<span>significant figures: 7, decimals: 3</span>
Chlorite is ClO2^-1, and potassium is K with a charge of +1. So, we would write is as KClO2, since the charges cancel out with each other.
Answer:
Lead (II) iodide
Explanation:
The reaction of lead (II) nitrate, Pb(NO₃)₂ with KI is:
Pb(NO₃)₂(aq) + 2KI(aq) → KNO₃(aq) + PbI₂(s)
This is a typical double-replacement reaction where anions and cations exchange its couple.
All nitrates are solubles, thus, KNO₃ is not the precipitate.
The only possibility of precipitate is PbI₂,
Lead (II) iodide, a yellow and insoluble solid...
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.
Sorry, but where are the ‘items’?
