Below is a program that takes an integer as input- Write a for loop wh.docx

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Below is a program that takes an integer as input. Write a for loop which prints a countdown from that number till 1 eg: if the inputted number is 4, then output should be: 4 3 2 1 Notes: 1. Use only for loop 2. Write your code only inside the placeholders provided. Do not edit the remaining part of the code. ------------------------------- import java.util.Scanner; class countdown{ public static void main(String args[]){ int startNum; Scanner scanner =new Scanner(System.in); System.out.println(\"Enter the number:\"); startNum=scanner.nextInt(); ///{ ///} } } Solution Answer import java.util.Scanner; class countdown { public static void main(String args[]) { int startNum; Scanner scanner =new Scanner(System.in); System.out.println(\"Enter the number:\"); startNum=scanner.nextInt(); for(int i=startNum;i>0;i--) //for loop starts from number entered { System.out.println(i); //prints the value of i } } } .

Below is a program that takes an integer as input. Write a for loop which prints a countdown
from that number till 1 eg: if the inputted number is 4, then output should be: 4 3 2 1 Notes: 1.
Use only for loop 2. Write your code only inside the placeholders provided. Do not edit the
remaining part of the code.
-------------------------------
import java.util.Scanner;
class countdown{
public static void main(String args[]){
int startNum;
Scanner scanner =new Scanner(System.in);
System.out.println("Enter the number:");
startNum=scanner.nextInt();
///{
///}
}
}
Solution
Answer
import java.util.Scanner;
class countdown
{
public static void main(String args[])
{

int startNum;
Scanner scanner =new Scanner(System.in);
System.out.println("Enter the number:");
startNum=scanner.nextInt();
for(int i=startNum;i>0;i--) //for loop starts from number entered
{
System.out.println(i); //prints the value of i
}
}
}

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Q1. Print all the odd numbers from 1 to a user specifiable upper limit (inclusive) using both while and for loops Example, Please enter a number: 9 Odd numbers from 1 to 9 are: 1 3 5 7 9 HINT: The number 9 should be included!!! 2. Write a program called CheckerBoard that displays the following n×n (n=7) checkerboard pattern using two nested for-loops. Solution Q)1) Using While Loop: DisplayOddNumbers.java import java.util.Scanner; public class DisplayOddNumbers { public static void main(String[] args) { //Declaring variables int number; int i=1; //Scanner object is used to get the inputs entered by the user Scanner sc=new Scanner(System.in); //getting the number entered by the user System.out.println(\"Please enter a number:\"); number=sc.nextInt(); //Displaying the odd numbers using while loop System.out.println(\"Odd numbers from 1 to 9 are:\"); while(number>0) { //Checking the number is even or odd if(i%2!=0) { System.out.println(i); } i++; number--; } } } _____________________ Output: Please enter a number: 9 Odd numbers from 1 to 9 are: 1 3 5 7 9 _______________ using for loop: DisplayOddNosUsingForLoop.java import java.util.Scanner; public class DisplayOddNosUsingForLoop { public static void main(String[] args) { //Declaring variables int number; //Scanner object is used to get the inputs entered by the user Scanner sc=new Scanner(System.in); //getting the number entered by the user System.out.println(\"Please enter a number:\"); number=sc.nextInt(); //Displaying the odd numbers using while loop System.out.println(\"Odd numbers from 1 to 9 are:\"); for(int i=1;i<=number;i++) { //Checking the number is even or odd if(i%2!=0) { System.out.println(i); } } } } ____________________ Output: Please enter a number: 9 Odd numbers from 1 to 9 are: 1 3 5 7 9 _____________________ 2)Q) package org.students; import java.util.Scanner; public class CheckerBoard { public static void main(String[] args) { //Displaying the checker board using nested for loop for(int i=1;i<=7;i++) { for(int j=1;j<=7;j++) { if(i%2==0) System.out.print(\" #\"); else System.out.print(\"# \"); } System.out.println(\" \"); } } } __________________ output: # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # ____________Thank You.

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Both these questions go together Solution When we bring a magnet near a metal, the metal enters a magnetic field, all of the electrons inside the metal \"line up,\" causing a temporary magnetic alignment that is attracted to the magnet . That alignment dissipates once the magnetic field is removed and therefore, the only way for a metal to be repelled by a magnet is if it\'s first magnetized to the opposite pole. so if the metal is magnetised and the a magnet with the same pole is brought near it will repel Note:- its easy to magnetise a iron piece just rub the magnet over the metal in one direction many, many times. By rubbing a magnet over your metal continuously, you are aligning the electrons in the metal in a certain way, toward a certain polarity. The more that you do this, the longer the effect will last, taking more time for the electrons to return to their random or unpolarized state. so the answer to the question is 1) c 2) d .

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6.3. (a) In which form does Cd(II) occur mainly in a water of the following composition: pH = 7.8 Solution (a) pH=7.8=-log [H+] [H+]=10^-7.8=1.585*10^-8M Also, [H+][OH-]=1*10^-14=Ionic product of water [OH-]=(1*10^-14)/[H+]=(1*10^-14)/(1.585*10^-8)=6.309*10^-7M Alkalinity=[OH-]+[CO3 2- ]=1.3*10^-3 eq/L [CO32-]=(1.3*10^-3 eq/L)-[OH-]=(1.3*10^-3 eq/L)-(6.309*10^-7M)=(1.3*10^-3 eq/L)=1/2*(1.3*10^-3 eq/L)=0.65*10^-3mol/L [For [OH-]eq=[OH]M ,but for Carbonate, 2eq=1mol] Thus,[CO32-]>>>[OH-] Also,Cd2+ +CO32- <--->CdCO3 ,log K=4.5,K=10^-4.5 But,Ca2+ +CO32- <--->CdCO3,log K=8.5 So,CdCO3 will precipitate first and CaCO3 later, Also, Cd T =1*10^-9 M<<<Ca T =1*10^-3 M And Cd 2+ +2OH- <--->Cd(OH)2 ,log K=14.3 The value of log K is much higher, so Cd(II) must be predominant as Cd(CO3) b) yes, CdCO3 precipitates due to very high equilibrium constant for this reaction compared to that for Cd(OH)2 c)[EDTA]=10^-7 mol/L Cd 2+ +EDTA 4- <--->CdEDTA 2- , log K=16.5,K=10^-16.5 K=[CdEDTA 2- ]/[Cd 2+ ][EDTA 4- ] or,[CdEDTA 2- ]=K*[Cd 2+ ][EDTA 4- ]=10^-16.5 *(10^-7)*[Cd2+]=10^-23.5 [Cd2+] But ,Cd 2+ +CO3 2- <--->CdCO3 ,log K=4.5,K=10^-4.5 Similarly,[CdCO3]=10^-4.5 *(6.309*10^-7)*[Cd2+]=(6.309*10^-11.5) [Cd2+] where., [OH-]=6.309*10^-7M Hence ,[CdEDTA 2- ]<[CdCO3] CdCO3 remains the predominant form for Cd(II) .

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6. A solution of an acid, with a general formula HA, contains 0.400 moles of H\"\'. The acid solution is neutralized by 43.4 mL of sodium hydroxide. What is the molarity of the sodium hydroxide solution? 7. Write the balanced chemical equations for the removal of water from the following hydrated crystals. a) Copper(Il) sulfate pentahydrate b) Tin(II) chloride dehydrate c) Cobalt(II) chloride hexahydrate d) Magnesium sulfate heptahydrate oint of solution made hy dissolving 6.000 q of urea Solution 6) HA + NaOH ------------------> NaA + H2O moles of HA = moles of NaOH = 0.400 moles = 0.400 volume = 43.4 mL = 0.0434 L molarity of NaOH = moles / volume = 0.400 / 0.0434 = 9.21 M 7) a) CuSO4 .5H2O ------------------> CuSO4 + 5 H2O b) SnCl2 H2O ----------------------> SnCl2 + H2O c) CoCl2.6H2O --------------------> CoCl2 + 6H2O d) MgSO4 7H2O -------------------> MgSO4 + 7H2O .

6- A solution of an acid- with a general formula HA- contains 0-400 mo.docx

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6. What are the differences between objects and abstract data types Solution Differences between Objects and Abstract data types: ______________________________________________ Objects: Object is an instance of a class and it has states and behaviours of class. Abstract data type: It is a data type of abstract and it hides the representation of class, and it is typically implemented as a class and each instance of Abstract data type is usually an object of that class. Objects and Abstract data types are not the same. But these are not different with each other. Objects are a kind of Abstract data types. An Abstract data type is an interface.But objects are instance of a class, with its own copy of any non-static variables. Abstract data type is a collection of methods but in case of objects it is not an interface.It is an representation of a class with new operator. A class can have one or more Abstract data types in the same way class can have one or more objects also. To understand you with clear for Example, Collection interface is Abstract data type, ArrayList is a class and you can make ArrayList object with new operator. .

6- What are the differences between objects and abstract data typesSol.docx

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6. What do the bubbles in boiling water principally c a. air b. O2 (e and H2e c. H20 d. vacuum e. nitrogen What is the correct name for Co3(PO4)4? a. cobalt phosphate b. tricobalt tetraphosphate c. cobalt (II) phosphate d. cobalt (III) phosphate e. cobalt (IV) phosphate 7. 8. How many neutrons are contained in *He? a. 0 b. 1 c. 2 d. 3 e. 4 9. How many electrons are contained in He 2 2 a. d. 3 e. 4 Solution 6) the bubbles in boiling water contains water vapours, hence H2O(g), as the vapour pressur eof li 7) The name for anion is phosphate , there are four phosphate groups The oxidation state of cobalt is 3x + 4(-3) = 0 x = +4 So the correct name is Cobal(IV)phosphate 8) neutrons = mass number - atomic number = 4-2 = 2 9) Electrons = 0 as the number of electrons in Helium in its ground state = 2 The charge is +2, which represents removal of two electrons .

6- What do the bubbles in boiling water principally c a- air b- O2 (e.docx

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6. Both solutions and colloids are mixtures. Differentiate between them. Solution Although solutions and colloid both are mixtures but we can differentiate solution and colloid on the basis of some parameter given below: 1) Particle size: Particle size of solution is less than 1nm or (10 -9 ) m Particles size of colloid are between 1nm to 1000 nm 2) Visibility In solutions the visibility of substances dissolved in it are not possible even with optical means. While in colloid it may be visible with ultra microscope. 3) Diffusion Diffusion is very rapid in solution while very slow in colloid 4) setting Solution doesn\'t settle while colloid may settle down under centrifuge . 5) Nature: Solution is Homogeneous in nature while colloid is heterogeneous in nature. 6) Tyndall effect Solution doesn\'t show Tyndall effect Colloid shows Tyndall effect So we can say that solution and colloid both are mixture but there are a lot of difference between them. .

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6). The Town of Elmwood applied for a competitive grant from the state government for providing an after- school recreational program for at-risk children from low-income families. On 2/1/2005, Elmwood was notified that it had been awarded a grant of $100,000 for the program. The program provides up to $1,000 per child served provided that family annual income is less than $25,000. The grant provides for reimbursement of all allowable costs of conducting the program on a twice-monthly basis. No time limit is imposed for completion of the program. Any unused grant award may be carried forward to the next fiscal year. By 12/31/2005, Elmwood?s fiscal year-end, $74,000 had been expended in providing recreational services for eligible beneficiaries. Reimbursement of $71,000 had been received from the state by this date. Required: [12 pts.] a). What are the eligibility requirements of this grant? b). As of 12/31/2005, identify the following: (1). Cash Received (2). Program Revenue (3). Expenditures Solution Total Expenditure = $ 74000 Cash Received = $ 71000 Revenue = 71000-74000 = $ 3000 The program must be carried on & also the expenditure will be for the children whose family income is less than $ 25000/-. .

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