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fenix001 [56]
3 years ago
8

If x is a string, then x = new String("OH"); and x = "OH"; will accomplish the same thing. Group of answer choices True False

Chemistry
2 answers:
Vinvika [58]3 years ago
5 0

Answer:

True is the correct answer to the above question.

Explanation:

  • If x is a string then it can be assigned by the help of two ways in java:
  • By the help of constructor:- When we write " x = new String("OH");", then it will create a pass a string "OH" into the constructor. It is because the String is a class in java and x is an object created by the constructor of the String class.
  • With the help of assigning: The "x= OH", which assigns the value of x which is an object of String class it can also use the constructor to initialize the "OH" string on the class.
  • The above question states the two scenarios which are defined above. Hence the question statement is true.
Akimi4 [234]3 years ago
5 0

Answer:

"True" is the correct answer to this question.

Explanation:

The program to the given question as follows:

Program:

public class data //defining class

{

  public static void main (String [] aw)//defining the main method

  {

String x="OH"; //defining string variable x and assign value

System.out.println("assign value: "+x); //print value

x = new String("OH"); //defining instance variable and assign value

System.out.println("assign value by creating instance: "+x); //print value

  }

}

Output:

assign value: OH

assign value by creating instance: OH

Explanation of the program:

In the above java program, a class data is defined, inside the class the main method is declared, In this main method a string variable "x" is defined that holds a value "OH", then we the print function to print this variable value.

In the next line, An instance of variable x is created, which holds a value "OH" in its parameter. In this question, both are correct because both hold the same value.

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Explanation:

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Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu
Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (<em>M)</em> is calculated from the following equation:

<em>M</em> = <em>n</em> ÷ <em>V</em>

where <em>n</em> is the number of moles of the solute in the solution, and <em>V</em> is the volume of the solution.

Accordingly, the number of moles of the solute is given by

<em>n</em> = <em>M</em> x <em>V</em>

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

<em>n </em>(1) = <em>M </em>(1) x <em>V </em>(1)

<em>n </em>(2) = <em>M </em>(2) x <em>V </em>(2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

<em>n</em> (1) = <em>n</em> (2)

and

<em>M</em> (1) x <em>V</em> (1)<em> </em>= <em>M</em> (2) x <em>V</em> (2)

For the stock solution, <em>M</em> (1) = 1.244 M, and <em>V</em> (1) is the volume needed. For the dilute solution, <em>M</em> (2) = 0,1281 M, and <em>V</em> (2) = 250.0 mL.

The volume of the stock solution needed, <em>V</em> (1), is calculated as follows:

<em>V</em> (1) = <em>M</em> (2) x <em>V</em> (2) ÷ <em>M</em> (1)

<em>V</em> (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

<em>V </em>(1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0
3 years ago
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