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):
<span>Hydroxide ions as the only negative ions</span>
Answer: With the cold front, warm air is rapidly forced upward (like the shavings) in advance of the actual front (the “cutter”), creating towering cumulus clouds, some hard showers and quite possibly a few gusty thunderstorms followed by a push of cooler and drier air in its wake.
Explanation:
Please be more specific with your question
Answer:
The answer to your question is: 43 %
Explanation:
Data
NO = 7 mol
O2 = 5 mol
NO2 = 3 mol
percent yield = ?
Reaction
2NO(g) + O2(g) ⇒ 2NO2(g)
Proportion of reactants
From the reaction 2 moles of NO / 1 moles of O2 = 2
From the experiment 7 moles of NO / 5 moles of O2 = 1.4
Then, the limiting reactant is NO.
Rule of three
2 moles of NO -------------- 2 moles of NO2
7 moles of NO -------------- x
x = (7 x 2) / 2
x = 7 moles of NO2
% yield = experimental/ theoretical x 100
% yield = 3/7 x 100
% yield = 42.9 ≈ 43
