Answer:
Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1
Explanation:
The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:
Vrms = [tex]\sqrt{3RT/M}[/tex]
where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol
For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R
Vrms = √(3 * R *4T)/0.04 = √300RT
For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R
Vrms = √(3 * R *T)/0.001 = √3000RT
Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316
Ratio of Vrms of argon to that of hydrogen = 0.316 : 1
How many neutrons does Carbon- 14 and Carbon -15 have? *
Answer: 8 for both
Explanation:
According to the following reaction, how many grams of ammonia will be formed upon the complete reaction of 31.2 grams of hydrogen gas with excess nitrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answer:
176.8 g of ammonia, NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar mass of H2 = 2x1 = 2 g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6 g
Molar mass of NH3 = 14 + (3x1) = 17 g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34 g.
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.
This can be obtained as follow:
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.
Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.
20. Stoichiometry is based on
A. molecular weight.
B. temperature.
C. conservation of matter.
D. pressure.
Answer:
The correct option is (c)
Answer:
the law of conservation of mass
what is the bond energy required to break one mole of carbon-carbon bonds
Answer:
100 kcal of bond energy
Which of the following is the balanced reaction, given the rate relationships below.
a. rate = − 13 Δ[CH4] Δt = − 12 Δ[H2O] Δt = − Δ[CO2] Δt = 14 Δ[CH3OH] Δt
b. rate = − 12 Δ[N2O5] Δt = 12 Δ[N2] Δt = 15 Δ[O2] Δt
c. rate = − 12 Δ[H2] Δt = − 12 Δ[CO2] Δt = − Δ[O2] Δt = 12 Δ[H2CO3] Δt
Answer:
a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]
b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]
c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]
Explanation:
Hello,
In this case, since those rate relationships have the stoichiometric coefficient at the denominators of the fractions preceding each ratio and the negative terms account for reactants and positive for products, we have:
a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]
b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]
c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]
Best regards.
A 1.0 L buffer solution is 0.250 M HC2H3O2 and 0.050 M LiC2H3O2. Which of the following actions will destroy the buffer?
A. adding 0.050 moles of NaOH
B. adding 0.050 moles of LiC2H3O2
C. adding 0.050 moles of HC2H3O2
D. adding 0.050 moles of HCl
E. None of the above will destroy the buffer.
Answer:
D
Explanation:
Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.
The action that destroys the buffer is option c. adding 0.050 moles of HCl.
What is acid buffer?It is a solution of a weak acid and salt.
Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.
The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this, there will be only acid in the solution.
Since
moles of HC2H3O2 = 1*0.250 = 0.250
moles of NaC2H3O2 = 1*0.050 = 0.050.
moles of HCl is added = 0.050
Now
The reaction between HCl and NaC2H3O2
[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]
Now
BCA table is
NaC2H3O2 HCl HC2H3O2
Before 0.050 0.050 0.250
Change -0.050 -0.050 +0.050
After 0 0 0.300
Now, the solution contains the acid (HC2H3O2 ) only.
Therefore addition of 0.050 moles of HCl will destroy the buffer.
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If the amount of radioactive iodine-123, used to treat thyroid cancer, in a sample decreases from 3.2 to 0.4 mg in 39.6 h, what is the half-life of iodine-123?
Answer:
Half life = 13.197 hour
Explanation:
Given:
Old amount (A₀) = 3.2
New amount (A) = 0.4
Radiation decay time (t) = 39.6 hour
Half life = T(1/2)
Find:
Half life = T(1/2) = T
Computation:
A = A₀[tex]e^{-(\frac{0.693t}{T} )}[/tex]
[tex]e^{-(\frac{0.693t}{T} )}[/tex] = 0.4 / 3.2
-[27.4428 / T] = In (0.125)
-[27.4428 / T] = -2.0794
[27.4428 / T] = 2.0794
T = 13.197
Half life = 13.197 hour
Which option is an isotope with 6 protons and 8 neutrons? (1 point)
O oxygen-6
o carbon-14
o oxygen-14
o carbon-8
Answer:
Carbon-14
Explanation:
It is because it have 6 protons and 8 neutrons which are added together to form an atomic mass of 14.
The mass number of an isotope is the sum of the number of protons and number of neutrons hence the correct answer is carbon-14.
Isotopes are atoms of the same element that has the same number of protons but different number of neutrons.
The atomic number is the same as the number of protons in the atom. Since there are six protons in the atom, the element must be carbon.
There are eight neutrons meaning that the mass number is 8 + 6 = 14
Therefore, the isotope is carbon-14.
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The substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an:_______
A) vicinal diol
B) geminal diol
C) acetal
D) ketal
Answer:
B) geminal diol
Explanation:
Hello,
In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.
Regards.
Hello, I am a bit stuck on this. Could someone help?
A 1.362 g sample of an iron ore that contained Fe3O4 was dissolved in acid with all of the iron being reduced to iron (II). The solution was acidified with sulfuric acid and titrated with 39.42 mL of 0.0281 M KMnO4, which oxidized the iron (II) to iron (III) while reducing the permanganate to manganese (II). Generate the balanced net ionic equation for the reaction. What is the mass percent of iron in this iron ore sample?
Answer:
a. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
b. 18.17% of Fe in the sample
Explanation:
a. In the reaction, Fe²⁺ is oxidized to Fe³⁺ and permanganate, MnO₄⁺ reduced to Mn²⁺, thus:
Fe²⁺ → Fe³⁺ + 1e⁻
MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O
5 times the iron and suming the manganese reaction:
MnO₄⁻ + 5e⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂Ob. Moles of permanganate in the titration are:
0.03942L × (0.0281 moles / L) = 1.108x10⁻³ moles of MnO₄⁻
Based on the reaction, 1 mole of permanganate reacts with 5 moles of iron, if 1.108x10⁻³ moles of MnO₄⁻ reacts, moles of iron are:
1.108x10⁻³ moles of MnO₄⁻ × (5 moles Fe²⁺ / 1 mole MnO₄⁻) =
4.431x10⁻³ moles of Fe²⁺. Molar mass of Fe is 55.845g/mol. 4.431x10⁻³ moles of Fe²⁺ are:
4.431x10⁻³ moles of Fe²⁺ ₓ (55.845g / mol) =
0.2474g of Fe you have in your sample.Percent mass is:
0.2474g Fe / 1.362g sample ₓ 100 =
18.17% of Fe in the sampleThe mass percent of iron in the sample is 22.6%.
The net ionic equation of the reaction is;
5Fe^2+(aq) + 8H^+(aq) + MnO4^- -----> 5Fe^3+(aq) + Mn^2+(aq) + 4H2O(l)
Number of moles of MnO4^- = 39.42/1000 L × 0.0281 M = 0.0011 moles
If 5 moles of Fe^2+ reacts with 1 mole of MnO4^-
x moles of Fe^2+ reacts with 0.0011 moles
x = 5 moles × 0.0011 moles/1 mole
x = 0.0055 moles
Mass of Fe^2+ = 0.0055 moles × 56 g/mol = 0.308 g
Mass percent of iron = 0.308 g/ 1.362 g × 100/1
= 22.6%
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What is the molecular mass of this substance? a. 31.02 u b. 63.02 u c. 47.02 u d. 126.04 u e. 110.01 u
Answer:
Molecular mass of HNO₃ = 63.015 g/mol
Explanation:
Note: The given question is incomplete , the given substance is nitric acid
Given:
Nitric acid (HNO₃)
Find:
Molecular mass
Computation:
Molecular mass of HNO₃ = 1.008 + 14.007 + 3(16)
Molecular mass of HNO₃ = 1.008 + 14.007 + 48
Molecular mass of HNO₃ = 63.015 g/mol
If a salt is formed by combining NH3 (Kb=1.8×10−5) and CH3COOH (Ka=1.8×10−5), an aqueous solution of this salt would be:
Answer:
Neutral
Explanation:
pKa of acid = -log Ka
= -log (1.8 x 10^-5)
= 4.74
pKb of base = -log Kb
= 4.74
pKa of acid = pKb of base
salt pH formula : pH = 7 + 1/2 [pKa -pKb ]
here pKa = pKb
so pH = 7
the salt it is CH3COONH4 exactly neutral solution .
If a salt is formed by combining NH₃ (Kb=1.8×10⁻⁵) and CH₃COOH (Ka=1.8×10⁻⁵), an aqueous solution of this salt would be neutral.
What information does pH convey?pH of any solution tells about the acidity or basicity or neutral nature of the solution.
pH of any solution is directly proportional to the acid dissociation constant value (Ka) and base dissociation constant (Kb). In the question it is given that,
Value of Kb for NH₃ = 1.8×10⁻⁵
Value of Ka for CH₃COOH = 1.8×10⁻⁵
Ka & Kb values for the base and acid is same means it dissociates with same extent. So the aqueous solution of this acid and base is a neutral in nature as they have same number of acid and base ions in it.
Hence resultant solution will be a neutral solution .
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Name the alkyne. Spelling and punctuation count.
The name of a compound is determined by IUPAC nomenclature
The International Union of Pure and Applied Chemistry have designated a universal rule for the nomenclature of organic compounds generally known as IUPAC nomenclature.
The whole idea of the IUPAC nomenclature is to have a universally agreed pattern of naming organic compounds according to their structure.
The compound is always named in such a way that the substituents and the functional group receives the lowest number.
In this case, the functional group is alkyne hence the name of the compound ends in -yne. The methyl group is substituted on position four.
Putting all these together the compound should be named as; 4-methylhept-2-yne
For IUPAC nomenclature of an inorganic compound
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What is the pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid with 15.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. K a = 1.8 ×× 10-5 for CH3CO2H.
Answer:
pH = 8.72
Explanation:
This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
As this compound acts like a base, we propose this equilibrium:
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰
Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M
So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
We can avoid the quadractic equation because Kb is so small
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH = 8.72
The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
Calculation of the pH of the solution:Since the following equation should be used.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
Now
(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
So,
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
Now
Kw = Ka. Kb
Kb = Kw/Ka
And,
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵
= 5.55×10⁻¹⁰
Now
[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
Now
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH
= 8.72
Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
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For the following reaction, 22.8 grams of diphosphorus pentoxide are allowed to react with 13.5 grams of water . diphosphorus pentoxide(s) water(l) phosphoric acid(aq) What is the maximum mass of phosphoric acid that can be formed
Answer:
[tex]m_{H_3PO_4}=31.5gH_3PO_4[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]P_2O_5(s)+3H_2O(l)\rightarrow 2H_3PO_4(aq)[/tex]
Thus, since the diphosphorus pentoxide to water molar ratio is 1:3 and we are given the mass of both of them, for the calculation of the maximum mass phosphoric acid that is yielded, one could first identify the limiting reactant, for which we compute the available moles of diphosphorus pentoxide (molar mass 142 g/mol):
[tex]n_{P_2O_5}=22.8gP_2O_5*\frac{1molP_2O_5}{142gP_2O_5}=0.161molP_2O_5[/tex]
And the moles of diphosphorus pentoxide that are consumed by 13.5 g of water (molar mass 18 g/mol):
[tex]n_{P_2O_5}^{consumed}=13.5gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molP_2O_5}{3molH_2O} =0.25molP_2O_5[/tex]
Hence, since less moles of diphosphorus pentoxide are available, we sum up it is the limiting reactant, therefore, the maximum mass of phosphoric acid (molar mass 98 g/mol) is computed by considering the 1:2 molar ratio between them as follows:
[tex]m_{H_3PO_4}=0.161molP_2O_5*\frac{2molH_3PO_4}{1molP_2O_5} *\frac{98gH_3PO_4}{1molH_3PO_4} \\\\m_{H_3PO_4}=31.5gH_3PO_4[/tex]
Regards.
Janet observes that bubbles rise inside water when water is heated. Which of the following best names and explains the change that causes bubbles to rise?
Answer:
Boiling
Explanation:
When a liquid is heated, the vapor pressure rises steadily. When water attains a temperature of 100°C or 212°F its vapor pressure is now equal to the atmospheric pressure at sea level, this is what we mean by boiling.
When this occurs, water continues to evaporate untill the vapor pressure inside the bubbles becomes high enough to stop water bubbles from collapsing again from the pressure of the water around it so the bubbles rise and break the surface.
When cyclopentane undergo free-radical substitution with bromine (Br2 /Heat) the product:
a. Bromo cyclopentene
b. Bromo cyclopentane + HBr
c. Bromo cyclopentene + HBr
Explanation:
b. Bromo cyclopentane + HBr
Which solution has the greatest buffer capacity? Select the correct answer below: 1 mole of acid and 1 mole of base in a 1.0 L solution
Answer:
The answer is
Explanation:
1 mole of acid.
Hope this helps....
Have a nice day!!!!
A buffer that is 1 M in acid and base will have the greatest capacity of buffer, and therefore the greatest buffer capacity.
What do you mean by the buffer solution ?A weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base, are combined to form the buffer solution, a water-based solvent solution.
In a biological system, a buffer's keep intracellular and extracellular pH levels within a relatively small range and to withstand pH fluctuations brought on by both internal and external factors.
A buffer is a substance that can withstand a pH shift when acidic or basic substances are added. It may balance out little quantities of additional acid or base, keeping the pH stable.
Thus, 1 M in acid and base solution has the greatest buffer capacity.
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9. In order to determine the melting point of tetradecanol, it was important that _____. the thermometer was attached to the capillary tube the water was frozen the capillary tube was immersed in the water without the thermometer
To record the melting point of tetradecanol, it is important the temperature is increase by little degrees after each trial.
The thermometer attached to the capillary tube was immersed in water without thermometer.
Usually heating curve describes the temperature change for a sample which passes through different states of matter.
The temperature than decreases and the solution in beaker cools down. To determine the melting point, it is important to keep the temperature consistent.
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in non- equilibrium cooling, carbon is allowed to diffuse its end position. true /false
Answer:
False because it is false
PLEASE HELP!!
You are performing an experiment that involves the electrolysis of gold (I) bromide, also know as AuBr. You know that gold is less reactive than hydrogen. Which of the following would be the product of the reaction?
A. Hydrogen gas
B. Gold bromide
C. Oxygen gas
D. Pure gold
Answer:
D. Pure gold
Explanation:
Hello,
In this case, since gold, as a heavy metal, is said to be less reactive than hydrogen, when it undergoes electrolysis process when forming a salt, due to the action of the electric current, we can appreciate the formation of a layer of gold on the surface of the cathode via a reduction half-reaction from gold (I) to metallic gold:
[tex]Au^++1e^-\rightarrow Au^0[/tex]
Thereby, D. Pure gold is formed as the product of the reaction.
In contrast, more reactive metals than hydrogen such as sodium or potassium, will remain in solution so the hydrogen converted to hydrogen gas.
Best regards-
2) What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 230 mL of solution?
Provide two real world examples of habitats and niches within ecosystems? Explain the interdependence that exists within those ecosystems.
Answer:
An ecosystem encompasses living organisms and the nonliving elements of their environments. Hence, the components of an ecosystem include animals, plants, microorganisms, rocks, soil, minerals, atmosphere, and the surrounding water masses. An ecosystem can be huge, cutting across several nations, or it can be relatively small, such as the body of an animal, which is home to numerous microorganisms.
We can describe ecosystems under two headings – natural ecosystems and unnatural ecosystems. Unnatural ecosystems, which include agricultural and urban areas, are greatly modified and maintained by human activity. Conversely, natural ecosystems are self-sufficient, balanced ecological units, with a high proportion of native biodiversity and minimal human disruption.
The natural ecosystem is broad. It is divided into two major groups – terrestrial and aquatic ecosystems. These are further divided into many other smaller types of ecosystems as outlined in this article.
Explanation:
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g Which ONE of the following pure substances will exhibit hydrogen bonding? A) methyl fluoride, FCH3 B) dimethyl ether, CH3C–O–CH3 C) formaldehyde, H2C=O D) trimethylamine, N(CH3)3 E) hydrazine, H2N-NH2
Answer:
C) formaldehyde, H2C=O.
Explanation:
Hello,
In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).
Best regards.
What is the maximum number of electrons in the following energy level?
n = 4
Answer: 32 electrons
Explanation:
What must happen to uranium before it can be used as a fuel source?
Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.
Determine what product will be produced at the negative electrode for the following reaction:
2KCl(aq) + 2H20(1) -> H2(g) + Cl2(g) + 2KOH(aq)
A. H2
B. Cl2
с. КОН
D. K
Answer:
Choice A. [tex]\rm H_{2}[/tex] would be produced at the negative electrode.
Explanation:
Ionic equation for this reaction:
[tex]2\, {\rm K^{+}} + 2\, {\rm Cl^{-}} + {2\, \rm H_{2} O} \to {\rm H_{2}} + {\rm Cl_{2}} + 2\, {\rm K^{+}} + {\rm 2\, OH^{-}}[/tex].
Net ionic equation:
[tex]2\, {\rm Cl^{-}} + 2\, \rm H_{2} O} \to {\rm H_{2}} + {\rm Cl_{2}} + 2\, {\rm OH^{-}}[/tex].
Half-equations:
[tex]2\, {\rm Cl^{-}} \to {\rm Cl_{2}} + 2\, {e^{-}}[/tex].
(Electrons travel from the solution to an electrode.)
[tex]2\, {\rm \overset{+1}{H}_{2} O} + 2\, {e^{-}} \to \overset{0}{\rm H}_{2} + 2\, {\rm O\overset{+1}{H}\!^{-}}[/tex].
(An electrode supply electrons to the solution to reduce some of the [tex]\rm H[/tex] atoms from [tex]\rm H_{2}O[/tex].)
In a DC circuit, electrons always enter the circuit from the negative terminal of the power supply and return to the power supply at the positive terminal.
The negative electrode is connected to the negative terminal of the power supply. Electrons from the power supply would flow into the solution through this electrode.
This continuous supply of electrons at the negative electrode would drive a reduction half-reaction. In this question, that corresponds to the reduction of water: [tex]2\, {\rm \overset{+1}{H}_{2} O} + 2\, {\rm e^{-}} \to \overset{0}{\rm H}_{2} + 2\, {\rm O\overset{+1}{H}\!^{-}}[/tex]. Hence, [tex]\rm H_{2}[/tex] would be produced at the negative electrode.
. You have two solutions, both with a concentration of 0.1M. Solution A contains a weak acid with a pKa of 5. ThepH of solution A is 3. Solution B contains a weak acid with a pKa of 9. The pH of solution B is:
Answer:
pH of solution B is 5
Explanation:
A weak acid, HA, is in equilibrium with water as follows:
HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)
Where Ka (10^-pKa = 1x10⁻⁹) is:
Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
Where concentrations of this species are equilibrium concentrations
As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:
[HA] = 0.1M - X
[A⁻] = X
[H₃O⁺] = X
Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.
Replacing in Ka expression:
1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
1x10⁻⁹ = [X] [X] / [0.1 - X]
1x10⁻¹⁰ - 1x10⁻⁹X = X²
1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0
Solving for X:
X = -0.00001 → False solution, there is no negative concentrations.
X = 1x10⁻⁵ → Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 1x10⁻⁵M
And pH = -log[H₃O⁺]
pH = 5
pH of solution B is 5
For the following set of volume/temperature data, calculate the missing quantity after the change is made. Assume that the pressure and the amount of gas remain constant.
V=2.91 L at 23.0 °C
V= 4.20 L at ? °C
Answer:
155 °C
Explanation:
Step 1: Given data
Initial volume (V₁): 2.91 LInitial temperature (T₁): 23.0°CFinal volume (V₂): 4.20 LFinal temperature (T₂): ?Step 2: Convert the initial temperature to Kelvin
We will use the following expression.
K = °C + 273.15 = 23.0°C + 273.15 = 296.2 K
Step 3: Calculate the final temperature
Assuming an ideal gas behavior, we can calculate the final temperature using Charles' law.
V₁/T₁ = V₂/T₂
T₂ = V₂ × T₁/V₁
T₂ = 4.20 L × 296.2 K/2.91 L
T₂ = 428 K
Step 4: Convert the final temperature to Celsius
We will use the following expression.
°C = K - 273.15 = 428 - 273.15 = 155 °C