Answer:
glycerol-3-phosphate, ADP, H⁺
Explanation:
The reaction of converting glycerol to glycerol-3-phosphate which makes is unfavorable and is coupled with the second reaction which involves conversion of ATP to ADP which is high energetically favorable.
Reaction 1: Glycerol + HPO₄²⁻ ⇒ Glycerol-3-phosphate + water
Reaction 2: ATP + H₂O ⇒ ADP + HPO₄²⁻ + H⁺
The coupled reaction of both the reactions become favorable. Thus, the overall coupled reaction is:
Glycerol + ATP ⇒ Glycerol-3-phosphate + ADP + H⁺
The net products are = glycerol-3-phosphate, ADP, H⁺
The net products of the coupled reactions as described are; glycerol-3-phosphate, ADP and H+.
According to the reaction;
According to the reaction;We are required to determine the net products of the coupled reactions above.The reactions given is as follows;
glycerol + (HPO4)²− + ATP+H2O ⟶⟶ glycerol-3-phosphate+H2O ADP+(HPO4)²- + H+
Since, the (HPO4)²- cancels out on both sides of the equation; the net products of the coupled reactions above are; glycerol-3-phosphate, ADP and H+.
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An adiabatic Turbine has R-134a as the working fluid. The inlet has a pressure of 200 kPa and a temperature of 20°C while the outlet temperature is -10°C. If the mass flow rate is 0.5 kg/s and the rate of entropy generation is 0.005 kW/K, determine the power output in kW
Hey there!:
Detailed solution is shown below ask if any doubt :
A futile cycle is ________. A futile cycle is ________. Two opposing cellular reactions that are independently controlled Two reactions or pathways that share substrates and products, and result in no net gain of ATP The citric acid cycle An endergonic reaction that uses up ATP
Answer:
The suitable answer to the given blank is
Two reactions or pathways that share the substrate or products and result in no net gain of ATP.
Explanation:
A futile cycle can be defined as those cycles which are involved in metabolism at cellular level to control or regulate biochemical pathways.
It is also known as substrate cycle and is when two metabolic pathways follow directions opposite to each other so that the overall effect is zero other than to dissipate energy.
Therefore, the result is zero net gain.
A futile cycle refers to simultaneous opposing cellular reactions resulting in no net gain of ATP. The citric acid cycle, in contrast, is a crucial pathway in cellular respiration that effectively produces energy for the cell.
Explanation:A futile cycle refers to a process where two opposing cellular reactions or metabolic pathways that share substrates and products occur simultaneously, resulting in no net gain of ATP or any other relevant output. It involves a cycle of reactions that essentially cancel each other out and can be a waste of energy if not properly regulated within the cell.
In contrast, the citric acid cycle (also known as the Krebs cycle or tricarboxylic acid cycle), is a crucial step in the cellular respiration process where carbohydrates, proteins, and fats are oxidized to produce energy. This cycle is an efficient process that leads to the removal of high-energy electrons and carbon dioxide, with the high-energy electrons then used to generate ATP (adenosine triphosphate), the cell's main energy source. Unlike a futile cycle, the citric acid cycle is not a wasteful process as it generates energy that the cell can use.
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At 85°C, the vapor pressure of A is 566 torr and that of B is 250 torr. Calculate the composition of a mixture of A and B that boils at 85°C when the pressure is 0.60 atm. Also, calculate the composition of the vapor mixture. Assume ideal behavior.
Answer:
Composition of the mixture:
[tex]x_{A} =0.652=65.2[/tex] %
[tex]x_{B} =0.348=34.8[/tex] %
Composition of the vapor mixture:
[tex]y_{A} =0.809=80.9[/tex]%
[tex]y_{B} =0.191=19.1[/tex]%
Explanation:
If the ideal solution model is assumed, and the vapor phase is modeled as an ideal gas, the vapor pressure of a binary mixture with [tex]x_{A}[/tex] and [tex]x_{B}[/tex] molar fractions can be calculated as:
[tex]P_{vap}=x_{A}P_{A}+x_{B}P_{B}[/tex]
Where [tex]P_{A}[/tex] and [tex]P_{B}[/tex] are the vapor pressures of the pure compounds. A substance boils when its vapor pressure is equal to the pressure under it is; so it boils when [tex]P_{vap}=P[/tex]. When the pressure is 0.60 atm, the vapor pressure has to be the same if the mixture is boiling, so:
[tex]0.60*760=P_{vap}=x_{A}P_{A}+x_{B}P_{B}\\456=x_{A}P_{A}+(1-x_{A})P_{B}\\456=x_{A}*(P_{A}-P_{B})+P_{B}\\\frac{456-P_{B}}{P_{A}-P_{B}}=x_{A}\\\\\frac{456-250}{566-250}=x_{A}=0.652[/tex]
With the same assumptions, the vapor mixture may obey to the equation:
[tex]x_{A}P_{A}=y_{A}P[/tex], where P is the total pressure and y is the fraction in the vapor phase, so:
[tex]y_{A} =\frac{x_{A}P_{A}}{P}=\frac{0.652*566}{456} =0.809=80.9[/tex] %
The fractions of B can be calculated according to the fact that the sum of the molar fractions is equal to 1.
Consider the following reaction at equilibrium. What effect will adding more SO3 have on the system? SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) A) The reaction will shift in the direction of products. B) The reaction will shift to decrease the pressure. C) No change will occur since SO3 is not included in the equilibrium expression. D) The reaction will shift in the direction of reactants. E) The equilibrium constant will decrease.
Answer:
D) "The reaction will shift in the direction of the reactants."
Explanation:
Consider the Le Chatelier's Principle. When there's a change in the conditions of an equilibrium, the equilibrium will shift in a way that minimizes the impact of those changes.
In this question, the change is an increase in the partial pressure of [tex]\mathrm{SO_3\;(g)}[/tex] (analog to concentration in a solution.) [tex]\mathrm{SO_3\;(g)}[/tex] is a product of the forward reaction and is consumed in the reverse reaction. Shifting the equilibrium towards the reactants will consume some of the additional [tex]\mathrm{SO_3\;(g)}[/tex] and reduce its partial pressure.
Alternatively, think about equilibrium as a balance between the forward and the backward reaction. When the system is at equilibrium, the two reaction rates are equal, so overall the composition will stay the same. However, when more of the product [tex]\mathrm{SO_3\;(g)}[/tex] is suddenly added to the system, the rate of the reverse reaction will jump upwards while the rate of the forward reaction will only gradually increase. Before the system reach a new equilibrium position, the reverse reaction will prevail and shift the equilibrium towards the reactants.
Either way, adding the product [tex]\mathrm{SO_3\;(g)}[/tex] to the system will shift the equilibrium towards the reactants. The Le Chatelier's principle might be easier to memorize. However, keep in mind that the Le Chatelier's principle is only a generalization of the observations; only the second explanation describes what's actually going on in the equilibrium.
Choice B) is not what will happen since there's an equal number of gas particles on both sides of this reaction. If all four gases behave like ideal gases, the shift in equilibrium position will not change the pressure if temperature stays the same.
As a side note on choice E), for a certain reaction, the equilibrium constant depends only on the temperature. In other words, adding or removing a reactant or a product will not change the equilibrium constant.
Final answer:
Adding more SO3 to the system SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) will cause the reaction to shift towards the reactants in accordance with Le Chatelier's principle. Therefore, the correct answer is D) The reaction will shift in the direction of reactants.
Explanation:
The student's question involves applying Le Chatelier's principle to a chemical equilibrium system. When additional SO3 is added to the equilibrium system SO2(g) + NO2(g) ⇌ SO3(g) + NO(g), this represents an increase in the concentration of a product.
According to Le Chatelier's principle, the system will counteract this change by shifting the equilibrium toward the reactants to reduce the added SO3. Therefore, the correct answer is D) The reaction will shift in the direction of reactants.
Use the following table to answer the question: Element Mass Number Atomic Number Aluminum 27 13 Nitrogen 14 7 Helium 4 2 Fluorine 19 9 Which element has the most neutrons in the nucleus? Aluminum Nitrogen Helium Fluorine
Answer:
Aluminium have the most neutrons in the nucleus.
Explanation:
The number of neutrons in the nucleus can be calculated by subtraction of the atomic number from the mass number.
neutrons in aluminium nucleus = 27 - 13 = 14
neutrons in nitrogen nucleus = 14 - 7 = 7
neutrons in helium nucleus = 4 - 2 = 2
neutrons in fluorine nucleus = 19 - 9 = 10
Answer:
aluminum has the most atomic number
Explanation:
the atomic number is equal to the protons and neutrons
the protons in aluminum in this case is 13, so the neutrons are 14.
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How much heat will be absorbed by a 55.5 g piece of aluminum (specific heat 0.930 J/g.°C) as it changes temperature from 23.0°C to 67.0°C? kJ
Answer: The amount of heat absorbed by aluminium is 2.271 kJ.
Explanation:
To calculate the amount of heat absorbed or released, we use the equation:
[tex]Q= m\times c\times \Delta T[/tex]
Q = heat absorbed = ?
m = mass of aluminium = 55.5 g
c = specific heat capacity of aluminium = 0.930 J/g ° C
Putting values in above equation, we get:
[tex]\Delta T={\text{Change in temperature}}=(67-23)^oC=44^oC[/tex]
[tex]Q=55.5g\times 0.930J/g^oC\times 44^oC[/tex]
Q = 2271.06 Joules
Converting this into kilo joules, we use the conversion factor:
1 kJ = 1000 J
So, 2271.06 J will be equal to 2.271 kJ
Hence, the amount of heat absorbed by aluminium is 2.271 kJ.
Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines. The decomposition of nitric oxide (NO) to N2 and O2 is second order with a rate constant of 0.0796 M−1⋅s−1 at 737∘C and 0.0815 M−1⋅s−1 at 947∘C. Calculate the activation energy for the reaction in kJ/mol
Answer : The activation energy for the reaction is, 1.151 KJ
Explanation :
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = rate constant at [tex]737^oC[/tex] = [tex]0.0796M^{-1}s^{-1}[/tex]
[tex]K_2[/tex] = rate constant at [tex]947^oC[/tex] = [tex]0.0815M^{-1}s^{-1}[/tex]
[tex]Ea[/tex] = activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = [tex]737^oC=273+737=1010K[/tex]
[tex]T_2[/tex] = final temperature = [tex]947^oC=273+947=1220K[/tex]
Now put all the given values in this formula, we get:
[tex]\log (\frac{0.0815M^{-1}s^{-1}}{0.0796M^{-1}s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{1010K}-\frac{1}{1220K}][/tex]
[tex]Ea=1151.072J/mole=1.151KJ[/tex]
Therefore, the activation energy for the reaction is, 1.151 KJ
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
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.
Write the balanced equation for the reaction of aluminum with copper(II) sulfate solution.
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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PH of Substances in Chemistry: In chemistry, the of a substance, a measure of acidity of a substance, is defined as PH-log[H0 ], where H30+] is the hydrogen ion concentration (which characterizes an acid solution), in moles per liter. a. Find the PH of each substance: [H30+ 3.2 10-5 b. Find the hydrogen ion concentration of each substance, given the PH-
Answer: A. PH=4.50
Explanation:
As we know that pH=-log(H3O+)
pH=-log(3.2x10^-5)
PH= -log(3.2)+log(10^-5)
=-(0.50+(-5))
PH= -(0.5-5)=-(-4.5)
pH=4.50
B. If we know pH , then [H3O+]= 10^-pH
To find the pH of a substance, use pH = -log[H3O+]. Given [H3O+] = 3.2 × 10-5, the pH ≈ 4.49. To find hydrogen ion concentration from pH, use [H3O+] = 10^-pH.
pH is a measure of the acidity or alkalinity of a solution, defined as the negative logarithm of the hydrogen ion concentration (expressed in molarity) of a solution.
To find the pH of a substance, you can use the formula: pH = -log[H3O+].
a. Given [H3O+] = 3.2 × 10-5, the pH is calculated as follows:
pH = -log(3.2 × 10-5)pH ≈ 4.49b. To find the hydrogen ion concentration given a pH value, you rearrange the formula as: [H3O+] = 10-pH.
For example, if the pH of a solution is 3:
[H3O+] = 10-3[H3O+] = 0.001 MTherefore, solutions with a pH of less than seven are acidic, while those with a pH greater than seven are basic (alkaline). The pH scale ranges from 0 to 14 and a pH of 7 is considered neutral.
Write the balanced chemical equation for the complete, stoichiometric combustion of ethylene in (a) nitrous oxide and (b) air. Compare the required number of moles and the oxidizer mass using each of the two oxidizers for the complete, stoichiometric combustion of ethylene.
Answer: Mass of nitrous oxide used is 264 grams and mass of oxygen gas used is 96 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
For a: Combustion of ethylene in nitrous oxideThe chemical equation for the combustion of ethylene in nitrous oxide follows the equation:
[tex]C_2H_4+N_2O\rightarrow 2CO_2+2H_2O+6N_2[/tex]
By stoichiometry of the reaction;
6 moles of nitrous oxide are required for the complete combustion of ethylene molecule.
Calculating the mass of nitrous oxide using equation 1:
Molar mass of nitrous oxide = 44 g/mol
Moles of nitrous oxide = 6 moles
Putting values in equation 1, we get:
[tex]6mol=\frac{\text{Mass of nitrous oxide}}{44g/mol}\\\\\text{Mass of nitrous oxide}=264g[/tex]
Thus, 264 grams of nitrous oxide are required for the complete combustion of ethylene.
For b: Combustion of ethylene in airThe chemical equation for the combustion of ethylene in air follows the equation:
[tex]C_2H_4+3O_2\rightarrow 2CO_2+2H_2O[/tex]
By stoichiometry of the reaction;
3 moles of oxygen are required for the complete combustion of ethylene molecule.
Calculating the mass of oxygen using equation 1:
Molar mass of oxygen = 32 g/mol
Moles of oxygen = 3 moles
Putting values in equation 1, we get:
[tex]3mol=\frac{\text{Mass of oxygen}}{32g/mol}\\\\\text{Mass of oxygen}=96g[/tex]
Thus, 96 grams of oxygen are required for the complete combustion of ethylene.
Cyclic compound A has molecular formula C5H10 and undergoes monochlorination to yield exactly three different constitutional isomers. Identify compound A and show the monochlorination products.
Answer:
A: 1,2-dimethylcyclopropane
Explanation:
The possible cyclic structure with formula C₅H₁₀ are shown in the image.
A is a cyclic compound. On monochlorination, A yields 3 products.
To have 3 products on monochlorination, there should be three different carbon atoms.
Considering structure 1, all carbons have same nature, thus only one product will be formed and thus not a structure of A.
Considering structure 2, there are two different carbon atoms, thus two different structure are formed and thus not a structure of A.
Considering structures 3 and 4 , there are four different carbon atoms, thus four products will be formed and either of them are not a structure of A.
Considering structure 5, there are three different carbon atoms, thus three different structure are formed and thus the A is structure 5.
The monochlorination products are 1,2-dimethylcyclopropane, 1,3-dichlorocyclopentane, and 1,1-dichlorocyclopentane.
What is Monochlorination?Monochlorination is the process in which substitution of one atom with chlorine atoms occur.
Given,
When [tex]C_5H_1_0[/tex] undergoes chlorination, it yields three different constitutional isomers.
So, the possible molecules are pentene and cyclopentane.
By monochlorination we get only one product that is 1-chlorocyclepentane.
By dichlorination of the compound, we get three products that are 1,2 -dichlorocyclopentane, 1,3-dichlorocyclopentane, and 1,1-dichlorocyclopentane respectively as given in the diagram.
Thus, the products are 1,2 -dichlorocyclopentane, 1,3-dichlorocyclopentane, and 1,1-dichlorocyclopentane.
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According to the following reaction, how many grams of oxygen gas are required for the complete reaction of 30.3 grams of hydrochloric acid? hydrochloric acid (aq) + oxygen (g) water (l) + chlorine (g) grams oxygen gas
Answer: The mass of oxygen gas required will be 6.648 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
Given mass of hydrochloric acid = 30.3 g
Molar mass of hydrochloric acid = 36.46 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of }HCl=\frac{30.3g}{36.46g/mol}=0.831mol[/tex]
For the given chemical reaction, the balanced equation follows:
[tex]4HCl+O_2\rightarrow 2H_2O+2Cl_2[/tex]
By Stoichiometry of the reaction:
4 moles of HCl reacts with 1 mole of oxygen gas.
So, 0.831 moles of HCl will react with = [tex]\frac{1}{4}\times 8.31=0.20775mol[/tex] of oxygen gas.
Now, calculating the mass of oxygen gas by using equation 1, we get:
Moles of oxygen gas = 0.20775 moles
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]0.20775mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=6.648g[/tex]
Hence, the mass of oxygen gas required will be 6.648 grams.
1. What volume of toluene do you need to add to 1 mL of ethyl acetate to make an equi-molar mixture? The density of ethyl acetate (C4H8O2) is 0.898 g/mL and the density of toluene (C7H8) is 0.867 g/mL.
Answer:
1.06 mL of toluene will be needed.
Explanation:
Equi-molar mixture means equal moles of all the components.
as given the volume of ethyl acetate = 1mL
Density of ethyl acetate = 0.898 g/mL
The relation between density, mass and volume is :
[tex]density=\frac{mass}{volume}[/tex]
mass=volumeXdensity
mass of ethyl acetate present = 1mL X 0.898g/mL = 0.598 grams
the moles are related to mass as:
[tex]moles=\frac{mass}{molarmass}[/tex]
For ethyl acetate molar mass = 4X12+8X1+2X16= 88g/mol
moles of ethyl acetate will be:
[tex]moles=\frac{0.898}{88}= 0.01mol[/tex]
So we need 0.01 moles of toluene also
For 0.01 moles the mass of toluene required = 0.01 X molar mass of toluene
mass required = 0.01 X 92=0.92grams
for 0.92 grams of toluene volume required will be:
[tex]volume=\frac{mass}{density}=\frac{0.92}{0.867}= 1.06mL[/tex]
To make an equi-molar mixture, you need to calculate the moles of ethyl acetate in 1 mL using its density and molar mass. Then, using the moles of ethyl acetate, find the volume of toluene needed that has the same number of moles. You'll need 1.081 mL of toluene to make this equi-molar mixture with 1 mL of ethyl acetate.
The question asks about making an equi-molar mixture of two solvents: ethyl acetate and toluene. To answer this question, we need to calculate the moles of ethyl acetate in 1 mL and then calculate the volume of toluene that contains the same number of moles.
First, we calculate the moles of ethyl acetate using its density and molar mass (88.11 g/mol):
Moles of ethyl acetate = (Density of ethyl acetate × volume in mL) / Molar mass of ethyl acetate
= (0.898 g/mL × 1 mL) / 88.11 g/mol
= 0.01019 mol
Next, we use the moles of ethyl acetate to determine the volume of toluene required, using the molar mass of toluene (92.14 g/mol) and its density:
Moles of toluene = Moles of ethyl acetate (since we want an equi-molar mixture)
Volume of toluene = Moles of toluene × Molar mass of toluene / Density of toluene
= 0.01019 mol × 92.14 g/mol / 0.867 g/mL
= 1.081 mL
Therefore, you would need to add 1.081 mL of toluene to 1 mL of ethyl acetate to make an equi-molar mixture.
During heat transfer, thermal energy always moves in the same direction: A. COLDHOT B. HOT COLD C. No movement D. HOTHOT
Answer: Option (B) is the correct answer.
Explanation:
Thermal energy is defined as the energy present within the molecules of a substance.
Also, when two objects that have different temperature and they are in contact with each other then heat will always flow from hot object to cold object.
For example, if a metal spoon is placed in a hot cup of coffee then heat will flow from hot coffee to the metal spoon.
Therefore, we can conclude that during heat transfer, thermal energy always moves in the same direction: HOT COLD.
Final answer:
Thermal energy during heat transfer always moves from a higher-temperature object to a lower-temperature object, as per the second law of thermodynamics.
Explanation:
When considering heat transfer, energy typically moves from a higher-temperature object to a lower-temperature object. This movement is adequately described by the laws of thermodynamics, particularly the second law of thermodynamics, which states that heat transfer flows from a hotter object to a cooler one and that heat energy, in any process, is lost to available work in a cyclical process. For example, in the scenario of heating food in a pot on a stove, the energy in the form of heat is transferred from the hot stove element (higher temperature) to the pot and its contents (lower temperature).
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 _____.
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.
Please explain and rank the order of oxidation of methanol, ethanol, 2-propanol, 2-methyl-2propanol reacting with potassium permanganate.
Answer:Methanol>Ethanol>2-Propanol>2-methyl-2Propanol
Explanation:The mechanism of oxidation using Potassium permanganate
involves two steps :
In the first step permanganate ion abstracts a alpha-hydrogen as hydride ion available at alcohol.
Alpha hydrogen is the hydrogen attached to the carbon bearing functional group.
In the second step the permanganate ion subsequently is reduced from +7 oxidation state of Manganese to +5 oxidation state of Manganese.
So for the oxidation of alcohol using potassium permanganate , there must be availability of alpha hydrogens.
we can also relate the order of oxidation with reference to number of alpha hydrogens present on the substrate.
so
Methanol has 3 available alpha hydrogen
Ethanol has 2 available alpha hydrogen
2-propanol has 1 available alpha hydrogen as it is a secondary alcohol
2-methyl-2propanol has 0 available alpha hydrogens as it is a tertiary alcohol.
Greater the number of available hydrogens easier would be oxidation of alcohols so the order of oxidation would be the following:
Methanol>Ethanol>2-Propanol>2-methyl-2Propanol
please refer the attachment for the structures of following compounds.
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.
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.
For which of the following aqueous solutions would one expect to have the largest van’t Hoff factor (i)? a. 0.050 m NaCl b. 0.500 m K2SO4 c. 0.050 m K2SO4 d. 0.500 m NaCl e. 0.500 m C6H12O6 (glucose)
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
The largest van't Hoff factor would be expected for the 0.500 m K2SO4 solution. This is because the van't Hoff factor depends on both the concentration and the degree of ionization, which are highest in this case.
Explanation:The van't Hoff factor (i) is a measure of the number of actual particles (ions or molecules) in solution after a compound has been dissolved compared to the number of formula units originally dissolved. It is used to predict various colligative properties of solutions such as freezing point depression and boiling point elevation.
In general, substances that do not ionize in solution, such as glucose (C6H12O6), have a van’t Hoff factor of 1. Sodium chloride (NaCl), when fully dissociated in water, yields two ions (Na+ and Cl-) and so has a van’t Hoff factor of 2 under ideal conditions. Potassium sulfate (K2SO4), when fully dissociated, yields three ions (2K+ and SO4--) and therefore has a van't Hoff factor of 3 under ideal conditions.
Given the choices provided, the 0.500 m K2SO4 solution would be expected to have the largest van't Hoff factor because both the concentration and degree of ionization are highest in this case.
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At 298 K, Kc = 1.45 for the following reaction 2 BrCl (g) Br2(g) + Cl2(g) A reaction mixture was prepared with the following initial concentrations. [BrCl] = 0.0400 M, [Br2] = 0.0300 M and [Cl2] = 0.0300 M Calculate their equilibrium concentrations.
Answer:
[BrCl]=0.02934M
[Br2]=[Cl2]=0.03533M
Explanation:
First, consider the Kc definition:
[tex]1.5=K_{c}=\frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}[/tex]
It is necessary to define a new variable, 'x', as the amount of moles of Br2 that are produced. If 'x' moles of Br2 are produced, the moles of the compounds will be calculated as;
[tex]n_{Br_{2}}=n_{{Br_{2}}^{0}} +x\\n_{Cl_{2}}=n_{{Cl_{2}}^{0}} +x\\n_{BrCl}=n_{{BrCl}}^{0}} -2x\\[/tex]
Where the zero superscript means the initial moles.
Dividing the last equations by the volume, which is constant because the reaction does not change the total moles number (2 moles of BrCl produce 2 moles, one of Cl2 and another of Br2), we have the molarity equations for all species:
[tex]M_{Br_{2}}=M_{{Br_{2}}^{0}} +x\\M_{Cl_{2}}=\\M_{BrCl}=M_{{BrCl}}^{0}} -2x\\[/tex]
And now 'x' is a change in molarity.
Replacing these in the Kc equation we have:
[tex]1.45=\frac{({M_{{Br_{2}}}^{0}}+x)({M_{{Cl_{2}}}^{0}}+x)}{({M_{{BrCl}}^{0}}-2x)^{2}}[/tex]
Where the only unknown is 'x'. So, let's solve the equation:
[tex]1.45=\frac{(0.03+x)(0.03+x)}{(0.04-2x)^{2} } \\1.45(0.04-2x)^2=(0.03+x)^2\\1.45(0.0016-0.16x+4x^2)=0.0009+0.06x+x^2\\4.8x^2-0.292x+0.00142=0\\x_{1}=0.0555\\x_{2}=0.00533[/tex]
The result [tex]x_{1}=0.0555[/tex] lacks of sense because it will give a negative concentration for BrCl, so the result is [tex]x_{2}=0.00533[/tex].
Applying the result, the concentrations at equilibrium are:
[tex]M_{Br_{2}}=M_{Cl_{2}}=0.03+0.00533=0.03533M\\M_{BrCl}=0.04-2*0.00533=0.02934M[/tex]
If you calculate Kc with this concentrations it will give 1.45 as a result.
Greets, I will be happy to solve any doubt you have.
To calculate the equilibrium concentrations for a given reaction with initial concentrations provided, use the equilibrium constant Kc to determine the equilibrium concentrations of the reactants and products.
Equilibrium concentrations calculation:
Given initial concentrations: [BrCl] = 0.0400 M, [Br2] = 0.0300 M, and [Cl2] = 0.0300 M, use the equilibrium constant Kc = 1.45 to calculate the equilibrium concentrations of Br2, Cl2, and BrCl in the reaction equation BrCl(g) = Br2(g) + Cl2(g). Work out the concentrations using the stoichiometry of the reaction.
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
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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Formic acid (HCO2H, Ka = 1.8 × 10-4) is the principal component in the venom of stinging ants. What is the molarity of a formic acid solution if 25.00 mL of the formic acid solution requires 39.80 mL of 0.0567 M NaOH to reach the equivalence point?
Answer : The molarity of a formic acid solution is, 0.0903 M
Explanation :
To calculate the molarity of formic acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of formic acid which is [tex]CH_3COOH[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of sodium hydroxide base which is NaOH.
As we are given:
[tex]n_1=1\\M_1=?\\V_1=25.00ml\\n_2=1\\M_2=0.0567M\\V_2=39.80ml[/tex]
Now put all the given values in above equation, we get:
[tex]1\times M_1\times 25.0ml=1\times 0.0567M\times 39.80ml\\\\M_1=0.0903M[/tex]
Hence, the molarity of a formic acid solution is, 0.0903 M
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.
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.
A 10.1 g sample of NaOH is dissolved in 250.0 g of water in a coffee-cup calorimeter. The temperature increases from 23.0 °C to ________°C. Specific heat of liquid water is 4.18 J/g-K and ΔH for the dissolution of sodium hydroxide in water is 44.4 kJ/mol.
Hey there!:
moles of NaOH = 10.1 / 40 = 0.2525
heat = ΔH x moles
= 44.4 x 0.2525
= 11.21 kJ
total mass = 10.1 + 250 = 260.1 g
Q = m Cp dT
11211 = 260.1 x 4.18 x dT
dT = 10.3
T2 = 10.3 + 23 = 33.3 °C
temperature = 33.3 ºC°
Hope this helps!
Answer:
33.3 °C
Explanation:
You have two heat flows in this experiment.
Heat from solution of NaOH + heat to warm water = 0
q1 + q2 = 0
nΔH + mCΔT = 0
Data:
m(NaOH) = 10.1 g
ΔH = -44.4 kJ/mol
m(H2O) = 250.0 g
C = 4.18 J/(K·mol)
Ti = 23.0 °C
Calculation:
n = 10.1 g NaOH × (1 mol NaOH/40.00g NaOH = 0.2525 mol NaOH
q1 = 0.2525 mol × (-44 400 J/mol) = -11 210 J
m(solution) = m(NaOH) + m(water) = 10.1 + 250.0 = 260.1g
q2 = 260.1 × 4.18 × ΔT = 1087ΔT J
-11 210 + 1087ΔT = 0
1087ΔT = 11 210
ΔT = 11 210/108745 = 10.31 °C
ΔT = T2 - T1 = T2 - 23.0 = 10.73
T2 = 23.0 + 10.31 = 33.3 °C
The temperature increases to 33.3 °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
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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You need to produce a buffer solution that has a pH of 5.03. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74.
Explanation:
We have given the acid that is, acetic acid. It is known that acetic acid is a weak acid.
Therefore, formula that depicts relationship between pH, [tex]pK_{a}[/tex] and weak acid is as follows.
pH = [tex]pK_{a}[/tex] + log[tex]\frac{[A^{-}]}{[HA]}[/tex]
where, [HA] = concentration of weak acid
[tex][A^{-}][/tex] = concentration of conjugate base of given weak acid
Since, we have to calculate the concentration of conjugate base of acetic acid. So, let it be equal to x. Whereas concentration of acetic acid is 10 mmol, [tex]pK_{a}[/tex] is 4.74 and pH is 5.03.
Hence, putting these values in the above formula as follows.
pH = [tex]pK_{a}[/tex] + log[tex]\frac{[A^{-}]}{[HA]}[/tex]
5.03 = 4.74 + log[tex]\frac{x}{10}[/tex]
[tex]\frac{x}{10}[/tex] = antilog (0.29)
x = 19.4 mmol
Thus, we can conclude that there is 19.4 mmol acetate (the conjugate base of acetic acid) will be needed to add to this given solution.
To produce a buffer solution with a pH of 5.03 using acetic acid, you need to calculate the concentration of acetic acid and acetate ion. By applying the Henderson-Hasselbalch equation, you can determine the concentration of acetate needed in the buffer solution.
Explanation:To produce a buffer solution with a pH of 5.03, you already have a solution containing 10 mmol of acetic acid. The first step is to calculate the concentration of acetic acid in the initial solution. Using the formula [H3O+] = √(√Ka x [CH3CO2H]), we find that [H3O+] = √(√1.8 x 10^-5 x 0.01) = 0.00113 M.
Next, we need to determine the concentration of acetate required. Since pH changes by 1 unit when the acetic acid concentration is reduced to 11% of the acetate ion concentration, we can calculate the concentration of acetate using the equation 0.11 = (0.00113-x)/(0.01+x), where x is the concentration of acetate.
Solving this equation, we can find x, which represents the concentration of acetate needed in the buffer solution.
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Draw a structure of glycogen with two alpha (1>4) linkages and two alpha (1>6) 3 linkages between glucose molecules What advantage does a branched structure have over a straight chain polysaccharide? b. Give the primary location of glycogen in the body.
Answer:
Here's what I get
Explanation:
a. Structure
The structure of glycogen is shown below.
The bonds between the glucose units in the horizontal chain are α(1⟶4) linkages.
The upper glucose units are linked to the main chain by α(1⟶6) linkages.
Enzymes attack the end units of a polysaccharide. The branched structure of glycogen provides more end units, so enzymes can break it down into glucose more quickly.
b. Storage site
The primary site for the storage if glycogen is the liver.
The secondary site is muscle tissue. The concentration of glycogen in the liver is five times that in muscle, but there is more glycogen in muscle because the body has a much greater muscle mass.
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
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 system absorbs 12 J of heat from the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the internal energy ΔEth of the system?
Answer: The value of change in internal energy of the system is, 40 J.
Explanation : Given,
Heat absorb from the surroundings = 12 J
Work done on the system = 28 J
First law of thermodynamic : It is a law of conservation of energy in which the total mass and the energy of an isolated system remains constant.
As per first law of thermodynamic,
[tex]\Delta U=q+w[/tex]
where,
[tex]\Delta U[/tex] = internal energy = ?
q = heat absorb from the surroundings
w = work done on the system
Now put all the given values in this formula, we get the change in internal energy of the system.
[tex]\Delta U=12J+28J[/tex]
[tex]\Delta U=40J[/tex]
Therefore, the value of change in internal energy of the system is, 40J.
Each of the following substances is a liquid at –50°C. Place these liquids in order of increasing vapor pressure: dimethyl ether (CH3OCH3), propane (C3H8), and ethanol (CH3CH2OH).a) propane < ethanol < dimethyl etherb)ethanol < propane < dimethyl etherc)propane < dimethyl ether < ethanold)ethanol < dimethyl ether < propanee)dimethyl ether < ethanol < propane
Answer:
Increasing vapor pressure order -
propane > dimethyl ether > ethanol
Explanation:
Vapor pressure - the pressure exerted the gaseous molecules , on the walls of the container is called the vapor pressure.
Boiling point - the temperature at which the the vapor pressure of the liquid equals the external atmospheric pressure.
Both boiling point and vapor pressure are linked by the inter molecular forces between the atoms.
The compound with stronger inter molecular forces are tightly held , hence more amount of energy is required to vaporize. Therefore, higher boiling point , and in turn the vapor pressure will be lower .
And the compound with weaker inter molecular forces are loosely held , hence less amount of energy is required to vaporize. Therefore, lower boiling point , and in turn the vapor pressure will be higher .
Therefore,
Ethanol has lowest vapor pressure , because at lower temperature , it has H - bonding , hence, its boiling point is more, therefore, less vapor pressure.
Dimethyl ether has vapor pressure more than ethanol as it has weaker dipole - dipole interactions as compared to strong H - bonding .
Propane has maximum vapor pressure among all , because of lower molecular mass than others , it can easily vaporize into vapors.
Hence,
Increasing vapor pressure order -
propane > dimethyl ether > ethanol
The Increasing vapor pressure order of vapor pressure for the liquids listed in the question is; propane > dimethyl ether > ethanol.
The vapor pressure of a liquid shows how easily the liquid is converted to vapor. Liquids that are easily converted to vapor are said to have a low vapor pressure.
The Increasing vapor pressure order of vapor pressure for the liquids listed in the question is; propane > dimethyl ether > ethanol. This is because, propane has the least intermolecular interaction and ethanol has the highest degree of intermolecular interaction among the molecules in the list.
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