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:
The correct answer is A Energy leaves the iron bar and enters the wood until the temperature are equal.
Explanation:
According to the law of conservation of energy or the first law thermodynamics energy neither be created nor destroyed, energy is transferred from one form to another form.
Here iron bar is placed in wood block energy is transferred from iron bar to wood until the temperatures are equal.
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
To learn more about pH here
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Potassium oxide: K₂O.
There's no need for prefixes since K₂O is an ionic compound.
<h3>Explanation</h3>
Find the two elements on a periodic table:
- Potassium- K- on the left end of period four.
- Oxygen- O- near the right end of periodic two.
Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.
- Potassium is a metal,
- Oxygen is a nonmetal.
A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:
- Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.
- Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion
when it combines with metals.
The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each ion is twice that on a 
ion. Each would pair up with two . Hence the subscript in the formula: 
.
There are two classes of compounds:
- Covalent compounds, which need prefixes, and
- Ionic compounds, which need no prefix.
Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide 
, the prefix di- indicates that there are two oxygen atoms in the formula . However, there's no need for prefix in ionic compounds such as 
.
