(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen.pdf
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(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these reactions. The result is a mild solution of sulfuric acid and nitric acid. Acid rain causes acidification of lakes and streams and contributes to damage of trees at high elevations and many sensitive forest soils. (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) Solution (1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these reactions. The result is a mild solution of sulfuric acid and nitric acid. Acid rain causes acidification of lakes and streams and contributes to damage of trees at high elevations and many sensitive forest soils. (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Vol.
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![(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in
the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these
reactions. The result is a mild solution of sulfuric acid and nitric acid.
Acid rain causes acidification of lakes and streams and contributes to damage of trees at high
elevations and many sensitive forest soils.
(2) 2NO2 + H2O ---> 2HNO3
Given mass of NO2 is , m = 5.47 mg
= 5.47 x 10 -3 g Since 1 mg = 10 -3 g
Volume of the solution , V = 1.0 L
Molar mass of NO2 is = 14 + (2x16) = 46 g/mol
So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution
= ( 5.47 x 10 -3 / 46 ) / 1.0
= 1.189 x 10 -4 M
According to the balanced equation,
2 mole of NO2 produces 2 moles of HNO3
So the concentration of HNO3 is = 1.189 x 10 -4 M
HNO3 ---> H+ + NO3-
1 mole of HNO3 produces 1 mole of H+
So [H+] = 1.189 x 10 -4 M
pH = - log [H+]
= -log ( 1.189 x 10 -4 )
= 4 - log 1.189
= 3.92
(3) HNO3 + H2O ----> HNO3 (aq)
H2SO4 + H2O ---> H2SO4 (aq)
HCl + H2O ---> HCl (aq)
(2) 2NO2 + H2O ---> 2HNO3
Given mass of NO2 is , m = 5.47 mg
= 5.47 x 10 -3 g Since 1 mg = 10 -3 g
Volume of the solution , V = 1.0 L
Molar mass of NO2 is = 14 + (2x16) = 46 g/mol
So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution
= ( 5.47 x 10 -3 / 46 ) / 1.0
= 1.189 x 10 -4 M
According to the balanced equation,](https://image.slidesharecdn.com/1acidrainoccurswhenthegasesso2noxreactwithwateroxygen-230409115518-e37168d5/85/1-Acid-rain-occurs-when-the-gases-SO2-NOx-react-with-water-oxygen-pdf-1-320.jpg)
![2 mole of NO2 produces 2 moles of HNO3
So the concentration of HNO3 is = 1.189 x 10 -4 M
HNO3 ---> H+ + NO3-
1 mole of HNO3 produces 1 mole of H+
So [H+] = 1.189 x 10 -4 M
pH = - log [H+]
= -log ( 1.189 x 10 -4 )
= 4 - log 1.189
= 3.92
(3) HNO3 + H2O ----> HNO3 (aq)
H2SO4 + H2O ---> H2SO4 (aq)
HCl + H2O ---> HCl (aq)
Solution
(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in
the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these
reactions. The result is a mild solution of sulfuric acid and nitric acid.
Acid rain causes acidification of lakes and streams and contributes to damage of trees at high
elevations and many sensitive forest soils.
(2) 2NO2 + H2O ---> 2HNO3
Given mass of NO2 is , m = 5.47 mg
= 5.47 x 10 -3 g Since 1 mg = 10 -3 g
Volume of the solution , V = 1.0 L
Molar mass of NO2 is = 14 + (2x16) = 46 g/mol
So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution
= ( 5.47 x 10 -3 / 46 ) / 1.0
= 1.189 x 10 -4 M
According to the balanced equation,
2 mole of NO2 produces 2 moles of HNO3
So the concentration of HNO3 is = 1.189 x 10 -4 M
HNO3 ---> H+ + NO3-
1 mole of HNO3 produces 1 mole of H+
So [H+] = 1.189 x 10 -4 M
pH = - log [H+]
= -log ( 1.189 x 10 -4 )](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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dont confuse this with MOLECULES which are composed of different atoms together..
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#include
#include
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class twoStacks
{
int *arr;
int size;
int top1, top2;
public:
twoStacks(int n) // constructor
{
size = n;
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top1 = -1;
top2 = size;
}
// Method to push an element x to stack1
void push1(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top1++;
arr[top1] = x;
}
else
{
cout << \"Stack Overflow\";
exit(1);
}
}
// Method to push an element x to stack2
void push2(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top2--;
arr[top2] = x;
}
else
{
cout << \"Stack Overflow\";
exit(1);
}
}
// Method to pop an element from first stack
int pop1()
{
if (top1 >= 0 )
{
int x = arr[top1];
top1--;
return x;
}
else
{
cout << \"Stack UnderFlow\";
exit(1);
}
}
// Method to pop an element from second stack
int pop2()
{
if (top2 < size)
{
int x = arr[top2];
top2++;
return x;
}
else
{
cout << \"Stack UnderFlow\";
exit(1);
}
}
};
/* Driver program to test twStacks class */
int main()
{
twoStacks ts(5);
ts.push1(5);
ts.push2(10);
ts.push2(15);
ts.push1(11);
ts.push2(7);
cout << \"Popped element from stack1 is \" << ts.pop1();
ts.push2(40);
cout << \"\ Popped element from stack2 is \" << ts.pop2();
return 0;
}
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#include
using namespace std;
class twoStacks
{
int *arr;
int size;
int top1, top2;
public:
twoStacks(int n) // constructor
{
size = n;
arr = new int[n];
top1 = -1;
top2 = size;
}
// Method to push an element x to stack1
void push1(int x)
{
// There is at least one empty space for new element
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top1++;
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cout << \"Stack Overflow\";
exit(1);
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void push2(int x)
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// There is at least one empty space for new element
if (top1 < top2 - 1)
{
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cout << \"Stack Overflow\";
exit(1);
}
}
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if (top1 >= 0 )
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int x = arr[top1];
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int x = arr[top2];
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Selection rule of Raman spectra: During the vibration polarizibility change.
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The main class of the tictoe game looks like.
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System.out.println();
}
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System.out.println();
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String choice = game.getPrompt();
if (!choice.equalsIgnoreCase(\"Y\")) {
continuePlaying = false;
}
}
}
public static void main(String[] args) {
Main main = new Main();
main.play();
}
}
The TicTacToe class code will be like
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/**
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int counter = 0;
for (int i = 0; i < 3; i++) {
for (int i1 = 0; i1 < 3; i1++) {.
the inner electronsSolutionthe inner electrons.pdf
This very short document does not provide enough context or information to generate a meaningful 3 sentence summary. The document consists of only two words repeated, without any surrounding context.
SolutionThe investees net income should be recorded as an increa.pdf
Solution
The investee\'s net income should be recorded as an increase to the investment account.
False
Explanation
Not all the net income of investee is recorded as an increase to investment account. Only the part
of net income proportionate to investor share is recorded as increase in investment account. For
example if investor have 30% share in investee company and the net income of investee is
100000 then only 30000 will be recorded as an increase in investment account.
QuestionAnswer1TrueThe corpus callosum , otherwise called th.pdf
Question
Answer
1
True
The corpus callosum , otherwise called the callosal commissure, is a wide, level heap of neural
filaments around 10 cm long underneath the cortex in the eutherian cerebrum at the longitudinal
gap. It associates the left and right cerebral sides of the equator and encourages interhemispheric
correspondence. It is the biggest white matter structure in the mind, comprising of 200–250
million contralateral axonal projections.
The (back) bit of the corpus callosum is known as the splenium; the foremost (front) is known as
the genu (or \"knee\"); between the two is the truncus, or \"body\", of the corpus callosum. The
part between the body and the splenium is frequently extraordinarily contracted and along these
lines alluded to as the \"isthmus\". The platform is the part of the corpus callosum that ventures
posteriorly and poorly from the anteriormost genu, as can be seen on the sagittal picture of the
cerebrum showed on the privilege. The platform is so named for its likeness to a winged
creature\'s mouth.
Corpus callosum
On either side of the corpus callosum, the filaments transmit in the white matter and go to the
different parts of the cerebral cortex; those bending forward from the genu into the frontal flap
constitute the forceps foremost, and those bending in reverse into the occipital projection, the
forceps back. Between these two sections is the fundamental body of the filaments which
constitute the tapetum and expand along the side on either side into the transient projection, and
cover in the focal part of the horizontal ventricle.
2
False
Ramus communicans (plural rami communicantes) is the Latin expression utilized for a nerve
which associates two different nerves, and can be deciphered as \"conveying branch\".When
utilized without further definition, it quite often alludes to an imparting branch between a spinal
nerve and the thoughtful trunk. All the more particularly, it generally alludes to one of the
accompanying :
Dark ramus communicans
White ramus communicans
The dark and white rami communicantes are in charge of passing on autonomic signs,
particularly for the thoughtful sensory system. Their distinction in tinge is brought about by
contrasts in myelination of the nerve filaments contained inside, i.e. there are more myelinated
than unmyelinated filaments in the white rami communicantes while the opposite is valid for the
dim rami communicantes.
3
False
tentorium cerebelli isolates the occipital flaps of the cerebral sides of the equator from the
cerebellum. Its inward, sunken, free fringe adds to the tentorial indent. (See underneath.) The
outer, arched fringe encases the transverse sinus where it appends to the dura over within the
occiput. Past the \"petrous edge,\" the tentorium is tied down to the foremost and back clinoid
forms
4
False
myelin sheath(deep) and Neurilemma (superficial)
5
False
Cranial nerves are the nerves that develop straightforwardly from the cerebrum (counting the
brainstem), a.
No. of Moles = Given MassMolecular Mass=3.0 x 101398= 3.06101.pdf
No. of Moles = Given Mass/Molecular Mass
=3.0 x 1013/98
= 3.06*1011 Moles
Solution
No. of Moles = Given Mass/Molecular Mass
=3.0 x 1013/98
= 3.06*1011 Moles.
Let the sample size be n ,Mean of the sample = 27 (same as pop.pdf
Let the sample size be n ,
Mean of the sample = 27 (same as population mean)
Standard deviation of sample = population standard deviation / sqrt(n)
= 2.5/sqrt(n)
Solution
Let the sample size be n ,
Mean of the sample = 27 (same as population mean)
Standard deviation of sample = population standard deviation / sqrt(n)
= 2.5/sqrt(n).
JeffSolutionJeff.pdf
This very short document appears to be a communication between two individuals, Jeff and Solution. It contains the name "Jeff" repeated twice without any other words or context, making it difficult to determine the intent or meaning based on the limited information provided.
import java.util.Scanner;public class Assignment4 { public st.pdf
import java.util.Scanner;
public class Assignment4 {
public static void main(String args[]) {
final int LEN = 5;
int[] hand = new int[LEN];
Scanner input = new Scanner(System.in);
//input the hand
System.out.println(\"Enter five numeric cards, no face cards. Use 2-9.\");
for (int i = 0; i < hand.length; i++) {
hand[i] = input.nextInt();
}
//sort the collection
bubbleSortCards(hand);
//determine players hand type
//flow of evaluation -> checking complex hands first
if (containsFullHouse(hand)) {
System.out.println(\"Full House!\");
} else if (containsStraight(hand)) {
System.out.println(\"Straight!\");
} else if (containsFourOfaKind(hand)) {
System.out.println(\"Four of a Kind!\");
} else if (containsThreeOfaKind(hand)) {
System.out.println(\"Three of a Kind!\");
} else if (containsTwoPair(hand)) {
System.out.println(\"Two Pair!\");
} else if (containsPair(hand)) {
System.out.println(\"Pair!\");
} else
System.out.println(\"High Card!\");
}
//cardInSets is called with hand[] and pair to check if a pair is found.
public static boolean containsPair(int hand[]) {
int pair = 2;
return cardInSets(hand, pair) != NOT_FOUND;
}
//cardInSets is called with hand[] and pair to check if a pair is found.
//Then calls back again with fcard result to check for another pair.
public static boolean containsTwoPair(int hand[]) {
int pair = 2;
int fcard = cardInSets(hand, pair);
int scard = cardInSets(hand, pair, fcard);
return fcard!=NOT_FOUND && scard!=NOT_FOUND;
}
//cardInSets is called with hand[] and threeaKind to check if three of a kind has been found.
public static boolean containsThreeOfaKind(int hand[]) {
int threeaKind = 3;
return cardInSets(hand, threeaKind) != NOT_FOUND;
}
//hasStraight will resolve to true, unless the previous value++ is not equivalent.
//The loop ends as soon as hasStaight is false or the end of the hand is reached.
public static boolean containsStraight(int hand[]) {
boolean hasStraight = true;
int i=1;
do {
hasStraight = (hand[i] == (hand[i-1] + 1));
} while(hasStraight && ++i < hand.length);
return hasStraight;
}
//cardInSets is called with hand[] and threeKinds to check if three of a kind is found.
//Then calls back again with pair and tcard result to check for a pair of another card type.
public static boolean containsFullHouse(int hand[]) {
int threeKinds = 3;
int pair = 2;
int tcard = cardInSets(hand, threeKinds);
int pcard = cardInSets(hand, pair, tcard);
return tcard!=NOT_FOUND && pcard!=NOT_FOUND;
}
//cardInSets is called with hand[] and fouraKind to check if four of a kind is found.
public static boolean containsFourOfaKind(int hand[]) {
int fouraKind = 4;
return cardInSets(hand, fouraKind) != NOT_FOUND;
}
//When ignoreCard isn\'t needed, it overloads with a sentinel 0.
public static int cardInSets(int hand[], int numKinds) {
return cardInSets(hand, numKinds, 0);
}
//The method cardInSets is a generic evaluation that returns if the parameter specified
numKinds(like 2 for pair) has been found in the hand.
//The ignoreCard d.
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
Walmart Business+ and Spark Good for Nonprofits.pdf
"Learn about all the ways Walmart supports nonprofit organizations.
You will hear from Liz Willett, the Head of Nonprofits, and hear about what Walmart is doing to help nonprofits, including Walmart Business and Spark Good. Walmart Business+ is a new offer for nonprofits that offers discounts and also streamlines nonprofits order and expense tracking, saving time and money.
The webinar may also give some examples on how nonprofits can best leverage Walmart Business+.
The event will cover the following::
Walmart Business + (https://business.walmart.com/plus) is a new shopping experience for nonprofits, schools, and local business customers that connects an exclusive online shopping experience to stores. Benefits include free delivery and shipping, a 'Spend Analytics” feature, special discounts, deals and tax-exempt shopping.
Special TechSoup offer for a free 180 days membership, and up to $150 in discounts on eligible orders.
Spark Good (walmart.com/sparkgood) is a charitable platform that enables nonprofits to receive donations directly from customers and associates.
Answers about how you can do more with Walmart!"
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the reaction of A to B since its Ea is the highes.pdf
the reaction of A to B since its Ea is the highest.
Solution
the reaction of A to B since its Ea is the highest..
the one thing that comes to mind for me is about .pdf
the one thing that comes to mind for me is about the diatomic elements
H,N,F,O,I,Cl,B they\'re all really the ones that could form a molecule since they are composed of
more than one element (all of them are subscript 2, as in H2). most other elements exists as 1.
dont confuse this with MOLECULES which are composed of different atoms together.
Solution
the one thing that comes to mind for me is about the diatomic elements
H,N,F,O,I,Cl,B they\'re all really the ones that could form a molecule since they are composed of
more than one element (all of them are subscript 2, as in H2). most other elements exists as 1.
dont confuse this with MOLECULES which are composed of different atoms together..
#includeiostream#includestdlib.husing namespace std;class .pdf
#include
#include
using namespace std;
class twoStacks
{
int *arr;
int size;
int top1, top2;
public:
twoStacks(int n) // constructor
{
size = n;
arr = new int[n];
top1 = -1;
top2 = size;
}
// Method to push an element x to stack1
void push1(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top1++;
arr[top1] = x;
}
else
{
cout << \"Stack Overflow\";
exit(1);
}
}
// Method to push an element x to stack2
void push2(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top2--;
arr[top2] = x;
}
else
{
cout << \"Stack Overflow\";
exit(1);
}
}
// Method to pop an element from first stack
int pop1()
{
if (top1 >= 0 )
{
int x = arr[top1];
top1--;
return x;
}
else
{
cout << \"Stack UnderFlow\";
exit(1);
}
}
// Method to pop an element from second stack
int pop2()
{
if (top2 < size)
{
int x = arr[top2];
top2++;
return x;
}
else
{
cout << \"Stack UnderFlow\";
exit(1);
}
}
};
/* Driver program to test twStacks class */
int main()
{
twoStacks ts(5);
ts.push1(5);
ts.push2(10);
ts.push2(15);
ts.push1(11);
ts.push2(7);
cout << \"Popped element from stack1 is \" << ts.pop1();
ts.push2(40);
cout << \"\ Popped element from stack2 is \" << ts.pop2();
return 0;
}
#include
#include
using namespace std;
class twoStacks
{
int *arr;
int size;
int top1, top2;
public:
twoStacks(int n) // constructor
{
size = n;
arr = new int[n];
top1 = -1;
top2 = size;
}
// Method to push an element x to stack1
void push1(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top1++;
arr[top1] = x;
}
else
{
cout << \"Stack Overflow\";
exit(1);
}
}
// Method to push an element x to stack2
void push2(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top2--;
arr[top2] = x;
}
else
{
cout << \"Stack Overflow\";
exit(1);
}
}
// Method to pop an element from first stack
int pop1()
{
if (top1 >= 0 )
{
int x = arr[top1];
top1--;
return x;
}
else
{
cout << \"Stack UnderFlow\";
exit(1);
}
}
// Method to pop an element from second stack
int pop2()
{
if (top2 < size)
{
int x = arr[top2];
top2++;
return x;
}
else
{
cout << \"Stack UnderFlow\";
exit(1);
}
}
};
/* Driver program to test twStacks class */
int main()
{
twoStacks ts(5);
ts.push1(5);
ts.push2(10);
ts.push2(15);
ts.push1(11);
ts.push2(7);
cout << \"Popped element from stack1 is \" << ts.pop1();
ts.push2(40);
cout << \"\ Popped element from stack2 is \" << ts.pop2();
return 0;
}
Solution
#include
#include
using namespace std;
class twoStacks
{
int *arr;
int size;
int top1, top2;
public:
twoStacks(int n) // constructor
{
size = n;
arr = new int[n];
top1 = -1;
top2 = size;
}
// Method to push an element x to stack1
void push1(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top1++;
arr[top1] = x;
}
else
{
cout << \"Stack Overflow\";
exit(1);
}
}
// Method to push an element x to stack2
void push2(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{.
Iodine will replace both Cl and Br It will be SN2.pdf
Iodine will replace both Cl and Br It will be SN2 reaction and hence there will be
Walden Inversion.
Solution
Iodine will replace both Cl and Br It will be SN2 reaction and hence there will be
Walden Inversion..
The number of outcomes in the event A is 4 Yes event A is a s.pdf
The number of outcomes in the event A is 4
Yes event A is a sample event
Solution
The number of outcomes in the event A is 4
Yes event A is a sample event.
There were very few planetary leftovers in this r.pdf
There were very few planetary leftovers in this region, because most of the solid
material was accreted by the terrestrial planets as the planets formed.
Solution
There were very few planetary leftovers in this region, because most of the solid
material was accreted by the terrestrial planets as the planets formed..
i have no idea. im just seeing if i get karma p.pdf
i have no idea. i\'m just seeing if i get karma points for a half ass answer.
Solution
i have no idea. i\'m just seeing if i get karma points for a half ass answer..
Fe+3 solution is more acidic because it produces .pdf
Fe+3 solution is more acidic because it produces H+ ion in the aqueous solution,
but Fe+2 is not. The following reaction represents this- [Fe(H2O)6]3+H2O ---------->
[Fe(H2O)5OH]+2+ + H3O+(aq) Therefore - a) Fe+3 b) ii
Solution
Fe+3 solution is more acidic because it produces H+ ion in the aqueous solution,
but Fe+2 is not. The following reaction represents this- [Fe(H2O)6]3+H2O ---------->
[Fe(H2O)5OH]+2+ + H3O+(aq) Therefore - a) Fe+3 b) ii.
Copper chloride. Copper ions will form precipitat.pdf
Copper chloride. Copper ions will form precipitated with sulfide, Coppersulfide.
For other cations, there are solubile when there form with eithersulfide ions or acetate ions. So,
the only way to separate is using copper ions.
Solution
Copper chloride. Copper ions will form precipitated with sulfide, Coppersulfide.
For other cations, there are solubile when there form with eithersulfide ions or acetate ions. So,
the only way to separate is using copper ions..
CCl4 molecule can give IR and Raman spectroscopy. Selection r.pdf
CCl4 molecule can give IR and Raman spectroscopy.
Selection rule of IR spectra: During the vibration dipolemoment change.
Selection rule of Raman spectra: During the vibration polarizibility change.
CCl4 molecule centrosymmetric molecule, so it can exibit Raman spectroscopy
According to mutual exclusion principle, IR inactive modes are raman active
Raman inactive modes are IR active modes.
Therefore, CCl4 molecule can give multiple peaks.
Solution
CCl4 molecule can give IR and Raman spectroscopy.
Selection rule of IR spectra: During the vibration dipolemoment change.
Selection rule of Raman spectra: During the vibration polarizibility change.
CCl4 molecule centrosymmetric molecule, so it can exibit Raman spectroscopy
According to mutual exclusion principle, IR inactive modes are raman active
Raman inactive modes are IR active modes.
Therefore, CCl4 molecule can give multiple peaks..
User interface design processNavigation DesignInput DesignOu.pdf
User interface design process
Navigation Design
Input Design
Output Design
Solution
User interface design process
Navigation Design
Input Design
Output Design.
True.it is an example of an independent-measures designSolution.pdf
True.it is an example of an independent-measures design
Solution
True.it is an example of an independent-measures design.
The main class of the tictoe game looks like.public class Main {.pdf
The main class of the tictoe game looks like.
public class Main {
public void play() {
TicTacToe game = new TicTacToe();
System.out.println(\"Welcome! Tic Tac Toe is a two player game.\");
System.out.print(\"Enter player one\'s name: \");
game.setPlayer1(game.getPrompt());
System.out.print(\"Enter player two\'s name: \");
game.setPlayer2(game.getPrompt());
boolean markerOk = false;
while (!markerOk) {
System.out.print(\"Select any letter as \" + game.getPlayer1() + \"\'s marker: \");
String marker = game.getPrompt();
if (marker.length() == 1 &&
Character.isLetter(marker.toCharArray()[0])) {
markerOk = true;
game.setMarker1(marker.toCharArray()[0]);
} else {
System.out.println(\"Invalid marker, try again\");
}
}
markerOk = false;
while (!markerOk) {
System.out.print(\"Select any letter as \" + game.getPlayer2() + \"\'s marker: \");
String marker = game.getPrompt();
if (marker.length() == 1 &&
Character.isLetter(marker.toCharArray()[0])) {
markerOk = true;
game.setMarker2(marker.toCharArray()[0]);
} else {
System.out.println(\"Invalid marker, try again\");
}
}
boolean continuePlaying = true;
while (continuePlaying) {
game.init();
System.out.println();
System.out.println(game.getRules());
System.out.println();
System.out.println(game.drawBoard());
System.out.println();
String player = null;
while (!game.winner() && game.getPlays() < 9) {
player = game.getCurrentPlayer() == 1 ? game.getPlayer1() : game.getPlayer2();
boolean validPick = false;
while (!validPick) {
System.out.print(\"It is \" + player + \"\'s turn. Pick a square: \");
String square = game.getPrompt();
if (square.length() == 1 && Character.isDigit(square.toCharArray()[0])) {
int pick = 0;
try {
pick = Integer.parseInt(square);
} catch (NumberFormatException e) {
//Do nothing here, it\'ll evaluate as an invalid pick on the next row.
}
validPick = game.placeMarker(pick);
}
if (!validPick) {
System.out.println(\"Square can not be selected. Retry\");
}
}
game.switchPlayers();
System.out.println();
System.out.println(game.drawBoard());
System.out.println();
}
if (game.winner()) {
System.out.println(\"Game Over - \" + player + \" WINS!!!\");
} else {
System.out.println(\"Game Over - Draw\");
}
System.out.println();
System.out.print(\"Play again? (Y/N): \");
String choice = game.getPrompt();
if (!choice.equalsIgnoreCase(\"Y\")) {
continuePlaying = false;
}
}
}
public static void main(String[] args) {
Main main = new Main();
main.play();
}
}
The TicTacToe class code will be like
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
/**
*
* author Iamavinashkotina
*/
public class TicTacToe {
private char[][] board = new char[3][3];
private String player1;
private String player2;
private int currentPlayer;
private char marker1;
private char marker2;
private int plays;
private BufferedReader reader =
new BufferedReader(new InputStreamReader(System.in));
protected void init() {
int counter = 0;
for (int i = 0; i < 3; i++) {
for (int i1 = 0; i1 < 3; i1++) {.
the inner electronsSolutionthe inner electrons.pdf
This very short document does not provide enough context or information to generate a meaningful 3 sentence summary. The document consists of only two words repeated, without any surrounding context.
SolutionThe investees net income should be recorded as an increa.pdf
Solution
The investee\'s net income should be recorded as an increase to the investment account.
False
Explanation
Not all the net income of investee is recorded as an increase to investment account. Only the part
of net income proportionate to investor share is recorded as increase in investment account. For
example if investor have 30% share in investee company and the net income of investee is
100000 then only 30000 will be recorded as an increase in investment account.
QuestionAnswer1TrueThe corpus callosum , otherwise called th.pdf
Question
Answer
1
True
The corpus callosum , otherwise called the callosal commissure, is a wide, level heap of neural
filaments around 10 cm long underneath the cortex in the eutherian cerebrum at the longitudinal
gap. It associates the left and right cerebral sides of the equator and encourages interhemispheric
correspondence. It is the biggest white matter structure in the mind, comprising of 200–250
million contralateral axonal projections.
The (back) bit of the corpus callosum is known as the splenium; the foremost (front) is known as
the genu (or \"knee\"); between the two is the truncus, or \"body\", of the corpus callosum. The
part between the body and the splenium is frequently extraordinarily contracted and along these
lines alluded to as the \"isthmus\". The platform is the part of the corpus callosum that ventures
posteriorly and poorly from the anteriormost genu, as can be seen on the sagittal picture of the
cerebrum showed on the privilege. The platform is so named for its likeness to a winged
creature\'s mouth.
Corpus callosum
On either side of the corpus callosum, the filaments transmit in the white matter and go to the
different parts of the cerebral cortex; those bending forward from the genu into the frontal flap
constitute the forceps foremost, and those bending in reverse into the occipital projection, the
forceps back. Between these two sections is the fundamental body of the filaments which
constitute the tapetum and expand along the side on either side into the transient projection, and
cover in the focal part of the horizontal ventricle.
2
False
Ramus communicans (plural rami communicantes) is the Latin expression utilized for a nerve
which associates two different nerves, and can be deciphered as \"conveying branch\".When
utilized without further definition, it quite often alludes to an imparting branch between a spinal
nerve and the thoughtful trunk. All the more particularly, it generally alludes to one of the
accompanying :
Dark ramus communicans
White ramus communicans
The dark and white rami communicantes are in charge of passing on autonomic signs,
particularly for the thoughtful sensory system. Their distinction in tinge is brought about by
contrasts in myelination of the nerve filaments contained inside, i.e. there are more myelinated
than unmyelinated filaments in the white rami communicantes while the opposite is valid for the
dim rami communicantes.
3
False
tentorium cerebelli isolates the occipital flaps of the cerebral sides of the equator from the
cerebellum. Its inward, sunken, free fringe adds to the tentorial indent. (See underneath.) The
outer, arched fringe encases the transverse sinus where it appends to the dura over within the
occiput. Past the \"petrous edge,\" the tentorium is tied down to the foremost and back clinoid
forms
4
False
myelin sheath(deep) and Neurilemma (superficial)
5
False
Cranial nerves are the nerves that develop straightforwardly from the cerebrum (counting the
brainstem), a.
No. of Moles = Given MassMolecular Mass=3.0 x 101398= 3.06101.pdf
No. of Moles = Given Mass/Molecular Mass
=3.0 x 1013/98
= 3.06*1011 Moles
Solution
No. of Moles = Given Mass/Molecular Mass
=3.0 x 1013/98
= 3.06*1011 Moles.
Let the sample size be n ,Mean of the sample = 27 (same as pop.pdf
Let the sample size be n ,
Mean of the sample = 27 (same as population mean)
Standard deviation of sample = population standard deviation / sqrt(n)
= 2.5/sqrt(n)
Solution
Let the sample size be n ,
Mean of the sample = 27 (same as population mean)
Standard deviation of sample = population standard deviation / sqrt(n)
= 2.5/sqrt(n).
JeffSolutionJeff.pdf
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import java.util.Scanner;public class Assignment4 { public st.pdf
import java.util.Scanner;
public class Assignment4 {
public static void main(String args[]) {
final int LEN = 5;
int[] hand = new int[LEN];
Scanner input = new Scanner(System.in);
//input the hand
System.out.println(\"Enter five numeric cards, no face cards. Use 2-9.\");
for (int i = 0; i < hand.length; i++) {
hand[i] = input.nextInt();
}
//sort the collection
bubbleSortCards(hand);
//determine players hand type
//flow of evaluation -> checking complex hands first
if (containsFullHouse(hand)) {
System.out.println(\"Full House!\");
} else if (containsStraight(hand)) {
System.out.println(\"Straight!\");
} else if (containsFourOfaKind(hand)) {
System.out.println(\"Four of a Kind!\");
} else if (containsThreeOfaKind(hand)) {
System.out.println(\"Three of a Kind!\");
} else if (containsTwoPair(hand)) {
System.out.println(\"Two Pair!\");
} else if (containsPair(hand)) {
System.out.println(\"Pair!\");
} else
System.out.println(\"High Card!\");
}
//cardInSets is called with hand[] and pair to check if a pair is found.
public static boolean containsPair(int hand[]) {
int pair = 2;
return cardInSets(hand, pair) != NOT_FOUND;
}
//cardInSets is called with hand[] and pair to check if a pair is found.
//Then calls back again with fcard result to check for another pair.
public static boolean containsTwoPair(int hand[]) {
int pair = 2;
int fcard = cardInSets(hand, pair);
int scard = cardInSets(hand, pair, fcard);
return fcard!=NOT_FOUND && scard!=NOT_FOUND;
}
//cardInSets is called with hand[] and threeaKind to check if three of a kind has been found.
public static boolean containsThreeOfaKind(int hand[]) {
int threeaKind = 3;
return cardInSets(hand, threeaKind) != NOT_FOUND;
}
//hasStraight will resolve to true, unless the previous value++ is not equivalent.
//The loop ends as soon as hasStaight is false or the end of the hand is reached.
public static boolean containsStraight(int hand[]) {
boolean hasStraight = true;
int i=1;
do {
hasStraight = (hand[i] == (hand[i-1] + 1));
} while(hasStraight && ++i < hand.length);
return hasStraight;
}
//cardInSets is called with hand[] and threeKinds to check if three of a kind is found.
//Then calls back again with pair and tcard result to check for a pair of another card type.
public static boolean containsFullHouse(int hand[]) {
int threeKinds = 3;
int pair = 2;
int tcard = cardInSets(hand, threeKinds);
int pcard = cardInSets(hand, pair, tcard);
return tcard!=NOT_FOUND && pcard!=NOT_FOUND;
}
//cardInSets is called with hand[] and fouraKind to check if four of a kind is found.
public static boolean containsFourOfaKind(int hand[]) {
int fouraKind = 4;
return cardInSets(hand, fouraKind) != NOT_FOUND;
}
//When ignoreCard isn\'t needed, it overloads with a sentinel 0.
public static int cardInSets(int hand[], int numKinds) {
return cardInSets(hand, numKinds, 0);
}
//The method cardInSets is a generic evaluation that returns if the parameter specified
numKinds(like 2 for pair) has been found in the hand.
//The ignoreCard d.
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- 2. 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) Solution (1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these reactions. The result is a mild solution of sulfuric acid and nitric acid. Acid rain causes acidification of lakes and streams and contributes to damage of trees at high elevations and many sensitive forest soils. (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 )
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