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):
Answer:
Chemical change
Explanation:
The rusting of iron/a nail is a chemical change
Iron (Fe) and Oxygen (O) combine to create the compound Iron Oxide (Fe2O3), which is rust.
First, We have to write the equation for neutralization:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
so, from the equation of neutralization, we can get the ratio between Ba(OH)2 and HCl. Ba(OH)2 : HCl = 1:2
- We have to get the no.of moles of Ba(OH)2 to do the neutralization as we have 25.9ml of 3.4 x 10^-3 M Ba(OH)2.
So no.of moles of Ba(OH)2 = (25.9ml/1000) * 3.4x10^-3 = 8.8 x 10^-5 mol
and when Ba(OH)2 : HCl = 1: 2
So the no.of moles of HCl = 2 * ( 8.8x10^-5) = 1.76 x 10^-4 mol
So when we have 1.76X10^-4 Mol in 16.6 ml (and we need to get it per liter)
∴ the molarity = no.of moles / mass weight
= (1.76 x 10^-4 / 16.6ml)* (1000ml/L) = 0.0106 M Hcl
Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
<span>It is known that amu is equal to the molar mass of 1 mole of substance. Therefore,
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Weight of one mole of uranium = 238 g/mol
The molar mass or molecular weight for U205 is 556 g/mol.
(238 x 2) + (16 x 5) = 556 g/mol
