K + I - > KI
Potassium (needs to lose 1 electron) responds with Iodine (needs to pick up 1 electron) to fulfill both component's octet, shaping a salt, potassium iodide
This is a similar case for NaCl, simply unique components. Trust this made a difference.
Given:
A compound with:
Number of carbon atoms = 9
Number of double bonds = 1
A double bond between 5th and 6th carbon
A propyl group (CH2CH2CH3) branching off the 3rd carbon from the left
Try to illustrate the given and observe the formation of the atoms. Now, follow the correct IUPAC naming system. The name of the compound is
4-propyl-1-hexene
Count from the right to the left, the double bond is between the 1st and 2nd carbon, thus, 1-hexene. The propyl branches out the 4th carbon from the right, thus 4-propyl.
Answer:
Pb(NO3)2(aq) + 2NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)
Explanation:
Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)
This is how it starts out.
Left:
Right
So the place to start with this equation is to bring the Cls up to 2
Pb(NO3)2(aq)+2NaCl(aq) -> NaNO3(aq)+PbCl2(s)
But the Nas are now out of kilter.
Pb(NO3)2(aq)+ 2NaCl(aq) -> NaNO3(aq)+PbCl2(s)
Now the right has a problem. There's only 1 Na
Pb(NO3)2(aq) + 2 NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)
Check it out. It looks like we are done.