Igneous <span>rock ! Of course oh how i miss middle school.........</span>
Answer:

Explanation:
Hello,
In this case, the undergoing chemical reaction is:

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

Finally, the resulting molarity in 30.8 mL (0.0308 L):

Regards.
Answer:
Molarity = 0.4M
Explanation:
Molar mass of NaOH (M)= 40
m= 8g, V= 500ml=0.5L
n= m/M=[8/40]= 0.2mol
Applying
n= CV
0.2= C×0.5
C= 0.4M