Answer:
AC₄ will precipitate out first.
Explanation:
A solid will precipitate out if the ionic product of the solution exceeds the solubility product.
Let us check the ionic product
a) A₂B₃
Ionic product = [A]²[B]³
[A] = say "s"
[B] = 0.05 , [B]³ = (0.05)³ = 0.000125
2.3 X 10⁻⁸ = [A]²(0.000125)
[A] = 0.0136
b) AC₄
Ionic product = [A] [C]⁴
[A] = "s"
[A][0.05]⁴ = 4.10 X 10⁻⁸
[A]=0.00656 M
So for ionic product to exceed solubility product, we need less concentration of A in case of AC₄.
Answer:
Explanation:
Hello,
In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior as a directly proportional relationship:
Thus, we solve for the final pressure P2 to obtain it as shown below:
Hence, we notice that the temperature doubles as well as the pressure.
Best regards.
Answer:
The number of moles of potassium hydroxide, KOH required to make 4 moles of K₂SO₄ is 8 moles of KOH
Explanation:
2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
From the above reaction, we have 2 moles of KOH combining with 1 mole of H₂SO₄ to produce 1 mole of K₂SO₄ and 2 moles of H₂O.
Therefore the number of moles of potassium hydroxide that will be needed to make 4 moles of K₂SO₄ is;
8KOH + 4H₂SO₄ → 4K₂SO₄ + 8H₂O
8 moles of KOH is required to make 4 moles of K₂SO₄.
Actually since Bromine is located at the 1 Carbon, so we
can say that this is a primary alkyl halide and which undergoes SN2 or E2
reactions. This reaction is a bimolecular, single step process because it is a
primary.
<span>The substitution product formed will be 1-ethoxybutane
(main product) and sodium bromide (side product).</span>
Answer:
There were 0.00735 moles Pb^2+ in the solution
Explanation:
Step 1: Data given
Volume of the KI solution = 73.5 mL = 0.0735 L
Molarity of the KI solution = 0.200 M
Step 2: The balanced equation
2KI + Pb2+ → PbI2 + 2K+
Step 3: Calculate moles KI
moles = Molarity * volume
moles KI = 0.200M * 0.0735L = 0.0147 moles KI
Ste p 4: Calculate moles Pb^2+
For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+
For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+
There were 0.00735 moles Pb^2+ in the solution
