Free Essay

Join

In:

Submitted By minter21
Words 1422
Pages 6
Unit: Reaction Rates
This Unit Activity will help you meet these educational goals: • Science Inquiry—You will explore the ozone layer by doing online research and collecting information. You will then communicate your results in written form. • STEM—You will use your knowledge of kinetics to analyze scientific investigations. • 21st Century Skills—You will employ online tools for research and analysis.

Introduction
You have probably heard about the ozone layer and/or the hole in the ozone layer. Ozone is an allotrope of oxygen (O3) which is a very important part of our lives. In this activity, you will learn more about ozone and how it is formed. You will also investigate what causes the ozone layer to be depleted over time in the upper atmosphere.
________________________________________________________________________________

Directions and Analysis

Task 1: The Ozone Layer

Read about the ozone layer and answer the following questions. You can use the key word “ozone” or “ozone layer” in a search engine to learn more about it.

Key search terms: ozone, ozone layer, benefits of ozone

1. Describe a few of the properties of ozone and compare these to the properties of diatomic oxygen.

Type your response here:

2. Describe the process by which ozone can be formed in large cities.

Type your response here:

3. Even though ozone may be dangerous to our health here on Earth, how can it be beneficial in the upper atmosphere?

Type your response here:

4. How is ozone formed in the upper atmosphere?

Type your response here:

5. How does the ozone protect us from harmful radiation?

Type your response here:

Task 2: Kinetics of ozone

As you have seen, ozone can be a very beneficial part of our world. It is required to shield us from the harmful rays that are given off by the sun. Let’s take a look at some kinetics involved with ozone.

1. As you have seen, ozone is formed in the upper atmosphere through the reaction: O(g) + O2 (g) ( O3(g) Data has shown that this is first order with respect to oxygen atoms and first order with respect to diatomic oxygen. Write the rate law equation for this reaction.

Type your response here:

2. The rate constant is found to be 3.9 x 10-5 M-1 s-1. Calculate the rate of formation of ozone when the concentration of oxygen atoms is 3.0 x 10-14 molar and the concentration of diatomic oxygen is 1.3 x 10-14 molar.

Type your response here:

3. In your reading, perhaps you saw that in 1995, three chemists were awarded the Nobel Prize for their studies involving the ozone layer. First, in 1970, scientist, Paul Cruzen, hypothesized that nitrogen oxides could participate in the destruction of ozone. He was concerned that high-flying jet aircraft, such as the Concorde, could be the culprit that created the nitrogen oxide. He proposed the following mechanism to show how these compounds were involved in the destruction of ozone:

Step 1 - NO(g) + O3(g) ( NO2(g) + O2(g) Step 2 - NO2(g) + O(g) ( NO(g) + O2(g)

The rate equation for this reaction was found to be: Rate of reaction = k[NO][O3] Looking at this rate law, which of the above steps would be the rate determining step and why?

Type your response here:

4. What is the overall reaction for the above mechanism?

Type your response here:

5. Is the NO shown acting as a catalyst or an intermediate in the reaction? Why?

Type your response here:

6. It was found that the effect of jet aircraft on the ozone layer was too insignificant to be the cause of ozone depletion. There were simply not enough aircraft to inflict lasting harm on the ozone layer. So, in 1973, Professor Sherry Rowland and his colleague Dr. Mario Molina began to investigate chlorofluorocarbons, or CFCs. Because they are such stable compounds, CFCs were used extensively in industry as aerosol propellants, refrigerants and in insulation foams.

So Rowland and Molina had two questions – first, how persistent are these stable CFCs, meaning what is the amount of time CFC molecules could remain in the lower atmosphere without reacting? Second, the researchers also wanted to find out what might happen if these CFCs made their way into the upper atmosphere and got subjected to high-energy ultraviolet radiation there. Their investigation found that, because of their stability, CFCs could remain in the air for many centuries without reacting. It was the answer to their second question that really got their interest.

They found that the breakdown of CFCs by ultraviolet radiation produces chlorine atoms:

CF2Cl2(g) ( CF2Cl(g) + Cl(g)

Their next question was what would happen to the chlorine atoms that would be floating around in the atmosphere. They concluded that the chlorine could potentially react with ozone using the following reaction mechanism:

Step 1 – Cl(g) + O3(g) ( ClO(g) + O2(g) (slow step) Step 2 – ClO(g) + O(g) ( Cl(g) + O2(g) (fast step)

Predict the rate equation for the mechanism for the reaction between chlorine atoms and ozone molecules and write the overall reaction.

Type your response here:

7. So, it is seen that the destruction of ozone can be catalyzed by NO or by Cl. It has been found that the activation energy for the reaction catalyzed by Cl is 2 kJ/mol and for the reaction catalyzed by NO is 12 kJ/mol. Which is the more effective catalyst? Why?

Type your response here:

8. Which of the reactions will have the greater rate constant? Why?

Type your response here:

9. On average, one chlorine atom can destroy 100,000 ozone molecules. Such destructive power is tremendous when it is related to the millions of tons of CFCs that had been produced up until that time.

When Rowland and Molina calculated the likely effect on the ozone layer, they could not believe their figures. They consulted another scientist Hal Johnson who confirmed their results. In 1974, Rowland and Molina published their predictions about ozone depletion in the scientific journal Nature. Although the United States was quick to ban the production of CFCs, many other countries including Britain, China, France and Japan did not. Look at the ozone depletion images from 1979-1994. These images show the amounts of ozone in the atmosphere above Antarctica in October. The colors indicate the relative amounts of ozone present. The “red” color indicates the highest amounts and the “purple” indicates the smallest amounts. In 1994, the “ozone hole” showed to have a loss of 65% of its original ozone amounts.

Using the images collected on the site, describe the trends you see over time in the strength of the ozone coverage.

Type your response here:

_______________________________________________________________________________

Resources • ozone layer • ozone depletion

________________________________________________________________________________

Evaluation
This project will be evaluated on a rubric that is based on the completeness, clarity, and thinking you exhibit in the Directions and Analysis section above.

10 points possible: • The ozone layer: 5 points maximum • Kinetics of ozone: 5 points maximum

|the ozone layer |kinetics of ozone |
|5 points |5 points |
|Properties of ozone (1 point) |Rate law and reaction rate of ozone (1 point) |
|Process by which ozone can be formed in large cities (1 point) |NO catalyzed ozone destruction (1 point) |
|Benefits of ozone in the upper atmosphere (1 point) |Cl catalyzed destruction (1 point) |
|Process by which ozone is formed in the upper atmosphere (2 points) |Conclusions drawn from the two different reactions |
| |(2 points) |

Self-Checked Activity: For Advanced Understanding
Note: This section may be optional; follow your instructor’s directions.

The free response test items accessed below are related to this unit’s concepts and content, but will draw from your prior chemistry knowledge as well. They will challenge your knowledge of the unit concepts, help you tie different concepts together, and help you gain confidence in addressing written response test items.

Download: Free Response Test

• Respond to question 3. • Read the directions section carefully. • Answer all parts of the question thoughtfully and concisely, to the extent of your ability. • Unless directed otherwise, answer the question in essay form; do not use bullets, outline format, etc.

For Scoring: Scoring Guidelines

• See Question 3

Download: Free Response Test

• Respond to question 2. • Read the directions section carefully. • Answer all parts of the question thoughtfully and concisely, to the extent of your ability. • Unless directed otherwise, answer the question in essay form; do not use bullets, outline format, etc.

For Scoring: Scoring Guidelines

• See Question 2

[pic]

-----------------------
Chemistry

Unit Activity

Similar Documents

Free Essay

Joins in Sql

...Join in SQL SQL Join is used to fetch data from two or more tables, which is joined to appear as single set of data. SQL Join is used for combining column from two or more tables by using values common to both tables. JoinKeyword is used in SQL queries for joining two or more tables. Minimum required condition for joining table, is(n-1) where n, is number of tables. A table can also join to itself known as, Self Join. Types of Join The following are the types of JOIN that we can use in SQL. * Inner * Outer * Left * Right Cross JOIN or Cartesian Product This type of JOIN returns the cartesian product of rows from the tables in Join. It will return a table which consists of records which combines each row from the first table with each row of the second table. Cross JOIN Syntax is, SELECT column-name-list from table-name1 CROSS JOIN table-name2; Example of Cross JOIN The class table, ID | NAME | 1 | abhi | 2 | adam | 4 | alex | The class_info table, ID | Address | 1 | DELHI | 2 | MUMBAI | 3 | CHENNAI | Cross JOIN query will be, SELECT * from class, cross JOIN class_info; The result table will look like, ID | NAME | ID | Address | 1 | abhi | 1 | DELHI | 2 | adam | 1 | DELHI | 4 | alex | 1 | DELHI | 1 | abhi | 2 | MUMBAI | 2 | adam | 2 | MUMBAI | 4 | alex | 2 | MUMBAI | 1 | abhi | 3 | CHENNAI | ...

Words: 1005 - Pages: 5

Free Essay

Outer Join and Scalar Queries

...and the number of sections that they have taught. Make sure to show the number of sections as 0 for instructors who have not taught any section. Your query should use an outer join, and should not use scalar subqueries. By using the university schema provided by db-book.com the following queries were done on the university database. The first query uses an outer join which works similar to the join operation but it keeps the rows that don’t match between the two tables that would be lost in a join operation. There are three forms of outer join: a) Left outer join displays the results from the left table even if the condition does not find any matching record in the right table. b) Right outer join will displays the results from the right table regardless if there is matching data in the left table. c) Full outer join will retain all rows from both tables, regardless if the data matches or not. The group by clause when used in a select statement collects data from multiple records and groups the results into one or more columns. The below is the query using left outer join operation: select ID, name, count(sec_id) as Number_of_sections from instructor natural left outer join teaches group by ID, name; The next query was written by using a scalar subquery without using an outer join operation. A scalar subquery is where the output of a subquery returns only one row which contains one single column/attribute. By using the select statement with count (*)...

Words: 414 - Pages: 2

Premium Essay

Sbsehe

...Table Name : Employee EMPLOYEE _ID FIRST_NA ME LAST_NA ME SALA RY JOINING_D ATE DEPARTME NT 1 John Abraham 1000000 01-JAN-13 12.00.00 AM Banking 2 Michael Clarke 800000 01-JAN-13 12.00.00 AM Insurance 3 Roy Thomas 700000 01-FEB-13 12.00.00 AM Banking 4 Tom Jose 600000 01-FEB-13 12.00.00 AM Insurance 5 Jerry Pinto 650000 01-FEB-13 12.00.00 AM Insurance 6 Philip Mathew 750000 01-JAN-13 12.00.00 AM Services 7 TestName1 123 650000 01-JAN-13 12.00.00 AM Services 8 TestName2 Lname% 600000 01-FEB-13 12.00.00 AM Insurance Table Name : Incentives EMPLOYEE_REF_ID INCENTIVE_DATE INCENTIVE_AMOUNT 1 01-FEB-13 5000 2 01-FEB-13 3000 3 01-FEB-13 4000 1 01-JAN-13 4500 2 01-JAN-13 3500 SQL Queries Interview Questions and Answers on "SQL Select" 1. Get all employee details from the employee table Select * from employee 2. Get First_Name,Last_Name from employee table Select first_name, Last_Name from employee 3. Get First_Name from employee table using alias name “Employee Name” Select first_name Employee Name from employee 4. Get First_Name from employee table in upper case Select upper(FIRST_NAME) from EMPLOYEE 5. Get First_Name from employee table in lower case Select lower(FIRST_NAME) from EMPLOYEE 6. Get unique DEPARTMENT from employee table select...

Words: 4444 - Pages: 18

Premium Essay

Note

...In SQL, a join is used to compare and combine rows from two or more table. Inner Join: An inner join returns records only those records are common in both tables. Outer Join An outer join returns a set of records (or rows) that include what an inner join would return but also includes other rows for which no corresponding match is found in the other table. There are three types of outer joins: * Left Outer Join (or Left Join) * Right Outer Join (or Right Join) * Full Outer Join (or Full Join) Each of these outer joins refers to the part of the data that is being compared, combined, and returned. Sometimes nulls will be produced in this process as some data is shared while other data is not. Left Outer Join A left outer join will return all the data in Table 1 and all the shared data (so, the inner part of the Venn diagram example), but only corresponding data from Table 2, which is the right join. Right Outer Join A right outer join returns Table 2's data and all the shared data, but only corresponding data from Table 1, which is the left join. Full Outer Join A full outer join, or full join, which is not supported by the popular MySQL database management system, combines and returns all data from two or more tables, regardless of whether there is shared information. Think of a full join as simply duplicating all the specified information, but in one table, rather than multiple tables. Where matching data is missing, nulls will be produced. These...

Words: 288 - Pages: 2

Free Essay

Mist Mysql Queries

...Project 6 Required Queries: *  Create a query that includes at least two INNER JOINS (i.e. spans at least three tables), a search criteria, and orders the resulting data. * >select M.MID, M.MFName * >from MEMBER as M * >inner join LINEITEM as LI on LI.MID=M.MID * >inner join ITEM as I on I.IID=LI.IID * >inner join DVD as D on D.IID=I.IID; * This shows which members have checked out DVDs * Create a query that includes a calculation (e.g. average, sum, etc.). * >Select M.MID, M.MFName, count(B.IID) as “Books Checked Out” * >from MEMBER as M * >inner join LINEITEM as LI on LI.MID=M.MID * >inner join ITEM as I on I.IID=LI.IID * >inner join BOOK as B on B.IID=I.IID; * This query shows how many books a member has checked out * Create a query that includes at least one RIGHT or LEFT outer join. * >select MEMBER.MID, MEMBER.MFName, LIBRARIAN.LFName * >from MEMBER * >left outer join LINEITEM on LINEITEM.MID=MEMBER.MID * >left outer join LIBRARIAN on LINEITEM.LID=LIBRARIAN.LID * >order by MEMBER.MID; * This query uses a left join to show the relations between which librarians have checked out which members * Create a query that uses a GROUP BY to perform a calculation on information spanning at least two tables. * Create a query that includes a GROUP BY, but selects items from...

Words: 289 - Pages: 2

Premium Essay

Programming

...Which is not true about the USING keyword? A) you use it to simplify the syntax for joining tables B) you code a USING clause in addition to the ON clause C) it can be used with inner or outer joins D) the join must be an equijoin, meaning the equals operator is used to compare the two columns Points Earned: 0.0/2.0 Correct Answer(s): B Correct 2. When you code a SELECT statement, you must code the four main clauses in the following order A) SELECT, FROM, ORDER BY, WHERE B) SELECT, ORDER BY, FROM, WHERE C) SELECT, WHERE, ORDER BY, FROM D) SELECT, FROM, WHERE, ORDER BY Points Earned: 2.0/2.0 Correct Answer(s): D Correct 3. When coded in a WHERE clause, which search condition will return invoices when payment_date isn’t null and invoice_total is greater than or equal to $500? A) payment_date IS NULL AND invoice_total > 500 B) payment_date IS NOT NULL OR invoice_total >= 500 C) NOT (payment_date IS NULL AND invoice_total = 500 Points Earned: 2.0/2.0 Correct Answer(s): D Correct 4. The order of precedence for the logical operators in a WHERE clause is A) Not, And, Or B) And, Or, Not C) Or, And, Not D) Not, Or, And Points Earned: 2.0/2.0 Correct Answer(s): A Correct 5. Which of the following types of SQL statements is not a DML statement? A) INSERT B) UPDATE C) SELECT D) CREATE TABLE Points Earned: 2.0/2.0 Correct Answer(s): D Incorrect 6. Which of the following is not a common error when entering and...

Words: 1309 - Pages: 6

Premium Essay

Data Base

... SELECT o.OrderID, n.ContactName, e.FirstName AS EmployeeFirstName, e.LastName AS EmployeeLastName, o.OrderDate FROM Orders o JOIN Employees e ON (e.EmployeeID = o.EmployeeID) JOIN Customers n ON (n.CustomerID = o.CustomerID) WHERE o.OrderDate > '1998-03-01' ORDER BY n.ContactName; 2. SELECT COUNT(DISTINCT Employees.EmployeeID) AS NumberOfEmployees, ISNULL(Employees.Country, 'None.') AS Country, COUNT(DISTINCT Customers.CustomerID) AS NumberOfCustomers, Customers.Country FROM Employees RIGHT JOIN Customers ON Employees.Country = Customers.Country GROUP BY Customers.Country, Employees.Country ORDER BY NumberOfEmployees DESC, NumberOfCustomers DESC; ΕΡΩΤΗΜΑ Δ 1. SELECT Address, Code, Country FROM Suppliers UNION SELECT Address, Code, Country FROM Customers UNION SELECT Address, Code, Country FROM Employees ORDER BY Country; 2. SELECT AVG(OD.Quantity) AS AverageQuantity, SUM(OD.Quantity * OD.UnitPrice) AS TotalOrderAmount, DATEDIFF(yy, e.BirthDate, e.HireDate) AS HireAge FROM [Order Details] OD JOIN Orders o ON ( OD.OrderID =O.OrderID) JOIN Employees E ON ( E.EmployeeID =O.EmployeeID) WHERE DATEDIFF(yy, e.BirthDate, e.HireDate) > = 30 GROUP BY DATEDIFF(yy, e.BirthDate, e.HireDate) ORDER BY DATEDIFF(yy, e.BirthDate, e.HireDate) DESC; ΕΡΩΤΗΜΑ Ε 1. SELECT o.ShippedDate, o.OrderDate FROM Orders o JOIN Suppliers sp ON (o.EmployeeID = sp.SupplierID) WHERE sp.CompanyName LIKE '%Pavlova%' ; 2. SELECT c.CompanyName,...

Words: 298 - Pages: 2

Free Essay

Saaadfasfwaeraerasfasdfaesfasdfs

...name who are treated by Doctor Dr Bill. Sol.select p.fname,p.lname from patients p inner join doctors d on p.docid=d.doctorid where d.fname='dr.bill'; Q2. Find Patient Details occupied in room 1100. Sol.select p.patientid,p.fname,p.lname,p.address from patients p inner join room r on p.patientid=r.patientid where r.roomno=1100; Q3. Nurse Sue checked which all patients on 09-09-2015. Sol.select p.fname,p.lname from patients p where p.patientid in(select r.patientid from room r inner join nursecheckingdetails n on n.roomno=r.roomno where n.daate='09-09-2015'); Q4. List Names of Nurses who work with Dr Bill. Sol.select n.fname,n.lname from nurses n inner join doctors d on n.doctorid=d.doctorid and d.fname='dr.bill'; Q5. List Names of Doctors and number of patients they are treating or have treated. Sol.select d.fname as Doctorfname,d.lname as DoctorLName,count(p.patientid) as NoofPatients from patients p inner join doctors d on p.docid=d.doctorid group by d.fname,d.lname; Q6. List Names of Doctors and number of nurses they have with them to help. Sol.select d.fname as Doctorfname,d.lname as DoctorLname,count(n.nurseno) from doctors d left outer join nurses n on d.doctorid=n.doctorid group by d.fname,d.lname; Q7. How many rooms to type Basic are occupied? Sol.select count(r.roomno) as NoOfRooms from room r inner join roomdetails rd on r.typeofroom=rd.roomtypeid where rd.type='basic' and r.status='occupied'; ...

Words: 461 - Pages: 2

Free Essay

Enginerring Manager

...SELECT column_list FROM table-name  [WHERE Clause] [GROUP BY clause] [HAVING clause] [ORDER BY clause]; SELECT name, salary, salary*1.2 AS new_salary FROM employee  WHERE new_salary /salary*1.2 > 30000; SELECT first_name, last_name, subject  FROM student_details  WHERE subject IN/NOT IN ('Maths', 'Science');  Select * from product p  where EXISTS (select * from order_items o  where o.product_id = p.product_id) ------------------------------------------------- SELECT user.name, course.name ------------------------------------------------- FROM `user` ------------------------------------------------- RIGHT JOIN `course` on user.course = course.id SELECT name, salary FROM employee ORDER BY 1, 2; ELECT name, salary  FROM employee  ORDER BY name DESC, salary DESC; SELECT COUNT (DISTINCT name) FROM employee; HAVING Group functions (MAX,MIN) cannot be used in WHERE Clause but can be used in HAVING clause. SELECT dept, SUM (salary)  FROM employee  GROUP BY dept  HAVING SUM (salary) > 25000  INSERT INTO TABLE_NAME  [ (col1, col2, col3,...colN)]  VALUES (value1, value2, value3,...valueN); INSERT INTO employee (id, name, dept, age, salary location) SELECT emp_id, emp_name, dept, age, salary, location  FROM temp_employee; Copy data from A to B. Select first then insert If you are inserting data to all the columns, the above insert statement can also be written as, INSERT INTO employee  SELECT...

Words: 652 - Pages: 3

Free Essay

Test

...ALTER PROCEDURE [dbo].[ASP_Weekly_Lift_Report] @Voyage_Start Varchar(3) = Null, @Voyage_End Varchar(3) = Null, @Sales_Rep Varchar(25) = Null, @Service Varchar(4) = Null AS SET @Voyage_Start = IsNull(@Voyage_Start, '%') SET @Voyage_End = IsNull(@Voyage_End, '%') SET @Sales_Rep = IsNull(@Sales_Rep, '%') SET @Service = IsNull(@Service, '%') -- SET NOCOUNT ON added to prevent extra result sets from -- interfering with SELECT statements. SET NOCOUNT ON; -- Insert statements for procedure here SELECT dbo.tbl_Lift_Report_PayorList.[Rep Name], dbo.tbl_Lift_Report_PayorList.Parent, dbo.tbl_Lift_Report_PayorList.Business_Type, dbo.tbl_Lift_Report_PayorList.PayorID, dbo.tbl_Lift_Report_PayorList.Payor, dbo.tbl_Lift_Report_PayorList.[Payor Name - ID], dbo.tbl_Lift_Report_PayorList.[Booking Payor], dbo.tbl_Lift_Report_PayorList.ShipperID, dbo.tbl_Lift_Report_PayorList.Shipper, dbo.tbl_Lift_Report_PayorList.ConsigneeID, dbo.tbl_Lift_Report_PayorList.Consignee, dbo.tbl_Lift_Report_PayorList.[Service], dbo.tbl_Lift_Report_PayorList.[Svc Desc], dbo.tbl_Lift_Report_PayorList.Vessel, dbo.tbl_Lift_Report_PayorList.Voyage, dbo.tbl_Lift_Report_PayorList.[Voyage No], dbo.tbl_Lift_Report_PayorList.[Cargo Type], dbo.tbl_Lift_Report_PayorList.Booking, dbo.tbl_Lift_Report_PayorList.[Profit Year], dbo.tbl_Lift_Report_PayorList...

Words: 356 - Pages: 2

Free Essay

Estadistica

...Estándares Transact-SQL Buenas Prácticas de Programación 1. Convenciones y Estándares de Nombres Nota: Los términos “notación Pascal” y “notación de Camell” son usados a través de este documento. Notación Pascal – El primer carácter de todas las palabras se escribe en Mayúsculas y los otros caracteres en minúsculas. Ejemplo: ColorDeFondo Notación de Camell – El primer carácter de todas las palabras, excepto de la primera palabra se escribe en Mayúsculas y los otros caracteres en minúsculas. Ejemplo: colorDeFondo 1. Usa notación Pascal para el nombre de las Tablas CREATE TABLE dbo.Employee 2. Usa notación Pascal para el nombre de los campos de tabla CREATE TABLE dbo.Employee ( IdEmployee INT, FirstName VARCHAR(20), LastName VARCHAR(20) ) 3. NO usar nunca “sp_”, La razón es que: SQL Server reconoce el prefijo “sp_” como “System Stored Procedure”, es decir, un procedimiento almacenado de Sistema y lo buscaría en la Base de Datos. Usa la siguiente sintaxis para los nombres de los Stored procedures: Usp_<Nombre Esquema>_<Nombre Tabla> _<Accion> Ejemplo: usp_GEN_Employee_Insert usp_GEN_Employee_GetAll 4. Usa esquemas para agrupar los objetos como tablas, los nombres deben de ser Abreviados. Incorrecto: GEN_Employee Correcto: GEN.Employee 2. Consideraciones en el diseño de base de datos 1. El nombre de la base de datos debe de asemejarse al nombre de la aplicación, no deberá de contener...

Words: 2608 - Pages: 11

Free Essay

Database Management

...Written Assignment 3 Explain the SQL commands Union, Intersect, and Minus with concepts like union-compatibility, and syntax alternatives such as IN/NOT IN and various JOIN options Database Management CIS-311 There are many different SQL commands. This assignment will focus on UNION, INTERSECT, and MINUS. UNION combines unique rows returned by two SELECT statements. UNION ALL functions in the same way as UNION except that it also returns duplicates. INTERSECT gives you rows that are found in both queries by eliminating rows that are only found in one or the other. An INTERSECT is simply an inner join where we compare the tuples of one table with those of the other, and select those that appear in both while weeding out duplicates. MINUS returns the rows that are in the first query but not the second by removing the rows that are only found in the second query. There are three primary SQL commands involved when implementing a Union, Intersection and difference relational operators. As you may know, SQL data manipulation commands are set-oriented which are involved in operating over entire sets of rows and columns in tables at once. The UNION, INTERSECT, and MINUS statements make sure these operations occur. Union, Intersect and Minus only work properly if relations are Union-Compatible, which is based on the names of the relation attributes that must be the same and their data types must be alike. Being compatible does not mean the data types have to be exactly the...

Words: 673 - Pages: 3

Free Essay

Join

...Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join join Join...

Words: 768 - Pages: 4

Premium Essay

Join

...NO IMPACT MAN This film shows us that you can change almost everything, even your way of life, every day we are surrounded of items and thing that we think are necessary for us, and some of this things are harmful for our environment. Why person can not change his or her footprint? A couple years ago Manhattanite author Colin Beavan with his wife Michele and their 2-years old daughter brake down this stereotype, they embody a solution to this issue in the life. It was Beavan's idea because he was interested in environment issue. Also he was keenly interested in getting a book and movie deal. As a result we can see this informative and interesting reality show/ documentary that reflects events in the life of the whole family.It was nice seeing how a family could completely change their lifestyle for a year and watching the progress of their journey seemed heartfelt and honest. The whole process was not just a straight up breeze in the park. It took effort and lots of adaptation. Colin and Michelle seemed real and maybe even relatable. It was very interesting to see family from huge city such as New York living without car, subway,TV,taxi,refrigerator far from civilization. "No impact man" means no air conditioning, no newspapers, no shopping for new clothes, no disposable diapers,no coffee, no toilet paper because Colin was thinking what if he actually try not to impact environment. From the first time it was hard to live without impact, and Michelle was arguing with Colin, because...

Words: 1154 - Pages: 5

Free Essay

Join

...Chapter 7 MULTIPLE CHOICE 1. This indicates the number of elements, or values, the array can hold. a.|the new operator| b.|the array’s size declarator| c.|the array’s data type| d.|the version of Java| ANS: B 2. What does the following statement do? double[] array1 = new double[10]; a.|Declares array1 to be a reference to an array of double values| b.|Creates an instance of an array of 10 double values| c.|Will allow valid subscripts in the range of 0 - 9| d.|All of the above| ANS: D 3. It is common practice to use a ____________ variable as a size declarator. a.|static| b.|reference| c.|final| d.|boolean| ANS: C 4. What do you call the number that is used as an index to pinpoint a specific element within an array? a.|subscript| b.|global unique identifier| c.|element| d.|argument| ANS: A 5. Subscript numbering always starts at what value? a.|0| b.|1| c.|-1| d.|None of the above| ANS: A 6. By default, Java initializes array elements with what value? a.|0| b.|100| c.|1| d.|-1| ANS: A 7. What will be the value of x[8] after the following code has been executed? final int SUB = 12; int[] x = new int[SUB]; int y = 100; for(int i = 0; i < SUB; i++) { x[i] = y; y += 10; } a.|170| b.|180| c.|190| d.|200| ANS: B 8. Java performs ____________, which means that it does not allow a statement to use a subscript that is outside the range of valid subscripts for the array...

Words: 1531 - Pages: 7