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):
The answer is liquid or solid.
That is a substance that forms a vapor is generally in liquid or solid physical state.
The substance that is in liquid or solid physical state, forms vapor.
When heated solid and liquid can be converted into vapor so the answer is solid or liquid is the physical state that forms vapors.
There are 5.43 x 10²³ present in 187 grams of XeF₄
<h3>Further explanation </h3>
The mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
Moles can also be determined from the amount of substance mass and its molar mass
mass of XeF₄ = 187 g
mol XeF₄ (MW=207,2836 g/mol) :
Number of molecules :
