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:
equivalent exchange forces cancel out but the substances are affected around the area
Explanation:
I think it’s c but I could be wrong
You have molarity and you have volume. Use the formula :
Molarity(M)= Moles(N)/Liter(L) to get the solution.
150 ml= .150 L
7.7 = N/.150
N=.1.155 moles of NaOH.
And since you know the moles, use the molar mass to figure out the grams.
<span> (40g/mol NaOH) x (1.155mol) =
46.2 g of NaOH.</span>
