Answer:
2KClO3 -------> 2KCl + 3O2
Explanation:
First, in balancing a chemical reaction such as the one given in the question, you should understand that for an equation to be balanced, the number of atoms and ions on both sides of the equation that is the right and left side must be equal. This follows the law of conservation of mass which tells us that matter can neither be created nor destroyed but can be changed into another form.
Next is to begin balancing the equation by identifying and writing down the substances given:
KCl03 ---------> KCl + O2
Next is to count he number of the individual atoms on each side and find out if they are the same on both sides and if not you must follow the next step.
Add a corresponding number and use it to multiply the atoms involved
KClO3 ---------> KCl + O2
Oxygen is 3 on the left side and two on the other side, so we multiply the left hand side by 2 and the right hand side by 3
2KClO3 -----> KCl + 3O2
The potassium and Chlorine are no longer balanced, so you multiply the right had=nd side of KCl by 2.
2KClO3 -----> 2KCl + 3O2
The reactionis herefore balanced as both sides have equal number of atoms and ions.
This says four figures. The 4 figures you should use are 4546 with a peek at 7 to see what effect it will have on the 4 main figures.
45.467 rounds to 45.47
You round the 4th figure up one. You are not concerned about the decimal places in this question.
Recall; pH + pOH = 14
In this case [OH-] =0.100 m
therefore;
pOH = -LOG[OH-]
= - Log (0.100)
= 1.00
Therefore; the pOH is 1.00
And since, pH +pOH = 14
Then pH = 14-pOH
= 14 -1
= 13
Thus the pH is 13.00
Two changes would make this reaction reactant-favored
C. Increasing the temperature
D. Reducing the pressure
<h3>Further explanation</h3>
Given
Reaction
2H₂ + O₂ ⇒ 2H₂0 + energy
Required
Two changes would make this reaction reactant-favored
Solution
The formation of H₂O is an exothermic reaction (releases heat)
If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)
While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient
in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2
As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)
And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater
