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MArishka [77]
3 years ago
14

Design and implement a program (name it SumValue) that reads three integers (say X, Y, and Z) and prints out their values on sep

arate lines with proper labels, followed by their average with proper label. Comment your code properly. Format the outputs following the sample runs below.
Sample run 1:
X = 3
Y = 2
Z = 4
Average = 3.0
Sample run 1:
X = 7
Y = 8
Z = 10
Average = 8.333333333333334
Sample run 1:
X = 7
Y = -3
Z = -5
Average = -0.3333333333333333
Chemistry
1 answer:
rjkz [21]3 years ago
3 0

Answer:

class sum (

public static void sumofvalue (int m, int n, int p)

{

System.out.println(m);

System.out.println(n);

System.out.println(p);

int SumValue=m+n+p;

System.out.println("Average="+Sumvalue/3);

}

)

Public class XYZ

(

public static void main(String [] args)

{

sum ob=new sum();

int X=3;

int X=4;

int X=5;

ob.sumofvalue(X,Y,Z);

int X=7;

int X=8;

int X=10;

ob.sumofvalue(X,Y,Z);

}

)

Explanation:

The above program is made in Java, in which first we have printed value in a separate line. After that, the average value of those three values has been printed according to the question.

The processing of the program is given below in detail

* The first one class named 'sum' has been created which contains the function to print individual value and the average of those three values.

* In seconds main class named 'XYZ', the object of that the above class had been created which call the method of the above class to perform functions.

* In the main class values are assigned to variables X, Y, Z.

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When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
Reika [66]

Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

Half-life

{t_{1/2}}A=2\ \text{min}

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

The order of the reaction is 2.

t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}

The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

3 0
3 years ago
I need your help!! about Chemistry (15 point!!)
irakobra [83]
1. I think it is true?
2. Low melting points
3. True
4. Atomic number, I think it’s periods?
5. Groups?


Sorry, I might not get all of them right :(


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6 0
3 years ago
How many molecules are in 3.6 grams of NaCl?
raketka [301]

Answer:

\boxed{\text{None}}

Explanation:

There are no molecules in NaCl, because it consists only of ions.

However, we can calculate the number of formula units (FU) of NaCl.

Step 1. Calculate the moles of NaCl

\text{No. of moles}=\text{3.6 g NaCl}\times \dfrac{\text{1 mol NaCl}}{\text{63.54 g NaCl}} = \text{0.0567 mol NaCl}

Step 2. Convert moles to formula units

\text{FU} = \text{0.0567 mol NaCl} \times \dfrac{6.022 \times 10^{23}\text{ FU NaCl}}{\text{1 mol NaCl}}\\\\= 3.4 \times 10^{22} \text{ FU NaCl}

There are \boxed{3.4 \times 10^{22} \text{ FU NaCl}} in 3.6 g of NaCl.

6 0
3 years ago
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