B and D is out. It cant be A because heat of combustion is substance not compound. So the answer is D.
Answer:
Peatification and coalification
Explanation:
https://energyeducation.ca/encyclopedia/Coal_formation
Answer:
A. There was still 140 ml of volume available for the reaction
Explanation:
According to Avogadro's law, we have that equal volumes of all gases contains equal number of molecules
According to the ideal gas law, we have;
The pressure exerted by a gas, P = n·R·T/V
Where;
n = The number of moles
T = The temperature of the gas
R = The universal gas constant
V = The volume of the gas
Therefore, given that the volumes and number of moles of the removed air and added HCl are the same, the pressure and therefore, the volume available for the reaction will remain the same
There will still be the same volume available for the reaction.
<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹)
and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³
n = ?
<span>
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
n = 4.0359 x 10</span>⁻³ mol
<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>
Moles = mass / molar
mass
</span>Mass of the gas = 0.529 g
<span>Molar mass of the gas</span> = mass / number of moles<span>
= </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span> = </span>131.07 g mol</span>⁻¹<span>
Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹
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.
