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Maru [420]
2 years ago
12

If A is slowly added to a solution containing 0.0500 M of B and 0.0500 M of C, which solid will precipitate first? The solubilit

y product constant for A2B3 is 2.30×10−8. The solubility product constant for AC4 is 4.10×10−8.
Chemistry
1 answer:
OleMash [197]2 years ago
5 0

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.

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You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2 that has a worn label mak
Orlov [11]

Answer:

Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL

Explanation:

Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃

For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.

Knowing that Concentration in mol/L = (number of moles)/(volume in L)

Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole

According to the reaction,

1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄

0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄

Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄

So, the concentration of commercial H₂SO₄ is usually 18.0M, using this as the assumed value.

Volume of H₂SO₄ = (number of H₂SO₄ required for it to be in excess)/(concentration of H₂SO₄)

Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.

QED!!!

5 0
2 years ago
) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H
Elanso [62]

Answer:

1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).

2. The difference in pH values is 4.95.

Explanation:

1. The pH of a compound can be found using the following equation:

pH = -log([H_{3}O^{+}])

First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.

<u>Trimethyl ammonium</u>:

We can calculate [H₃O⁺] using the Ka as follows:

(CH₃)₃NH⁺ + H₂O  →  (CH₃)₃N + H₃O⁺    

1.0 - x                               x           x  

Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}

10^{-pKa} = \frac{x*x}{1.0 - x}

10^{-9.80}(1.0 - x) - x^{2} = 0    

By solving the above equation for x we have:  

x = 0.097 = [H₃O⁺]

pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01                                      

<u>Phenol</u>:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x                        x             x

Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}

10^{-10} = \frac{x^{2}}{1.0 - x}

1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00

Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.

2. The difference in pH values for the two acids is:

\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95

Therefore, the difference in pH values is 4.95.

I hope it helps you!

7 0
2 years ago
The number of p atoms in 1.0 g of ba3(po4)2 is:
marta [7]
First, we determine the number of moles of barium phosphate. This is done using:

Moles = mass / Mr
Moles = 1 / 602
Moles = 0.002

Now, we see from the formula of the compound that each mole of barium phosphate has 2 moles of phosphorus (P). Therefore, the moles of phosphorus are:
0.002 * 2 = 0.004

The number of particles in one mole of substance is 6.02 x 10²³. The number of phosphorus atoms will be:
0.004 * 6.02 x 10²³

2.41 x 10²¹ atoms of phosphorus are present
3 0
2 years ago
Which part of experimental design is most important to a scientist when replicating an experiment? Having exactly the same data
yarga [219]

Answer:

The answer to your question is below:

Explanation:

Having exactly the same data as the previous experiment I think that having the same data as the previous experiment is extremely important but not the most important, for me is the second most important.

Using the same procedure and variables as the previous experiment For me, this is the most importan thing when a scientist is designing an experiment, because if he or she follow exactly the same procedure and variables, then the results will be very close.

Conducting an experiment similar to the previous experiment  This characteristic is important but not the most important.

Using the same laboratory that was used in the previous experiment It is not important the laboratory, if the procedure and variables are the same, your experiment must give the same results in whatever laboratory.

5 0
2 years ago
Read 2 more answers
Fluorine gas is placed in contact with calcium metal at high temperatures to produce calcium fluoride powder. What is the formul
Mars2501 [29]

Answer:

Upper F subscript 2 (g) plus upper C a (s) right arrow with delta above upper C a upper F subscript 2 (s).

Explanation:

This is a chemical reaction problem.

In expressing any chemical reaction, we need to understand that there are reactants and products.

  • The reactants are the species on the left hand side that are combining.
  • The products are the species on the right hand side that are formed.
  • Every chemical reaction is obeys the law of conservation of matter i.e equal number of matter on both sides.

Using the statement of this problem, we can deduce that;

 Reactants are Fluorine gas and Calcium metal

  Product is Calcium Fluoride

Note: A metal is a solid(s) and powder is a solid(s). A gas is denoted as (g). They depict the state of the species reacting.

                    F₂_{(g)}     +    Ca_{(s)}               →           CaF₂_{(s)}

We can see that equal number of atoms are on both sides of the expression.

7 0
2 years ago
Read 2 more answers
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