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Travka [436]
4 years ago
12

The solubility of lead (ii) chloride (pbcl2) is 1.6 ⋅ 10-2 m. what is the ksp of pbcl2? the solubility of lead (ii) chloride (pb

cl2) is 1.6 10-2 m. what is the ksp of pbcl2? 3.1 ⋅ 10-7 5.0 ⋅ 10-4 1.6 ⋅ 10-5 1.6 ⋅ 10-2 4.1 ⋅ 10-6
Chemistry
2 answers:
Yakvenalex [24]4 years ago
8 0
According to the equilibrium equation for this reaction:

PbCl2(s)  ↔ Pb2+(aq)   +  2Cl-(aq)

So when Ksp is the solubility product constant for a solid substance when it

dissolved in the solution. and measure how a solute dissolves in the solution 

So, Ksp expression = [Pb2+] [Cl-]^2

and when the solubility is the maximum quantity of solute which can dissolve in a certain solute.
 
So, we assume the solubility = X 

∴[Pb2+] = X = 1.6 x 10^-2 M

[Cl-] = 2X = 2 * 1.6 x 10^-2 = 0.032 M

by substitution:

∴ Ksp = (1.6 x 10^-2) * (0.032)^2

           = 1.64 x 10^-5  
DedPeter [7]4 years ago
8 0

Answer:The Solubility product of PbCl_2 is 1.63\times 10^{-5}.

Explanation;

Solubility of PbCl_2,S=1.6\times 10^{-2} M

                         PbCl_2\rightleftharpoons Pb^{2+}+2Cl^-

                                                 S        2S

The expression of K_{sp} is given as:

K_{sp}=S\times (2S)^2=4S^3

K_{sp}=4\times (1.6\times 10^{-2})^3=1.63\times 10^{-5}

The Solubility product of PbCl_2 is 1.63\times 10^{-5}.

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Barbara is converting 78°F to degrees Celsius. First, she subtracts 32 from 78. What is the next step?
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3 years ago
28 ml of 0.10 m hcl is added to 60 ml of 0.10 m sr(oh)2. determine the concentration of oh− in the resulting solution.
Blababa [14]
Here is an acid-base reaction. Hydrochloric acid (HCl) reacts with strontium hydroxide [ Sr(OH)2 ]

Ions H+ and OH- neutralize each other. If the amounts are not equal, one of them will be in excess. 
Follow the steps as

1. Find moles of ions: mole= Molarity * Volume (in liter) ; n= M * V OR millimole = Molarity * Volume (in ml) ;
2. Write the equation
3. Find out excess ion
4. Use final volume (V acid + V base ) to calculate concentration of excess ion.

n HCI = 28 ml * 0.10 M = 0.28 mmol, releases 0.28 mmol H+ ions
n Sr(OH)2= 60 ml * 0.10 M= 0.60 mmol, releases 2* 0.60=1.20 mmol OH- ions
since Sr(OH)2⇒ Sr2+ + 2OH-

Neutralization reaction is OH- + H+ ---> H2O. The ratio is 1:1. That means 1 mmol hydroxide ions will neutralize 1 mmol hydrogen ions. Since OH- ions are greater in amount, they will be in excess

n(OH-) - n(H+)= 1.20 - 0.28 = 0.92 mmol OH- ions UNREACTED.

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3 years ago
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Explanation:

6 0
3 years ago
How many grams are in 1.23 x 1020 atoms of arsenic?
Alina [70]

Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g

<h3>Avogadro's hypothesis </h3>

6.02×10²³ atoms = 1 mole of arsenic

But

1 mole of arsenic = 75 g

Thus, we can say that:

6.02×10²³ atoms = 75 g of arsenic

<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>

6.02×10²³ atoms = 75 g of arsenic

Therefore,

1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)

1.23×10²⁰ atoms = 0.0153 g of arsenic

Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic

Learn more about Avogadro's number:

brainly.com/question/26141731

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