The substituent groups make up the following ether are the substituent groups that make up ethyl phenyl ether are A. ethyl and phenyl, and the substituent groups that make up propyl benzyl ether are B. propyl and benzyl.
An ether is a class of organic compounds that contain an oxygen atom that is bonded to two alkyl or aryl groups. Ethyl phenyl ether is an ether that is formed when phenol is treated with ethyl iodide in the presence of a base such as sodium hydroxide. Ethyl phenyl ether has a sweet, fruity odor and is used as a solvent and a flavoring agent in the food industry.
Propyl benzyl ether, on the other hand, is formed by the reaction of benzyl chloride with n-propyl alcohol. It is used as a solvent for resins, lacquers, and oils. The given ethers are ethyl phenyl ether, propyl benzyl ether, ethyl benzyl ether, and propyl phenyl ether. Among these, the substituent groups that make up ethyl phenyl ether are A. ethyl and phenyl, and the substituent groups that make up propyl benzyl ether are B. propyl and benzyl.
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106 Many drugs decompose in blood by a first-order process. (a) Two tablets of aspirin supply 0.60g of the active compound, After 30 min, this compound reaches a maximum concentration of 2mg/100 mL of blood. If the halftife for its breakdown is 90 min, what is its concentration (in mg/100 mL ) 2.5 h after it reaches its maximum concentration? (b) For the decomposition of an antibictic in a person with a normal temperature (98.6
∘
F),k=3.1×10
−5
s
−1
; for a person with a fever (temperature of 101.9
∘
F),k=3.9×10
−5
s
−1
. If the person with the fever must take another pill when
3
2
of the first pill has decomposed. how many hours should she wait to take a second pill? A third pill? (Assume that the pill is effective immediately.) (c) Calculate E
2
for decomposition of the antibiotic in part (b).
(a). The concentration of the compound after 2.5 h is 0.45mg/100mL.
(b). The person with the fever must wait 7.4 h to take the next pill and 14 h to take the third pill.
(c). The activation energy is 134.5 kJ/mol.
(a). First, let's calculate the rate constant, k. The half-life is given as 90 min.
For a first-order reaction, the half-life is given as:
t1/2 = (0.693/k)
Rearranging, we get:
k = (0.693/t1/2)
Substituting the values:
k = (0.693/90)
k = 7.7×10⁻³/min.
Now, let's use the equation for the first-order rate law for the decomposition of aspirin in blood:
C = C₀e−kt
Where:
C₀ is the concentration of the compound initially.
C is the concentration of the compound at any given time t.
e is the mathematical constant approximately equal to 2.71828.
t is the time elapsed.
Substituting the values:
C = 2mg/100mL
C₀ = 0.60g/(2 tablets×100)
= 0.003g/mL
= 3mg/ mL
Thus:
C = 3×2.71828−7.7×10−3×30
= 1.9mg/100mL
We want to calculate the concentration after 2.5 hours. 2.5 hours is 150 min.
Thus:
C = 2mg/100mL×2.71828−7.7×10⁻³×150
=0.45mg/100mL
(b). We can calculate the rate constant for both normal temperature and fever temperature, k₁ and k₂, respectively:
k₁ = 3.1 × 10⁻⁵ s⁻¹
k₂ = 3.9 × 10⁻⁵ s⁻¹
To find out how long the person with the fever should wait before taking the next pill, we can use the equation for the first-order rate law:
N/N₀ = e−kt
Where:
N₀ is the initial number of molecules.
N is the number of molecules at any given time t.
e is the mathematical constant approximately equal to 2.71828.
t is the time elapsed.
We know that the person with the fever should wait for 32 of the first pill to decompose.
Thus,
N/N₀ = 0.68.
Substituting the values for k₂:
N/N₀ = e−3.9×10−5 tln(0.68)
= −3.9×10−5 tln(1)−ln(0.68)
= −3.9×10−5 t
Thus, t = 7.4 hours.
Similarly, we can find the time for the third pill. If the first pill has decomposed by 64, then N/N₀ = 0.32.
Substituting the values for k₂:
N/N₀ = e−3.9×10⁻⁵ tln(0.32)
=−3.9×10⁻⁵ t
Thus, t = 14 hours
(c). The activation energy can be calculated using the Arrhenius equation:
k = Ae−Ea/RT
Where:
k is the rate constant
A is the pre-exponential factor
Ea is the activation energy
R is the gas constant (8.314 J/K·mol)
T is the temperature in Kelvin.
Rearranging the equation:
lnk = lnA−Ea/RT
Thus, the plot of ln k versus 1/T will be a straight line with slope −Ea/R.
Using the values for k₁ and k₂ from part (b), we can calculate the activation energy.
Ea = (−R)(slope of the plot)
= 8.314×(3.9×10⁻⁵−3.1×10⁻⁵)/(1/311−1/309.7)
= 134.5 kJ/mol (approx)
Thus,
E2−E1 = 134.5−92.3
= 42.2 kJ/mol.
Therefore, the concentration of the compound after 2.5 h is 0.45mg/100mL, the person with the fever must wait 7.4 h to take the next pill and 14 h to take the third pill and the activation energy is 134.5 kJ/mol.
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Complete question is,
106 Many drugs decompose in blood by a first-order process.
(a) Two tablets of aspirin supply 0.60g of the active compound, After 30 min, this compound reaches a maximum concentration of 2mg/100 mL of blood. If the halftime for its breakdown is 90 min, what is its concentration (in mg/100 mL ) 2.5 h after it reaches its maximum concentration?
(b) For the decomposition of an antibiotic in a person with a normal temperature (98.6 ∘ F),k=3.1×10 −5s −1 ; for a person with a fever (temperature of 101.9 ∘F),k=3.9×10 −5 s −1 . If the person with the fever must take another pill when 32 of the first pill has decomposed. how many hours should she wait to take a second pill? A third pill? (Assume that the pill is effective immediately.)
(c) Calculate E 2for decomposition of the antibiotic in part (b).
Draw (R)-2,3-dimethylheptane in a structural condensed format. Use a dash or wedge bond to indicate stereochemistry of substituents on asymmetric centers, where applicable.
Draw (S)-2,4-dimethylheptane in a structural condensed format. Use a dash or wedge bond to indicate stereochemistry of substituents on asymmetric centers, where applicable.
The structural condensed format of (R)-2,3-dimethylheptane and (S)-2,4-dimethylheptane can be represented as follows: (R)-2,3-dimethylheptane: CH3CH(CH3)CH(CH3)CH2CH2CH3 (S)-2,4-dimethylheptane: CH3CH(CH3)CH(CH3)CH(CH3)CH2CH3
In (R)-2,3-dimethylheptane, the "R" configuration indicates that the priority groups are arranged clockwise around the asymmetric carbon atom at position 2. The structural condensed formula is CH3CH(CH3)CH(CH3)CH2CH2CH3, where the methyl groups (CH3) are attached to the second carbon atom (numbered as 2 and 3).
In (S)-2,4-dimethylheptane, the "S" configuration indicates that the priority groups are arranged counterclockwise around the asymmetric carbon atom at position 2. The structural condensed formula is CH3CH(CH3)CH(CH3)CH(CH3)CH2CH3, where the methyl groups (CH3) are attached to the second and fourth carbon atoms (numbered as 2 and 4).
In both structures, the dash or wedge bonds are not necessary as there are no chiral centers or asymmetric carbon atoms present. The structures represent the arrangement of atoms in a simplified condensed form, where the carbon atoms are represented by vertical lines, and the hydrogen and methyl groups are implied but not explicitly shown.
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The decomposition of nitroxyl bromide is second-order. If it takes 0.20 min to decompose 15% of a 0.300M solution of nitrosyl chloride, what is k for the reaction, including units? Your Answer: 2.95M
∧
−1 min
∧
−1
To determine the rate constant, k, for the second-order decomposition reaction of nitroxyl bromide, we can use the integrated rate law for a second-order reaction. The value of k for the reaction is 94.445 M^(-1) min^(-1).
To determine the rate constant, k, for the second-order decomposition reaction of nitroxyl bromide, we can use the integrated rate law for a second-order reaction:
1/[A]t - 1/[A]0 = kt
Where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
In this case, we are given that it takes 0.20 min to decompose 15% of the initial concentration of nitroxyl bromide, which is 0.300 M. Therefore, [A]t = 0.15 * 0.300 M = 0.045 M, and [A]0 = 0.300 M.
Substituting these values into the integrated rate law:
1/0.045 - 1/0.300 = k * 0.20
Simplifying the equation:
22.222 - 3.333 = 0.2k
18.889 = 0.2k
Dividing both sides by 0.2:
k = 18.889 / 0.2
k = 94.445 M^(-1) min^(-1)
Therefore, the value of k for the reaction is 94.445 M^(-1) min^(-1).
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what is the standard heat of formation of a compound
The standard heat of formation of a compound refers to the enthalpy change of the reaction which forms 1 mole of the compound from its elements in their standard states. It is denoted by ΔH°f. The standard states of elements refer to their most stable state at a given temperature and pressure.
When any substance is formed from its constituent elements, heat is either released or absorbed. The heat change of a reaction is measured under standard conditions, and the energy required to make or break bonds is described in terms of enthalpy changes. Standard heat of formation of a compound, ΔHf°, is the heat absorbed or released when one mole of a substance is formed from its constituent elements, in their stable state, at standard temperature and pressure (STP).It is considered to be zero for all the elements in their standard state.
A compound's standard heat of formation is a thermodynamic property of a substance. It is described in terms of enthalpy changes, and its sign (+/-) gives information about the heat absorbed or released during the formation of the compound.
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In an oxidation-reduction reaction, the substance reduced always...
a. Gains electrons
b. Loses electrons
c. Gains protons
d. Loses protons
In an oxidation-reduction reaction, the substance reduced always gains electrons.
Oxidation is a process that involves a loss of electrons, while reduction involves a gain of electrons. Oxidation-reduction reactions, often known as redox reactions, involve a transfer of electrons from one atom to another.
An oxidation reaction involves the loss of electrons, whereas a reduction reaction involves the gain of electrons. In a redox reaction, both of these reactions happen at the same time, resulting in the transfer of electrons.
In this case, the answer is option A, which is "gains electrons." In an oxidation-reduction reaction, the substance that is reduced gains electrons.
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How would you prepare 250 mL of 0.40M solution of copper sulfate solution (MM of CuSO4 is 159.609 g/mol
To prepare 250 mL of 0.40M solution of copper sulfate solution, 15.961 grams of copper sulfate must be dissolve in enough water to make 250mL solution.
To prepare 250mL of 0.4M solution of Copper Sulfate, the number of moles of the compound needed and mass of the compound should be calculated.
The number of moles (n) can be calculated using the formula;
n = M x V
Where;
n = number of moles
M = molarity (0.4M)
V = volume (250 mL or 0.25 L)
Thus;
n = 0.4 x 0.25
n = 0.1 moles of CuSO₄
The mass (m) of CuSO₄ can be calculated using the formula;
m = n x MM
Where;
m = mass
n = number of moles
MM = molar mass
Thus;
m = 0.1 x 159.609
m = 15.961 grams of CuSO₄
Hence, you have to dissolve 15.961 grams of copper sulfate in enough water to make 250mL solution.
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5. Draw structural formulas for the following names. Identify compounds that can exist as cis-trans isomers. Draw structural formulas for both isomers. a. 2-heptene b. 2,3-dibromo-2-butene c. propene d. 1,1-dichloroethene 6. Write equations for each of the following, using structural formulas for the reactants and products. Name the product(s) and supply any missing reactant(s) and or catalyst(s). a. combustion of 3,3-dimethylnonane b. halogenation (fluorine) of m-terbutyltoluene c. halogenation (chlorine) of 2-methylpentane d. halogenation (iodine) of 4 -ethyl-2-hexyne c. Write a short summary about the similarities and differences between halogenation reactions in alkanes, alkenes, alkynes and aromatics.
Structural formula: CH3CH=CHCH2CH2CH2CH3
To draw the structural formula for 2-heptene, we start with the parent chain containing 7 carbon atoms. We number the carbons so that the double bond is between the 2nd and 3rd carbons. We then add the appropriate number of hydrogen atoms to complete the structure.
b. 2,3-dibromo-2-butene:
Structural formula for cis isomer: BrCH2CH=CHCH2Br
Structural formula for trans isomer: BrCH2CH=CHCH2BrTo draw the structural formulas for the cis and trans isomers of 2,3-dibromo-2-butene, we start with the parent chaincontaining 4 carbon atoms. We number the carbons so that the double bond is between the 2nd and 3rd carbons. We then add the bromine atoms on the 2nd and 3rd carbons. For the cis isomer, the bromine atoms are on the same side of the double bond, while for the trans isomer, they are on opposite sides.
c. propene:
Structural formula: CH3CH=CH2
To draw the structural formula for propene, we start with the parent chain containing 3 carbon atoms. We number the carbons so that the double bond is between the 1st and 2nd carbons. We then add the appropriate number of hydrogen atoms to complete the structure.
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Which is defined as the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature?
A. precipitation
B. super saturation
C. dilution
D. solubility
E. combustion
The correct answer is option (D). The term that is defined as the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature is solubility.
Solubility refers to the ability of a substance, known as the solute, to dissolve in a solvent and form a homogeneous mixture called a solution. The solubility of a solute is typically expressed in terms of the maximum amount of solute that can dissolve in a given amount of solvent at a particular temperature.
It is often measured in units such as grams per liter or moles per liter. The solubility of a substance is influenced by factors such as temperature, pressure, and the nature of the solute and solvent.
When the amount of solute dissolved in the solvent reaches its maximum solubility at a particular temperature, the solution is said to be saturated. If additional solute is added to a saturated solution, it will not dissolve and will instead precipitate out.
This is different from super saturation, which occurs when a solution contains more solute than what is normally soluble at that temperature. Dilution, on the other hand, involves adding more solvent to a solution, reducing the concentration of the solute present. Combustion, while relevant in chemistry, is not directly related to the concept of solubility.
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compared to the mass of a proton, the mass of an electron is
The mass of an electron is significantly less compared to the mass of a proton.
]Electrons and protons are two of the most important subatomic particles. They have different masses. The electron is a negatively charged subatomic particle while the proton is a positively charged subatomic particle. The mass of an electron is approximately 1/1836th of the mass of a proton. The electron has a mass of 9.10938356 × 10^-31 kg, while the proton has a mass of 1.6726219 × 10^-27 kg.
The mass of a proton is 1,836 times greater than that of an electron. The proton, on the other hand, is nearly 2,000 times heavier than the electron. Despite their small mass, electrons play a significant role in chemical reactions. They are involved in the chemical bonding between atoms, determining the electronic structure of atoms and molecules, and in electrical conductivity.
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The d
z
2
orbital falls under which of the following irreducible representations in the C
4v
point group? a. B
2
b. E c. B
1
d. A
1
What is the order of the D
3
point group?
The [tex]dz^2[/tex] orbital belongs to the E irreducible representation in the C4v point group and the order is given by symmetry operations, which is 5.
The character table for the C4v point group can help us determine irreducible representation.
The C4v point group has irreducible representations such as,
A1, A2, B1, B2, and E.
The [tex]dz^2[/tex] orbital belongs to the E irreducible representation in the C4v point group.
The order of the D3 point group refers to the number of symmetry operations present in that specific group.
The D3 point group has the symmetry operations given by,
E (identity)
[tex]C3[/tex] (rotation by 120 degrees counterclockwise)
[tex]C3^2[/tex] (rotation by 240 degrees counterclockwise)
[tex]S3[/tex] (rotation-reflection axis)
σh (horizontal mirror plane)
The total number of these symmetry operations is 5.
Therefore, the order of the D3 point group is 5, and E is an irreducible representation in the C4v point group.
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Which of the following fatty acids would you expect to have a melting point lower than that of myristic acid (14 :0)
18:2, Delta^ 9.12
16: 0
18 : 0
22:0
All are correct
None are correct
The options 18:2, Delta^ 9.12, and 16:0 would be expected to have a lower melting point than myristic acid (14:0). Therefore, the correct answer is "Both 18:2, Delta^ 9.12 and 16:0 are correct."
The fatty acid with a lower melting point than myristic acid (14:0) would typically have a shorter carbon chain length or contain double bonds. Let's analyze the given options:
18:2, Delta^ 9.12 - This fatty acid has 18 carbon atoms and 2 double bonds. The presence of double bonds tends to lower the melting point, so this fatty acid would be expected to have a lower melting point than myristic acid.
16:0 - This fatty acid has 16 carbon atoms and no double bonds. It has a shorter carbon chain length compared to myristic acid, so it would also be expected to have a lower melting point.
18:0 - This fatty acid has 18 carbon atoms and no double bonds. It has the same carbon chain length as the first option (18:2), but it lacks the double bonds. Therefore, its melting point would be higher than myristic acid.
22:0 - This fatty acid has 22 carbon atoms and no double bonds. It has a longer carbon chain length compared to myristic acid, so its melting point would also be higher.
Based on the analysis, the options 18:2, Delta^ 9.12 and 16:0 would be expected to have a lower melting point than myristic acid (14:0). Therefore, the correct answer is "Both 18:2, Delta^ 9.12 and 16:0 are correct."
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The normal boiling point of a certain liquid X is 133.40∘C, but when 81.7 g of glycine (C2H5NO2) are dissolved in 550 . g of X the solution boils at 138.0∘C instead. Use this information to calculate the molal bolling point elevation constant Kb of X. Round your answer to 2 significant digits. Kb=□mol∘C−kg
Using the information stated, the molal boiling point elevation constant Kb of X is 2.32 mol °C⁻¹ kg.
The normal boiling point of a solution is greater than that of a solvent. The boiling point elevation is directly proportional to the molality of the solute present in the solution. This constant of proportionality is known as the molal boiling point elevation constant (Kb).
The boiling point elevation (ΔTb) is calculated as:
ΔTb = Kb × m × i
where ΔTb = change in boiling point temperature, Kb = molal boiling point elevation constant, m = molality of the solution, and i = the van't Hoff factor, which accounts for the degree of dissociation of the solute in the solvent. When glycine (C₂H₅NO₂) is added to a certain liquid X, the boiling point increases from 133.40°C to 138.0°C. Let's solve for Kb:
ΔTb = 138.0°C - 133.40°C = 4.6°C
mass of glycine (C₂H₅NO₂) = 81.7 g
mass of solvent X = 550 g
number of moles of C₂H₅NO₂: n(C₂H₅NO₂) = mass/molar mass = 81.7 g/(75.07 g/mol) = 1.09 mol
The molality of the solution is:
m = (n of solute)/(mass of solvent in kg) = 1.09 mol/0.55 kg = 1.98 mol/kg
Now we have all the variables to solve for Kb:
ΔTb = Kb × m × i
note that i = 1 for glycine since it does not dissociate
Kb = ΔTb/(m × i) = (4.6∘C)/(1.98 mol/kg) = 2.32 mol °C⁻¹ kg⁻¹
Therefore, the value of Kb is 2.32 mol °C⁻¹ kg.
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Which of the following tripeptides has a net charge of +1 at pH8 ? SIT DIG PEN WTG AKT The amino and carboxyl groups of amino acids are bonded to which carbon? The amino is bonded to the α-carbon, and the carboxyl is bonded to the β-carbon. The amino is bonded to the β-carbon, and the carboxyl is bonded to the α-carbon. Both are bonded to the β-carbon. Both are bonded to the a-carbon. Question 7 1 pts Which of the following amino acids would you expect to carry an overall positive charge at pH5.0 ? Glutamine Glutamic acid Tyrosine Glutamic acid and Tyrosine Histidine
The tripeptide that has a net charge of +1 at pH 8 is AKT To determine the net charge of a tripeptide at a specific pH, we need to consider the charges of the amino acids it contains and the p Ka values of their side chains. At pH 8, which is more basic
the carboxyl groups (-COOH) of the amino acids will be deprotonated and carry a negative charge (-COO-). The amino groups (-NH2) will remain protonated and carry a positive charge (+NH3+). Analyzing the given options, AKT contains the amino acid lysine (K), which has a side chain with a pKa value of around 10.5. At pH 8, lysine will be mostly protonated and carry a positive charge, contributing to a net charge of +1 for the tripeptide AKT.
The other options, SIT, DIG, PEN, and WTG, do not contain amino acids with side chains that have pKa values in the relevant pH range. Therefore, they will not contribute to the net charge at pH 8. The amino and carboxyl groups of amino acids are bonded to which carbon The amino is bonded to the α-carbon, and the carboxyl is bonded to the β-carbon.
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Give the reaction equation: 1). KHP with NaOH : 2). HCl with Na
2
CO
3
: Find the mass of KHP (FW 204.223 g/mol ) to react with 30 mL of 0.1MNaOH ? (1 mole of KHP reacts with 1 mole of NaOH ) Show your work. Galculate the volume (mL) of 12MHCl that should be added to produce 2 L of 0.1MHCl ?
1) The mass of KHP required to react with 30 mL of 0.1M NaOH is 30 mL
2) The volume of 12M HCl that should be added to produce 2 L of 0.1M HCl is 33 mL
The reaction equation for KHP with NaOH is: KHP + NaOH → NaKP + H2O.
To find the mass of KHP needed to react with 30 mL of 0.1M NaOH, we can use the formula: M1V1 = M2V2.
M1 = 0.1 M (concentration of NaOH)
V1 = 30 mL (volume of NaOH)
M2 = unknown (concentration of KHP)
V2 = unknown (mass of KHP)
Rearranging the formula, we have: M2 = (M1 x V1) / V2.
Substituting the values, we get: M2 = (0.1 M x 30 mL) / V2.
Since 1 mole of KHP reacts with 1 mole of NaOH, the molar ratio is 1:1. So, the molarity of KHP is also 0.1 M.
Plugging in the values, we get: 0.1 M = (0.1 M x 30 mL) / V2.
Solving for V2, we get: V2 = (0.1 M x 30 mL) / 0.1 M.
Simplifying, V2 = 30 mL.
The mass of KHP required to react with 30 mL of 0.1M NaOH is 30 mL.
2) The reaction equation for HCl with Na2CO3 is: 2 HCl + Na2CO3 → 2 NaCl + CO2 + H2O.
To calculate the volume of 12M HCl needed to produce 2 L of 0.1M HCl, we can again use the formula: M1V1 = M2V2.
M1 = 12 M (concentration of HCl)
V1 = unknown (volume of HCl)
M2 = 0.1 M (concentration of desired HCl)
V2 = 2 L (volume of desired HCl)
Rearranging the formula, we have: V1 = (M2 x V2) / M1.
Substituting the values, we get: V1 = (0.1 M x 2 L) / 12 M.
Simplifying, V1 = 0.033 L, which is equal to 33 mL.
Therefore, the volume of 12M HCl that should be added to produce 2 L of 0.1M HCl is 33 mL.
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An isothermal pulse test on a piece of reaction equipment gave the following results: The output concentrations rose linearly from zero to 0.5 mmol/L in 5 min, and then fell linearly to zero in 10 min ager reaching the maximum value.
E(t) = 0.05t for 0 < t < 4 min
E(t) = (1–0.1t)/3 for 4 < t < 10 min
E(t) = 0 for t > 10 min
(a) What is the mean residence time tm?
The second order liquid phase reaction 2A à B + C is carried out in the system.
The entering concentration is 2 mol/L and the specific reaction rate is 0.2 L/(mol.s).
(b) What is the conversion after 30 seconds in a batch reactor?
(c) What is the conversion predicted by the segregation model?
a) The mean residence time is 7.5 minute.
b) The conversion after 30 seconds is 0.142 .
c) The segregation model posits that the mixture is divided into totally mixed and plug flow areas, with conversion based on mixing intensity.
(a) The mean residence time tm is the average time taken by the fluid to pass through the reactor and is calculated using the formula (time for concentration rise + time for concentration fall)/2:tm = (5 + 10)/2 = 7.5 min
(b) The conversion after 30 seconds in a batch reactor is calculated using the formula:X = (1 - [tex]e^-^k^t[/tex])/Cao = (1 - [tex]e^(^-^0^.^2^*^0^.^5)[/tex])/2 = 0.142
(c) The segregation model assumes that the mixture is segregated into completely mixed and plug flow regions, and that the conversion depends on the mixing intensity.
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Which of the following nuclides are most likely to decay via beta decay? Cs-137 Na-20 N-13 |-126 Al-24
Beta decay can be classified into three forms. These are the beta-minus decay, beta-plus decay, and electron capture decay.
Each of these decay modes occurs under certain circumstances. In the provided nuclides, Cs-137, Na-20, and N-13 are the most likely to decay via beta decay. Beta Decay: Beta decay occurs when a neutron in the nucleus of an atom is transformed into a proton or vice versa. This occurs due to the presence of weak nuclear forces. Beta decay can occur in three ways: beta-minus decay, beta-plus decay, and electron capture decay.
Beta-Minus Decay:When a neutron decays into a proton, a high-energy electron (beta particle) and an antineutrino are emitted. This is called beta-minus decay or negatron decay. Beta-minus decay can occur when the neutron-to-proton ratio in a nucleus is too high.
Cs-137 and Na-20 undergo beta-minus decay. N-13 can undergo either beta-plus decay or positron emission.Electron Capture Decay: When a proton in the nucleus combines with an electron, electron capture decay occurs. This results in the conversion of a proton into a neutron, which lowers the neutron-to-proton ratio. Al-24 can undergo electron capture decay.
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Consider this reaction: H
2
CO
3
(aq)→H
2
O(aq)+CO
2
(aq) At a certain temperature it obeys this rate law. rate =(0.35M
−1
⋅s
−1
)[H
2
CO
3
]
2
Suppose a vessel contains H
2
CO
3
at a concentration of 0.630M. Calculate how long it takes for the concentration of Hi CO, to decrease to 9.095 of its initial value, You may assume no other reaction is important. Round your answer to 2 significant digits: Consider this reaction: ClCH
2
CH
2
Cl(s)→CH
2
CHCl(e)+HCl( e) At a certain temperature it obeys this rate law. rate =(3,00=−1)[ClCH
2
CH
2
Cl] 0.0378M. You may assume no other reaction is important. Round your answer to 2 significant digits.
We can use the integrated rate law for a second-order reaction. It takes approximately 4.92 seconds for the concentration of H2CO3 to decrease to 9.095% of its initial value.
To calculate the time it takes for the concentration of H2CO3 to decrease to 9.095% of its initial value, we can use the integrated rate law for a second-order reaction:
t = 1 / (k * [H2CO3])
Given that the initial concentration of H2CO3 is 0.630 M and the final concentration is 9.095% of the initial value, we can substitute these values into the equation:
[Initial] = 0.630 M
[Final] = 0.09095 * [Initial] = 0.09095 * 0.630 M = 0.05731 M
Now we need to determine the rate constant (k). From the rate law, we can see that the rate constant is equal to 0.35 M^(-1)⋅s^(-1). Therefore, k = 0.35 M^(-1)⋅s^(-1).
Substituting the values into the equation, we can solve for time (t):
t = 1 / (k * [H2CO3])
t = 1 / (0.35 M^(-1)⋅s^(-1) * 0.05731 M)
t ≈ 4.92 seconds
Therefore, it takes approximately 4.92 seconds for the concentration of H2CO3 to decrease to 9.095% of its initial value.
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Explain why copper is formed at the cathode during the electrolysis of its 2
salts. *
Your answer
T
What product(s) would you expect from the following substitution reaction of
14
C-labeled propene? A. CH
2
=CH−
14
CH
2
Br alone B.
14
CH
2
=CH−CH
2
Br alone C. More CH
2
=CH
14
CH
2
Br but a little
14
CH
2
=CHCH
2
Br D. More
14
CH
2
=CHCH
2
Br but a little CH
2
=CH
14
14
CH
2
Br E.
14
CH
2
=CH−CH
2
Br and CH
2
=CH−
14
CH
2
Br in equal amounts
The substitution reaction of 14C-labeled propene can result in various products, including CH2=CH-14CH2Br, 14CH2=CH-CH2Br, and CH2=CH-14CH2CH2Br. The specific product(s) formed depends on the conditions and reactants used.
In a substitution reaction, the halogen (Br) replaces a hydrogen atom in the propene molecule. When CH2=CH-14CH2Br is the only product formed, it suggests that the substitution occurred at one specific position, resulting in a single product with a 14C-labeled carbon attached to the CH2 group. Similarly, if 14CH2=CH-CH2Br is the sole product, it indicates substitution at a different position in the propene molecule, resulting in the 14C label on the CH2 group next to the double bond.
However, it is also possible to obtain a mixture of products. For instance, when there is more 14CH2=CH-CH2Br but a small amount of CH2=CH-14CH2Br, it suggests that the substitution occurs predominantly at one position but with some minor substitution at a different position. The same principle applies to the scenario where there is more CH2=CH-14CH2CH2Br but a little 14CH2=CH-CH2Br, indicating that substitution mainly occurs at one position, with minor substitution at another position.
Lastly, if both 14CH2=CH-CH2Br and CH2=CH-14CH2Br are formed in equal amounts, it implies that the substitution reaction occurs at two different positions in propene, resulting in a mixture of products with the 14C label attached to different carbon atoms. The distribution of products will depend on factors such as reaction conditions and the specific reagents used.
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A cube of butter weights 0260lb and has a voluse of 1303mI. Express the density in grams per minhider to three significant figures. Part C A gom heys a mass of 600 g. When the gem is placed in a grafuased cylinder containing a o00 Express the density in grams per millititer to three significant figures.
3 g/mL is the density in grams per millititer to three significant figures.
To express the density of a cube of butter in grams per milliliter, we need to convert the weight from pounds to grams and the volume from cubic inches to milliliters.
1 pound is approximately equal to 453.592 grams, so the weight of the cube of butter in grams would be:
0.260 lb * 453.592 g/lb = 117.81992 g (rounded to three significant figures as 118 g)
1 cubic inch is equal to approximately 16.387 milliliters, so the volume of the cube of butter in milliliters would be:
[tex]1303 cubic inches * 16.387 mL/in^3 = 21,365.861 mL[/tex] (rounded to three significant figures as 21,400 mL)
Therefore, the density of the cube of butter would be:
Density = Mass / Volume
= 118 g / 21,400 mL
= 0.00551 g/mL (rounded to three significant figures as 0.005 g/mL)
For the gem, we already have the mass as 600 g and the volume as 200 mL.
Therefore, the density of the gem would be:
Density = Mass / Volume
= 600 g / 200 mL
= 3 g/mL (rounded to three significant figures)
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A complete combustion of octane, C8H18, a component of gasoline, proceeds as C8H18(I)+O2( g)→CO2( g)+H2O(I) (unbalanced equation) Atomic weights (g/mol)⋅C=12,H=1,O=16 How many grams of carbon dioxide will be produced from 101 moles of oclane? Enter final answer with hWO decimal places (e.g 37.18 or 35.00 or 9820 ). numenc answers only, do not include the units
The amount of carbon dioxide produced from 101 moles of octane is 3193.82 grams.
To determine the grams of carbon dioxide produced from 101 moles of octane (C8H18), we need to balance the given chemical equation and calculate the molar ratio between octane and carbon dioxide. The balanced equati
on for the complete combustion of octane is:
C8H18 + 12.5O2 → 8CO2 + 9H2O
From the balanced equation, we can see that for every 1 mole of octane, 8 moles of carbon dioxide are produced. Therefore, to find the grams of carbon dioxide produced from 101 moles of octane, we use the following calculation:
101 moles octane × (8 moles CO2 / 1 mole octane) × (44 g CO2 / 1 mole CO2) = 3193.82 g CO2
Therefore, 101 moles of octane will produce 3193.82 grams of carbon dioxide.
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what volume is occupied by 1 mole of any ideal gas at stp?
One mole of any ideal gas at STP occupies 22.4 L of volume.
Standard temperature and pressure (STP) is defined as a temperature of 273.15 K (0 °C, 32 °F) and a pressure of exactly 100 kPa (1 bar).The volume of one mole of an ideal gas at STP can be found using Avogadro's Law, which states that equal volumes of gases at the same temperature and pressure contain the same number of molecules or moles.
The volume of 1 mole of any ideal gas at STP is equal to 22.4 L, according to Avogadro's Law. This volume is also known as the molar volume of a gas at STP, and it applies to all gases regardless of their chemical properties.
The formula to find the volume of any number of moles of a gas at STP is:
Volume = (number of moles) x (molar volume at STP)
Therefore, if we have 2 moles of an ideal gas at STP, then the volume occupied by this gas is given by:
Volume = 2 moles x 22.4 L/mole = 44.8 L
This is how we can calculate the volume of any ideal gas at STP.
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[11 points] A tank in the Unit Operations Lab in the department initially containș 1,000 kg brine containing 12% salt by mass. An inlet stream of brine containing 20% salt by mass flows into the tank at a rate of 22 kg/min. If the solution in the tank is kept well mixed by stirring, and brine is removed from the tank at the bottom at a rate of 12 kg/min, find a). [2 points] how the total mass in the tank changes with time t;b ). [3 points] the linear ODE that relates the amount of salt in the tank at any time t; and c). [6 points] the expression for the amount of salt as a function of time t.
a) The total mass in the tank changes by 10 kg/min.
b) This is the linear ODE that relates the amount of salt in the tank at any time (t).
a) To determine how the total mass in the tank changes with time (t), we need to consider the inflow and outflow rates of brine.
The rate of brine entering the tank is 22 kg/min, and the rate of brine leaving the tank is 12 kg/min.
Therefore, the net change in mass per unit time (dm/dt) can be calculated as:
dm/dt = (rate of inflow) - (rate of outflow)
dm/dt = 22 kg/min - 12 kg/min
dm/dt = 10 kg/min
So, the total mass in the tank changes by 10 kg/min.
b) To derive the linear ordinary differential equation (ODE) that relates the amount of salt in the tank at any time (t), we need to consider the rate of change of salt mass in the tank.
Let's denote the amount of salt in the tank as S(t) (in kg), where t represents time.
The rate of change of salt mass (dS/dt) can be expressed as:
dS/dt = (rate of salt inflow) - (rate of salt outflow)
The rate of salt inflow is given by the concentration of salt in the incoming brine multiplied by the inflow rate:
(rate of salt inflow) = (concentration of salt in incoming brine) × (rate of inflow)
(rate of salt inflow) = 0.20 × 22 kg/min
(rate of salt inflow) = 4.4 kg/min
The rate of salt outflow is given by the concentration of salt in the tank multiplied by the outflow rate:
(rate of salt outflow) = (concentration of salt in the tank) × (rate of outflow)
Since the concentration of salt in the tank is not given, we can express it in terms of the total mass in the tank (M) and the amount of salt in the tank (S):
(concentration of salt in the tank) = S(t) / M(t)
Substituting the values, we have:
(rate of salt outflow) = (S(t) / M(t)) × (rate of outflow)
(rate of salt outflow) = (S(t) / M(t)) × 12 kg/min
Now, we can substitute the expressions for the rates of salt inflow and outflow into the rate of change of salt mass equation:
dS/dt = 4.4 kg/min - (S(t) / M(t)) × 12 kg/min
This is the linear ODE that relates the amount of salt in the tank at any time (t).
c) To obtain the expression for the amount of salt (S) as a function of time (t), we need to solve the ODE:
dS/dt = 4.4 - (12/M) × S
This is a first-order linear ODE, which can be solved using various techniques such as separation of variables, integrating factors, or the method of exact equations.
The specific solution will depend on the initial conditions (i.e., the amount of salt in the tank at t=0).
By solving the ODE with the appropriate initial conditions, you can obtain the expression for the amount of salt as a function of time.
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Which one of the following is NOT an allowable set of quantum numbers? n=3ℓ=2mt=−2ms=1/2 n=2ℓ=1mt=−1ms=1/2 n=3ℓ=3mt=−1ms=1/2 n=2ℓ=0ml=0ms=1/2 n=1mt=0ms=−1/2
One of the following is not an allowable set of quantum numbers is D. n = 2, l = 0, ml = 0, ms = 1/2.
The allowed values of quantum numbers are as follows: - Principle quantum number (n): the permitted values of n are positive whole numbers starting from 1. The average separation between the nucleus and the electron shell is determined by the value of n. - Azimuthal quantum number (l): the permitted values of l are all positive integers ranging from 0 to n - 1, it is known as the orbital angular momentum quantum number. - Magnetic quantum number (m): the permitted values of m are integers ranging from -l to l.
The magnetic field affects the orientation of the atom. - Spin quantum number (s): The permitted values of s are +1/2 or -1/2. It distinguishes between two electrons in the same orbital. Therefore, the fourth set of quantum numbers, n = 2, l = 0, ml = 0, ms = 1/2, is not allowed. The permitted values of ml are integers ranging from -l to l; thus, the allowed value for ml is 0 for l = 0. So the correct answer is D. n = 2, l = 0, ml = 0, ms = 1/2.
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Calculate the volume in cubic nanometers of the cobalt crystal
structure unit cell (use the larger cell). Cobalt is HCP at 20°C
with a = 0.2507 nm and c = 0.4069 nm.
The volume of the cobalt crystal structure unit cell is approximately 0.03381 cubic nanometers
To calculate the volume of the cobalt crystal structure unit cell, we need to consider the hexagonal close-packed (HCP) arrangement. The HCP structure consists of three layers: two hexagonal layers and one triangular layer in between.
The volume of the HCP unit cell can be calculated using the formula:
V = (3√2) * a^2 * c / 2
where a is the lattice parameter in the basal plane and c is the height of the unit cell.
Given that a = 0.2507 nm and c = 0.4069 nm, we can substitute these values into the formula:
V = (3√2) * (0.2507 nm)^2 * 0.4069 nm / 2
Calculating this expression gives us the volume of the unit cell in cubic nanometers.
V ≈ 0.03381 nm^3
Therefore, the volume of the cobalt crystal structure unit cell is approximately 0.03381 cubic nanometers.
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Is density an intensive or extensive property?
2. The following data was collected to determine the density of water. Use the data to
calculate the density of water.
Mass of flask and stopper: 62.468 g
Mass of flask, water, and stopper: 87.418 g
Volume of water: 25.00 mL
3. If 50.00 mL of an unknown liquid was pipetted into a flask with a mass of 124.011 g and
the mass of the liquid and flask was 180.011 g, what is the density of the unknown liquid?
Is the unknown liquid water? How do you know?
4. The density of gallium triiodide, GaI , is 4.15 g/cm3. If the initial volume of water in a
graduated cylinder is 45.6 mL, what will the volume be after adding a crystal of gallium
triiodide which has a mass of 24.659 g? (The crystal is not soluble in water.)
5. Copper has a density of 8.92 g/cm3. What is the mass of a sample of copper with a volume
of 33.252 mL?
41
6. Lead has a density of 11.344 g/cm3. A lead brick has a length of 22.86 cm, a width of 7.62 cm,
and a mass of 20.07 kg. Determine the height of the brick in cm.
7. If a piece of aluminum foil has a length of 27.94 cm, a width of 22.86 cm, and a thickness
of 2.54 × 10–3 cm, what is the mass of the foil?
Density is an intensive property. Intensive properties are those that do not depend on the amount or size of the sample but rather describe the inherent characteristics of the substance.
Density is defined as the ratio of mass to volume and remains constant regardless of the size or amount of the substance.
Answer 2. The density of water is 0.998 g/mL.
To calculate the density of water using the given data, we need to divide the mass of water by its volume. The mass of water can be obtained by subtracting the mass of the empty flask and stopper from the mass of the flask, water, and stopper. In this case:
Mass of water = (Mass of flask, water, and stopper) - (Mass of flask and stopper)
= 87.418 g - 62.468 g
= 24.95 g
The volume of water is given as 25.00 mL.
Density of water = Mass of water / Volume of water
= 24.95 g / 25.00 mL
= 0.998 g/mL
Therefore, the density of water is 0.998 g/mL.
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How many grams of carbon dioxide (CO
2
) are there in 0.634 mol of the compound?
To determine the number of grams of CO2 present in 0.634 moles of CO2, the molar mass of CO2 must be determined.
According to the given information:Therefore, the molar mass of CO2 is 12.01 + 2 (16.00) = 44.01 g/mol.
Then, multiply the molar mass of CO2 with the number of moles of CO2 to find the number of grams of CO2.
Mass of CO2 = Number of moles of CO2 × Molar mass of CO2
Mass of CO2 = 0.634 mol = 44.01 g/mol
Mass of CO2 = 27.88 g
Therefore, there are 27.88 grams of CO2 in 0.634 moles of CO2.
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calculate the [h3o+] and ph of each polyprotic acid solution.
The H₃O⁺ is 0.380 M and the pH of this solution is 0.420.
To calculate the [H₃O⁺] (hydronium ion concentration) of a polyprotic acid solution, we need to consider the ionization steps of the acid.
H₃PO₄ is a polyprotic acid that ionizes in multiple steps:
H₃PO₄ ⇌ H⁺ + H₂PO₄⁻
H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻
HPO₄²- ⇌ H⁺ + PO₄³⁻
Since we are given the concentration of H₃PO₄, we can assume that it is fully ionized in the first step.
Therefore, the concentration of H⁺ from H₃PO₄ is equal to the concentration of H₃PO₄ itself.
[H₃O⁺] = [H⁺] = 0.380 M
So, the [H₃O⁺] is 0.380 M.
To calculate the pH of the solution, we can use the formula:
pH = -log[H₃O⁺]
pH = -log(0.380)
= -(-0.420)
= 0.420
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Calculate the [H3O+] of the following polyprotic acid solution: 0.380 M H3PO4. Express your answer using two significant figures. [H3O+] =
Calculate the pH of this solution. Express your answer using one decimal place. pH =
Be sure to answer all parts. Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with superheated steam: C(s)+H
2
O(g)→CO(g)+H
2
( g) ΔH
rxB
0
=129.7 kJ (a) Combine the reaction above with the following two to write an overall reaction for the production of methane:
CO(g)+H
2
O(g)→CO
2
( g)+H
2
( g)
CO(g)+3H
2
( g)→CH
4
(g)+H
2
O(g)
ΔH
rnn
0
=−4 kJ
ΔH
rxn
0
=−206 kJ
In the overall reaction, include the physical states of each product and reactant. In the overall reaction, include the physical states of each product and reactant. (b) Calculate ΔH
rxn
0
for this overall clange. (c) Using the value in (b) and calculating ΔH
rxn
0
for the combustion of methane, find the total heat for gasifying 1.97 kg of coal and burning the methane formed. Assume water forms as a gas and the molar mass of coal is 12.00 g/mol. kJ
(a) The given reactions are as follows:
CO(g) + H2O(g) → CO2(g) + H2(g)
CO(g) + 3H2(g) → CH4(g) + H2O(g)
The reaction for the production of methane can be found by adding these two equations:
CO(g) + H2O(g) → CO2(g) + H2(g) .... (1)
CO(g) + 3H2(g) → CH4(g) + H2O(g) .... (2)
On addition, we get:
CO(g) + 3H2(g) + CO(g) + H2O(g) → CH4(g) + CO2(g) + 2H2(g)
The overall balanced reaction for the production of methane is:
2CO(g) + 4H2(g) + O2(g) → 2CH4(g) + 2CO2(g)
(b) Calculation of ΔH°rxnThe given data is:
C(s) + H2O(g) → CO(g) + H2(g);
ΔH°rxn = 129.7 kJ (Exothermic)
CO(g) + 3H2(g) → CH4(g) + H2O(g);
ΔH°rxn = -206 kJ (Exothermic)
We have to find the enthalpy change for the overall reaction, which can be found by combining the given equations:
On reversing the first equation, we get:
CO(g) + H2(g) → C(s) + H2O(g);
ΔH°rxn = -129.7 kJ
On multiplying the second equation by 2, we get:
2CO(g) + 6H2(g) → 2CH4(g) + 2H2O(g);
ΔH°rxn = -412 kJ
Now, we have to cancel the reactants and products common to both the reactions:
C(s) + 2H2(g) → CH4(g)
ΔH°rxn = [-206 - (-129.7) - 2(412)] kJ
= -235.6 kJ (Endothermic)
Thus, the enthalpy change for the overall reaction is -235.6 kJ.
(c) Calculation of total heat for gasifying 1.97 kg of coal and burning the methane formed
The given data is:
Mass of coal = 1.97 kg
The molar mass of coal = 12.00 g/mol
The moles of coal can be calculated as:
n = mass/molar mass= 1970 g / 12.00 g/mol
= 164.2 mol
The enthalpy change for the overall reaction is -235.6 kJ.
As per the balanced equation:
2CO(g) + 4H2(g) + O2(g) → 2CH4(g) + 2CO2(g)
The enthalpy change for the combustion of methane:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g);
ΔH°rxn = -891 kJ (Exothermic)
Thus, the enthalpy change for 2 moles of methane is -891 x 2 kJ= -1782 kJ.
The heat required for gasifying 1.97 kg of coal and burning the methane formed is given by the sum of enthalpy changes of gasification and combustion processes.
∆H = 164.2 mol x [-235.6 kJ + (-1782 kJ)] = -313.6 MJ
Therefore, the total heat required is -313.6 MJ.
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A sample consisting of six moles of Cl2 gas that was initially confined in a 30.0 L vessel is allowed to adiabatically expand against a constant pressure of 750 torr to a final volume that has increased by a factor of 2. Calculate: heat (q), work (w), and the changes in temperature, internal energy, and enthalpy (DT, DU, DH). Assume the gas sample behaves ideally.
For a sample of 6 moles of Cl₂ gas adiabatically expanding against constant pressure of 750 torr to a final volume that is doubled, q = 0, w = -22,500 torr L, ΔT = 0, ΔU = 0, ΔH = -22,500 torr L.
The heat (q), work (w), changes in temperature (ΔT), internal energy (ΔU), and enthalpy (ΔH) can be calculated using the first law of thermodynamics and ideal gas law.
To calculate heat (q), we can use the formula [tex]q = \triangle U + w[/tex], where ΔU is the change in internal energy and w is the work done by the system.
Since the process is adiabatic, [tex]q = 0[/tex] (no heat exchange with the surroundings).
The work (w) can be calculated using the formula [tex]w = -P\triangle V[/tex], where P is the constant pressure and ΔV is the change in volume.
Since the volume increased by a factor of 2, the final volume is 60.0 L.
Therefore, [tex]w = -(750 torr)(60.0 L - 30.0 L)[/tex]
[tex]= -22,500 torr L[/tex]
To calculate [tex]\triangle T[/tex], we can use the formula [tex]\triangle T = q / nC_v[/tex], where[tex]n[/tex] is the number of moles and [tex]C_v[/tex] is the molar heat capacity at constant volume.
For an ideal gas, [tex]C_v= (3/2)R[/tex], where R is the gas constant.
[tex]\triangle T = q / (6 moles)(3/2)R[/tex]
To calculate ΔU, we can use the formula [tex]\triangle U = nC_v\triangle T[/tex]
[tex]\triangle U = (6 moles)(3/2)R\triangle T[/tex]
To calculate ΔH, we can use the formula [tex]\triangle H = \triangle U + P\triangle V[/tex]
[tex]\triangle H = \triangle U + w[/tex]
The changes in temperature (ΔT), internal energy (ΔU), and enthalpy (ΔH) are all zero in this case, as there is no heat exchange and the volume change does not affect them. Thus, [tex]\triangle T = \triangle U = \triangle H = 0[/tex]
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