From the balanced equation 2KClO3 → 2KCl + 3O2, the coefficients are the following:
coefficient 2 in front of potassium chlorate KClO3
coefficient 2 in front of potassium chloride KCl
coefficient 3 in front of oxygen molecule O2
We got this balanced equation by identifying the number of atoms of each element that we have in the given equation KClO3 → KCl + O2.
Looking at the subscripts of each atom on the reactant side and on the product side, we have
KClO3 → KCl + O2
K=1 K=1
Cl=1 Cl=1
O=3 O=2
We can see that the oxygens are not balanced. We add a coefficient 2 to the 3 oxygen atoms on the left side and another coefficient 3 to the 2 oxygen
atoms on the right side to balance the oxygens:
2KClO3 → KCl + 3O2
The coefficient 2 in front of potassium chlorate KClO3 multiplied by the subscript 3 of the oxygen atoms on the left side indicates 6 oxygen atoms just as the coefficient 3 multiplied by the subscript 2 on the right side indicates 6 oxygen atoms.
The number of potassium K atoms and chloride Cl atoms have changed as well:
2KClO3 → KCl + 3O2
K=2 K=1
Cl=2 Cl=1
O=6 O=6
We now have two potassium K atoms and two chloride Cl atoms on the reactant side, so we add a coefficient 2 to the potassium chloride KCl on the product side:
2KClO3 → 2KCl + 3O2, which is our final balanced equation.
K=2 K=2
Cl=2 Cl=2
O=6 O=6
The potassium, chlorine, and oxygen atoms are now balanced.
2.3 mols of Al*26.98 molar mass of Al=62.054g
Answer: Zinc nitride (Zn3N2) is an inorganic compound of zinc and nitrogen, usually obtained as (blue)grey crystals. It is a semiconductor. In pure form, it has the anti-bixbyite structure.
Chemical formula: Zn3N2
Space group: Ia-3, No. 206
EC Number: 215-207-3
Appearance: blue-gray cubic crystals
Explanation:
Boyle’s Law illustrates the inverse relationship of volume and pressure. It follows the formula p1V1 = P2V2 , where P1V1 denotes initial pressure and volume and P2V2 denotes values of pressure and volume.
Now, let us work out for what is asked above.
a. if the pressure is doubled
50.0 p = V x 2p
V = 50.0 p / 2p
= 50.0 /2
= 25.0 m^3
b. if the pressure is cut in half
50.0 p = V x p/2
100 p = V x p
V = 100 m^3
c. if the pressure is tripled
50.0 p = V x 3p
V = 50.0 p / 3p
= 50.0 /3
=16.7 m^3
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