Answer:
[tex][Ag^+]=4.82x10^{-5}M[/tex]
Explanation:
Hello,
In this case, the dissociation reaction for silver phosphate is:
[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4^{3-}(aq)[/tex]
Therefore, the equilibrium expression is:
[tex]Ksp=[Ag^+]^3[PO_4^{3-}][/tex]
And in terms of the reaction extent [tex]x[/tex] is:
[tex]Ksp=1.8x10^{-18}=(3x)^3(x)[/tex]
Thus, [tex]x[/tex] turns out:
[tex]1.8x10^{-18}=27x^4\\\\x=\sqrt[4]{\frac{1.8x10^{-18}}{27} } \\\\x=1.61x10^{-5}M[/tex]
In such a way, the concentration of the silver ion is:
[tex][Ag^+]=3x=3*1.61x10^{-5}M=4.82x10^{-5}M[/tex]
Best regards.
Which of the following is not the same as 1,400 mL? a. 1.4 cm³ b 1.4 L c. 1,400 cm³ d. 140 cL
answer should be 1.4 cm³
1 L = 10 and so
dL = 100 and then
cL = 1,000
mL = 0.001 m³
1 m³ = 1,000
dm³ = 1,000,000
cm³ = 1,000,000,000
mm³ = 1,000 L
So, 1 mL = 1 cm³ = 0.001 L = 0.1 cL
1,400 mL = 1,400 cm³ = 1.4 L = 140 cL
Answer:
1.4 cm^3
Explanation:
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.
A scientist observes that the electrical resistance of a superconducting material drops to zero when the material is cooled to very low temperatures. Which of the following statements best describes what the scientist is observing?
The scientist is observing the electrical power of a superconductor.
The scientist is observing the temperature of a superconductor.
The scientist is observing an intensive property of a superconductor.
The scientist is observing an extensive property of a superconductor
Answer:
The scientist is observing an intensive property of a superconductor.
Explanation:
An intensive property is a bulk property of matter. This means that an intensive property does not depend on the amount of substance present in the material under study. Typical examples of intensive properties include; conductivity, resistivity, density, hardness, etc.
An extensive property is a property that depends on the amount of substance present in a sample. Extensive properties depend on the quantity of matter present in the sample under study. Examples of extensive properties include, mass and volume.
Resistance of a superconducting material has nothing to do with the amount of the material present hence it is an intensive property of the superconductor.
Answer:
The scientist is observing an intensive property of a superconductor.
Explanation:
its the only one that makes logical sense
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
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:
What is the primary source of energy in most living communities?
Answer:
The sun
Explanation:
The sun is the primary source of energy in most living communities. The producers or the green plants that prepare their own food by the use of sunlight and other natural resources. Carbon dioxide, water, and other minerals are used by the plants to make their food in the presence of chlorophyll. Plants are then consumed by the consumers. This chain helps in forming the food chain and the food web.
A galvanic cell is powered by the following redox reaction:
2Zn2+(aq) + N2H4(aq) 4OH-zn2+ right arrow(aq) 2Zn(s) + N2(g) + 4H2O(I)
1. Write a balanced equation for the half-reaction that takes place at the cathode.
2. Write a balanced equation for the half-reaction that takes place at the anode.
3. Calculate the cell voltage under standard conditions.
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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Assume you dissolve 0.235 g of the weak benzoic acid, C6H5CO2H in enough water to make 100.0 mL of the solution and then titrate the solution with 0.108 M NaOH. Benzoic acid is a monoprotic acid.
1. What is the pH of the original benzoic acid solution before the titration is started?
2. What is the pH when 7.00 mL of the base is added? (Hint: This is in the buffer region.)
3. What is the pH at the equivalence point?
Answer:
1. pH = 2.98
2. pH = 4.02
3. pH = 8.12
Explanation:
1. Initial molarity of benzoic acid (Molar mass: 122.12g/mol; Ka = 6.14x10⁻⁵) is:
0.235 ₓ (1mol / 122.12g) = 1.92x10⁻³ moles / 0.100L = 0.01924M
The equilibrium of benzoic acid with water is:
C6H5CO2H(aq) + H2O(l) → C6H5O-(aq) + H3O+(aq)
And Ka is defined as the ratio between equilibrium concentrations of products over reactants, thus:
Ka = 6.14x10⁻⁵ = [C6H5O⁻] [H3O⁺] / [C6H5CO2H]
The benzoic acid will react with water until reach equilibrium. And equilibrium concentrations will be:
[C6H5CO2H] = 0.01924 - X
[C6H5O⁻] = X
[H3O⁺] = X
Replacing in Ka:
6.14x10⁻⁵ = [X] [X] / [0.01924 - X]
1.1815x10⁻⁶ - 6.14x10⁻⁵X = X²
1.1815x10⁻⁶ - 6.14x10⁻⁵X - X² = 0
Solving for X:
X = -0.0010→ False solution. There is no negative concentrations
X = 0.0010567M → Right solution.
pH = - log [H3O⁺] and as [H3O⁺] = X:
pH = - log [0.0010567M]
pH = 2.982.
pH of a buffer is determined using H-H equation (For benzoic acid:
pH = pka + log [C6H5O⁻] / [C6H5OH]
pKa = -log Ka = 4.21 and [] could be understood as moles of each chemical
The benzoic acid reacts with NaOH as follows:
C6H5OH + NaOH → C6H5O⁻ + Na⁺ + H₂O
That means NaOH added = Moles C6H5O⁻ And C6H5OH = Initial moles (1.92x10⁻³ moles - Moles NaOH added)
7.00mL of NaOH 0.108M are:
7x10⁻³L ₓ (0.108 mol / L) = 7.56x10⁻⁴ moles NaOH = Moles C₆H₅O⁻
And moles C6H5OH = 1.92x10⁻³ moles - 7.56x10⁻⁴ moles = 1.164x10⁻³ moles C₆H₅OH
Replacing in H-H equation:
pH = 4.21 + log [7.56x10⁻⁴ moles] / [ 1.164x10⁻³ moles]
pH = 4.023. At equivalence point, all C6H5OH reacts producing C6H5O⁻. The moles are 1.164x10⁻³ moles
Volume of NaOH to reach equivalence point:
1.164x10⁻³ moles ₓ (1L / 0.108mol) = 0.011L. As initial volume was 0.100L, In equivalence point volume is 0.111L and concentration of C₆H₅O⁻ is:
1.164x10⁻³ moles / 0.111L = 0.01049M
Equilibrium of C₆H₅O⁻ with water is:
C₆H₅O⁻(aq) + H₂O(l) ⇄ C₆H₅OH(aq) + OH⁻(aq)
Kb = [C₆H₅OH] [OH⁻]/ [C₆H₅O⁻]
Kb = kw / Ka = 1x10⁻¹⁴ / 6.14x10⁻⁵ = 1.63x10⁻¹⁰
Equilibrium concentrations of the species are:
C₆H₅O⁻ = 0.01049M - X
C₆H₅OH = X
OH⁻ = X
Replacing in Kb expression:
1.63x10⁻¹⁰ = X² / 0.01049- X
1.71x10⁻¹² - 1.63x10⁻¹⁰X - X² = 0
Solving for X:
X = -1.3x10⁻⁶ → False solution
X = 1.3076x10⁻⁶ → Right solution
[OH⁻] = 1.3076x10⁻⁶
as pOH = -log [OH⁻]
pOH = 5.88
And pH = 14 - pOH
pH = 8.12what is the valency of element sulphur
sorry not sure wish I can help u
If 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:
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
"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.What happens if we put raw eggs in a pot full of hot oil?
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
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]
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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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
In the laboratory, a general chemistry student measured the pH of a 0.425 M aqueous solution of benzoic acid, C6H5COOH to be 2.270.
Use the information she obtained to determine the Ka for this acid.
Ka(experiment) = _____
Answer:
Ka = 6.87x10⁻⁵
Explanation:
The equilibrium of benzoic acid in water is:
C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)
The equilibrium constant, Ka, is:
Ka = [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH]
The initial concentration of benzoic acid is 0.425M. In equilibrium its concentration is 0.425M - X and [C₆H₅COO⁻] [H₃O⁺] = X.
X is the reaction coordinate. How many acid produce C₆H₅COO⁻ and H₃O⁺ until reach equilibrium.
Concentrations in equilibrium are:
[C₆H₅COOH] = 0.425M - X[C₆H₅COO⁻] = X [H₃O⁺] = XpH is defined as -log [H₃O⁺]. As pH = 2.270
2.270 = -log [H₃O⁺]
10^-2.270 = [H₃O⁺]
5.37x10⁻³M = [H₃O⁺] = X.
Replacing, concentrations in equilibrium are:
[C₆H₅COOH] = 0.425M - 5.37x10⁻³M = 0.4196M
[C₆H₅COO⁻] = 5.37x10⁻³M
[H₃O⁺] = 5.37x10⁻³M
Ka = [5.37x10⁻³M] [5.37x10⁻³M] / [0.4196M]
Ka = 6.87x10⁻⁵By 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
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.
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
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 .....................................
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)
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!
When the equation MnO₄⁻ + I⁻ + H₂O → MnO₂ + IO₃⁻ is balanced in basic solution, what is the smallest whole-number coefficient for OH⁻?
Answer:
The smallest whole-number coefficient for OH⁻ is 2
Explanation:
Step 1: The equation redox reaction is divided into two half equations
Reduction half equation: MnO₄⁻ ----> MnO₂
Oxidation half-equation: I⁻ ---> IO₃⁻
Step 2: Next the atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation;
MnO₄⁻ + 2H₂O ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O
Step 3 : The charges are then balanced by adding electrons to the appropriate sides of each half equation
MnO₄⁻ + 2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
Step 4: Oxidation half equation is multiplied by 2 while reduction half equation is multiplied by 1 to balance the number of electrons gained and lost for the reaction
2MnO₄⁻ + 4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
Step 5 : addition of the two half equations to yield a net ionic equation
2MnO₄⁻ + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻
The smallest whole number coefficient for OH⁻ is 2
A redox reaction is divided into two half equations which are shown below:
Reduction half equation: MnO₄⁻ ----> MnO₂
Oxidation half-equation: I⁻ ---> IO₃⁻
Atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation to make the equation complete ;
MnO₄⁻ + 2H₂O ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O
The charges needs to be balanced and this is done by adding electrons to the appropriate sides of each half equation
MnO₄⁻ + 2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
The equation needs to be balanced by multiplying the oxidation half equation by 2 while reduction half equation is multiplied by 1 to balance the number of electrons on both sides of the equations.
2MnO₄⁻ + 4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
The two half equations are then added and written together to form a net ionic equation
2MnO₄⁻ + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻
The smallest whole-number coefficient for OH⁻ is therefore 2.
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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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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/molWhat 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.
Write the IUPAC and common names, if any, of the carboxylate salts produced in the reaction of each of the following carboxylic acids with NaOH: 2-methylhexanoic acid
Part A
2-bromopropanoic acid Spell out the full name of the compound. If there is a common name separate your answers by a comma.
Part B
2-methylhexanoic acid Spell out the full name of the compound. If there is a common name separate your answers by a comma.
Answer:
Following are the explanation to this question:
Explanation:
The salts of carboxylate are named by the writing, which is also named as the creation of the first, which is followed by the name of the carboxylic acid were '-ic' of the acid end and replaced by the 'ate'.
Following are the description of the given reaction:
In reaction A:
2-Bromopropanoic acid= [tex]C_3H_5BrO_2[/tex]
[tex]C_3H_5BrO_2+NaOH[/tex]⇄ [tex]C_3H_4BrNaO_2 +H_2O[/tex]
The IUPAC name is Sodium-2-Bromopropanate
In reaction B:
2-Methylhexanoic acid= [tex]C_7H_{14}O_2[/tex]
[tex]C_7H_{14}O_2+NaOH[/tex]⇄ [tex]C_7H_{13}NaO_{2}+H_2O[/tex]
The IUPAC name is Sodium-2-Methyl hexanoate
The IUPAC name of compound is Sodium-2-Bromopropanate in reaction 1 and in reaction 2 it is 2-Methylhexanoic acid.
The salts of carboxylate are named by the writing, which is also named as the creation of the first, which is followed by the name of the compound carboxylic acid were '-ic' of the acid end and replaced by the 'ate'.
Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.
Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.
Thus, the IUPAC name of compound is Sodium-2-Bromopropanate in reaction 1 and in reaction 2 it is 2-Methylhexanoic acid.
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