Producer. Hope this helps!
Answer:

Explanation:
Hello there!
In this case, since perchloric acid is HClO4 and is a strong acid and calcium hypochlorite is Ca(ClO)2, the undergoing molecular chemical reaction turns out:

Thus, since the resulting hypochlorous acid is weak, it does not fully ionize, so it remains unionized, however, we can write the ions for the other species:

Now, we can cancel out the spectator ions, calcium and perchlorate, to obtain:

Best regards!
The grams of NaCl that are required to make 150.0 ml of a 5.000 M solution is 43.875 g
calculation
Step 1:calculate the number of moles
moles = molarity x volume in L
volume = 150 ml / 1000 = 0.15 L
= 0.15 L x 5.000 M = 0.75 moles
Step 2: calculate mass
mass = moles x molar mass
molar mass of NaCl = 23 + 35.5 = 58.5 mol /L
mass is therefore =0.75 moles x 58.5 mol /l =43.875 g
Precipitation occurs when the product of the ion concentration exceeds the Ksp.