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.
Answer:
There are 0.219 mol of LINO3
This problem requires our calculation to undergo the dimensional analysis approach. In this approach, you disregard the actual quantity and focus on the units of measurement. This helps us know the units of our final answer.
First, let's ignore 16. Let's focus on converting the units kPa-mm³/s to mJ/s. The unit kPa stands for kiloPascals which is 1000 times greater than 1 Pa. The unit mJ, on the other hand, stands for millijoules, which is 1000 times lesser than Joules. The relationship between the two is that, Joules = Pa × m³. But since we want our final answer to be mJ, that would be equal to Pa×mm³. Since the original unit already contains mm³, all we have to do is convert kPa to Pa.
16 kPa-mm³/s * (1000 Pa/1 kPa) = 16,000 Pa-mm³/s
Since Pa-mm³ is equal to mJ, the final conversion yields to 16,000 Pa-mm³/s.
