplied by [H+][OH-] , which is Kwater, and divided by [H+] the result is
[Al(OH-)4][H+][OH-]/[OH-][H+] = [Al(OH-)]
adding the first and seond terms gives
S = [Al3+] + [Al(OH-)]
Solution
This is a fun problem. How often do you get to think about a compound that is more soluble in
strong base and strong acid? Pretty nifty!
a. In very acidic solution La Chateliers shows that reaction will shift to products
Al(OH)3(s) + 3H+ ----> Al3+ + 3H2O EQN 1
or if you start with acidic solution and then add ions
Al3+ + OH- + H+ ----> Al3+ + H2O
In very basic solution there are so many OH- ions that probability is high that four will be near
each AL3+
Al 3+ +4OH- -----> Al(OH)4-
b. S has two terms, the first is large when [H+] is high the second one is large when [H+] is low.
For the first term using chemical EQN 1
[Al-] [H2O]3
Ksp = ----------------
[H+]3
To get [Al-] all alone in the numerator we multiply both sides by con of H+ cubed and divide by
water cubed gets [Al-], so
[Al-] = [H+]^3 * Ksp/Kwater^3
For the second term
K is given as[Al(OH-)4]/[OH-]
If this is multiplied by [H+][OH-] , which is Kwater, and divided by [H+] the result is
[Al(OH-)4][H+][OH-]/[OH-][H+] = [Al(OH-)]
adding the first and seond terms gives
S = [Al3+] + [Al(OH-)].
there are several theory on defining acid and base, Lewis acids and .pdf
there are several theory on defining acid and base, Lewis acids and bases is the broadest. The
species donate electron pair is lewis base and the species accept electron pair is lewis acid. so for
the second one. the oxygen in OH- has 3 pair of electrons. It donate one pair to the Al which has
empty p-orbits thus Al accept one pair of electons. So OH- is lewis base, Al(OH)3 is lewis acid.
For the first one, it would be easier when you think their product in water solution. SO2 will give
H2SO3 and CaO will give Ca(OH)2, H2SO3 is clearly a brownsted acid(also lewis acid) and
Ca(OH)2 is a brownsted base(also lewis base).
Solution
there are several theory on defining acid and base, Lewis acids and bases is the broadest. The
species donate electron pair is lewis base and the species accept electron pair is lewis acid. so for
the second one. the oxygen in OH- has 3 pair of electrons. It donate one pair to the Al which has
empty p-orbits thus Al accept one pair of electons. So OH- is lewis base, Al(OH)3 is lewis acid.
For the first one, it would be easier when you think their product in water solution. SO2 will give
H2SO3 and CaO will give Ca(OH)2, H2SO3 is clearly a brownsted acid(also lewis acid) and
Ca(OH)2 is a brownsted base(also lewis base)..
Acid base concept
An acid is any substance that in water solution tastes sour, changes blue litmus paper to red, reacts with some metals to liberate hydrogen, reacts with bases to form salts, and promotes chemical reactions (acid catalysis).
A base is a substance that can neutralize the acid by reacting with hydrogen ions. Most bases are minerals that react with acids to form water and salts.
Salt is a chemical compound consisting of an ionic assembly of cations and anions.
1) The Arrhenius Theory of acids and bases Acids are substances wh.pdf
1) The Arrhenius Theory of acids and bases: Acids are substances which produce hydrogen
ions in solution. Bases are substances which produce hydroxide ions in solution. Neutralisation
happens because hydrogen ions and hydroxide ions react to produce water. HCl(acid) + NaOH
(base) = H2O + NaCl 2) The Bronsted-Lowry Theory of acids and bases An acid is a proton
(hydrogen ion) donor. A base is a proton (hydrogen ion) acceptor. H2O (base) + HCl(acid) =
H3O^+ + Cl^- 3) The Lewis Theory of acids and bases: An acid is an electron pair acceptor. A
base is an electron pair donor. BH3(acid) + NH3(base) = H3B:NH3 (where NH3 donotes one
pair of electrons to the empty orbital of BH3 to form H3BNH3 complex.)
Solution
1) The Arrhenius Theory of acids and bases: Acids are substances which produce hydrogen
ions in solution. Bases are substances which produce hydroxide ions in solution. Neutralisation
happens because hydrogen ions and hydroxide ions react to produce water. HCl(acid) + NaOH
(base) = H2O + NaCl 2) The Bronsted-Lowry Theory of acids and bases An acid is a proton
(hydrogen ion) donor. A base is a proton (hydrogen ion) acceptor. H2O (base) + HCl(acid) =
H3O^+ + Cl^- 3) The Lewis Theory of acids and bases: An acid is an electron pair acceptor. A
base is an electron pair donor. BH3(acid) + NH3(base) = H3B:NH3 (where NH3 donotes one
pair of electrons to the empty orbital of BH3 to form H3BNH3 complex.).
#includestdio.h#includestdlib.h structure of a stack node .pdf
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct sNode* new_node =
(struct sNode*) malloc(sizeof(struct sNode));
if(new_node == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*top_ref);
/* move the head to point to the new node */
(*top_ref) = new_node;
}
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref)
{
int res;
struct sNode *top;
/*If stack is empty then error */
if(*top_ref == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
else
{
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}
/* Driver function to test anove functions */
int main()
{
/* Create a queue with items 1 2 3*/
struct queue *q = (struct queue*)malloc(sizeof(struct queue));
q->stack1 = NULL;
q->stack2 = NULL;
enQueue(q, 1);
enQueue(q, 2);
enQueue(q, 3);
/* Dequeue items */
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
getchar();
}
Solution
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct s.
Solution Since we know that the subshell with .pdf
Solution
: Since we know that the subshell with : n=2 and l=1 is the 2p subshell; Similarly : if n=3 and
l=0, it is the 3s subshell and so on that\'s why it is n=3, l=2 and m_l=0 Please rate.
let V be nonempty open subset of X then V inters.pdf
let V be nonempty open subset of X then V intersection U is nonempty because U
is dense in X therefore V is not contained in X/U so X/U is nowhere dense in X
Solution
let V be nonempty open subset of X then V intersection U is nonempty because U
is dense in X therefore V is not contained in X/U so X/U is nowhere dense in X.
Look at the periodic table. When Be loses its f.pdf
Look at the periodic table. When Be loses it\'s first electron it now only has 3
electroms (Just like Li) when it loses it\'s second electron in only has 2 electrons (just like He)
when it loses it\'s third electron it goes from 1s2 to 1s1 therefore the 1s orbital loses an electron
in the third ionization step for Be
Solution
Look at the periodic table. When Be loses it\'s first electron it now only has 3
electroms (Just like Li) when it loses it\'s second electron in only has 2 electrons (just like He)
when it loses it\'s third electron it goes from 1s2 to 1s1 therefore the 1s orbital loses an electron
in the third ionization step for Be.
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))....pdf
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))...........for periodic functions the trans
form is given as ........inte(0 to T)(T(t))exp(-st)*dt/[1-exp(-sT)].........solving we get the the
integral as [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]...........putting it in the equation
we get [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]/([s^2+4s+14])
Solution
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))...........for periodic functions the trans
form is given as ........inte(0 to T)(T(t))exp(-st)*dt/[1-exp(-sT)].........solving we get the the
integral as [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]...........putting it in the equation
we get [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]/([s^2+4s+14]).
the salt is neutral in charge. OS of Ca is +2, so.pdf
the salt is neutral in charge. OS of Ca is +2, so (NO3)2 bears -2 charges, so NO3
bears -1 charge, we know O holds -2 charge Thus, (the charge of N) + -2x3 =-1 so, N = +5 Ca
+2, N +5, O -2
Solution
the salt is neutral in charge. OS of Ca is +2, so (NO3)2 bears -2 charges, so NO3
bears -1 charge, we know O holds -2 charge Thus, (the charge of N) + -2x3 =-1 so, N = +5 Ca
+2, N +5, O -2.
nonmetals, metals, metalloids, and gases. They ar.pdf
nonmetals, metals, metalloids, and gases. They are important because they are
involved in everyday life
Solution
nonmetals, metals, metalloids, and gases. They are important because they are
involved in everyday life.
there are two types that are presented as ch3 and ch...Solution.pdf
there are two types that are presented as ch3 and ch...
Solution
there are two types that are presented as ch3 and ch....
In thermodynamics, a state function, function of .pdf
In thermodynamics, a state function, function of state, state quantity, or state
variable is a property of a system that depends only on the current state of the system, not on the
way in which the system acquired that state (independent of path). A state function describes the
equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state
quantities because they describe quantitatively an equilibrium state of a thermodynamic system,
irrespective of how the system arrived in that state. In contrast, mechanical work and heat are
process quantities because their values depend on the specific transition (or path) between two
equilibrium states. The following are considered to be state functions in thermodynamics: Mass
(m) Energy (E) Enthalpy (H) Internal energy (U) Gibbs free energy (G) Helmholtz free energy
(A) Exergy Entropy (S) Pressure (p) Temperature (T) Volume (V) Chemical composition
Specific volume (v) or its reciprocal Density (?) Fugacity Altitude Particle number (ni)
Solution
In thermodynamics, a state function, function of state, state quantity, or state
variable is a property of a system that depends only on the current state of the system, not on the
way in which the system acquired that state (independent of path). A state function describes the
equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state
quantities because they describe quantitatively an equilibrium state of a thermodynamic system,
irrespective of how the system arrived in that state. In contrast, mechanical work and heat are
process quantities because their values depend on the specific transition (or path) between two
equilibrium states. The following are considered to be state functions in thermodynamics: Mass
(m) Energy (E) Enthalpy (H) Internal energy (U) Gibbs free energy (G) Helmholtz free energy
(A) Exergy Entropy (S) Pressure (p) Temperature (T) Volume (V) Chemical composition
Specific volume (v) or its reciprocal Density (?) Fugacity Altitude Particle number (ni).
The process of protein synthesis and secretion starts from the endop.pdf
The process of protein synthesis and secretion starts from the endoplasmic reticulum. The
polypeptide chain is formed in the ribosomes, and then transferred to the Golgi apparatus, where
the final modifications are done. The Golgi apparatus sorts the modified poly peptide molecules
and packages into new transport vesicles which reach their particular destinations. The
tranlocons transport the polypeptides with a signal sequence to the interior of the endoplasmic
reticulum. Multiple control mechanisms of the endoplasmic reticulum ensures that only properly
folded proteins to export towards the Golgi apparatus.
The plasma membrane proteins synthesized in the endoplasmic reticulum (ER). The proteins
generally contain endoplasmic reticulum signal sequence at their N-terminus, which directs them
to the rough endoplasmic reticulum. After completion of translation process in the RER, the
polypeptide chains get inserted into the ER membrane or into the lumen of ER.
Some proteins move to the transport vesicles, which fuse together to form cis- Golgi vesicles, the
cis-form again moves to trans-form. During this process, many proteins undergo modifications
particularly at oligosaccharide chains. From the trans-Golgi vesicles, proteins move to cell
membrane through small transportation vesicles (exocytosis).
Solution
The process of protein synthesis and secretion starts from the endoplasmic reticulum. The
polypeptide chain is formed in the ribosomes, and then transferred to the Golgi apparatus, where
the final modifications are done. The Golgi apparatus sorts the modified poly peptide molecules
and packages into new transport vesicles which reach their particular destinations. The
tranlocons transport the polypeptides with a signal sequence to the interior of the endoplasmic
reticulum. Multiple control mechanisms of the endoplasmic reticulum ensures that only properly
folded proteins to export towards the Golgi apparatus.
The plasma membrane proteins synthesized in the endoplasmic reticulum (ER). The proteins
generally contain endoplasmic reticulum signal sequence at their N-terminus, which directs them
to the rough endoplasmic reticulum. After completion of translation process in the RER, the
polypeptide chains get inserted into the ER membrane or into the lumen of ER.
Some proteins move to the transport vesicles, which fuse together to form cis- Golgi vesicles, the
cis-form again moves to trans-form. During this process, many proteins undergo modifications
particularly at oligosaccharide chains. From the trans-Golgi vesicles, proteins move to cell
membrane through small transportation vesicles (exocytosis)..
The IUPAC names of the given compounds is 2-chloro-1-butanol.pdf
The IUPAC names of the given compounds is:
2-chloro-1-butanol
2-methoxycyclohexanol
Solution
The IUPAC names of the given compounds is:
2-chloro-1-butanol
2-methoxycyclohexanol.
Solution - (c) - 20001 SharesCalculation of Number of shares to be.pdf
Solution
- (c) - 20001 Shares
Calculation of Number of shares to be controlled to guarantee the election to be board.
= [1/(Number of Seats of Board of Directors + 1)*Shares Outstanding] + 1
= [1/(5+1)*120000 + 1]
= [(1/6)*120000] + 1
= 20001 Shares..
NOBecause here we are not working on past experience of data.S.pdf
NO
Because here we are not working on past experience of data.
Solution
NO
Because here we are not working on past experience of data..
Move decimal to behind the first number4.71Round4.714.71 X 1.pdf
Move decimal to behind the first number
4.71
Round
4.71
4.71 X 10^2 (2=the number of spaces you had to move the decimal)
Solution
Move decimal to behind the first number
4.71
Round
4.71
4.71 X 10^2 (2=the number of spaces you had to move the decimal).
More Related Content
Similar to Lewis acid is the electron pair acceptorLewis base is the electron.pdf
Acid base concept
An acid is any substance that in water solution tastes sour, changes blue litmus paper to red, reacts with some metals to liberate hydrogen, reacts with bases to form salts, and promotes chemical reactions (acid catalysis).
A base is a substance that can neutralize the acid by reacting with hydrogen ions. Most bases are minerals that react with acids to form water and salts.
Salt is a chemical compound consisting of an ionic assembly of cations and anions.
1) The Arrhenius Theory of acids and bases Acids are substances wh.pdf
1) The Arrhenius Theory of acids and bases: Acids are substances which produce hydrogen
ions in solution. Bases are substances which produce hydroxide ions in solution. Neutralisation
happens because hydrogen ions and hydroxide ions react to produce water. HCl(acid) + NaOH
(base) = H2O + NaCl 2) The Bronsted-Lowry Theory of acids and bases An acid is a proton
(hydrogen ion) donor. A base is a proton (hydrogen ion) acceptor. H2O (base) + HCl(acid) =
H3O^+ + Cl^- 3) The Lewis Theory of acids and bases: An acid is an electron pair acceptor. A
base is an electron pair donor. BH3(acid) + NH3(base) = H3B:NH3 (where NH3 donotes one
pair of electrons to the empty orbital of BH3 to form H3BNH3 complex.)
Solution
1) The Arrhenius Theory of acids and bases: Acids are substances which produce hydrogen
ions in solution. Bases are substances which produce hydroxide ions in solution. Neutralisation
happens because hydrogen ions and hydroxide ions react to produce water. HCl(acid) + NaOH
(base) = H2O + NaCl 2) The Bronsted-Lowry Theory of acids and bases An acid is a proton
(hydrogen ion) donor. A base is a proton (hydrogen ion) acceptor. H2O (base) + HCl(acid) =
H3O^+ + Cl^- 3) The Lewis Theory of acids and bases: An acid is an electron pair acceptor. A
base is an electron pair donor. BH3(acid) + NH3(base) = H3B:NH3 (where NH3 donotes one
pair of electrons to the empty orbital of BH3 to form H3BNH3 complex.).
Similar to Lewis acid is the electron pair acceptorLewis base is the electron.pdf (7)
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#include
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struct sNode
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int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
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struct sNode *stack2;
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getchar();
exit(0);
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{
/* allocate node */
struct sNode* new_node =
(struct sNode*) malloc(sizeof(struct sNode));
if(new_node == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*top_ref);
/* move the head to point to the new node */
(*top_ref) = new_node;
}
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref)
{
int res;
struct sNode *top;
/*If stack is empty then error */
if(*top_ref == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
else
{
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}
/* Driver function to test anove functions */
int main()
{
/* Create a queue with items 1 2 3*/
struct queue *q = (struct queue*)malloc(sizeof(struct queue));
q->stack1 = NULL;
q->stack2 = NULL;
enQueue(q, 1);
enQueue(q, 2);
enQueue(q, 3);
/* Dequeue items */
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
getchar();
}
Solution
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct s.
Solution Since we know that the subshell with .pdf
Solution
: Since we know that the subshell with : n=2 and l=1 is the 2p subshell; Similarly : if n=3 and
l=0, it is the 3s subshell and so on that\'s why it is n=3, l=2 and m_l=0 Please rate.
let V be nonempty open subset of X then V inters.pdf
let V be nonempty open subset of X then V intersection U is nonempty because U
is dense in X therefore V is not contained in X/U so X/U is nowhere dense in X
Solution
let V be nonempty open subset of X then V intersection U is nonempty because U
is dense in X therefore V is not contained in X/U so X/U is nowhere dense in X.
Look at the periodic table. When Be loses its f.pdf
Look at the periodic table. When Be loses it\'s first electron it now only has 3
electroms (Just like Li) when it loses it\'s second electron in only has 2 electrons (just like He)
when it loses it\'s third electron it goes from 1s2 to 1s1 therefore the 1s orbital loses an electron
in the third ionization step for Be
Solution
Look at the periodic table. When Be loses it\'s first electron it now only has 3
electroms (Just like Li) when it loses it\'s second electron in only has 2 electrons (just like He)
when it loses it\'s third electron it goes from 1s2 to 1s1 therefore the 1s orbital loses an electron
in the third ionization step for Be.
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))....pdf
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))...........for periodic functions the trans
form is given as ........inte(0 to T)(T(t))exp(-st)*dt/[1-exp(-sT)].........solving we get the the
integral as [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]...........putting it in the equation
we get [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]/([s^2+4s+14])
Solution
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))...........for periodic functions the trans
form is given as ........inte(0 to T)(T(t))exp(-st)*dt/[1-exp(-sT)].........solving we get the the
integral as [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]...........putting it in the equation
we get [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]/([s^2+4s+14]).
the salt is neutral in charge. OS of Ca is +2, so.pdf
the salt is neutral in charge. OS of Ca is +2, so (NO3)2 bears -2 charges, so NO3
bears -1 charge, we know O holds -2 charge Thus, (the charge of N) + -2x3 =-1 so, N = +5 Ca
+2, N +5, O -2
Solution
the salt is neutral in charge. OS of Ca is +2, so (NO3)2 bears -2 charges, so NO3
bears -1 charge, we know O holds -2 charge Thus, (the charge of N) + -2x3 =-1 so, N = +5 Ca
+2, N +5, O -2.
nonmetals, metals, metalloids, and gases. They ar.pdf
nonmetals, metals, metalloids, and gases. They are important because they are
involved in everyday life
Solution
nonmetals, metals, metalloids, and gases. They are important because they are
involved in everyday life.
there are two types that are presented as ch3 and ch...Solution.pdf
there are two types that are presented as ch3 and ch...
Solution
there are two types that are presented as ch3 and ch....
In thermodynamics, a state function, function of .pdf
In thermodynamics, a state function, function of state, state quantity, or state
variable is a property of a system that depends only on the current state of the system, not on the
way in which the system acquired that state (independent of path). A state function describes the
equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state
quantities because they describe quantitatively an equilibrium state of a thermodynamic system,
irrespective of how the system arrived in that state. In contrast, mechanical work and heat are
process quantities because their values depend on the specific transition (or path) between two
equilibrium states. The following are considered to be state functions in thermodynamics: Mass
(m) Energy (E) Enthalpy (H) Internal energy (U) Gibbs free energy (G) Helmholtz free energy
(A) Exergy Entropy (S) Pressure (p) Temperature (T) Volume (V) Chemical composition
Specific volume (v) or its reciprocal Density (?) Fugacity Altitude Particle number (ni)
Solution
In thermodynamics, a state function, function of state, state quantity, or state
variable is a property of a system that depends only on the current state of the system, not on the
way in which the system acquired that state (independent of path). A state function describes the
equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state
quantities because they describe quantitatively an equilibrium state of a thermodynamic system,
irrespective of how the system arrived in that state. In contrast, mechanical work and heat are
process quantities because their values depend on the specific transition (or path) between two
equilibrium states. The following are considered to be state functions in thermodynamics: Mass
(m) Energy (E) Enthalpy (H) Internal energy (U) Gibbs free energy (G) Helmholtz free energy
(A) Exergy Entropy (S) Pressure (p) Temperature (T) Volume (V) Chemical composition
Specific volume (v) or its reciprocal Density (?) Fugacity Altitude Particle number (ni).
The process of protein synthesis and secretion starts from the endop.pdf
The process of protein synthesis and secretion starts from the endoplasmic reticulum. The
polypeptide chain is formed in the ribosomes, and then transferred to the Golgi apparatus, where
the final modifications are done. The Golgi apparatus sorts the modified poly peptide molecules
and packages into new transport vesicles which reach their particular destinations. The
tranlocons transport the polypeptides with a signal sequence to the interior of the endoplasmic
reticulum. Multiple control mechanisms of the endoplasmic reticulum ensures that only properly
folded proteins to export towards the Golgi apparatus.
The plasma membrane proteins synthesized in the endoplasmic reticulum (ER). The proteins
generally contain endoplasmic reticulum signal sequence at their N-terminus, which directs them
to the rough endoplasmic reticulum. After completion of translation process in the RER, the
polypeptide chains get inserted into the ER membrane or into the lumen of ER.
Some proteins move to the transport vesicles, which fuse together to form cis- Golgi vesicles, the
cis-form again moves to trans-form. During this process, many proteins undergo modifications
particularly at oligosaccharide chains. From the trans-Golgi vesicles, proteins move to cell
membrane through small transportation vesicles (exocytosis).
Solution
The process of protein synthesis and secretion starts from the endoplasmic reticulum. The
polypeptide chain is formed in the ribosomes, and then transferred to the Golgi apparatus, where
the final modifications are done. The Golgi apparatus sorts the modified poly peptide molecules
and packages into new transport vesicles which reach their particular destinations. The
tranlocons transport the polypeptides with a signal sequence to the interior of the endoplasmic
reticulum. Multiple control mechanisms of the endoplasmic reticulum ensures that only properly
folded proteins to export towards the Golgi apparatus.
The plasma membrane proteins synthesized in the endoplasmic reticulum (ER). The proteins
generally contain endoplasmic reticulum signal sequence at their N-terminus, which directs them
to the rough endoplasmic reticulum. After completion of translation process in the RER, the
polypeptide chains get inserted into the ER membrane or into the lumen of ER.
Some proteins move to the transport vesicles, which fuse together to form cis- Golgi vesicles, the
cis-form again moves to trans-form. During this process, many proteins undergo modifications
particularly at oligosaccharide chains. From the trans-Golgi vesicles, proteins move to cell
membrane through small transportation vesicles (exocytosis)..
The IUPAC names of the given compounds is 2-chloro-1-butanol.pdf
The IUPAC names of the given compounds is:
2-chloro-1-butanol
2-methoxycyclohexanol
Solution
The IUPAC names of the given compounds is:
2-chloro-1-butanol
2-methoxycyclohexanol.
Solution - (c) - 20001 SharesCalculation of Number of shares to be.pdf
Solution
- (c) - 20001 Shares
Calculation of Number of shares to be controlled to guarantee the election to be board.
= [1/(Number of Seats of Board of Directors + 1)*Shares Outstanding] + 1
= [1/(5+1)*120000 + 1]
= [(1/6)*120000] + 1
= 20001 Shares..
NOBecause here we are not working on past experience of data.S.pdf
NO
Because here we are not working on past experience of data.
Solution
NO
Because here we are not working on past experience of data..
Move decimal to behind the first number4.71Round4.714.71 X 1.pdf
Move decimal to behind the first number
4.71
Round
4.71
4.71 X 10^2 (2=the number of spaces you had to move the decimal)
Solution
Move decimal to behind the first number
4.71
Round
4.71
4.71 X 10^2 (2=the number of spaces you had to move the decimal).
first treat with alcoholic KOH to get cyclo hexene and then treat it.pdf
first treat with alcoholic KOH to get cyclo hexene and then treat it with alkaline KMnO4 to get
trans cyclohxane -1,2 -diol
Solution
first treat with alcoholic KOH to get cyclo hexene and then treat it with alkaline KMnO4 to get
trans cyclohxane -1,2 -diol.
An azeotrope is a mixture of two or more liquids .pdf
An azeotrope is a mixture of two or more liquids in such a ratio that its composition
cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the
resulting vapor has the same ratio of constituents as the original mixture. Because their
composition is unchanged by distillation, azeotropes are also called constant boiling mixtures.
Common drying agenets for organic compounds : Alcohols : Anhydrous K2CO3, CaSO4 or
Mg2SO4 and CaO Alkylhalides/aryl halides : Anhydrous CaCl2,CaSO4 or Mg2SO4 and
phosphoric acid Saturated aromatic hydrocarbonds : CaCl2,CaSO4 and phosphoric oxide
Aldehydes : CaSO4,Mg2SO4 Na2SO4 Ketones : CaSO4,MgSO4 ,NaSO4 or K2CO3 Organic
bases : Solid KOH/NaOH ,CaO or Barium oxide Organic acids : Calcium sulphate, Mg2SO4
Solution
An azeotrope is a mixture of two or more liquids in such a ratio that its composition
cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the
resulting vapor has the same ratio of constituents as the original mixture. Because their
composition is unchanged by distillation, azeotropes are also called constant boiling mixtures.
Common drying agenets for organic compounds : Alcohols : Anhydrous K2CO3, CaSO4 or
Mg2SO4 and CaO Alkylhalides/aryl halides : Anhydrous CaCl2,CaSO4 or Mg2SO4 and
phosphoric acid Saturated aromatic hydrocarbonds : CaCl2,CaSO4 and phosphoric oxide
Aldehydes : CaSO4,Mg2SO4 Na2SO4 Ketones : CaSO4,MgSO4 ,NaSO4 or K2CO3 Organic
bases : Solid KOH/NaOH ,CaO or Barium oxide Organic acids : Calcium sulphate, Mg2SO4.
Here we have sodium bisulfate NaHSO4, as you have shown the ions pr.pdf
Here we have sodium bisulfate: NaHSO4, as you have shown the ions present are of equal
charge. Thus, there are equal amounts of both ions.
Recall molarity,
molarity = mole / volume
Thus, we can calculate the number of moles:
M = mol / V
0.55 = mol / 1.32L
Solving for moles gives: mole = 0.726 moles.
Since we established earlier, both species (Na+ and HSO4- are present in equal amounts), thus
moles (Na+) = moles (HSO4-) = 0.726 moles
Please don\'t forget to rate, I hope this helped! :)
Solution
Here we have sodium bisulfate: NaHSO4, as you have shown the ions present are of equal
charge. Thus, there are equal amounts of both ions.
Recall molarity,
molarity = mole / volume
Thus, we can calculate the number of moles:
M = mol / V
0.55 = mol / 1.32L
Solving for moles gives: mole = 0.726 moles.
Since we established earlier, both species (Na+ and HSO4- are present in equal amounts), thus
moles (Na+) = moles (HSO4-) = 0.726 moles
Please don\'t forget to rate, I hope this helped! :).
Frontal lobe is the region of forebrain which comprises of various c.pdf
Frontal lobe is the region of forebrain which comprises of various centres for motor and concious
functions, including intelligence and decision making. Some specific areas are defined in the
frontal lobe according to their specifc functions. These are:
1. The motor strip: It is a part of the primary motor cortex of the frontal and prefrontal cortex of
the brain and maintains essential functions related to motion and movement. It is also known as
the precentral gyrus.
2. The precentral gyrus: This area structurally forms the posterior lobes of the frontal area cortex
and is important for the activity of motor neurons of limbs.
3. Broca\'s speech area: This region resides in a major part of the frontal lobe and is essential for
learning ability of languages and speech.
Importantly, Wernicke\'s speech area is an integral part of the temporal lobe of the brain near the
ears. It is not a part of the frontal lobe. Thus, the correct answers is choice A, WERNICKE\'S
SPEECH AREA.
Solution
Frontal lobe is the region of forebrain which comprises of various centres for motor and concious
functions, including intelligence and decision making. Some specific areas are defined in the
frontal lobe according to their specifc functions. These are:
1. The motor strip: It is a part of the primary motor cortex of the frontal and prefrontal cortex of
the brain and maintains essential functions related to motion and movement. It is also known as
the precentral gyrus.
2. The precentral gyrus: This area structurally forms the posterior lobes of the frontal area cortex
and is important for the activity of motor neurons of limbs.
3. Broca\'s speech area: This region resides in a major part of the frontal lobe and is essential for
learning ability of languages and speech.
Importantly, Wernicke\'s speech area is an integral part of the temporal lobe of the brain near the
ears. It is not a part of the frontal lobe. Thus, the correct answers is choice A, WERNICKE\'S
SPEECH AREA..
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NOBecause here we are not working on past experience of data.S.pdf
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Lewis acid is the electron pair acceptorLewis base is the electron.pdf
- 1. Lewis acid is the electron pair acceptor Lewis base is the electron pair donar a) OH- + Al(OH)3 -----> Al(OH)4- Al(OH)3 is the lewis acid OH- is the lewis base b) SO3 + H2O --------> H2SO4 H2O is the lewis acid SO3 is the lewis base H2O is the lewis acid SO3 is the lewis base c) Co3+ + 6NH3 -----> Co(NH3)63+ Co 3+ is the lewis acid NH3 is the lewis base Co 3+ is the lewis acid NH3 is the lewis base Solution Lewis acid is the electron pair acceptor Lewis base is the electron pair donar a) OH- + Al(OH)3 -----> Al(OH)4- Al(OH)3 is the lewis acid OH- is the lewis base b) SO3 + H2O --------> H2SO4 H2O is the lewis acid SO3 is the lewis base H2O is the lewis acid SO3 is the lewis base c) Co3+ + 6NH3 -----> Co(NH3)63+
- 2. Co 3+ is the lewis acid NH3 is the lewis base Co 3+ is the lewis acid NH3 is the lewis base