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.
Answer:
[KBr] = 454.5 m
Explanation:
m is a sort of concentration that indicates the moles of solute which are contianed in 1kg of solvent.
In this case, the moles of solute are 0.25 moles.
Let's determine the mass of solvent in kg.
Density of heavy water, solvent, is 1.1 g/L and our volume is 0.5L.
1.1 g = mass of solvent / 0.5L, according to density.
mass of solvent = 0.5L . 1.1g/L = 0.55 g
We convert the mass to kg → 0.55 g . 1kg /1000g = 5.5×10⁻⁴ kg
m = mol/kg → 0.25 mol /5.5×10⁻⁴ kg = 454.5 m
Energy lost to condense = 803.4 kJ
<h3>Further explanation</h3>
Condensation of steam through 2 stages:
1. phase change(steam to water)
2. cool down(100 to 0 C)
1. phase change(condensation)
Lv==latent heat of vaporization for water=2260 J/g
2. cool down
c=specific heat for water=4.18 J/g C
Total heat =
Answer: Thus concentration of 
in 
is 0.011 and in 
is 0.814
Explanation:
To calculate the concentration of , we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is
We are given:
Putting values in above equation, we get:
The concentration in is 
Thus concentration of is 
and 
Answer:
Moles of 
= 6 moles
Explanation:
The reaction of 
and 
to make is:
⇒
The above reaction shows that 2 moles of Sc can react with 3 moles of to form 
Mole Ratio= 2:3
For 10 moles of Sc we need:
Moles of = 
Moles of = 
Moles of =15 moles
So 15 moles of are required to react with 10 moles of but we have 9 moles of , it means is limiting reactant.
Moles of ScCl_3= 6 moles
