The number of liters of 3.00 M lead (II) iodide : 0.277 L
<h3>Further explanation</h3>
Reaction(balanced)
Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)
moles of KI = 1.66
From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):

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Answer:
The answer is
<h3>6000 N/m² or 6000 Pa</h3>
Explanation:
The pressure exerted by an object given the force of the object and the area can be found by using the formula

where
P is the pressure
f is the force
a is the area
From the question
f = 2400 N
a = 0.4 m²
So we have

We have the final answer as
<h3>6000 N/m² or 6000 Pa</h3>
Hope this helps you