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):
(a) 33.6 L of oxygen would be produced.
(b) 106 grams of would be needed
1 mole of gas = 22.4 L
(a) From the equation, 2 moles of produce 3 moles of
. 1 mole of
will, therefore, produce 1.5 moles of
.
1.5 moles of oxygen = 22.4 x 1.5 = 33.6 L
(b) 22.4 L of is produced at STP. This means that 1 mole of the gas is produced.
From the equation, 1 mole of requires 1 mole of
.
Molar mass of = (23x2)+ (12)+(16x3) = 106 g/mol
Mass of 1 mole = 1 x 106 = 106 grams
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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Answer:
a) Sm
Explanation:
Hello,
This substantiation could be done by knowing how to compute the released energy for 1 mol the oxide. Thus, for Sm, the computation turns out into:
The sign just indicates that the heat is released due to the chemical reaction and it matches with the indicated heat in the statement.
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