Answers
The answer is in the photo
The molarity of the luminol stock solution is calculated by dividing the number of moles of luminol (0.08464 moles) by the volume of the solution in liters (0.075 L), resulting in a molarity of 1.13 M.
The ideal way to characterise stock solutions is as concentrated solutions with precise, known concentrations that will be diluted for later usage in laboratories. Although you have the option to forego preparing stock solutions, doing so can simplify your workflow and save you a significant amount of time and money. To solve the question we need to first determine the number of moles of luminol and then use the volume of the solution to find the molar concentration.
The molecular weight of luminol (C₈H₇N₃O₂) is approximately 177.16 g/mol. The number of moles of luminol can be calculated by dividing the mass of luminol by its molecular weight:
Number of moles of luminol = Mass of luminol / Molecular weight of luminol
Number of moles of luminol = 15.0 g / 177.16 g/mol = 0.08464 moles
Next, we convert the volume of water from milliliters to liters:
Volume in liters = 75.0 mL × (1 L / 1000 mL) = 0.075 L
Now we can calculate the molarity of the luminol solution:
Molarity (M) = Number of moles of solute / Volume of solution in liters
Molarity (M) = 0.08464 moles / 0.075 L
Molarity (M) = 1.13 M
The stock solution of luminol has a molarity of 1.13 moles per liter (1.13 M).
Related Questions
The 1H NMR spectrum of chloroethane (CH3CH2Cl) recorded on a 500 MHz NMR spectrometer consists of signals at chemical shifts of 1.48 ppm and 3.57 ppm. Calculate the frequency, downfield from TMS, of each absorption. Be sure to answer all parts.
Answers
Answer:The frequency of absorption for the proton having chemical shift 1.48 ppm is 740 Hz downfield from TMS.
The frequency of absorption for the proton having chemical shift 3.57 ppm is 1785 Hz downfield from TMS.
Explanation:
We are given with the following data:
Frequency of the instrument(NMR spectrometer)=500MHz=500×10⁶Hz
Chemical shift (δ) value for 1st proton=1.48PPM=1.48×10⁻⁶
Chemical shift (δ) value for 2nd proton=3.57PPM=3.57×10⁻⁶
We know that frequency of reference that is of TMS(Tetramethylsilane) is assumed to be 0.
We have to calculate the frequency of absorption of each protons downfield from TMS.
The formula for the chemical shift (δ) is:
δ=[Frequency of sample(νₐ)-Frequency of TMS(νₓ)]÷Frequency of instrument MHz(Mega Hertz)
So using the above formula we can calculate the frequency of absorption for the two protons whose δ value is given.
1. For the proton having δ value 1.48ppm:
1.48= [Frequency of sample(νₐ)-Frequency of TMS(νₓ)]÷Frequency of instrument MHz(Mega Hertz)
1.48×10⁻⁶=[Frequency of sample(νₐ)-0]÷[500×10⁶Hz]
1.48×10⁻⁶×500×10⁶Hz=[Frequency of sample(νₐ)]
740Hz=[Frequency of sample(νₐ)]
2. For the proton having δ value 3.57ppm:
3.57= [Frequency of sample(νₐ)-Frequency of TMS(νₓ)]÷Frequency of instrument MHz(Mega Hertz)
3.57×10⁻⁶=[Frequency of sample(νₐ)-0]÷[500×10⁶Hz]
3.57×10⁻⁶×500×10⁶Hz=[Frequency of sample(νₐ)]
1785Hz=[Frequency of sample(νₐ)]
So the frequency of absorption for the proton having δ value 1.48ppm is 740 Hz and for the proton having δ value 3.57 ppm is 1785Hz.
Final answer:
To calculate the NMR frequencies for chloroethane at 1.48 ppm and 3.57 ppm on a 500 MHz spectrometer, multiply each chemical shift by 500 MHz. The results are 740 Hz for 1.48 ppm and 1785 Hz for 3.57 ppm.
Explanation:
The question pertains to calculating the frequency of each absorption in the 1H NMR spectrum of chloroethane, observed at chemical shifts of 1.48 ppm and 3.57 ppm, on a 500 MHz NMR spectrometer. The frequency of an absorption in Hz can be calculated by multiplying the parts per million (ppm) chemical shift by the frequency of the spectrometer in MHz. To calculate the frequency downfield from TMS (tetramethylsilane), the chemical shifts in ppm are multiplied by the spectrometer frequency.
For the 1.48 ppm peak
500 MHz Spectrometer
= 1.48 ppm *500 MHz
= 740 Hz
For the 3.57 ppm peak:
= 3.57 ppm *500 MHz
= 1785 Hz
Therefore, the frequencies of the absorptions downfield from TMS for chloroethane on a 500 MHz NMR spectrometer are 740 Hz and 1785 Hz for the chemical shifts of 1.48 ppm and 3.57 ppm, respectively.
Pure HN3 (atom sequence HNNN) is explosive. In aqueous solution, it is a weak acid that yields the azide ion, N3−. Draw one resonance structure for N3− and one resonance structure for HN3. Include all lone pair electrons and nonzero formal charges in your structures.
Answers
Answer:
see explanation...
Explanation:
The central nitrogen atom in the azide ion has a formal charge of +1. The other nitrogen atoms have a formal charge of -1.
The resonance structure of azide ion and hydrazoic acid is shown in the image attached to this answer. We can see that the central nitrogen atom in the azide ion has a formal charge of +1. The other nitrogen atoms have a formal charge of -1.
In hydrazoic acid, hydrogen is attached to one of the end nitrogen atoms while the other nitrogen atoms have a formal of +1 and -1 respectively.
Learn more:
When 7.50 grams of Iodine (I2) is added to 48.7 grams of Carbon Tetrachloride (CCl4), the iodine dissolves and a solution is formed. What is the weight percent of the iodine in the solution?
Answers
Answer:
weight percent of Iodine in the solution is 13.34%
Explanation:
Weight percentage -
weight percentage of A is given as , the weight of the substance A by weight of the total solution multiplied by 100.
i.e.
weight % A = weight of A / weight of solution * 100
From the question ,
weight of Iodine = 7.50 g
weight of carbon tetrachloride = 48.7 g
iodine is the solute and carbon tetrachloride is the solvent ,
Since,
solution = solute + solvent.
Hence,
weight of solution = weight of solute ( iodine ) + weight of solvent ( carbon tetrachloride)
weight of solution = 7.50 g + 48.7 g = 56.20 g
now,
weight of Iodine is calculated as,
weight % iodine = weight of iodine / weight of solution * 100
weight % iodine = 7.50 g / 56.20 g * 100
weight % iodine = 13.34 %
The fraction of a radioactive isotope remaining at time t is (1/2)^t/t1/2 where t1/2 is the half-life. If the half-life of carbon−14 is 5,730 yr, what fraction of carbon−14 in a piece of charcoal remains after
(a) 14.0 yr?
(b) 1.900 × 10^4 yr? _________ × 10 (Enter your answer in scientific notation.)
(c) 1.0000 × 10^5 yr? __________× 10 (Enter your answer in scientific notation.)
Answers
Answer:
a) [tex]1.065\times 10^{-8} [/tex]fraction of carbon−14 in a piece of charcoal remains after 14.0 years.
b) [tex]0.000\times 10^{-3} [/tex]fraction of carbon−14 in a piece of charcoal remains after [tex]1.900\times 10^4 years[/tex]
c) [tex]0.0000\times 10^{-4} [/tex]fraction of carbon−14 in a piece of charcoal remains after [tex]1.0000\times 10^5 years[/tex].
Explanation:
The fraction of a radioactive isotope remaining at time t is given by:
[tex][A]=\frac{(\frac{1}{2})^t}{t_{\frac{1}{2}}}[/tex]
Taking log both sides:
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[A] = fraction at given time t
[tex]t_{\frac{1}{2}}[/tex] = half life of the carbon−14 =5,730 years
a)When , t = 14 years
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[tex]\log [A]= 14 years\times (-3010)-\log [5,730 years][/tex]
[tex][A]=1.065\times 10^{-8} [/tex]
b)When , t = [tex]1.900\times 10^4 years[/tex]
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[tex]\log [A]= 1.900\times 10^4 years\times (-3010)-\log [5,730 years][/tex]
[tex][A]=0.000\times 10^{-3} [/tex [/tex]
c)When , t = [tex]1.0000\times 10^5 years[/tex]
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[tex]\log [A]= 1.0000\times 10^5 years\times (-3010)-\log [5,730 years][/tex]
[tex][A]=0.0000\times 10^{-4} [/tex]
The equilibrium-constant of the reaction NO2(g)+NO3(g)⇌N2O5(g) is K=2.1×10−20. What can be said about this reaction? a. At equilibrium the concentration of products and reactants is about the same. b. At equilibrium the concentration of products is much greater than the concentration of reactants. c. At equilibrium the concentration of reactants is much greater than that of products. d. There are no reactants left over once the reaction reaches equilibrium.
Answers
Answer:
[tex]\boxed{\text{c. The concentration of reactants is much greater than that of products.}}[/tex]
Explanation:
NO₂ + NO₃ ⇌ N₂O₅; K = 2.1 × 10⁻²⁰
We often write K as
[tex]K = \dfrac{[\text{Products}]}{[\text{Reactants}]}[/tex]
If K is large, more of the molecules exist as products.
If K is small, more of the molecules exist as reactants.
[tex]\text{Since K is small, }\\\boxed{\textbf{the concentration of reactants is much greater than that of products.}}[/tex]
a. b., and d. are wrong. The concentration of reactants is greater.
Final answer:
The equilibrium constant of K = 2.1 × 10−20 suggests that at equilibrium, the reactants' concentration significantly surpasses the products', confirming option (c) as correct.
Explanation:
The equilibrium constant for the reaction NO2(g) + NO3(g) ⇌ N2O5(g) is K = 2.1 × 10−20. An equilibrium constant of such a small magnitude indicates that, at equilibrium, the concentration of reactants (NO2 and NO3) will be much greater than the concentration of products (N2O5). This means that option (c) is correct: At equilibrium the concentration of reactants is much greater than that of products. It does not mean reactions stop; both the forward and reverse reactions continue occurring at equal rates, maintaining the equilibrium concentrations of reactants and products.
If a solid line represents a covalent bond and a dotted line represents intermolecular attraction, which of the choices shows a hydrogen bond? H−H H4C⋅⋅⋅⋅⋅⋅H−F H3N⋅⋅⋅⋅⋅⋅H−O−H H2O⋅⋅⋅⋅⋅⋅H−CH3
Answers
Answer: [tex]H_{3}N-H-O-H[/tex] shows a hydrogen bond is the correct answer.
Explanation:
A hydrogen bond is defined as a weak bond that is formed between an electropositive atom (generally hydrogen atom) and an electronegative atom like oxygen, nitrogen and fluorine.
This bond is formed due to difference in the electronegativity of atoms present in a compound which contains a hydrogen bond. So, there occurs a partial positive charge on hydrogen atom and a partial negative charge on the electronegative atom.
For example, [tex]H_{3}N-H-O-H[/tex] molecule will have hydrogen bonds.
This is because both nitrogen and oxygen atoms are electronegative in nature. Therefore, partial charges will develop on the both electronegative and electropositive atoms of this compound.
Thus, we can conclude that out of the given options [tex]H_{3}N-H-O-H[/tex] shows a hydrogen bond.
A hydrogen bond is an intermolecular attractive force in which a hydrogen atom is attracted to a lone pair of electrons on an atom in a neighboring molecule. The correct choice that shows a hydrogen bond is H3N⋅⋅⋅⋅⋅⋅H−O−H.
Explanation:A hydrogen bond is an intermolecular attractive force in which a hydrogen atom, that is covalently bonded to a small, highly electronegative atom, is attracted to a lone pair of electrons on an atom in a neighboring molecule. Hydrogen bonds are very strong compared to other dipole-dipole interactions, but still much weaker than a covalent bond. A typical hydrogen bond is about 5% as strong as a covalent bond.
Looking at the choices, H−H and H2O⋅⋅⋅⋅⋅⋅H−CH3 do not involve hydrogen bonding, as they lack a small, highly electronegative atom with a lone pair of electrons. H4C⋅⋅⋅⋅⋅⋅H−F does not involve hydrogen bonding either, as the fluorine atom is not highly electronegative enough to form a hydrogen bond. Therefore, the correct choice that shows a hydrogen bond is H3N⋅⋅⋅⋅⋅⋅H−O−H.
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Metals often form several cations with different charges. Cerium, for example, forms Ce3+ and Ce4+ ions, and thallium forms Tl+ and Tl3+ ions. Cerium and thallium ions react as follows:
2Ce4+(aq)+Tl+(aq)→2Ce3+(aq)+Tl3+(aq)
This reaction is very slow and is thought to occur in a single elementary step. The reaction is catalyzed by the addition of Mn2+(aq) according to the following mechanism:
Ce4+(aq)+Mn2+(aq)Ce4+(aq)+Mn3+(aq)Mn4+(aq)+Tl+(aq)→→ →Ce3+(aq)+Mn3+(aq)Ce3+(aq)+Mn4+(aq)Mn2+(aq)+Tl3+(aq)
Part A
What is the rate law for the uncatalyzed reaction?
rate=k[Ce4+]2[Tl+]2
rate=k[Ce4+][Tl+]
rate=k[Ce4+]
rate=k[Tl+]
rate=k[Ce4+]2[Tl+]
rate=k[Ce4+][Tl+]2
If the uncatalyzed reaction occurs in a single elementary step, why is it a slow reaction?
Check all that apply.
All reactions that occur in one step are slow.
The reaction requires the collision of three particles with the correct energy and orientation.
The probability of an effective three-particle collision is low.
The transition state is low in energy.
The catalyzed reaction is first order in [Ce4+] and first order in [Mn2+]. Which of the steps in the catalyzed mechanism is rate determining?
The catalyzed reaction is first order in and first order in . Which of the steps in the catalyzed mechanism is rate determining?
Ce4+(aq)+Mn2+(aq)→Ce3+(aq)+Mn3+(aq)
Ce4+(aq)+Mn3+(aq)→Ce3+(aq)+Mn4+(aq)
Mn4+(aq)+Tl+(aq)→Mn2+(aq)+Tl3+(aq)
Answers
What is the rate law for the uncatalyzed reaction?
In this type of reactions the rate law is dependent on the concentration of the reactants. In this case Ce and Tl ions. The answer will be rate=k[Ce[tex]^{4+}[/tex]][tex]^{2}[/tex][Tl[tex]^{+}[/tex]] because according to the rate law the reactants concentration should be powered at the number of moles, and from the equation we deduce that Ce[tex]^{4+}[/tex] concentration will be powered to 2.
the other answers are wrong
If the uncatalyzed reaction occurs in a single elementary step, why is it a slow reaction?
“All reactions that occur in one step are slow” - it is not correct because there are one step reactions that are fast.
“The reaction requires the collision of three particles with the correct energy and orientation” - it is the correct answer because the particles should collide in exactly the same time with the required amount of energy for the reaction to take place. The probability of an effective three-particle collision is low so the reaction will be a slow one.
“The transition state is low in energy” - in this case the rate of the reaction should be fast because the energy for the reaction needed to take place is low.
The catalyzed reaction is first order in [Ce[tex]^{4+}[/tex]] and first order in [Mn[tex]^{2+}[/tex]]. Which of the steps in the catalyzed mechanism is rate determining?
rate =k[Ce[tex]^{4+}[/tex]] [Mn[tex]^{2+}[/tex]] because the rate of the reaction may be done by increasing the amount of the catalyst (Mn[tex]^{2+}[/tex]).
The catalyzed reaction is first order in which of the steps in the catalyzed mechanism is rate determining?
Ce[tex]^{4+}[/tex][tex]_{(aq)}[/tex]+Mn[tex]^{2+}[/tex][tex]_{(aq)}[/tex]→Ce[tex]^{3+}[/tex][tex]_{(aq)}[/tex]+Mn[tex]^{3+}[/tex][tex]_{(aq)}[/tex]
A piece of metal with a mass of 12.5 g is placed into a graduated cylinder that contains 25.00 mL of water, raising the water level to 27.5 mL. What is the density of the metal?0.45 g/mL2.23 g/mL4.0 g/mL5.0 g/mL5.50 g/mL
Answers
The answer is in the photo
7. A sample of hydrogen gas is mixed with water vapor. The mix has a total pressure of 725 torr and the water vapor has a partial pressure of 24 torr. What amount (in moles) of hydrogen gas is contained in 1.85 L of this mixture at 308 K?
Answers
Answer: 0.07 moles
Explanation:
According to Dalton's law, the total pressure of a mixture of gases is the sum of individual pressures exerted by the constituent gases.
Thus [tex]p_{total}=p_{H_2O}+p_{H_2}[/tex]
Given: [tex]p_{total}=725torr[/tex]
[tex]p_{H_2O}=24torr[/tex]
[tex]p_{H_2}=?[/tex]
Thus [tex]725torr=24torr+p_{H_2}[/tex]
[tex]p_{H_2}=701torr[/tex]
According to the ideal gas equation:'
[tex]PV=nRT[/tex]
P= Pressure of the gas = 701 torr = 0.92 atm (760torr=1atm)
V= Volume of the gas = 1.85 L
T= Temperature of the gas = 308 K
R= Value of gas constant = 0.082 Latm\K mol
[tex]n=\frac{PV}{RT}=\frac{0.92\times 1.85L}{0.0821 \times 308}=0.07moles[/tex]
Thus 0.07 moles of hydrogen gas is contained in 1.85 L of this mixture at 308 K.
Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.54 cm. (1 f-12 inches)
Answers
Answer: The density of gold in [tex]g/cm^3[/tex] is [tex]19.22g/cm^3[/tex]
Explanation:
Density is defined as the ratio of mass of the object and volume of the object. Mathematically,
[tex]\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}[/tex]
We are given:
Density of gold = [tex]1200lb/ft^3[/tex]
Using conversion factors:
1 lb = 453.6 g
1 feet = 12 inches
1 inch = 2.54 cm
Converting given quantity into [tex]g/cm^3[/tex], we get:
[tex]\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3[/tex]
Hence, the density of gold in [tex]g/cm^3[/tex] is [tex]19.22g/cm^3[/tex]
Write the structure of a D-aldohexose that gives a meso compound on treatment with NaBH4 in water.
Answers
Explanation:
Meso compounds are optically inactive stereoisomers. Despite of having chiral carbon, they do not show optical activity because it has a plane of symmetry in its structure itself. It is superposable on its mirror image.
Aldo sugars on reaction which reducing agents such as , NaBH₄ reduces the carbonyl group in the sugar to alcohol and gives corresponding alditol.
The D- aldohexose which on reduction gives a meso alditol are allose and galactose.
The structure is shown in the image below. Thus, in allitol and galactitol formed have internal plane of symmetry which makes the optically inactive.
What is the percent yield of a reaction in which 74.1 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 8.01 mL of water (d = 1.00 g/mL)?
Answers
Final answer:
The percent yield of the reaction producing tungsten and water is approximately 13.66%. To find this, the theoretical yield of tungsten was calculated using stoichiometry, and compared with the actual yield of water, assuming a 1:1 correspondence between tungsten and water produced by the reaction.
Explanation:
To calculate the percent yield of the reaction, we need to know the theoretical yield and the actual yield. The theoretical yield can be calculated using stoichiometry based on the balanced chemical equation for the reaction. In this case, we are given the production of water as a byproduct of the reaction between tungsten(VI) oxide (WO3) and hydrogen gas. We're also provided the mass of water produced (8.01 g, since the density of water is given as 1.00 g/mL and volume is 8.01 mL).
The balanced reaction is: WO3 + 3H2 → W + 3H2O
To find the theoretical yield of tungsten, we need the molar mass of WO3 and W:
Molar mass of WO3 = (1 x 183.84) + (3 x 16.00) = 231.84 g/molMolar mass of W = 183.84 g/molUsing the molar mass, we can calculate the moles of WO3 reacted:
74.1 g WO3 × (1 mol / 231.84 g) = 0.319 moles of WO3
Because the reaction stoichiometry is 1:1 for WO3 to W, the moles of tungsten produced should also be 0.319. Using the molar mass of tungsten, we can calculate the theoretical yield:
0.319 moles of W × 183.84 g/mol = 58.665 g of W
The actual yield given is 8.01 g of water, which we assume corresponds to the amount of tungsten produced since 3 moles of water are produced per mole of tungsten. Therefore, the percent yield is:
Percent Yield = (Actual Yield / Theoretical Yield) × 100%
Percent Yield = (8.01 g / 58.665 g) × 100% ≈ 13.66%
Which of the following is incorest regarding monosaccharides? b) smallest monosaccharides are composed of three carbons. d) must be either ether or ester a) aldoses and ketoses c) exist in many isomeric forms e) Two monosaccharides can combine to form a disaccharide
Answers
Answer:
Must be either ether or ester.
Explanation:
The incorrect statement is that the monosaccharides must ether or esters.
The main and compulsory functional group in a monosaccharide are aldehyde and ketones.
The other statements are true
1) smallest monosaccharides are composed of three carbons: the smallest monosaccharide is glyceraldehyde.
a) aldoses and ketoses : they have either aldehyde group or ketone group
example: glucose is aldose and fructose if keotse
c) exist in many isomeric forms: there can be many arrangement on chiral carbons thus many stereoisomers are possible
e) Two monosaccharides can combine to form a disaccharide: yes for example glucose combines with fructose to give sucrose.
Write the balanced equation for the reaction of aluminum with copper(II) sulfate solution.
Answers
Answer: [tex]2Al+3CuSO_4\rightarrow Al_2(SO_4)_3+3Cu[/tex]
Explanation:
A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution. Thus one element should be different from another element.
A general single displacement reaction can be represented as :
[tex]A+BC\rightarrow AC+B[/tex]
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The balanced equation for the reaction of aluminum with copper(II) sulfate solution is:
[tex]2Al+3CuSO_4\rightarrow Al_2(SO_4)_3+3Cu[/tex]
The balanced equation for the reaction between aluminum and copper(II) sulfate solution is: 2Al (s) + 3CuSO4 (aq) -> Al2(SO4)3 (aq) + 3Cu (s). In this reaction, aluminum displaces copper to form aluminum sulfate, and copper is precipitated out.
Explanation:The question asks for the balanced equation for the reaction of aluminum with copper(II) sulfate solution. When aluminum reacts with copper(II) sulfate solution, it displaces copper to form aluminum sulfate, and copper is precipitated out.
The unbalanced equation for the reaction is:
Al (s) + CuSO4 (aq) -> Al2(SO4)3 (aq) + Cu (s)
To balance it, add stoichiometric coefficients to the reactants and products:'
2Al (s) + 3CuSO4 (aq) -> Al2(SO4)3 (aq) + 3Cu (s)
This means that 2 moles of aluminum react with 3 moles of copper(II) sulfate to produce 1 mole of aluminum sulfate and 3 moles of copper.
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A mixture of 8.0 molHe and 8.0 molXe are at STP in a rigid container. Which of the following statements is TRUE? O Both gases have the same average kinetic energy O The mixture has a volume of 22.4 L O Both gases contribute equally to the density of the mixture under these conditions O Both gases have the same molecular speed O All of the above are TRUE
Answers
Answer : The correct option is, Both gases have the same average kinetic energy.
Explanation :
As we are given that there are 8 moles of 'He' and 'Xe'.
At STP condition,
The temperature and pressure are 273 K and 1 atm respectively.
The formula of average kinetic energy is, [tex]K.E=\frac{3}{2}RT[/tex]. From this we conclude that the average kinetic energy is depends on the temperature only. So, at same temperature the average kinetic energy will also be same for both the gases.
As we know that the molecular speed is inversely proportional to the square root of the molecular mass. That means, it depends on the molar mass of substance. So, both the gases have the different molecular speed.
As we know that the density is directly proportional to the mass of substance. That means, it depends on the mass of substance. So, both the gases have the different density.
At STP,
As, 1 mole of He gas contains 22.4 liter volume of He gas
So, 8 mole of He gas contains [tex]8\times 22.4=179.2[/tex] liter volume of He gas
As, 1 mole of Xe gas contains 22.4 liter volume of Xe gas
So, 8 mole of Xe gas contains [tex]8\times 22.4=179.2[/tex] liter volume of Xe gas
The mixture of has volume = 179.2 + 179.2 = 358.4 L
Hence, from the above we conclude that the correct option is, Both gases have the same average kinetic energy.
A rigid sphere is falling from a height of 1m in a fluid of density 1000 m3/kg. The time it took to fall is 500 seconds. If the density of the sphere is 1100 m3/kg and the viscosity of the fluid is 100 cP, what is the diameter of the sphere? (a. 6.1 m , b. 0.61 m , c. 0.061 m , d. 0.0061 m)
Answers
Answer:
0.00191 meter is the diameter of the sphere.
Explanation:
Velocity of the rigid sphere = [tex]v=\frac{1m}{500 s}[/tex]
Density of the fluid = [tex]\rho _f=1000 kg/m^3[/tex]
Density of the particle= [tex]\rho _p=1100 kg/m^3[/tex]
Viscosity of the fluid =[tex]\eta =100 centiPoise = 1 Poise=0.1 Kg/(m s)[/tex]
Radius of the sphere = r
[tex]v=\frac{2}{9}\times \frac{\rho _p-\rho _f}{\eta }gr^2[/tex]
(Terminal velocity)
[tex]\frac{1 m}{500 s}=\frac{2}{9}\times \frac{1100 kg/m^3-1000 kg/m^3}{0.1 kg/(m s)}\times 9.8 m/s^2\times r^2[/tex]
r = 0.00191 m
0.00191 meter is the diameter of the sphere.
Ethylene glycol (C2H4(OH)2) , when dissolved in water, provides the standard ‘anti-freeze’ coolant for water-cooled engines. In order to depress the freezing point of water by 20 °C, how many grams of ethylene glycol would need to be dissolved in 15 kg of pure water? (The molal freezing point depression constant for water Kf = 1.86 K mol-1 kg and the relevant atomic masses are: C = 12g, H = 1g and O = 16g.) Note: ethylene glycol is an organic compound and does NOT break up or dissociate when it dissolves in water. Hint: first calculate the molality (m) of the ethylene glycol solution.
What is the boiling point of this same solution at atmospheric pressure? (The molal boiling point elevation constant for water Kb = 0.52 K mol -1 kg.)
Answers
Answer:
146,575 grams of ethylene glycol would need to be dissolved in 15 kg of pure water
The boiling point of the solution is 181.95°C.
Explanation:
Mass of ethylene glycol = x
Molar mass of ethylene glycol = 2 × 12 g/mol + 2 × 32 g/mol+ 4 × 1= 62g/mol
Moles of ethylene glycol =[tex]\frac{x}{62 g/mol}[/tex]
Mass of solvent that is water = 15 kg
The molal freezing point depression constant for water [tex]K_f=1.86 K kg/mol[/tex]
Molality of the solution:
[tex]Molality=\frac{\text{Moles of compound}}{\text{Mass of solvent (kg)}}[/tex]
[tex]Molality=m=\frac{x}{62 g/mol\times 15 kg}[/tex]
Depression in freezing point of water =[tex]\Delta T_f=20^oC=293.15 K[/tex]
((T)°C =T+ 273.15 K)
[tex]\Delta T_f=K_f\times m[/tex]
[tex]293.15 K=1.86 K kg/mol\times \frac{x}{62 g/mol\times 15 kg}[/tex]
x = 146,575 g
Boiling point of this solution =[tex]T_b[/tex]
The molal boiling point elevation constant for water[tex]K_b = 0.52 K kg/mol[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]\Delta T_b=0.52 K kg/mol\times \frac{146,575 g}{62 g/mol\times 15 kg}[/tex]
[tex]\Delta T_b=81.95K[/tex]
Normal boiling point of water is = T = 373.15 K
[tex]\Delta T_b=T_b-T[/tex]
[tex]T_b=\Delta T_b+T=81.95K+373.15 K=455.1 K=181.95^oC[/tex]
The boiling point of the solution is 181.95°C.
Be sure to answer all parts. There are three different dichloroethylenes (molecular formula C2H2Cl2), which we can designate X, Y, and Z. Compound X has no dipole moment, but compound Z does. Compounds X and Z each combine with hydrogen to give the same product: C2H2Cl2(X or Z) + H2 → ClCH2―CH2Cl What are the structures of X, Y, and Z? Be sure to include lone pair electrons
Answers
Answer:
Explanation:Since the compound X has no net-dipole moment so we can ascertain that this compound is not associated with any polarity.
hence the compound must be overall non-polar. The net dipole moment of compound is zero means that the vector sum of individual dipoles are zero and hence the two individual bond dipoles associated with C-Cl bond must be oriented in the opposite directions with respect to each other.]
So we can propose that compound X must be trans alkene as only in trans compounds the individual bond dipoles cancel each other.
If one isomer of the alkene is trans then the other two isomers may be cis .
Since the two alkenes give the same molecular formula on hydrogenation which means they are quite similar and only slightly different.
The two possibility of cis structures are possible:
in the first way it is possible the one carbon has two chlorine substituents and the carbon has two hydrogens.
Or the other way could be that two chlorine atoms are present on the two carbon atoms in cis manner that is on the same side and two hydrogens are also present on the different carbon atoms in the same manner.
Kindly refer the attachments for the structure of compounds:
Compound X is a trans-form of dichloroethylene (ClCH=CHCl), compound Z, which has a dipole moment, is a cis-form of dichloroethylene (ClC=CH2) and, compound Y is ClCH=CH2.
Explanation:The structures of the three different dichloroethylenes designated as X, Y, and Z can be understood based on their dipole moments and their reactivity with hydrogen. Compound X, which has no dipole moment, must have a symmetrical structure with the chlorines on opposite sides of the double bond, resulting in a trans conformer. This could be represented as ClCH=CHCl. Since Compounds X and Z each combine with hydrogen to give the same product (ClCH2―CH2Cl), Z must also possess this structure; however, it has a dipole moment. Therefore, Z has the chlorines on the same side of the double bond, resulting in a cis conformer. This structure can be represented as ClC=CH2. The remaining isomer, Y, must therefore have one chlorine atom at each carbon, represented as ClCH=CH2.
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Which of the following is an acceptable IUPAC name?(A) 3,6-dimethylheptane (B) 2-ethyl-3-methylheptane (C) 5-methyl-3-ethylheptane (D) 4-ethyl-4-methylheptaneD
Answers
The acceptable IUPAC name among the given options is 3,6-dimethylheptane, as it correctly follows the IUPAC rules for nomenclature where similar substituents are denoted with prefixes like 'di-', 'tri-' etc.
Explanation:The correct answer is option (A) 3,6-dimethylheptane. The IUPAC rules for nomenclature of organic compounds mandate that when there are multiple substituents of the same type, as in this case where we have two methyl groups, the carbon atoms to which they are attached are indicated by commas between the numbers and preceded by prefixes such as 'di-', 'tri-' etc. Thus, 3,6-dimethylheptane follows these rules, making it an acceptable IUPAC name. The other options do not follow the IUPAC rules for nomenclature.
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According to IUPAC nomenclature rules for organic compounds, option (B) 2-ethyl-3-methylheptane is the correct name. This indicates a heptane chain with an ethyl group on the second carbon and a methyl group on the third carbon.
Explanation:The question pertains to the correct naming of organic compounds according to the system laid out by the International Union of Pure and Applied Chemistry (IUPAC). According to IUPAC nomenclature, the longest carbon chain in the molecule is identified first, and is named using Greek prefixes indicating the number of carbon atoms. All substituents (any atoms or groups of atoms) that are attached to the longest chain are then identified and designated with prefixes and the position on the carbon chain.
After examining the given options, Option (B) 2-ethyl-3-methylheptane is the correct IUPAC name. This indicates that the longest carbon chain in the molecule is heptane (7 carbons), with an ethyl group (2 carbons) attached to the second carbon and a methyl group (1 carbon) attached to the third carbon. The other choices (3,6-dimethylheptane; 5-methyl-3-ethylheptane; and 4-ethyl-4-methylheptane) do not follow the IUPAC rules of nomenclature appropriately, either because they involve numbering from the wrong end of the carbon chain or because identical groups on the same carbon are not correctly identified.
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A 55.0 L steel tank at 20.0 ∘C contains acetylene gas, C2H2, at a pressure of 1.39 atm. Assuming ideal behavior, how many grams of acetylene are in the tank?
Answers
Answer:
PV=nRT
n = PV/RT
n = m/Mm
m/Mm = PV/RT
m = MmPV/RT
T in kelvin = T Celsius + 273.15 = 293.15 K
m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)
mass in grams = 82.8 grams
Explanation:
Ideal gases formula is PV=nRT, where:
P is the pressure (1.39 atm in this case)
V is the volume (55.0 L in this case)
R is the gas constant (0.08206 L.atm/K.mole)
T is the temperature (20.0C) should be converted to Kelvin
all the unit should correspond to the one in the R.
we also know that to find the mass, we can use number mole with the formula number of mole(n) = mass (m) divided by the molar mass (Mm). therefore we substituted that in the formula and make (m) the subject of the formula.
we found the mass to be 82.8 grams
PV=nRT is the formula for ideal gases, where P is the pressure and R is the constant and T is the temperature (1.39 atm in this case)
The letter V stands for volume (55.0 L in this case)(0.08206 L.atm/K.mole) is the gas constant.T stands for temperature (20.0C) All units should be changed to Kelvin and correspond to the one in the R.We also know that we can apply the formula number of mole(n) = mass (m) divided by the molar mass to obtain the mass (Mm). As a result, we inserted that in the formula and made (m) the subject.[tex]m = MmPV/RT\\m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)[/tex]
Thus, the mass was discovered to be 82.8 grams.
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What is the IUPAC name for the following compound? Question 10 options: 1-methyl-3-cyclohexene methylcyclohexene 5-methylcyclohexene 1-methyl-4-cyclohexene 4-methylcyclohexene
Answers
Answer:
4-methylcyclohexene (the last choice)Explanation:
The compound is an alkene (cycloalkene), with a methyl group as substituent (it substitutes one hydrogen in the carbon chain).
The IUPAC's rules state that the location of the carbon-carbon double bond in the structure is indicated by specifying the number of the carbon atom at which the C=C bond starts, assigning the lowest possible number to the double bond: this in this case is number 1.
Cyclohexene is the main chain and mehtyl is a substituent, as already said.
So, the name 1-methyl-3-cychlohexene is not valid (position 1 is for the carbon-carbon double bond).
The name methycyclohexene is not valid because it is not telling the position of methylgroup.
The name 5-methylcyclohexene is not valid because the position five should be named 2 in the cyclohexene (you must use the smallest number), so the name should be 2-methyl... instead of 5 methyl...
1-methyl-4cyclohexene is not valid because, as said, the position 1 is reserved for the carbon-carbon double bond.
Only 4-methylcyclohexene is a valid name.
The file attached shows the structure. I have added numbers on the carbons of the main chain to show you how that the methyl group is in the positiion number 4.
.Entropy is a measure of_______
Answers
Answer:
Disorder
Explanation:
Entropy is a measure of disorder.
Entropy is a measure of disorder.
How does heat transfer takes place between any two systems?
Answers
Explanation:
For transfer of heat between any two substances or matter can be occurred by conduction or convection process.
Conduction is a process in which two objects of different temperature are placed in contact with each other. Therefore, heat transfers from hotter object to colder object until the temperature of both objects becomes equal.
For example, a metal spoon placed in a hot cup of tea.
Convection is a process in which a fluid generally liquid or gas is heated and denser material (colder) sinks at the bottom whereas less denser material (hotter) rises at the top. This causes conventional currents in the fluid.
For example, heating rice in a pot full of water.
Consider the neutralization reaction 2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq) A 0.105 L sample of an unknown HNO3 solution required 29.1 mL of 0.200 M Ba(OH)2 for complete neutralization. What is the concentration of the HNO3 solution?
Answers
Answer: 0.11 M
Explanation:
The equation for the reaction is given as:
[tex]2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O[/tex]
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of [tex]HNO_3[/tex] solution = ?
[tex]V_1[/tex] = volume of [tex]HNO_3[/tex] solution = 0.105 L= 105 ml
[tex]M_2[/tex] = molarity of [tex]Ba(OH)_2[/tex] solution = 0.2 M
[tex]V_2[/tex] = volume of [tex]Ba(OH)_2[/tex] solution = 29.1 ml
[tex]n_1[/tex] = valency of [tex]HNO_3[/tex] = 1
[tex]n_2[/tex] = valency of [tex]Ba(OH)_2[/tex] = 2
[tex]1\times M_1\times 105=2\times 0.2\times 29.1[/tex]
[tex]M_1=0.11M[/tex]
Therefore, the concentration of [tex]HNO_3[/tex] will be 0.11 M.
a good sample of benzoic acid melts at 121-122. however, a student had a sample of benzoic acid that melted over a range, 105-115. what could the student concllude about his sample?
Answers
Answer:
The student conclude that the sample of benzoic acid is impure.
Explanation:
The observed melting point of benzoic acid when a student melts his/her sample is low than the actual value. The reason for this might be:
(a) The most probable reason is that the sample is impure. Impurities in the sample leads to lowering the value of the melting point. The reason for the phenomenon is that when impurity is present in the compound, the pattern of the crystal lattice disturbs and thus it less amount of heat is require to break the lattice.
(b) There may be some experimental errors like:
Non-uniform heating of the sampleThe sample is not tightly filled in the capillary if the student is following Kjeldahl's flask method.If the student is using melting point machine, there there might be some instrumental errors.The student conclude that the sample is impure.
Final answer:
The student's benzoic acid sample is likely impure, as indicated by a melting point range significantly lower than the expected 121-122 degrees Celsius for a pure sample.
Explanation:
If a good sample of benzoic acid typically melts at 121-122 degrees Celsius, but a student's sample melted over a range of 105-115 degrees Celsius, it could indicate that their benzoic acid sample is impure. The presence of impurities in a compound will generally lower the melting point and cause the substance to melt over a broader temperature range. This can be seen in a MelTemp apparatus where samples with varying amounts of impurities, such as those containing acetanilide, melt at lower temperatures and have a wider melting range compared to pure benzoic acid.
A process has a AH of +22 kJ and a AS of -13 J/K. At which temperatures is the process spontaneous?
Answers
Answer:
At 1692.31 K.
Explanation:
The Gibbs free energy is a measurement of the energy available for the system to realize a change. And can be calculated by the relation of enthalpy, temperature and entropy by the Gibbs equation:
[tex]\Delta G = \Delta H + T\Delta S[/tex]
A procces will be spontaneous if the Gibbs free energy of the system is negative.
So, if we set the ΔG of the system as -.1 J to make it so that the proccess will be spontanteous and input the given values of entropy and enthalpy into the Gibbs equation, we get:
[tex]-.01 J = 22,000 J - T(-13 J/K)[/tex]
[tex]T = 1692.3 K[/tex]
A sample of gas in a cylinder as in the example in Part A has an initial volume of 52.0 L , and you have determined that it contains 1.40 moles of gas. The next day you notice that some of the gas has leaked out. The pressure and temperature remain the same, but the volume has changed to 13.0 L . How many moles of gas (n2) remain in the cylinder?
Answers
Answer: 0.35 moles
Explanation:
Avogadro's Law: This law states that volume is directly proportional to the number of moles of the gas at constant pressure and temperature.
[tex]V\propto n[/tex] (At constant temperature and pressure)
[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas = 52.0 L
[tex]V_2[/tex] = final volume of gas = 13.0 L
[tex]n_1[/tex] = initial number of moles = 1.40
[tex]n_2[/tex] = final number of moles = ?
Now put all the given values in the above equation, we get the final pressure of gas.
[tex]\frac{52.0}{1.40}=\frac{13.0}{n_2}[/tex]
[tex]n_2=0.35moles[/tex]
Therefore, the moles of the gas that will remain is 0.35 moles.
Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-chlorobutane. Based on this information, the major organic product(s) of this reaction are expected to be _____.
Answers
Answer:but-1-ene
Explanation:This is an E2 elimination reaction .
Kindly refer the attachment for complete reaction and products.
Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.
Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .
As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.
Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.
2-butene is more thermodynamically6 stable as compared to 1-butene
The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.
The reaction between sodium tert-butoxide and 2-chlorobutane is expected to yield 2-butene and 1-butene as the major organic products, with 2-butene being more predominant due to Hofmann Elimination with the bulky base.
When sodium tert-butoxide (NaOC(CH₃)₃) reacts with 2-chlorobutane, it acts as a bulky Bronsted-Lowry base. Due to its steric bulk, it tends to remove a hydrogen from the less hindered (more accessible) carbon adjacent to the carbon bearing the chlorine, which in this case is the carbon at the 3-position. As a result, the major organic product(s) of this reaction are expected to be alkenes, specifically 2-butene and, to a lesser extent, 1-butene due to the Hofmann Elimination favoring the less substituted alkene when considering a bulky base like sodium tert-butoxide.
Decide whether a chemical reaction happens in either of the following situations. If a reaction does happen, write the chemical equation for it. Be sure your chemical equation is balanced and has physical state symbols. situation chemical reaction? chemical equation A strip of solid lead metal is put into a beaker of 0.043M Fe(NO3)2 solution. yes no A strip of solid iron metal is put into a beaker of 0.044M Pb(NO3)2 solution.
Answers
Answer: The chemical equations are given below.
Explanation:
A balanced chemical reaction follows law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.
Single displacement reaction is defined as the reaction in which more reactive metal displaces a less reactive metal from its chemical reaction.
General equation representing single displacement reaction follows:
[tex]AB+C\rightarrow CB+A[/tex]
C is more reactive element than element A.
The reactivity of metals is judged by the series known as reactivity series. Elements lying above in the series are more reactive than the elements lying below in the series.
1. When solid lead metal is put in the beaker of 0.043 M [tex]Fe(NO_3)_2[/tex] solution.Lead lies below in the reactivity series than iron. Thus, it will not replace iron from its chemical reaction.
[tex]Pb(s)+Fe(NO_3)_2(aq.)\rightarrow \text{No reaction}[/tex]
2. When solid iron metal is put in the beaker of 0.044 M [tex]Pb(NO_3)_2[/tex] solution.Iron lies above in the reactivity series than lead. Thus, it will easily replace lead from its chemical reaction.
[tex]Fe(s)+Pb(NO_3)_2(aq.)\rightarrow Fe(NO_3)_2(aq.)+Pb(s)[/tex]
Hence, the chemical equations are given above.
In the first scenario, no reaction occurs when solid lead is placed into a solution of Fe(NO₃)₂ because lead is less reactive than iron. In the second scenario, a reaction occurs when iron is placed into a solution of Pb(NO₃)₂ because iron is more reactive than lead. The resulting balanced chemical equation is: Fe(s) + Pb(NO₃)₂(aq) -> Fe(NO₃)₂(aq) + Pb(s).
Explanation:In chemistry, a chemical reaction occurs when two or more atoms bond together to form molecules or when bonded atoms break apart. We refer to the substances used in the beginning of a reaction as the reactants and the substances found at the end of the reaction as the products.
The first scenario involves a strip of solid lead (Pb) placed into a solution of Fe(NO₃)₂. No reaction happens in this case because lead is less reactive than iron. In the second scenario, a strip of iron (Fe) is placed into a solution of Pb(NO₃)₂. A reaction occurs because iron is more reactive than lead. The balanced chemical equation for this reaction is: Fe(s) + Pb(NO₃)₂(aq) -> Fe(NO₃)₂(aq) + Pb(s).
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Chlorination of pentane gives a mixture of isomers having the molecular formula C5H11Cl. The percentage of 1-chloropentane is 22%. Assuming the secondary hydrogens in pentane are equally reactive to monochlorination, what is the percentage of 3-chloropentane in the mixture?
Answers
Hey there!:
Pentane is a 5 member hydrocarbon
1 chloropentane can be on either side and will still have thesame name because of the IUPAC rules .
To determine the relative amounts of products obtained from radical chlorination of an alkane, both probability (the number of hydrogens that can be abstracted that will lead to the formation of the particular product) and reactivity (the relative rate at which a particular hydrogen is abstracted) must be taken into account.
The precise ratios differ at different temperatures.
For 3- chloro pentane , There are two hydrogen only and the reactivity is same for all 2° hydrogen
The 2° hydrogen % should = 100-22 = 78% ( all remaining are 2° hydrogen only)
since all are equally recative 2° hydrogen , proportionate distribution will be there.
% Yield = {( Total hydrogen which can give 3 choloropentane) / (all hydrogen of 2° ) } * 78
= (2/6)* 78 = 26%
Hope this helps!
In the chlorination of pentane, if 1-chloropentane is 22% of the mixture and secondary hydrogens are equally reactive as in monochlorination, the percentage of 3-chloropentane in the mixture is approximately 15%. The percentages are derived taking into consideration that the ratio of monochlorination among the different positions in pentane (primary, secondary, tertiary) is 3:2:0.
Explanation:In organic chemistry, different isomers of a compound can be formed via a process of substitution like chlorination. In the chlorination of pentane, we have a mixture of isomers, including 1-chloropentane and 3-chloropentane, among others. The question stated that the given percentage of 1-chloropentane is 22%. If the reactivity of secondary hydrogens in pentane is equal to that of monochlorination, then it implies the formation of 2-chloropentane and 3-chloropentane have the same ratio because both have their Cl bound to a secondary carbon. Thus, the distribution of monochlorination among 3 positions (primary, secondary, and tertiary) is in a ratio 3:2:0 for 1,2 and 3-chloropentane respectively.
To derive the individual rates of formation, we'd have to divide each by the total, e.g., for 1-chloropentane (primary position), it would be 3/5 = 0.60 or 60%. However, the question gives us a value of 22% for the 1-chloropentane, which suggests our chlorination is only 22/60 = 0.37 or 37% effective. Applying this scale factor to 2,3 chloropentane (secondary position), we get 0.37*40 = 14.8%. Therefore, 3-chloropentane will constitute around 15% of the mixture.
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