bonde order of NO in nitrate 1.33 in nitrite is 1.pdf
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bonde order of NO in nitrate 1.33 in nitrite is 1.5 in NO+ is 3.0 in NO- is 2.0 j. NO+ > NO- > nitrite > nitrate is correct. Solution bonde order of NO in nitrate 1.33 in nitrite is 1.5 in NO+ is 3.0 in NO- is 2.0 j. NO+ > NO- > nitrite > nitrate is correct..
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(deftemp Animal (slot loc (default greencouch)) (slot.pdf
(deftemp Animal
(slot loc
(default greencouch))
(slot on-top-of
(default roof))
(slot hold
(default nothing)))
(deftemp thing
(slot name
(default NONE))
(slot loc
(default NONE))
(slot on-top-of
(default roof))
(slot mass
(default light weight)))
(deftemp chest
(slot name
(default NONE))
(slot contents
(default NONE))
(slot unlocked-by
(default NONE)))
(deftemp goal-was
(slot act
(default NONE))
(multislot args))
(defrule hold chest to get of roof \"\"
(goal-was (act unlock) (args ?chest))
(thing (name ?chest) (on-top-of ~roof) (mass light weight))
(Animal (hold ~?chest))
(not (goal-was (act hold) (args ?chest)))
=>
(assert (goal-was (act hold) (args ?chest))))
(defrule put-chest-on-roof \"\"
(goal-was (act unlock) (args ?chest))
?Animal <- (Animal (loc ?place) (on-top-of ?on) (hold ?chest))
?thing <- (thing (name ?chest))
=>
(printout t \"Animal throws the \" ?chest \" off the \"
?on \" onto the roof.\" crlf)
(modify ?Animal (hold blank))
(modify ?thing (loc ?place) (on-top-of roof)))
(defrule get-key-to-unlock \"\"
(goal-was (act unlock) (args ?obj))
(thing (name ?obj) (on-top-of roof))
(chest (name ?obj) (unlocked-by ?key))
(Animal (hold ~?key))
(not (goal-was (act hold) (args ?key)))
=>
(assert (goal-was (act hold) (args ?key))))
(defrule move-to-chest-with-key \"\"
(goal-was (act unlock) (args ?chest))
(Animal (loc ?mplace) (hold ?key))
(thing (name ?chest) (loc ?cplace&~?mplace) (on-top-of roof))
(chest (name ?chest) (unlocked-by ?key))
(not (goal-was (act walk-to) (args ?cplace)))
=>
(assert (goal-was (act walk-to) (args ?cplace))))
(defrule unlock-chest-with-key \"\"
?goal <- (goal-was (act unlock) (args ?name))
?chest <- (chest (name ?name) (contents ?contents) (unlocked-by ?key))
(thing (name ?name) (loc ?place) (on-top-of ?on))
(Animal (loc ?place) (on-top-of ?on) (hold ?key))
=>
(printout t \"Animal opens the \" ?name \" with the \" ?key
\" revealing the \" ?contents \".\" crlf)
(modify ?chest (contents nothing))
(assert (thing (name ?contents) (loc ?place) (on-top-of ?name)))
(retract ?goal))
;;;*********************
;;;* HOLD OBJECT RULES *
;;;*********************
(defrule unlock-chest-to-hold-object \"\"
(goal-was (act hold) (args ?obj))
(chest (name ?chest) (contents ?obj))
(not (goal-was (act unlock) (args ?chest)))
=>
(assert (goal-was (act unlock) (args ?chest))))
(defrule use-ladder-to-hold \"\"
(goal-was (act hold) (args ?obj))
(thing (name ?obj) (loc ?place) (on-top-of ceiling) (mass light weight))
(not (thing (name ladder) (loc ?place)))
(not (goal-was (act move) (args ladder ?place)))
=>
(assert (goal-was (act move) (args ladder ?place))))
(defrule climb-ladder-to-hold \"\"
(goal-was (act hold) (args ?obj))
(thing (name ?obj) (loc ?place) (on-top-of ceiling) (mass light weight))
(thing (name ladder) (loc ?place) (on-top-of roof))
(Animal (on-top-of ~ladder))
(not (goal-was (act on) (args ladder)))
=>
(assert (goal-was (act on) (args ladder))))
(defrule grab-object-from-ladder \"\"
?goal <- (goal-was (act hold) (args ?.
2. The pressures of O2 and NO will be negligibleSolution 2. T.pdf
2. The pressures of O2 and NO will be negligible
Solution
2. The pressures of O2 and NO will be negligible.
# Imports# Include your imports here, if any are used. import.pdf
# Imports
# Include your imports here, if any are used.
import math
import random
import copy
import Queue
from collections import deque
# Section 1: N-Queens
# N^2 choose N
def num_placements_all(n):
return math.factorial(n * n) / (math.factorial(n) * math.factorial((n * n) - n))
#N^N
def num_placements_one_per_row(n):
return n ** n
def n_queens_valid(board):
# set() will merge any row with the same column number
# hence if there are conflicts the length of the set will
# be smaller than the actual list
if len(set(board)) < len(board):
return False
# check if each pair of queen is on a diagonal
for i in range(len(board)) :
for j in range(i+1,len(board)) :
if j-i == abs(board[i]-board[j]) :
return False
# return true if every piece is in valid spot
return True
def n_queens_solutions(n):
for i in range(n):
for solution in n_queens_helper(n, [i]):
yield solution
def n_queens_helper(n, board):
if n_queens_valid(board):
if len(board) == n:
yield board
else:
#iterate through all col not added so far
for i in [col for col in range(n) if col not in board]:
#do a quick check if the piece we\'re adding is not directly below
#nor diagonal to the last piece - faster than n_queens_valid
if(i !=board[-1] and i !=board[-1]+1 and i!=board[-1]-1):
newBoard = list(board);
newBoard.append(i);
for solution in n_queens_helper(n, newBoard):
if solution:
yield solution
# Section 2: Lights Out
class LightsOutPuzzle(object):
def __init__(self, board):
self.board = board
self.rLength = len(board)-1
self.cLength = len(board[0])-1
def get_board(self):
#so we don\'t edit the internal representation by mistake on
#the outside
return list(self.board)
def perform_move(self, row, col):
self.board[row][col] = not self.board[row][col]
#bounds checking
if row - 1 >= 0:
self.board[row - 1][col] = not self.board[row - 1][col]
if row + 1 <= self.rLength :
self.board[row + 1][col] = not self.board[row + 1][col]
if col - 1 >= 0:
self.board[row][col - 1] = not self.board[row][col - 1]
if col + 1 <= self.cLength :
self.board[row][col + 1] = not self.board[row][col + 1]
def scramble(self):
for row in range(self.rLength+1):
for col in range(self.cLength+1):
if random.random() < 0.5:
self.perform_move(row, col)
def is_solved(self):
return not any([i for row in self.board for i in row])
def copy(self):
return copy.deepcopy(self)
def successors(self):
for row in range(self.rLength+1):
for col in range(self.cLength+1):
newBoard = self.copy()
newBoard.perform_move(row, col)
yield ((row, col), newBoard)
def find_solution(self):
q = Queue.Queue()
q.put(self)
explored = set()
parent = {}
parent[self] = None
moves = {}
moves[self] = None
solution = []
while not q.empty() :
board = q.get()
explored.add(board.toTup());
if board.is_solved():
node = board
while not parent[node] == None:
solution.append(tuple(moves[node]))
node = parent[node]
return list(reversed(solution))
else :
for move, nextBoard in board.successors() :
if nextBoard.toTup() not in explored :
q.put(nextBoard)
moves[nextBoard] = m.
The OMe is a strong electron withdrawing group an.pdf
The OMe is a strong electron withdrawing group and the OSiMe3 is a good electron
donating group. Together they provide good regioselectivity because they make their Cs more
reactive to the nucleophile and electrophile respectively of the dienophile. Also the OSiMe3 is a
large substituent and can use steric hinderance to direct the regioselectivity.
Solution
The OMe is a strong electron withdrawing group and the OSiMe3 is a good electron
donating group. Together they provide good regioselectivity because they make their Cs more
reactive to the nucleophile and electrophile respectively of the dienophile. Also the OSiMe3 is a
large substituent and can use steric hinderance to direct the regioselectivity..
O bears three lone pairs and is bound to a H, thu.pdf
O bears three lone pairs and is bound to a H, thus it is sp3-hybrized.
Solution
O bears three lone pairs and is bound to a H, thus it is sp3-hybrized..
Nacl and H20 so, Salt & water .pdf
Nacl and H20 so, Salt & water
Solution
Nacl and H20 so, Salt & water.
In terms of ion charge, we notice that the charge.pdf
In terms of ion charge, we notice that the charge of elements increases by one as
you move from left to right. For example, Sodium, Na, has a plus 1+ charge in solution and
Calcium, Ca, has a 2+ charge. On the opposite side of the table, the elements are more likely to
gain and electron and have a negative charge than lose an electron and have a positive charge in
solution, so starting from the noble gases and moving towards the center, the elements tend to
gain a negative charge for each column. The halides, which include Chlorine, Fluorine, Bromine
and Iodine all tend to have a 1- charge, while Oxygen has a 2- and nitrogen, a 3-. The transition
metals in the middle of the table tend to form positive charges of different values, known as
oxidation states. These do not necessarily follow left to right trend.
Solution
In terms of ion charge, we notice that the charge of elements increases by one as
you move from left to right. For example, Sodium, Na, has a plus 1+ charge in solution and
Calcium, Ca, has a 2+ charge. On the opposite side of the table, the elements are more likely to
gain and electron and have a negative charge than lose an electron and have a positive charge in
solution, so starting from the noble gases and moving towards the center, the elements tend to
gain a negative charge for each column. The halides, which include Chlorine, Fluorine, Bromine
and Iodine all tend to have a 1- charge, while Oxygen has a 2- and nitrogen, a 3-. The transition
metals in the middle of the table tend to form positive charges of different values, known as
oxidation states. These do not necessarily follow left to right trend..
Here are four characteristics of acid a) A char.pdf
Here are four characteristics of acid: a) A characteristic sour taste b) ability to
change the color of litmus paper from blue to red c) react with certain metals to produce gaseous
H2 d) react with bases to form a salt and water
Solution
Here are four characteristics of acid: a) A characteristic sour taste b) ability to
change the color of litmus paper from blue to red c) react with certain metals to produce gaseous
H2 d) react with bases to form a salt and water.
Recommended
(deftemp Animal (slot loc (default greencouch)) (slot.pdf
(deftemp Animal
(slot loc
(default greencouch))
(slot on-top-of
(default roof))
(slot hold
(default nothing)))
(deftemp thing
(slot name
(default NONE))
(slot loc
(default NONE))
(slot on-top-of
(default roof))
(slot mass
(default light weight)))
(deftemp chest
(slot name
(default NONE))
(slot contents
(default NONE))
(slot unlocked-by
(default NONE)))
(deftemp goal-was
(slot act
(default NONE))
(multislot args))
(defrule hold chest to get of roof \"\"
(goal-was (act unlock) (args ?chest))
(thing (name ?chest) (on-top-of ~roof) (mass light weight))
(Animal (hold ~?chest))
(not (goal-was (act hold) (args ?chest)))
=>
(assert (goal-was (act hold) (args ?chest))))
(defrule put-chest-on-roof \"\"
(goal-was (act unlock) (args ?chest))
?Animal <- (Animal (loc ?place) (on-top-of ?on) (hold ?chest))
?thing <- (thing (name ?chest))
=>
(printout t \"Animal throws the \" ?chest \" off the \"
?on \" onto the roof.\" crlf)
(modify ?Animal (hold blank))
(modify ?thing (loc ?place) (on-top-of roof)))
(defrule get-key-to-unlock \"\"
(goal-was (act unlock) (args ?obj))
(thing (name ?obj) (on-top-of roof))
(chest (name ?obj) (unlocked-by ?key))
(Animal (hold ~?key))
(not (goal-was (act hold) (args ?key)))
=>
(assert (goal-was (act hold) (args ?key))))
(defrule move-to-chest-with-key \"\"
(goal-was (act unlock) (args ?chest))
(Animal (loc ?mplace) (hold ?key))
(thing (name ?chest) (loc ?cplace&~?mplace) (on-top-of roof))
(chest (name ?chest) (unlocked-by ?key))
(not (goal-was (act walk-to) (args ?cplace)))
=>
(assert (goal-was (act walk-to) (args ?cplace))))
(defrule unlock-chest-with-key \"\"
?goal <- (goal-was (act unlock) (args ?name))
?chest <- (chest (name ?name) (contents ?contents) (unlocked-by ?key))
(thing (name ?name) (loc ?place) (on-top-of ?on))
(Animal (loc ?place) (on-top-of ?on) (hold ?key))
=>
(printout t \"Animal opens the \" ?name \" with the \" ?key
\" revealing the \" ?contents \".\" crlf)
(modify ?chest (contents nothing))
(assert (thing (name ?contents) (loc ?place) (on-top-of ?name)))
(retract ?goal))
;;;*********************
;;;* HOLD OBJECT RULES *
;;;*********************
(defrule unlock-chest-to-hold-object \"\"
(goal-was (act hold) (args ?obj))
(chest (name ?chest) (contents ?obj))
(not (goal-was (act unlock) (args ?chest)))
=>
(assert (goal-was (act unlock) (args ?chest))))
(defrule use-ladder-to-hold \"\"
(goal-was (act hold) (args ?obj))
(thing (name ?obj) (loc ?place) (on-top-of ceiling) (mass light weight))
(not (thing (name ladder) (loc ?place)))
(not (goal-was (act move) (args ladder ?place)))
=>
(assert (goal-was (act move) (args ladder ?place))))
(defrule climb-ladder-to-hold \"\"
(goal-was (act hold) (args ?obj))
(thing (name ?obj) (loc ?place) (on-top-of ceiling) (mass light weight))
(thing (name ladder) (loc ?place) (on-top-of roof))
(Animal (on-top-of ~ladder))
(not (goal-was (act on) (args ladder)))
=>
(assert (goal-was (act on) (args ladder))))
(defrule grab-object-from-ladder \"\"
?goal <- (goal-was (act hold) (args ?.
2. The pressures of O2 and NO will be negligibleSolution 2. T.pdf
2. The pressures of O2 and NO will be negligible
Solution
2. The pressures of O2 and NO will be negligible.
# Imports# Include your imports here, if any are used. import.pdf
# Imports
# Include your imports here, if any are used.
import math
import random
import copy
import Queue
from collections import deque
# Section 1: N-Queens
# N^2 choose N
def num_placements_all(n):
return math.factorial(n * n) / (math.factorial(n) * math.factorial((n * n) - n))
#N^N
def num_placements_one_per_row(n):
return n ** n
def n_queens_valid(board):
# set() will merge any row with the same column number
# hence if there are conflicts the length of the set will
# be smaller than the actual list
if len(set(board)) < len(board):
return False
# check if each pair of queen is on a diagonal
for i in range(len(board)) :
for j in range(i+1,len(board)) :
if j-i == abs(board[i]-board[j]) :
return False
# return true if every piece is in valid spot
return True
def n_queens_solutions(n):
for i in range(n):
for solution in n_queens_helper(n, [i]):
yield solution
def n_queens_helper(n, board):
if n_queens_valid(board):
if len(board) == n:
yield board
else:
#iterate through all col not added so far
for i in [col for col in range(n) if col not in board]:
#do a quick check if the piece we\'re adding is not directly below
#nor diagonal to the last piece - faster than n_queens_valid
if(i !=board[-1] and i !=board[-1]+1 and i!=board[-1]-1):
newBoard = list(board);
newBoard.append(i);
for solution in n_queens_helper(n, newBoard):
if solution:
yield solution
# Section 2: Lights Out
class LightsOutPuzzle(object):
def __init__(self, board):
self.board = board
self.rLength = len(board)-1
self.cLength = len(board[0])-1
def get_board(self):
#so we don\'t edit the internal representation by mistake on
#the outside
return list(self.board)
def perform_move(self, row, col):
self.board[row][col] = not self.board[row][col]
#bounds checking
if row - 1 >= 0:
self.board[row - 1][col] = not self.board[row - 1][col]
if row + 1 <= self.rLength :
self.board[row + 1][col] = not self.board[row + 1][col]
if col - 1 >= 0:
self.board[row][col - 1] = not self.board[row][col - 1]
if col + 1 <= self.cLength :
self.board[row][col + 1] = not self.board[row][col + 1]
def scramble(self):
for row in range(self.rLength+1):
for col in range(self.cLength+1):
if random.random() < 0.5:
self.perform_move(row, col)
def is_solved(self):
return not any([i for row in self.board for i in row])
def copy(self):
return copy.deepcopy(self)
def successors(self):
for row in range(self.rLength+1):
for col in range(self.cLength+1):
newBoard = self.copy()
newBoard.perform_move(row, col)
yield ((row, col), newBoard)
def find_solution(self):
q = Queue.Queue()
q.put(self)
explored = set()
parent = {}
parent[self] = None
moves = {}
moves[self] = None
solution = []
while not q.empty() :
board = q.get()
explored.add(board.toTup());
if board.is_solved():
node = board
while not parent[node] == None:
solution.append(tuple(moves[node]))
node = parent[node]
return list(reversed(solution))
else :
for move, nextBoard in board.successors() :
if nextBoard.toTup() not in explored :
q.put(nextBoard)
moves[nextBoard] = m.
The OMe is a strong electron withdrawing group an.pdf
The OMe is a strong electron withdrawing group and the OSiMe3 is a good electron
donating group. Together they provide good regioselectivity because they make their Cs more
reactive to the nucleophile and electrophile respectively of the dienophile. Also the OSiMe3 is a
large substituent and can use steric hinderance to direct the regioselectivity.
Solution
The OMe is a strong electron withdrawing group and the OSiMe3 is a good electron
donating group. Together they provide good regioselectivity because they make their Cs more
reactive to the nucleophile and electrophile respectively of the dienophile. Also the OSiMe3 is a
large substituent and can use steric hinderance to direct the regioselectivity..
O bears three lone pairs and is bound to a H, thu.pdf
O bears three lone pairs and is bound to a H, thus it is sp3-hybrized.
Solution
O bears three lone pairs and is bound to a H, thus it is sp3-hybrized..
Nacl and H20 so, Salt & water .pdf
Nacl and H20 so, Salt & water
Solution
Nacl and H20 so, Salt & water.
In terms of ion charge, we notice that the charge.pdf
In terms of ion charge, we notice that the charge of elements increases by one as
you move from left to right. For example, Sodium, Na, has a plus 1+ charge in solution and
Calcium, Ca, has a 2+ charge. On the opposite side of the table, the elements are more likely to
gain and electron and have a negative charge than lose an electron and have a positive charge in
solution, so starting from the noble gases and moving towards the center, the elements tend to
gain a negative charge for each column. The halides, which include Chlorine, Fluorine, Bromine
and Iodine all tend to have a 1- charge, while Oxygen has a 2- and nitrogen, a 3-. The transition
metals in the middle of the table tend to form positive charges of different values, known as
oxidation states. These do not necessarily follow left to right trend.
Solution
In terms of ion charge, we notice that the charge of elements increases by one as
you move from left to right. For example, Sodium, Na, has a plus 1+ charge in solution and
Calcium, Ca, has a 2+ charge. On the opposite side of the table, the elements are more likely to
gain and electron and have a negative charge than lose an electron and have a positive charge in
solution, so starting from the noble gases and moving towards the center, the elements tend to
gain a negative charge for each column. The halides, which include Chlorine, Fluorine, Bromine
and Iodine all tend to have a 1- charge, while Oxygen has a 2- and nitrogen, a 3-. The transition
metals in the middle of the table tend to form positive charges of different values, known as
oxidation states. These do not necessarily follow left to right trend..
Here are four characteristics of acid a) A char.pdf
Here are four characteristics of acid: a) A characteristic sour taste b) ability to
change the color of litmus paper from blue to red c) react with certain metals to produce gaseous
H2 d) react with bases to form a salt and water
Solution
Here are four characteristics of acid: a) A characteristic sour taste b) ability to
change the color of litmus paper from blue to red c) react with certain metals to produce gaseous
H2 d) react with bases to form a salt and water.
Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu .pdf
Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu
Solution
Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu.
You can use the below function for linear interpolation i.e.#inclu.pdf
You can use the below function for linear interpolation i.e.
#include
int NearestIndex( double value, double *x, int len )
{
double dist;
int index;
int i;
index = -1;
dist = Double_Max;
for ( i = 0; i < len; i++ ) {
double newDist = value - x[i];
if ( newDist > 0 && newDist < dist ) {
dist = newDist;
index = i;
}
}
return index;
}
void linearinterpolation(double *x, int x_tem, double *y, double *xx, int xx_tem, double *yy)
{
double da, db, *slp, *inter;
int i, Envectorindice;
slp=(double *)calloc(x_tem,sizeof(double));
inter=(double *)calloc(x_tem,sizeof(double));
for(i = 0; i < x_tem; i++){
if(i
Solution
You can use the below function for linear interpolation i.e.
#include
int NearestIndex( double value, double *x, int len )
{
double dist;
int index;
int i;
index = -1;
dist = Double_Max;
for ( i = 0; i < len; i++ ) {
double newDist = value - x[i];
if ( newDist > 0 && newDist < dist ) {
dist = newDist;
index = i;
}
}
return index;
}
void linearinterpolation(double *x, int x_tem, double *y, double *xx, int xx_tem, double *yy)
{
double da, db, *slp, *inter;
int i, Envectorindice;
slp=(double *)calloc(x_tem,sizeof(double));
inter=(double *)calloc(x_tem,sizeof(double));
for(i = 0; i < x_tem; i++){
if(i.
These are the following tool to build a data model for an applicatio.pdf
These are the following tool to build a data model for an application :
PowerDesigner
PowerDesigner is arguably the industry’s leading data modelling tool. Its features include: fully
integrated models, different modelling techniques that cater to both an IT-centric audience and
non IT-centric. It also supports a powerful metadata repository and various output formats. It has
a nice and polished user interface with easily readable help documentation aiding the user to
quickly solve ad hoc problems.
ER/Studio
ER/Studio is an intuitive data modelling tool that supports single and multi-platform
environments, with native integration for big data platforms such as – MongoDB and Hadoop
Hive. It can forward and reverse engineer models, includes a compare and merge function and is
able to create reports in various formats (XML, PNG, JPEG). Built-in features automate routine
tasks and supports the popular database platforms. ER/Studio is a great tool that that is easy to
start working with due to its intuitive design and good user support.
Sparx Enterprise Architect
Enterprise Architect is a feature rich data modelling tool that prides itself on being the cost-
efficient option. It helps business users build robust and maintainable systems quickly and can
easily scale to accommodate large teams collaborating on shared projects. Enterprise Architect
also has the capability of running a dynamic model simulations to verify the correctness of
models and provide better understanding of how specific business systems operate.
Oracle SQL Developer Data Modeler
Oracle data modeller is a free graphical tool to help business users with data modelling tasks.
This tool is robust, offering features and utilities centred around productivity. This includes
easily accessible report tool, DDL preview capabilities, built in quality check tool and a
sophisticated search engine.
CA ERwin
ERwin is one of the leading data modelling solutions that provides a simple, polished user
interface for a complex data environment. This solution provides business agility – models and
metadata can be managed in a common repository to ensure consistency and security. ERwin
supports high customisation and automation allowing macro language, custom datatypes, APIs
and much more. It also has an extensive user community enabling consumers to share knowledge
and expertise.
IBM - InfoSphere Data Architect
InfoSphere is an innovative data modelling tool that runs on an open-source platform – Eclipse.
Infopshere focuses on three key areas: efficiency, simplicity and integration. This tools helps
business users create logical and physical data model diagrams which can be used for a variety of
applications and systems. This end-to-end solution can be used to create, deploy and update data
models in a quick and efficient manner. It also provides easy integration with other related IBM
products.
Solution
These are the following tool to build a data model for an application :
PowerDesigner
PowerDesi.
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2.pdf
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2
Solution
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2.
Water is a good solvent due to its polarity. The oxygen atom in wate.pdf
Water is a good solvent due to its polarity. The oxygen atom in water attracts the electrons more
strongly than the hydrogen atoms. Hence the molecule has partial positive and partial negative
ends.
This allows water molecules to seperate polar solute molecules.
When a solute is dissolved in water, the solvent molecules surround the solute molecule. The
relative small size of water molecules enable many of them to surround one solute molecule.
Ionic substances dissociate to charged ions while covalent substances which are polar have
charged ends. This allows for water (a polar solvent) to easily dissolve such substances. Hence
ionic substance and polar covalent substances dissolve easily in water.
Solution
Water is a good solvent due to its polarity. The oxygen atom in water attracts the electrons more
strongly than the hydrogen atoms. Hence the molecule has partial positive and partial negative
ends.
This allows water molecules to seperate polar solute molecules.
When a solute is dissolved in water, the solvent molecules surround the solute molecule. The
relative small size of water molecules enable many of them to surround one solute molecule.
Ionic substances dissociate to charged ions while covalent substances which are polar have
charged ends. This allows for water (a polar solvent) to easily dissolve such substances. Hence
ionic substance and polar covalent substances dissolve easily in water..
V. The two types of larvae found in the life cycle of Diphyllobothri.pdf
V. The two types of larvae found in the life cycle of Diphyllobothrium spp are
1. Procercoid larva (tailed larva that develop in the body cavity of copepod)
2. Plerocercoid larvae (sparganum): occurs in the muscles of the fishes
VI The three types of larvae found in the life cycle of trematodes are:
1. Miracidium ( released from the egg)
2. Sporocyst ( forms inside the snail first intermediate host)
3. Redia ( also forms inside snail first intermediate host)
Solution
V. The two types of larvae found in the life cycle of Diphyllobothrium spp are
1. Procercoid larva (tailed larva that develop in the body cavity of copepod)
2. Plerocercoid larvae (sparganum): occurs in the muscles of the fishes
VI The three types of larvae found in the life cycle of trematodes are:
1. Miracidium ( released from the egg)
2. Sporocyst ( forms inside the snail first intermediate host)
3. Redia ( also forms inside snail first intermediate host).
The tolerance has been given as +-0.05 inchSolutionThe tolera.pdf
The tolerance has been given as +/-0.05 inch
Solution
The tolerance has been given as +/-0.05 inch.
The voltmeter reading indicates that Zn is the negative electrode an.pdf
The voltmeter reading indicates that Zn is the negative electrode and Fe is the positive electrode.
Current flows from Fe to Zn.
Zn is the anode (negative electrode where oxidation takes place)
Fe is the cathode (positive electrode where reduction takes place)
Electrons flow from Zn to Fe (opposite to current direction)
Zn is more active (Zn is oxidized more easily and electrons are produced at the Zn electrode)
Solution
The voltmeter reading indicates that Zn is the negative electrode and Fe is the positive electrode.
Current flows from Fe to Zn.
Zn is the anode (negative electrode where oxidation takes place)
Fe is the cathode (positive electrode where reduction takes place)
Electrons flow from Zn to Fe (opposite to current direction)
Zn is more active (Zn is oxidized more easily and electrons are produced at the Zn electrode).
The foreign exchange market facilitates the exchange of Currencies. .pdf
The foreign exchange market facilitates the exchange of Currencies. it facilitates the
international investment and capital flows.
Solution
The foreign exchange market facilitates the exchange of Currencies. it facilitates the
international investment and capital flows..
c. defendant that the complaint lacks true facts .pdf
c. defendant that the complaint lacks true facts
Solution
c. defendant that the complaint lacks true facts.
mean of sampling distribution = mean of the population from which th.pdf
mean of sampling distribution = mean of the population from which the sample is taken
Solution
mean of sampling distribution = mean of the population from which the sample is taken.
At the equivalence point all the HNO2 has been co.pdf
At the equivalence point all the HNO2 has been converted to NO2- so we can just
set up an ice chart but in the reverse direction: Oh and to make things simple just say there was
.5L of acid and 0.5L of NaOH were added for equivalence. NO2- + H2O <----> HNO2 + OH-
0.1 0 0 -x +x +x .1-x x x Now because we are going in the reverse direction we need the Kb not
the Ka. Using the identity Ka * Kb = Kw Kb = Kw/Ka = 1e-14/4.5e-4 = 2.2222e-11 The good
thing is that since the final volume is 1L the number of moles is now also the concentration :) Kb
= [OH-][HNO2]/[NO2-] = (x * x)/(0.1 - x) = 2.2222e-11 Let us assume that x is small compared
to 0.1 x^2/0.1 = 2.222e-11 x^2 = 2.222e-12 x = 1.49e-6 This is the concentration of OH- to get
pH we can just get the pOH and subtract it from 14: pOH = -log(1.49e-6) = 5.83 pH = 14 - pOH
= 8.17 and there you are Hope you understood it, please comment if you need further
explanation
Solution
At the equivalence point all the HNO2 has been converted to NO2- so we can just
set up an ice chart but in the reverse direction: Oh and to make things simple just say there was
.5L of acid and 0.5L of NaOH were added for equivalence. NO2- + H2O <----> HNO2 + OH-
0.1 0 0 -x +x +x .1-x x x Now because we are going in the reverse direction we need the Kb not
the Ka. Using the identity Ka * Kb = Kw Kb = Kw/Ka = 1e-14/4.5e-4 = 2.2222e-11 The good
thing is that since the final volume is 1L the number of moles is now also the concentration :) Kb
= [OH-][HNO2]/[NO2-] = (x * x)/(0.1 - x) = 2.2222e-11 Let us assume that x is small compared
to 0.1 x^2/0.1 = 2.222e-11 x^2 = 2.222e-12 x = 1.49e-6 This is the concentration of OH- to get
pH we can just get the pOH and subtract it from 14: pOH = -log(1.49e-6) = 5.83 pH = 14 - pOH
= 8.17 and there you are Hope you understood it, please comment if you need further
explanation.
isomerisms of groups are correct given isomerism.Solutionisome.pdf
isomerisms of groups are correct given isomerism.
Solution
isomerisms of groups are correct given isomerism..
let 10x+5 =z 10dx = dz I = int 110 dzz = 110 ln z 110 ln.pdf
let 10x+5 =z
10dx = dz
I = int 1/10 dz/z
= 1/10 ln z
1/10 ln (10x+5) + C
Solution
let 10x+5 =z
10dx = dz
I = int 1/10 dz/z
= 1/10 ln z
1/10 ln (10x+5) + C.
iii. Chi-square test of independence1. The sampling method is simp.pdf
iii. Chi-square test of independence
1. The sampling method is simple random sampling.
2. The variables under study are each categorical.
Solution
iii. Chi-square test of independence
1. The sampling method is simple random sampling.
2. The variables under study are each categorical..
earning can be done through two methods 1. simple interest - it is.pdf
earning can be done through two methods
1. simple interest - it is an interest income which is calculated on principle amount deposited . if
in our question it is a simple interest than we can calculate earning as follow
simple interest = principle * interest rate
=18000*16.6%
=2988
2. compounding interest is which along with interst on principle it also make interest on interest
.it is of different type like compunded annually, semi-annually,monthly etc
we assume it as compounded semi-annually
C.I = P (1+R/N)T*N
N=NO. OF TIMES COMPOUNDING = IN THIS QUESTION 2TIMES I.E SEMI ANNUALLY
T= YEAR
R=RATE OF INTEREST
CI= 18000(1+0.166/2)^1*2
=18000(1.083)^2
=21112.02
INT INCOME = 21112-18000
=3112
Solution
earning can be done through two methods
1. simple interest - it is an interest income which is calculated on principle amount deposited . if
in our question it is a simple interest than we can calculate earning as follow
simple interest = principle * interest rate
=18000*16.6%
=2988
2. compounding interest is which along with interst on principle it also make interest on interest
.it is of different type like compunded annually, semi-annually,monthly etc
we assume it as compounded semi-annually
C.I = P (1+R/N)T*N
N=NO. OF TIMES COMPOUNDING = IN THIS QUESTION 2TIMES I.E SEMI ANNUALLY
T= YEAR
R=RATE OF INTEREST
CI= 18000(1+0.166/2)^1*2
=18000(1.083)^2
=21112.02
INT INCOME = 21112-18000
=3112.
According to Bronsted-Lowry theory of acids and b.pdf
According to Bronsted-Lowry theory of acids and bases,weak acids have strong
conjugate bases and strong acids have weak conjugate bases. b,c,d are weak acids.
Solution
According to Bronsted-Lowry theory of acids and bases,weak acids have strong
conjugate bases and strong acids have weak conjugate bases. b,c,d are weak acids..
Unit 8 - Information and Communication Technology (Paper I).pdf
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
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Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu .pdf
Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu
Solution
Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu.
You can use the below function for linear interpolation i.e.#inclu.pdf
You can use the below function for linear interpolation i.e.
#include
int NearestIndex( double value, double *x, int len )
{
double dist;
int index;
int i;
index = -1;
dist = Double_Max;
for ( i = 0; i < len; i++ ) {
double newDist = value - x[i];
if ( newDist > 0 && newDist < dist ) {
dist = newDist;
index = i;
}
}
return index;
}
void linearinterpolation(double *x, int x_tem, double *y, double *xx, int xx_tem, double *yy)
{
double da, db, *slp, *inter;
int i, Envectorindice;
slp=(double *)calloc(x_tem,sizeof(double));
inter=(double *)calloc(x_tem,sizeof(double));
for(i = 0; i < x_tem; i++){
if(i
Solution
You can use the below function for linear interpolation i.e.
#include
int NearestIndex( double value, double *x, int len )
{
double dist;
int index;
int i;
index = -1;
dist = Double_Max;
for ( i = 0; i < len; i++ ) {
double newDist = value - x[i];
if ( newDist > 0 && newDist < dist ) {
dist = newDist;
index = i;
}
}
return index;
}
void linearinterpolation(double *x, int x_tem, double *y, double *xx, int xx_tem, double *yy)
{
double da, db, *slp, *inter;
int i, Envectorindice;
slp=(double *)calloc(x_tem,sizeof(double));
inter=(double *)calloc(x_tem,sizeof(double));
for(i = 0; i < x_tem; i++){
if(i.
These are the following tool to build a data model for an applicatio.pdf
These are the following tool to build a data model for an application :
PowerDesigner
PowerDesigner is arguably the industry’s leading data modelling tool. Its features include: fully
integrated models, different modelling techniques that cater to both an IT-centric audience and
non IT-centric. It also supports a powerful metadata repository and various output formats. It has
a nice and polished user interface with easily readable help documentation aiding the user to
quickly solve ad hoc problems.
ER/Studio
ER/Studio is an intuitive data modelling tool that supports single and multi-platform
environments, with native integration for big data platforms such as – MongoDB and Hadoop
Hive. It can forward and reverse engineer models, includes a compare and merge function and is
able to create reports in various formats (XML, PNG, JPEG). Built-in features automate routine
tasks and supports the popular database platforms. ER/Studio is a great tool that that is easy to
start working with due to its intuitive design and good user support.
Sparx Enterprise Architect
Enterprise Architect is a feature rich data modelling tool that prides itself on being the cost-
efficient option. It helps business users build robust and maintainable systems quickly and can
easily scale to accommodate large teams collaborating on shared projects. Enterprise Architect
also has the capability of running a dynamic model simulations to verify the correctness of
models and provide better understanding of how specific business systems operate.
Oracle SQL Developer Data Modeler
Oracle data modeller is a free graphical tool to help business users with data modelling tasks.
This tool is robust, offering features and utilities centred around productivity. This includes
easily accessible report tool, DDL preview capabilities, built in quality check tool and a
sophisticated search engine.
CA ERwin
ERwin is one of the leading data modelling solutions that provides a simple, polished user
interface for a complex data environment. This solution provides business agility – models and
metadata can be managed in a common repository to ensure consistency and security. ERwin
supports high customisation and automation allowing macro language, custom datatypes, APIs
and much more. It also has an extensive user community enabling consumers to share knowledge
and expertise.
IBM - InfoSphere Data Architect
InfoSphere is an innovative data modelling tool that runs on an open-source platform – Eclipse.
Infopshere focuses on three key areas: efficiency, simplicity and integration. This tools helps
business users create logical and physical data model diagrams which can be used for a variety of
applications and systems. This end-to-end solution can be used to create, deploy and update data
models in a quick and efficient manner. It also provides easy integration with other related IBM
products.
Solution
These are the following tool to build a data model for an application :
PowerDesigner
PowerDesi.
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2.pdf
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2
Solution
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2.
Water is a good solvent due to its polarity. The oxygen atom in wate.pdf
Water is a good solvent due to its polarity. The oxygen atom in water attracts the electrons more
strongly than the hydrogen atoms. Hence the molecule has partial positive and partial negative
ends.
This allows water molecules to seperate polar solute molecules.
When a solute is dissolved in water, the solvent molecules surround the solute molecule. The
relative small size of water molecules enable many of them to surround one solute molecule.
Ionic substances dissociate to charged ions while covalent substances which are polar have
charged ends. This allows for water (a polar solvent) to easily dissolve such substances. Hence
ionic substance and polar covalent substances dissolve easily in water.
Solution
Water is a good solvent due to its polarity. The oxygen atom in water attracts the electrons more
strongly than the hydrogen atoms. Hence the molecule has partial positive and partial negative
ends.
This allows water molecules to seperate polar solute molecules.
When a solute is dissolved in water, the solvent molecules surround the solute molecule. The
relative small size of water molecules enable many of them to surround one solute molecule.
Ionic substances dissociate to charged ions while covalent substances which are polar have
charged ends. This allows for water (a polar solvent) to easily dissolve such substances. Hence
ionic substance and polar covalent substances dissolve easily in water..
V. The two types of larvae found in the life cycle of Diphyllobothri.pdf
V. The two types of larvae found in the life cycle of Diphyllobothrium spp are
1. Procercoid larva (tailed larva that develop in the body cavity of copepod)
2. Plerocercoid larvae (sparganum): occurs in the muscles of the fishes
VI The three types of larvae found in the life cycle of trematodes are:
1. Miracidium ( released from the egg)
2. Sporocyst ( forms inside the snail first intermediate host)
3. Redia ( also forms inside snail first intermediate host)
Solution
V. The two types of larvae found in the life cycle of Diphyllobothrium spp are
1. Procercoid larva (tailed larva that develop in the body cavity of copepod)
2. Plerocercoid larvae (sparganum): occurs in the muscles of the fishes
VI The three types of larvae found in the life cycle of trematodes are:
1. Miracidium ( released from the egg)
2. Sporocyst ( forms inside the snail first intermediate host)
3. Redia ( also forms inside snail first intermediate host).
The tolerance has been given as +-0.05 inchSolutionThe tolera.pdf
The tolerance has been given as +/-0.05 inch
Solution
The tolerance has been given as +/-0.05 inch.
The voltmeter reading indicates that Zn is the negative electrode an.pdf
The voltmeter reading indicates that Zn is the negative electrode and Fe is the positive electrode.
Current flows from Fe to Zn.
Zn is the anode (negative electrode where oxidation takes place)
Fe is the cathode (positive electrode where reduction takes place)
Electrons flow from Zn to Fe (opposite to current direction)
Zn is more active (Zn is oxidized more easily and electrons are produced at the Zn electrode)
Solution
The voltmeter reading indicates that Zn is the negative electrode and Fe is the positive electrode.
Current flows from Fe to Zn.
Zn is the anode (negative electrode where oxidation takes place)
Fe is the cathode (positive electrode where reduction takes place)
Electrons flow from Zn to Fe (opposite to current direction)
Zn is more active (Zn is oxidized more easily and electrons are produced at the Zn electrode).
The foreign exchange market facilitates the exchange of Currencies. .pdf
The foreign exchange market facilitates the exchange of Currencies. it facilitates the
international investment and capital flows.
Solution
The foreign exchange market facilitates the exchange of Currencies. it facilitates the
international investment and capital flows..
c. defendant that the complaint lacks true facts .pdf
c. defendant that the complaint lacks true facts
Solution
c. defendant that the complaint lacks true facts.
mean of sampling distribution = mean of the population from which th.pdf
mean of sampling distribution = mean of the population from which the sample is taken
Solution
mean of sampling distribution = mean of the population from which the sample is taken.
At the equivalence point all the HNO2 has been co.pdf
At the equivalence point all the HNO2 has been converted to NO2- so we can just
set up an ice chart but in the reverse direction: Oh and to make things simple just say there was
.5L of acid and 0.5L of NaOH were added for equivalence. NO2- + H2O <----> HNO2 + OH-
0.1 0 0 -x +x +x .1-x x x Now because we are going in the reverse direction we need the Kb not
the Ka. Using the identity Ka * Kb = Kw Kb = Kw/Ka = 1e-14/4.5e-4 = 2.2222e-11 The good
thing is that since the final volume is 1L the number of moles is now also the concentration :) Kb
= [OH-][HNO2]/[NO2-] = (x * x)/(0.1 - x) = 2.2222e-11 Let us assume that x is small compared
to 0.1 x^2/0.1 = 2.222e-11 x^2 = 2.222e-12 x = 1.49e-6 This is the concentration of OH- to get
pH we can just get the pOH and subtract it from 14: pOH = -log(1.49e-6) = 5.83 pH = 14 - pOH
= 8.17 and there you are Hope you understood it, please comment if you need further
explanation
Solution
At the equivalence point all the HNO2 has been converted to NO2- so we can just
set up an ice chart but in the reverse direction: Oh and to make things simple just say there was
.5L of acid and 0.5L of NaOH were added for equivalence. NO2- + H2O <----> HNO2 + OH-
0.1 0 0 -x +x +x .1-x x x Now because we are going in the reverse direction we need the Kb not
the Ka. Using the identity Ka * Kb = Kw Kb = Kw/Ka = 1e-14/4.5e-4 = 2.2222e-11 The good
thing is that since the final volume is 1L the number of moles is now also the concentration :) Kb
= [OH-][HNO2]/[NO2-] = (x * x)/(0.1 - x) = 2.2222e-11 Let us assume that x is small compared
to 0.1 x^2/0.1 = 2.222e-11 x^2 = 2.222e-12 x = 1.49e-6 This is the concentration of OH- to get
pH we can just get the pOH and subtract it from 14: pOH = -log(1.49e-6) = 5.83 pH = 14 - pOH
= 8.17 and there you are Hope you understood it, please comment if you need further
explanation.
isomerisms of groups are correct given isomerism.Solutionisome.pdf
isomerisms of groups are correct given isomerism.
Solution
isomerisms of groups are correct given isomerism..
let 10x+5 =z 10dx = dz I = int 110 dzz = 110 ln z 110 ln.pdf
let 10x+5 =z
10dx = dz
I = int 1/10 dz/z
= 1/10 ln z
1/10 ln (10x+5) + C
Solution
let 10x+5 =z
10dx = dz
I = int 1/10 dz/z
= 1/10 ln z
1/10 ln (10x+5) + C.
iii. Chi-square test of independence1. The sampling method is simp.pdf
iii. Chi-square test of independence
1. The sampling method is simple random sampling.
2. The variables under study are each categorical.
Solution
iii. Chi-square test of independence
1. The sampling method is simple random sampling.
2. The variables under study are each categorical..
earning can be done through two methods 1. simple interest - it is.pdf
earning can be done through two methods
1. simple interest - it is an interest income which is calculated on principle amount deposited . if
in our question it is a simple interest than we can calculate earning as follow
simple interest = principle * interest rate
=18000*16.6%
=2988
2. compounding interest is which along with interst on principle it also make interest on interest
.it is of different type like compunded annually, semi-annually,monthly etc
we assume it as compounded semi-annually
C.I = P (1+R/N)T*N
N=NO. OF TIMES COMPOUNDING = IN THIS QUESTION 2TIMES I.E SEMI ANNUALLY
T= YEAR
R=RATE OF INTEREST
CI= 18000(1+0.166/2)^1*2
=18000(1.083)^2
=21112.02
INT INCOME = 21112-18000
=3112
Solution
earning can be done through two methods
1. simple interest - it is an interest income which is calculated on principle amount deposited . if
in our question it is a simple interest than we can calculate earning as follow
simple interest = principle * interest rate
=18000*16.6%
=2988
2. compounding interest is which along with interst on principle it also make interest on interest
.it is of different type like compunded annually, semi-annually,monthly etc
we assume it as compounded semi-annually
C.I = P (1+R/N)T*N
N=NO. OF TIMES COMPOUNDING = IN THIS QUESTION 2TIMES I.E SEMI ANNUALLY
T= YEAR
R=RATE OF INTEREST
CI= 18000(1+0.166/2)^1*2
=18000(1.083)^2
=21112.02
INT INCOME = 21112-18000
=3112.
According to Bronsted-Lowry theory of acids and b.pdf
According to Bronsted-Lowry theory of acids and bases,weak acids have strong
conjugate bases and strong acids have weak conjugate bases. b,c,d are weak acids.
Solution
According to Bronsted-Lowry theory of acids and bases,weak acids have strong
conjugate bases and strong acids have weak conjugate bases. b,c,d are weak acids..
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special B.ed 2nd year old paper_20240531.pdf
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Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
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The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
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- 1. bonde order of NO in nitrate 1.33 in nitrite is 1.5 in NO+ is 3.0 in NO- is 2.0 j. NO+ > NO- > nitrite > nitrate is correct. Solution bonde order of NO in nitrate 1.33 in nitrite is 1.5 in NO+ is 3.0 in NO- is 2.0 j. NO+ > NO- > nitrite > nitrate is correct.