Answer:
the mass does not change
Explanation:
For the reaction 3H 2(g) + N 2(g) 2NH 3(g), K c = 9.0 at 350°C. What is the value of ΔG at this temperature when 1.0 mol NH 3, 5.0 mol N 2, and 5.0 mol H 2 are mixed in a 2.5 L reactor?
Answer:
ΔG = - 31.7kJ/mol
Explanation:
It is possible to find ΔG of a reaction at certain temperature knowing Kc following the equation:
ΔG = ΔG° + RT ln Q
ΔG° = -RT lnKc
ΔG = -RT lnKc + RT ln Q (1)
Where R is gas constant (8.314J/molK), T absolute temperature (350°C + 273.15 = 623.15K) and Q reaction quotient
For the reaction,
3H₂(g) + N₂(g) ⇄ 2NH₃(g)
Q = [NH₃]² / [H₂]³[N₂]
Where the concentrations of each chemical are:
[NH₃] = 1.0mol / 2.5L = 0.4M
[H₂] = 5.0mol / 2.5L = 2M
[N₂] = 2.5mol / 2.5L = 1M}
Q = [0.4M]² / [2M]³[1M]
Q = 0.02
And replacing in (1):
ΔG = -RT lnKc + RT ln Q
ΔG = -8.314J/molK*623.15K ln 9 + 8.314J/molK*623.15K ln 0.02
ΔG = - 31651J/mol
ΔG = - 31.7kJ/molIf there are a 1000 ml per 1 L and a 1000g per kilogram
a. How many ml are there in 5.0 L?
b. How many kg are there in 230g?
Answer:
hbchbjH j jas a aa a s ds d as das
Explanation:
A sample of ice absorbs 15.6kJ of heat as it undergoes a reversible phase transition to form liquid water at 0∘C. What is the entropy change for this process in units of JK? Report your answer to three significant figures. Use −273.15∘C for absolute zero.
Answer:
Entropy change of ice changing to water at 0°C is equal to 57.1 J/K
Explanation:
When a substance undergoes a phase change, it occurs at constant temperature.
The entropy change Δs, is given by the formula below;
Δs = q/T
where q is the quantity of heat absorbed or evolved in Joules and T is temperature in Kelvin at which the phase change occur
From the given data, T = 0°C = 273.15 K, q = 15.6 KJ = 15600 J
Δs = 15600 J / 273.15 K
Δs = 57.111 J/K
Therefore, entropy change of ice changing to water at 0°C is equal to 57.1 J/K
The entropy change of ice changing to water will be "57.1 J/K".
Entropy changeThe shift in what seems like a thermodynamic system's condition of confusion is caused by the transformation of heat as well as enthalpy towards activity. Entropy seems to be greater mostly in a network with a high quantity or measure of chaos.
According to the question,
Temperature, T = 0°C or,
= 273.15 K
Heat, q = 15.6 KJ or,
= 15600 J
We know the formula,
Entropy change, Δs = [tex]\frac{q}{T}[/tex]
By substituting the values, we get
= [tex]\frac{15600}{273.15}[/tex]
= 57.11 J/K
Thus the above answer is correct.
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If one pound is the same as 454 grams, then convert the mass of 78 grams to pounds.
Answer:
0.17 lb
Explanation:
78 g * (1 lb/454 g)=0.17 lb
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A piece of solid Fe metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction that may be predicted. Assume that the oxidation state of in the resulted solution is 2 . (Use the lowest possible coefficients for the reaction. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)
Answer:
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
Explanation:
The ionic equation shows the actual reaction that took place. It excludes the spectator ions. Spectator ions are ions that do not really participate in the reaction even though they are present in the system.
For the reaction between iron and copper II nitrate, the molecular reaction equation is;
Fe(s) + Cu(NO3)2(aq)----> Fe(NO3)2(aq) +Cu(s)
Ionically;
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
what is the valency of element sulphur
sorry not sure wish I can help u
What is the pOH of a solution at 25.0∘C with [H3O+]=4.8×10−6 M?
Answer:
8.68
Explanation:
pOH = 8.68
all you need is contained in the sheet
Answer:
Approximately [tex]8.68[/tex].
Explanation:
The [tex]\rm pOH[/tex] of a solution can be found from the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] with the following equation:
[tex]\displaystyle \rm pOH = -\log_{10} \rm \left[OH^{-}\right][/tex].
On the other hand, the ion-product constant of water, [tex]K_{\text{w}}[/tex], relates the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] of a solution to its hydronium ion concentration [tex]\rm \left[{H_3O}^{+}\right][/tex]:
[tex]K_\text{w} = \rm \left[{H_3O}^{+}\right] \cdot \rm \left[OH^{-}\right][/tex].
At [tex]25 \; ^\circ \rm C[/tex], [tex]K_{\text{w}} \approx 1.0 \times 10^{-14}[/tex]. For this particular [tex]25 \; ^\circ \rm C[/tex] solution, [tex]\rm \left[{H_3O}^{+}\right] = 4.8 \times 10^{-6}\; \rm mol \cdot L^{-1}[/tex].Hence the [tex]\rm \left[OH^{-}\right][/tex] of this solution:
[tex]\begin{aligned}\left[\mathrm{OH}^{-}\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^{+}\right]} \\ &= \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-6}}\; \rm mol\cdot L^{-1} \approx 2.08333 \times 10^{-9}\; \rm mol\cdot L^{-1}\end{aligned}[/tex].
Therefore, the [tex]\rm pOH[/tex] of this solution would be:
[tex]\begin{aligned}\rm pOH &= -\log_{10} \rm \left[OH^{-}\right] \\ &\approx -\log_{10} \left(4.8 \times 10^{-6}\right) \approx 8.68\end{aligned}[/tex].
Note that by convention, the number of decimal places in [tex]\rm pOH[/tex] should be the same as the number of significant figures in [tex]\rm \left[OH^{-}\right][/tex].
For example, because the [tex]\rm \left[{H_3O}^{+}\right][/tex] from the question has two significant figures, the [tex]\rm \left[OH^{-}\right][/tex] here also has two significant figures. As a result, the [tex]\rm pOH[/tex] in the result should have two decimal places.
i) Briefly discuss the strengths and weaknesses of the four spectroscopy techniques listed below. Include in your answer the specific structural information you get from each method.
IR
UV-VIS
NMR
Mass Spec
delete please .....................................
An atom of 120In has a mass of 119.907890 amu. Calculate the mass defect (deficit) in amu/atom. Use the masses: mass of 1H atom
Answer:
a
Explanation:
answer is a on edg
Select the correct answer.
Which state of matter is highly compressible, is made of particles moving independently of each other, and is present in large quantities near Earth’s surface?
A.
solid
B.
liquid
C.
gas
D.
plasma
Answer:
C. Gas
Explanation:
2. Find the two generic molecules from Part 1 that are made of 3 atoms. a. Compare and contrast these two molecules by listing two similarities and two differences.
Answer:
hello the molecules are missing from your question below are the Generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]
answer : It can be determined that both generic molecules are polar
It can be determined that both generic molecules have similar molecular shape
They have different Geometry
They differ in bond angles as well
Explanation:
The two generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]
comparing(similarities) these two generic molecules
It can be determined that both generic molecules are polar
It can be determined that both generic molecules have similar molecular shape
differences between the generic molecules
They have different Geometry
They differ in bond angles as well
What happens if we put raw eggs in a pot full of hot oil?
For element radon, give the chemical symbol, atomic number, and group number.
Draw the product that valine forms when it reacts with excess CH3CH2OH and HCl followed by a wash with aqueous base.
Answer:
Product: ethyl L-valinate
Explanation:
If we want to understand what it is the molecule produced we have to analyze the reagents. We have valine an amino acid, in this kind of compounds we have an amine group ([tex]NH_2[/tex]) and a carboxylic acid group ([tex]COOH[/tex]). Additionally, we have an alcohol ([tex]CH_3CH_2OH[/tex]) in the presence of HCl (a strong acid) in the first step, and a base ([tex]OH^-[/tex]).
When we have an acid and an alcohol in a vessel we will have an esterification reaction. In other words, an ester is produced. As the first step, the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the second step, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In step 3, a proton is transferred to produce a better leaving group ([tex]H_2O[/tex]). In step 4, a water molecule leaves the main structure to produce again the double bond C=O. Finally, a base ([tex]OH^-[/tex]) removes the hydrogen from the C=O bond to produce ethyl L-valinate
See figure 1
I hope it helps!
Complete the following equation of nuclear transmutation.
23892U + 126C → 24498Cf + 6 ______
Complete the following equation of nuclear transmutation.
U + C → Cf + 6 ______
A) 1n
B) 0 e
C) 0 e
D) 1H
E) 0g 0 -1 +1 1 0
Answer:
Option A. 1 0n
Explanation:
Details on how to balanced the equation for the reaction given in the question above can be found in the attached photo.
The missing part of the transmutation equation as it has been shown is 1/o n. Option A
What is nuclear transmutation?Nuclear transmutation is the process of shifting the number of protons in an atom's nucleus to change one element into another. Nuclear processes that change one atomic nucleus into another with a different atomic number are involved.
The production of nuclear energy, radioactive decay, and the creation of new isotopes for use in science and industry all depend on nuclear transmutation, a fundamental idea in nuclear physics.
We have the equation as;
238/92 U + 12/6 C ----> 244/98 Cf + 6 1/0 n
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Enter your answer in the provided box.
The equilibrium constant KP for the reaction
CO(g) + Cl2(g) ⇌ COCl2(g)
is 5.62 × 1035 at 25°C. Calculate ΔG
o
f
for COCl2 at 25°C.
Answer:
The correct answer is -341.2 kJ per mole.
Explanation:
The reaction given is:
CO (g) + Cl₂ (g) ⇔ COCl₂ (g)
Kp = 5.62 × 10³⁵
T = 25 °C or 298 K
The formula for calculating ΔG is,
ΔG° = -RTlnKp
ΔG° = -8.314 × 298 ln (5.62 × 10^35)
ΔG° = -203.9 kJ/mol
ΔG° = ∑nΔG°f (products) -∑nΔG°f (reactants)
ΔG° = ΔG°f (COCl₂ (g)) - [ΔG°f (CO(g)) + ΔG°f (Cl₂(g))]
ΔG°f (COCl₂ (g)) = ΔG° + [ΔG°f (CO (g)) + ΔG°f (Cl₂(g))]
ΔG°f (COCl₂ (g)) = -203.9 + (-137.28 + 0.00)
ΔG°f (COCl₂ (g)) = -341.2 kJ/mol
The standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol
The given equation for the chemical reaction is
CO(g) + Cl2(g) ⇌ COCl2(g)
At the temperature of 25°C = (273 + 25) K, the equilibrium constant [tex]\mathbf{K_p = 5.62\times 10^{35}}[/tex]
Consider the expression for the relationship between [tex]\mathbf{\Delta G^o}[/tex] and [tex]\mathbf{K_p }[/tex] for the equilibrium reaction can be expressed as:
[tex]\mathbf{\Delta G^o = - RT In K_p}[/tex]
where;
gas constant (R) = 8.314 × 10⁻³ kJ/K.mol∴
[tex]\mathbf{\Delta G^o = - (8.314 \times 10^{-3}\ kJ/K.mol \times 298 \ K) \times In (5.62 \times 10^{35} )}[/tex]
[tex]\mathbf{\Delta G^o = -2.477572\ K \times 82.31680992}[/tex]
[tex]\mathbf{\Delta G^o = 203.95 \ kJ}[/tex]
Thus, the standard free energy for the reaction is 203.95 kJ/mol
For a given reaction, the standard Gibbs free energy can be calculated by using the formula:
[tex]\mathbf{\Delta G^o_{rxn} = \sum n \Delta G^o_f (products) - \sum m \Delta G^o_f (reactants) }[/tex]
[tex]\mathbf{\Delta G^o_{rxn} =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(\Delta G^o_{f} (CO)_{(g)} + \Delta G^o_{f} (Cl)_{2(g)} ) \Big ) \Big ] }[/tex]
replacing the values of and solving for COCl2 at standard free energy of formation of substances, we have:
[tex]\mathbf{-203.95 \ kJ/mol =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(-137.3 kJ/mol + 0 \ kJ/mol\Big ) \Big ] }[/tex]
Collecting like terms, we have:
[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -203.95 \ kJ/mol -137.3 kJ/mol }[/tex]
[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -341.25 \ kJ/mol }[/tex]
Therefore, we can conclude that the standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol
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By heating a 93% pure kclo3 sample, what percentage of its mass is reduced?
2KCLO3---->2KCL+3O2
Explanation:
free your mind drink water and go outside take fresh air you will get answers
"Aqueous solutions of lead nitrate and ammonium chloride are mixed" together. Which statement is correct
Answer:
PbCl₂ will precipitate from solution.
Explanation:
Statements are:
Insufficient information is given.
Both NH4NO3 and PbCl2 precipitate from solution.
No precipitate forms.
PbCl2 will precipitate from solution.
NH4NO3 will precipitate from solution.
The reaction of ammonium chloride (NH₄Cl) with lead nitrate (Pb(NO₃)₂) is:
Pb(NO₃)₂ + 2NH₄Cl → PbCl₂ + 2 NH₄NO₃
Talking of rules of solubility, all nitrates are soluble in water, that means NH₄NO₃ is soluble and no precipitate is formed.
In the same way, all chlorides are soluble except silver chloride and lead chloride. That means:
PbCl₂ (Lead chloride) will precipitate from solution.Consider the following reaction at 298.15 K: Co(s)+Fe2+(aq,1.47 M)⟶Co2+(aq,0.33 M)+Fe(s) If the standard reduction potential for cobalt(II) is −0.28 V and the standard reduction potential for iron(II) is −0.447 V, what is the cell potential in volts for this cell? Report your answer with two significant figures.
Answer:
The correct answer is 0.186 V
Explanation:
The two hemirreactions are:
Reduction: Fe²⁺ + 2 e- → Fe(s)
Oxidation : Co(s) → Co²⁺ + 2 e-
Thus, we calculate the standard cell potential (Eº) from the difference between the reduction potentials of cobalt and iron, respectively, as follows:
Eº = Eº(Fe²⁺/Fe(s)) - Eº(Co²⁺/Co(s)) = -0.28 V - (-0.447 V) = 0.167 V
Then, we use the Nernst equation to calculate the cell potential (E) at 298.15 K:
E= Eº - (0.0592 V/n) x log Q
Where:
n: number of electrons that are transferred in the reaction. In this case, n= 2.
Q: ratio between the concentrations of products over reactants, calculated as follows:
[tex]Q = \frac{ [Co^{2+} ]}{[Fe^{2+} ]} = \frac{0.33 M}{1.47 M} = 0.2244[/tex]
Finally, we introduce Eº= 0.167 V, n= 2, Q=0.2244, to obtain E:
E= 0.167 V - (0.0592 V/2) x log (0.2244) = 0.186 V
The compound methylamine, CH3NH2, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibrium that occurs in an aqueous solution of methylamine:
Answer:
Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]
Explanation:
According to Brönsted-Lowry acid-base theory:
An acid is a substance that donates H⁺.A base is a substance that accepts H⁺.When methylamine reacts with water, it behaves as a Brönsted-Lowry base, according to the following reaction.
CH₃NH₂(aq) + H₂O(l) ⇄ CH₃NH₃⁺(aq) + OH⁻(aq)
The basic equilibrium constant (Kb) is:
Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]
A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup (see sketch at right).
Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to 2 significant digits.
You'll need to know that the density of fresh water at the temperature of the sea around the sailor is 0.999/gcm3. You'll also want to remember Archimedes' Principle, that objects float when they displace a mass of water equal to their own mass.
Answer:
Amount of salt in 1 L seawater = 34 g
Explanation:
According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater
mass of freshwater = density * volume
1 cm³ = 1 mL
mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g
mass of freshwater + cup = 734.265 + 25 = 759.265 g
Therefore, mass of equal volume of seawater = 759.265 g
Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)
1 liter = 1000 cm³ = 1000 mL;
Density of seawater = mass / volume
Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L
Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L
mass of 1 Liter seawater = 1033.01 g
mass of 1 Liter freshwater = 999 g
mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g
Therefore, amount of salt in 1 L seawater = 34 g
2. In what part of an atom can protons be found?
a. Inside the electrons
b. Inside the neutrons
C. Inside the atomic nucleus
d. Inside the electron shells
Answer:
c
Explanation:
it's found inside the atomic nucleus
A galvanic cell consists of a Cu(s)|Cu2+(aq) half-cell and a Cd(s)|Cd2+(aq) half-cell connected by a salt bridge. Oxidation occurs in the cadmium half-cell. The cell can be represented in standard notation as
Answer:
[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]
Explanation:
A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.
The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.
As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.
[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]
Which of the following combinations will result in a reaction that is spontaneous at all temperatures?
Negative enthalpy change and negative entropy change
Negative enthalpy change and positive entropy change
Positive enthalpy change and negative entropy change
Positive enthalpy change and positive entropy change
PLS EXPLAIN WHAT EACH MEANS AND THE VARIABLES AND THE EXPLANATION BEHIND IT
Answer:
[tex]\huge\boxed{Option \ 2}[/tex]
Explanation:
A reaction is spontaneous at all temperatures by the following combinations:
=> A negative enthalpy change ( [tex]\triangle H < 0[/tex] )
=> A positive entropy change ( [tex]\triangle S > 0[/tex] )
See the attached file for more better understanding!
from Gibbs Equation, [tex] \Delta G = \Delta H - T\Delta S [/tex]
reaction is spontaneous if $\Delta G$ is negative.
so, first option is not valid at high temperature, ($-h+ts$)
second, is always a spontaneous reaction, ($-h-ts$)
third, is never spontaneous ($+h+ts$)
4th is similar to second, spontaneous at higher temperatures ($+h-ts$)
10. For the following isotopes that have missing information, fill in the missing informatic
complete the notation: 36P
Answer:
Krypton.
Explanation:
Krypton is an atom which has 36 protons in its nucleus. There are 31 isotopes of Krypton which have same number of protons i. e. 36, same number of electrons i. e. 36 but different number of neutrons. Isotope refers to those atoms having same atomic number i. e. number of proton but different mass number i. e. number of neutron. For example, in Krypton-78, there 36 protons and 42 neutrons.
Which of the following chemical equations corresponds to the standard molar enthalpy of formation of Na_2CO_3(s)?
a. 2 NA(s) + C(s) + 3 O(g) ------------> Na_2CO_3(s)
b. Na_2O(s) + CO_2(g) --------------->Na_2CO_3 (s)
c. Na_2(s) + C(s) + 3 O(g) -------------> Na_2CO_3 (s)
d. Na_2O(s) + CO(g) ---------------> Na_2CO_3(s)
e. 2 Na(s)+ C(s) + 3/2 O_2(g) ------------> Na_2CO_3(s)
Answer:
2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)
Explanation:
The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)
For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:
2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)Does a reaction occur when aqueous solutions of potassium hydroxide and chromium(III) bromide are combined
Explanation:
Potassium hydroxide = KOH
Chromium(iii)bromide = CrBr3
Yes! A reaction occurs. This is given by the balanced equation;
3 KOH + CrBr3 → 3 KBr + Cr(OH)3
A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an average volume of the titrant was 38.2 mL. The starting volume of the HCl solution was 20 mL. What's the concentration of the HCl? answer options: A) 0.788 M B) 0.284 M C) 3.34 M D) 0.191 M
Answer: it is A
Explanation: I am sure
Answer:
0.191 M
Explanation:
i took the test.
If the Ksp for Li3PO4 is 5.9×10−17, and the lithium ion concentration in solution is 0.0020 M, what does the phosphate concentration need to be for a precipitate to occur?
Answer:
7.4 × 10⁻⁹ M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷
Concentration of lithium ion: 0.0020 M
Step 2: Write the reaction for the solution of Li₃PO₄
Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
Step 3: Calculate the phosphate concentration required for a precipitate to occur
The solubility product constant is:
Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³
[PO₄³⁻] = 7.4 × 10⁻⁹ M
The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?
Answer:
[tex]Ksp=1.07x10^{-8}[/tex]
Explanation:
Hello,
In this case, the dissociation reaction is:
[tex]PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)[/tex]
For which the equilibrium expression is:
[tex]Ksp=[Pb^{2+}][I^-]^2[/tex]
Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)
[tex]Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M[/tex]
In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:
[tex][Pb^{2+}]=1.39x10^{-3}M[/tex]
[tex][I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M[/tex]
Thereby, the solubility product results:
[tex]Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}[/tex]
Regards.
Solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].
The dissociation reaction for lead (II) iodide
[tex]\bold {Pb I^2 (s) \leftrightharpoons Pb^2^+ + 2I^- }[/tex]
Solubility product constant at equilibrium.
[tex]\bold {Ksp = [Pb^2^++[I^-]^2}[/tex]
The molar solubility of the substance can be calculated by using the molar mass,
[tex]\bold {s = \dfrac {0.064}{100 mL} \times 461.2 g/mol = 1.39x10^-^3}[/tex]
Molar ratio between between PbI to lead and iodide ions is 1:1 and 1:2 respectively.
Thus Ksp will be,
[tex]\bold {Ksp =(1.39x10^-^3)(2.78x10^-^3 )^2}\\\\\bold {Ksp = 1.07x 10^-^8}[/tex]
Therefore, solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].
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