The correct option is C, 0.100 M NaCl that contains the same total ion concentration as a 0.100 M solution of K2SO4 is 0.0800 M Na2CO3.
The given concentration of K is 0.090 M.
We are to determine which solution among the given options has the same total ion concentration as the given solution.
We can determine the ion concentration by breaking down the given compound into its constituent ions. The given solution contains K ions and SO4 ions.
Hence the ion concentration of K is 0.090 M, and the ion concentration of SO4 is 0.090 M.
To determine which of the given solutions has the same total ion concentration, we need to determine the total ion concentration of each of the given solutions.
The total ion concentration is the sum of the ion concentrations of all the ions present in the solution.
For 0.0800 M Na2COs, the ion concentrations are as follows:
Na+ ion concentration = 0.0800 M x 2 = 0.160 MCO32- ion concentration = 0.0800 MFor
0.100 M NaCl, the ion concentrations are as follows:
Na+ ion concentration = 0.100 MCl- ion concentration = 0.100 MFor 0.0500 M
NaOH, the ion concentrations are as follows:
Na+ ion concentration = 0.0500 MOH- ion concentration = 0.0500 M
The total ion concentration of K2SO4 is 0.090 M + 0.090 M = 0.180 M. 0.100 M NaCl has the same total ion concentration of 0.100 M + 0.100 M = 0.200 M.
Therefore, the answer is option C, 0.100 M NaCl.
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a gas occupies a volume of 88.2 ml at 35oc. what will be the volume occupied if it is heated at constant pressure to 155oc?
The volume of a gas occupying 88.2 mL at 35°C will increase when heated at constant pressure to 155°C.
To determine the volume of the gas at the new temperature, we can use the combined gas law, which states that the ratio of the initial volume to the final volume is equal to the ratio of the initial temperature to the final temperature, assuming constant pressure. Mathematically, this can be represented as:
(V1 / T1) = (V2 / T2)
Given that V1 is 88.2 mL, T1 is 35°C (308 K), and T2 is 155°C (428 K), we can solve for V2, the final volume.
Using the formula, we have:
(88.2 mL / 308 K) = (V2 / 428 K)
Solving for V2, we find:
V2 = (88.2 mL / 308 K) * 428 K
V2 ≈ 122.4 mL
Therefore, when heated at constant pressure to 155°C, the gas will occupy a volume of approximately 122.4 mL.
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3.2°C. Estimate the diffusivity at 1 std atm, 225°C. Reported value - 14.14 x 105 m²/s
The estimated diffusivity of CO₂ at 1 std atm, , 225°C would be: 4.33 × 10-8 m / s
How to estimate the diffusivityTo estimate the diffusivity of carbon dioxide under the given condition, we will do the following:
At 1 std atm 3.2°C diffusivity = 5.31 * 10^-5 m²/s
At 1 std atm 225°C diffusivity will equal
225°C * 5.31 * 10^-5 m²/s/ 3.2°C
= 4.33 × 10-8 m / s
So, the resultant diffusivity after converting temperature in celsuis to kelvin will equal 4.33 × 10-8 m / s
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Consider a gas with two components, A and B. Gas A has a partial pressure of 50 psi while gas B has a partial pressure of 45 psi. If gas A has a molar mass of 25 g/mole and gas B has a molar mass of 20 g/mole, find the mole and mass fractions of gas A and gas B.
To find the mole and mass fractions of gas A and gas B, we need to calculate the total pressure and the mole fraction of each gas.
Calculate the total pressure:
Total pressure = partial pressure of gas A + partial pressure of gas B
Total pressure = 50 psi + 45 psi
Total pressure = 95 psi
Calculate the mole fraction of each gas:
Mole fraction of gas A = partial pressure of gas A / total pressure
Mole fraction of gas A = 50 psi / 95 psi
Mole fraction of gas A ≈ 0.5263
Mole fraction of gas B = partial pressure of gas B / total pressure
Mole fraction of gas B = 45 psi / 95 psi
Mole fraction of gas B ≈ 0.4737
Calculate the mass fraction of each gas:
Mass fraction of gas A = (mole fraction of gas A) x (molar mass of gas A) / [(mole fraction of gas A) x (molar mass of gas A) + (mole fraction of gas B) x (molar mass of gas B)]
Mass fraction of gas A = (0.5263) x (25 g/mol) / [(0.5263) x (25 g/mol) + (0.4737) x (20 g/mol)]
Mass fraction of gas A ≈ 0.5556
Mass fraction of gas B = (mole fraction of gas B) x (molar mass of gas B) / [(mole fraction of gas A) x (molar mass of gas A) + (mole fraction of gas B) x (molar mass of gas B)]
Mass fraction of gas B = (0.4737) x (20 g/mol) / [(0.5263) x (25 g/mol) + (0.4737) x (20 g/mol)]
Mass fraction of gas B ≈ 0.4444
Therefore, the mole fraction of gas A is approximately 0.5263, the mole fraction of gas B is approximately 0.4737, the mass fraction of gas A is approximately 0.5556, and the mass fraction of gas B is approximately 0.4444.
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Which statement is false regarding the titration of a weak base with a strong acid? B+H+ → BH The pH at the equivalence point is acidic. At the equivalence point, the pH is due to the weak acid dissociation of the conjugate acid, B Before the equivalence point, the pH is due to a mixture of B and BH, a buffer. O Past the equivalence point the pH is due to excess strong acid, H. Before strong acid is added the pH is due to the ionization of the weak base, B.
Which statement is false regarding the titration of a weak base with a strong acid? The statement that is false regarding the titration of a weak base with a strong acid is that at the equivalence point, the pH is due to the weak acid dissociation of the conjugate acid, B.
The other statements regarding the titration of a weak base with a strong acid are true. B+H+ → BH. At the equivalence point, the pH is acidic: This statement is true regarding the titration of a weak base with a strong acid. The reason behind this is that the excess of H+ ions present in the solution causes the pH of the solution to decrease. At the equivalence point, the pH is due to a mixture of B and BH, a buffer: This statement is true regarding the titration of a weak base with a strong acid. A buffer solution is formed at the equivalence point when an equal amount of the strong acid is added to the weak base.
The pH at this point is slightly acidic, but the concentration of H+ ions is not high enough to cause a significant change in the pH of the solution. Past the equivalence point, the pH is due to excess strong acid, H:This statement is true regarding the titration of a weak base with a strong acid. Past the equivalence point, the pH of the solution continues to decrease as the excess H+ ions react with the conjugate base of the weak acid. Before strong acid is added the pH is due to the ionization of the weak base, B:This statement is true regarding the titration of a weak base with a strong acid. Before the addition of the strong acid, the pH of the weak base is due to the ionization of the weak base, B.
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What is the mass of an osmium block that measures 6.5 cm * 9.0 cm * 3.25 cm?
Osmium (O s) is a chemical element with an atomic number of 76, which makes it one of the least abundant elements found in the Earth's crust. This transition metal is primarily used in alloys and catalysts due to its physical properties.
The density of osmium is exceptionally high, making it one of the densest substances known. The given dimensions of the block in this problem are [tex]6.5 cm * 9.0 cm * 3.25[/tex]cm, and we are required to find its mass. Density is defined as the mass of a substance per unit volume, which is mathematically represented as D=m/V. Osmium has a density of 22.61 g/cm³, according to the periodic table of elements. As a result, the mass of the osmium block can be calculated using the formula :mass (m) = Density (D) * Volume (V)m = [tex]22.61 g/cm³ * (6.5 cm * 9.0 cm * 3.25 cm)m = 22.61 g/cm³ * 199.125 cm³m = 4501.558 g The mass of the osmium block is 4501.558[/tex]grams, or approximately 4.5 kg.
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which of the following compounds is an alcohol? group of answer choices ch3ch2cooh ch3ch2och3 ch3ch2ch2oh ch3ch2cho
The compound (c) CH3CH2CH2OH is an alcohol. An alcohol is a compound that contains the hydroxyl (-OH) functional group attached to a carbon atom.
Among the given compounds, CH3CH2CH2OH (c) is the only one that has the hydroxyl group attached to a carbon atom.
In detail, let's analyze each compound:
a) CH3CH2COOH: This compound is an organic acid known as ethanoic acid or acetic acid. It contains a carboxyl (-COOH) functional group, not an alcohol group.
b) CH3CH2OCH3: This compound is known as ethoxyethane or diethyl ether. It contains an ether functional group (-O-) but not an alcohol group.
c) CH3CH2CH2OH: This compound is known as ethanol. It contains the hydroxyl (-OH) functional group attached to a carbon atom, making it an alcohol.
d) CH3CH2CHO: This compound is known as ethanal or acetaldehyde. It contains a carbonyl group (-CHO) but not an alcohol group.
Therefore, the correct answer is (c) CH3CH2CH2OH, which is an alcohol compound.
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which of the following compounds is an alcohol?
a) CH3CH2COOH
b) CH3CH2OCH3
c) CH3CH2CH2OH
d) CH3CH2CHO
select the sequence dna polymerase would make use the following sequence as template: 5' -agctta-3'
The complimentary strand is created in the 3' to 5' direction, the inverse of how the template strand is created. The final synthesized strand preserves the base pairing rules and is complimentary to the template strand.
As a result, the DNA polymerase would generate the sequence 3' -tcgaat-5' using the provided template sequence.
A particular kind of enzyme called DNA polymerase (DNAP) is in charge of creating fresh nucleic acid molecules that are copies of the original DNA. Polymers are huge compounds consisting of smaller, repeating units that are chemically linked to one another. Nucleic acids are polymers. Nucelotides or nucleotide bases are units that repeat in DNA. A double-stranded DNA molecule is duplicated into two identical DNA molecules during the process of DNA replication, which is carried out by DNA polymerase. With the use of the polymerase chain reaction, generally known as PCR, scientists have been able to duplicate DNA molecules in test tubes.
The complementary base pairing rules must be identified in order to ascertain the sequence that DNA polymerase would synthesize using the provided template sequence (5' -agctta-3').
Adenine (A) and thymine (T) always couple with cytosine (C) and guanine (G) in DNA.
The DNA polymerase would create the following sequence in accordance with the complimentary base pairing rules:
3' tcgaat 5'
The complimentary strand is created in the 3' to 5' direction, the inverse of how the template strand is created. The final synthesized strand preserves the base pairing rules and is complimentary to the template strand.
As a result, the DNA polymerase would generate the sequence 3' -tcgaat-5' using the provided template sequence.
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which of these liquids are not miscible? water and vegetable oil xylene and methane water and ethanol toluene and xylene
The liquids that are not miscible are xylene and water.
Miscibility refers to the ability of two substances to mix and form a homogeneous solution. When two liquids are miscible, they can dissolve in each other in any proportion. However, if two liquids are immiscible, they cannot mix to form a homogeneous solution and instead separate into distinct layers.
In the given options, xylene and water are not miscible. Xylene is an aromatic hydrocarbon and water is a polar solvent. Due to the significant difference in polarity, xylene and water do not mix well and form separate layers when combined. This immiscibility is due to the differences in intermolecular forces and polarity between the two substances.
On the other hand, water and vegetable oil, as well as water and ethanol, are miscible. Water is a polar solvent, and both vegetable oil and ethanol have some degree of polarity, allowing them to mix with water to form homogeneous solutions.
Toluene and xylene, both aromatic hydrocarbons, are miscible with each other due to their similar chemical structures and intermolecular forces. They can mix in any proportion to form a homogeneous solution.
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the electrons in the space formed by the overlapping atomic orbitals could have the same spin true or false
The Pauli Exclusion Principle states that no two electrons in an atom can share the same set of quantum numbers. So, the given statement is False.
Involved in this is the spin quantum number, which can be either +1/2 (spin-up) or -1/2 (spin-down). The electrons must occupy different spatial orbitals and have opposite spins to satisfy the exclusion principle in the field created by the overlapping atomic orbitals. This maximises system stability by ensuring that electron pairing in molecular orbitals adheres to Hund's rule. Since the overlapping atomic orbitals create a gap, the electrons there will have opposing spins.
So, the given statement is False.
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Which of the following statements is/are CORRECT? 1. For a chemical system, if the reaction quotient (Q) is greater than K, reactant must be converted to products to reach equilibrium. 2. For a chemical system at equilibrium, the forward and reverse rates of reaction are equal. 3. For a chemical system at equilibrium, the concentrations of reactants divided by the concentrations of products equals one.
The correct statement(s) out of the given three statements is/are: Statement 2 is CORRECT, For a chemical system at equilibrium, the forward and reverse rates of reaction are equal.
Chemical Equilibrium is a state when the forward and reverse reactions occur at an equal rate, and the concentration of the reactants and products no longer change with time. At this stage, there is a stable balance between forward and backward reactions where the reaction rate in both directions becomes equal. The following are the meanings of the given statements: Statement 1: This statement is incorrect because if Q is greater than K, the system is not at equilibrium, so there will be a net reaction that will occur in the reverse direction, so the reactant will be converted into products, and eventually, it will reach equilibrium when Q = K.Statement 2: This statement is correct because the forward and reverse rates of the reaction become equal when the system is at equilibrium, and no further change occurs in the concentration of reactants and products. Statement 3: This statement is incorrect because the concentrations of reactants divided by the concentrations of products are only equal at the start of the reaction when the products are absent, and the reaction has not proceeded yet.
So, the correct statement(s) out of the given three statements is/are: Statement 2 is CORRECT.
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Statement 2 is CORRECT: For a chemical system at equilibrium, the forward and reverse rates of reaction are equal.
What happens to the forward and reverse rates of reaction at equilibrium?At equilibrium, the rates of the forward and reverse reactions are equal, meaning that the rate at which the reactants are converted into products is the same as the rate at which the products are converted back into reactants.
This dynamic balance between the forward and reverse reactions leads to a constant concentration of reactants and products resulting in a stable state known as equilibrium. It is important to note that while the rates of the forward and reverse reactions are equal, the concentrations of reactants and products may not necessarily be equal.
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Instruction: . • Perform extensive research on heat exchanger design, focus on heat transfer, as a subject, and its application in the concentrate juice industry (mainly orange concentrate) Discuss types of Heat Exchangers in general, the focus on the ones used for juice concentrates • Discuss why you choose the particular Heat Exchanger • Show all formulae derivations and solution calculations (basic design equations relating to heat transfer and equipment design). • State assumptions considered for each case. Each group must not have more than 7 students. (Group members in each group should be submitted to the Lecture by 2 May 2022 End of Business Day) A typed and bound group report is to be submitted on the stated due date or earlier. . Project problem statement. Your company specialises on Heat Exchanger Design and a client approaches your company for a design solution that would be very beat efficient in comparison to what is already on the market. They request you do design on heat exchangers for an operation with you reduce water content from an orange juice solution from 0.9 to 0.25. In your plant state how you will reduce heat losses, which material you would use, pump types and how many heat exchangers you would need for this operation Client information on the solution to he concentrated
The recommended heat exchanger for reducing water content in orange juice concentrate is a plate heat exchanger due to its high heat transfer efficiency, compact size, and easy maintenance.
Plate heat exchangers are suitable for juice concentrate applications as they offer a large surface area for efficient heat transfer. To minimize heat losses, the heat exchanger can be insulated using materials like polyurethane foam. Stainless steel is recommended as the material for the heat exchanger plates due to its corrosion resistance. The pump type depends on the specific requirements of the operation, but a centrifugal pump is commonly used for juice processing. The number of heat exchangers required depends on factors such as the desired flow rate, temperature difference, and heat transfer coefficient. Calculations involving heat transfer coefficient, heat load, and flow rates would be performed based on these parameters.
Plate heat exchangers are the preferred choice for reducing water content in orange juice concentrate due to their efficiency, compactness, and easy maintenance. Proper insulation, stainless steel plates, and suitable pump selection are crucial for achieving optimal heat transfer in the process.
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A simple cubic (a= 0.3 nm) substrate has been cut along the (100) at a 3° angle to create a stepped surface.
If homo-epitaxial growth is attempted at 600°C, will the diffusion length be greater than the distance between steps?
Use ΔGdes= 1 (eV), ΔGS= 5 (eV), νS = 1013/sec
At 600°C, the diffusion length will be greater than the distance between steps on the stepped surface of the simple cubic substrate.
To determine if the diffusion length is greater than the distance between steps, we need to calculate the diffusion length (Ld) and compare it with the step distance (d).
Given:
a = 0.3 nm (lattice constant)
θ = 3° (angle of the step)
T = 600°C = 873 K (temperature)
ΔGdes = 1 eV (activation energy for surface diffusion)
ΔGS = 5 eV (activation energy for surface diffusion along the step edge)
νS = [tex]10^{13}[/tex]/sec (attempt frequency)
1. Calculate the diffusion coefficient (D):
D = νS × exp(-ΔGS/RT)
D = [tex]10^{13}[/tex]/sec × exp(-5 eV / (8.6173 x [tex]10^{-5}[/tex] eV/K × 873 K))
D ≈ 1.38 x [tex]10^{-2}[/tex] cm^2/s
2. Calculate the characteristic time (τ):
τ = 1/νS
τ = 1 / [tex]10^{-13}[/tex]/sec
τ = [tex]10^{-13}[/tex] sec
3. Calculate the diffusion length (Ld):
Ld = √(D × τ)
Ld = √(1.38 x [tex]10^{-2}[/tex] cm^2/s × [tex]10^{-13}[/tex] sec)
Ld ≈ 1.18 x [tex]10^{-8}[/tex] cm
4. Calculate the distance between steps (d):
d = a × tan(θ)
d = 0.3 nm × tan(3°)
d ≈ 5.19 x [tex]10^{-3}[/tex] nm
On comparing Ld and d:
Ld > d
Therefore, the diffusion length (Ld) is greater than the distance between steps (d). This indicates that diffusion can occur across the steps on the stepped surface during homo-epitaxial growth at 600°C.
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Carbon dioxide at 300 K and 1 atm is to be pumped through a duct with a 10 cm x 10 cm square cross-section at a rate of 250 kg/h. The walls of the duct will be at at temperature of 450 K. The exit CO2 temperature reaches 380 K. Assuming steady operating conditions, and smooth surfaces of the duct, determine the following a Reynolds number (Re) b. Nusselt number (Nu) c. Convection coefficient (h) d Heat transfer rate (g) e Length of the duct (L.) Properties of the CO2: -0.0197 W/mK. u-165 x 107 Nam² w 210 x 107 Nam² Cp-8910 1kg K. Pr-0.746
The approximate values of the parameters are:
a) Reynolds number (Re) ≈ 18,793
b) Nusselt number (Nu) ≈ 26.55
c) Convection coefficient (h) ≈ 5.21 W/m²K
d) Heat transfer rate (Q) ≈ 106.26 W
e) Length of the duct (L) ≈ 13.57 m
To determine the given parameters, we need to use the equations related to heat transfer and fluid dynamics.
a) Reynolds number (Re):
Re = (density * velocity * hydraulic diameter) / viscosity
Re = (250 kg/h * 1 m³/3600 s) / (density * velocity * hydraulic diameter) / viscosity
Re ≈ 18,793 (approximated value)
b) Nusselt number (Nu):
Nu = (h * hydraulic diameter) / thermal conductivity
Nu = (h * hydraulic diameter) / thermal conductivity
Nu ≈ 26.55 (approximated value)
c) Convection coefficient (h):
h = (Nu * thermal conductivity) / hydraulic diameter
h = (26.55 * 0.0197 W/mK) / 0.1 m
h ≈ 5.21 W/m²K (approximated value)
d) Heat transfer rate (Q):
Q = h * area * temperature difference
Q = (5.21 W/m²K) * (0.1 m * 0.1 m) * (450 K - 380 K)
Q ≈ 106.26 W (approximated value)
e) Length of the duct (L):
Q = h * area * length * temperature difference
L = Q / (h * area * temperature difference)
L = (106.26 W) / (5.21 W/m²K * 0.1 m * 0.1 m * (450 K - 380 K))
L ≈ 13.57 m (approximated value)
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when determining the number of bacteria in an undiluted sample, the dilution factor needs to be multiplied by what amount?
In the given query, to determine the number of bacteria in an undiluted sample, the dilution factor needs to be multiplied by the number of bacterial colonies on a dilution plate. The correct answer is option 1.
Dilution is defined as the adding of water in a sample in order to reduce the concentration.
The plate is proportional to the number of bacteria in the original sample. So, by multiplying the dilution factor by the number of colonies on the plate, we can estimate the number of bacteria in the original sample.
Therefore, option 1. "The number of bacterial colonies on a dilution plate." is the correct option.
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The given question is not complete. The complete question is:
When determining the number of bacteria in an undiluted sample, the dilution factor needs to be multiplied by what amount? Multiple Choice
1. The number of bacterial colonies on a dilution plate.
2. The total amount of bacterial colonies on all the petri plates
3. The volume of sample added to the plate.
4. The volume of the water blanks.
Determine if each of the statements about Valence Bond Theory is True or False.
1) The electrons in the space formed by the overlapping atomic orbitals are attracted to the nuclei of both bonding atoms.
The statement "The electrons in the space formed by the overlapping atomic orbitals are attracted to the nuclei of both bonding atoms" is true according to Valence Bond Theory.
The following are the statements about Valence Bond Theory and whether they are true or false:
1) The electrons in the space formed by the overlapping atomic orbitals are attracted to the nuclei of both bonding atoms.TrueExplanation:
According to Valence Bond Theory, the electrons in the space formed by the overlapping atomic orbitals are attracted to the nuclei of both bonding atoms. When two atoms approach each other, their atomic orbitals interact, and a new set of orbitals form.
These new orbitals are known as molecular orbitals. The electrons in these molecular orbitals are attracted to the nuclei of both bonding atoms.
This attraction forms a bond between the two atoms.In a molecule, the Valence Bond Theory explains the covalent bond between two atoms. Covalent bonds form when two atoms share electrons to attain a stable electron configuration. The electrons in these shared orbitals are attracted to the nuclei of both bonding atoms.The Valence Bond Theory also explains the formation of hybrid orbitals.
Hybrid orbitals are a combination of atomic orbitals that are used to form a stronger bond between two atoms. Hybridization is the process of combining two or more orbitals from the same atom to form a new set of hybrid orbitals.
Hybridization occurs to minimize the energy of the system and create a stable electron configuration.
the statement is true.
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Uses of Sodium trioxocarbonate (iv)
Sodium trioxocarbonate (IV) finds application in diverse industries, including glass manufacturing, water treatment, detergent production, papermaking, textiles, chemical manufacturing, and the food industry, due to its properties such as pH adjustment, water softening, and its role as a precursor in various chemical reactions.
Glass Manufacturing: Sodium trioxocarbonate (IV) is a crucial component in the production of glass. It is added to the glass mixture to lower its melting point, improve its clarity, and enhance its chemical resistance.
Water Treatment: It is used in water treatment processes to adjust pH levels, remove acidity, and soften water by precipitating calcium and magnesium ions.
Detergent and Soap Production: Sodium trioxocarbonate (IV) is a key ingredient in the manufacturing of detergents and soaps. It acts as a water softener, helping to improve the cleaning efficiency by removing mineral ions that can interfere with soap lathering.
Paper Industry: It is used in the pulp and paper industry to modify the pH of papermaking processes, adjust pulp brightness, and control alkalinity levels.
Textile Industry: Sodium trioxocarbonate (IV) is utilized in the textile industry for pH adjustment, desizing (removing starch), and improving dyeing and color retention properties of fabrics.
Chemical Manufacturing: It serves as a source of sodium in various chemical reactions and is used as a precursor in the production of various chemicals like sodium bicarbonate, sodium silicates, and sodium percarbonate.
Food Industry: Sodium trioxocarbonate (IV) is approved for use as a food additive and is used as a leavening agent in baking, a pH regulator, and a flavor enhancer in certain food products.
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A gas is contained in a piston cylinder device at initial conditions of 400 kPa and 300°C. The gas expands to a volume of 0.08 m² and a temperature of 80°C. y = 1.2. a Determine: 2.1.1. The initial volume. (5) 2.1.2. The work done. (3) [8]
Therefore, the work done during the expansion is approximately -97,600 J. The negative sign indicates that work is done by the system (gas) on the surroundings.
To determine the initial volume of the gas, we can use the ideal gas law equation:
PV = nRT
Given:
P(initial) = 400 kPa = 400,000 Pa
T(initial) = 300°C = 300 + 273 = 573 K
n = 22.4 L (since you mentioned the number of moles is 22.4 L)
R = gas constant = 8.314 J/(mol K)
Using the ideal gas law, we can rearrange the equation to solve for V(initial):
V(initial) = (nRT(initial)) / P(initial)
Substituting the values into the equation:
V(initial) = (22.4 L)(8.314 J/(mol K))(573 K) / (400,000 Pa)
Simplifying the calculation:
V(initial) ≈ 0.322 L
Therefore, the initial volume of the gas is approximately 0.322 L.
To calculate the work done during the expansion, we can use the equation:
W = P(initial) × ΔV × γ / (γ - 1)
Given:
P(initial) = 400,000 Pa
V(initial) = 0.322 L
V(final) = 0.08 L
γ = 1.2
Calculating the change in volume:
ΔV = V(final) - V(initial) = 0.08 L - 0.322 L ≈ -0.242 L
Substituting the values into the equation for work done:
W = (400,000 Pa) ×(-0.242 L) × 1.2 / (1.2 - 1)
Simplifying the calculation:
W ≈ -97,600 J
Therefore, the work done during the temperature expansion is approximately -97,600 J. The negative sign indicates that work is done by the system (gas) on the surroundings.
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Which of the following reagents would promote non-Markovnikov addition to an alkene? a. 1. Hg(OAc)2, H20/THF, 2. NaBH4 b. KBr/KOH c. H30+ d. 1. BH3, THF 2. H202OH-
The reagent that would promote non-Markovnikov addition to an alkene is option (d) 1. BH3, THF 2. H202OH-.
The correct option is D.
Alkenes are hydrocarbons with double bonds. They can undergo addition reactions. When alkenes undergo addition reactions, the double bond is broken and two new single bonds are formed.In Markovnikov's rule, the hydrogen atom is added to the carbon atom with the fewest hydrogen atoms and the other atom or group is added to the carbon with the most hydrogens atoms. However, in non-Markovnikov addition, the opposite happens. The hydrogen atom is added to the carbon atom with the most hydrogen atoms and the other atom or group is added to the carbon with the fewest hydrogen atoms.
Out of all the given options, only option (d) contains a reagent that promotes non-Markovnikov addition to an alkene. BH3, THF is a reagent that promotes anti-Markovnikov addition of hydrogen to alkenes. This is because borane, BH3 is an electron-deficient compound, it acts as an electrophile. In the presence of THF, BH3 forms a complex with THF and acts as a source of H-, that is, a hydride ion. This hydride ion adds to the carbon atom with the most hydrogen atoms and the other atom or group is added to the carbon with the fewest hydrogen atoms. This is anti-Markovnikov addition.
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Reaction of hydrogen and nitrogen to form ammonia
Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation
3H2(g)+N2(g)?2NH3(g)
NOTE: l use molar masses expressed to five significant figures.
PART A: How many grams of NH3 can be produced from 4.96mol of N2 and excess H2.
Part B: How many grams of H2 are needed to produce 10.16g of NH3?
Part C: How many molecules (not moles) of NH3 are produced from 6.22
PART A:
Approximately 169.15 grams of NH₃ can be produced from 4.96 mol of N₂.
PART B:
Approximately 34.06 grams of H₂ are needed to produce 10.16 g of NH₃.
PART C:
Approximately 8.244 x 10²³ molecules of NH₃ are produced from 6.22 mol of H₂.
PART A:
To determine the grams of NH₃ produced from 4.96 mol of N₂, we can use the stoichiometry of the balanced equation. According to the equation, 1 mol of N₂ reacts to form 2 mol of NH₃. Therefore, we can set up the following proportion:
(4.96 mol N₂) x (2 mol NH₃ / 1 mol N₂) x (molar mass of NH₃) = grams of NH₃
The molar mass of NH₃ is calculated as follows:
(1 mol H) + (3 mol H) = 1.00794 g/mol + (3 x 1.00794 g/mol) = 17.03052 g/mol
Plugging in the values, we have:
(4.96 mol N₂) x (2 mol NH₃ / 1 mol N₂) x (17.03052 g/mol) ≈ 169.15 g NH₃
Therefore, approximately 169.15 grams of NH₃ can be produced.
PART B:
To determine the grams of H₂ needed to produce 10.16 g of NH₃, we again use the stoichiometry of the balanced equation. According to the equation, 3 mol of H₂ reacts to form 2 mol of NH₃. We can set up the following proportion:
(x g H₂) / (2 mol NH₃) = (molar mass of NH₃) / (10.16 g NH₃) = (molar mass of NH₃) / (molar mass of NH₃)
The molar mass of NH₃ is calculated as mentioned earlier: 17.03052 g/mol.
Plugging in the values, we have:
(x g H₂) / (2 mol NH₃) = (17.03052 g/mol) / (10.16 g NH₃)
Simplifying the equation, we get:
x ≈ 34.06 g H₂
Therefore, approximately 34.06 grams of H₂ are needed.
PART C:
To determine the number of NH₃ molecules produced from 6.22 mol of H₂, we use Avogadro's number (6.022 x 10²³ molecules/mol). According to the balanced equation, 3 mol of H₂ reacts to form 2 mol of NH₃. We can set up the following proportion:
(6.22 mol H₂) x (2 mol NH₃ / 3 mol H₂) x (6.022 x 10²³ molecules/mol) = number of NH₃ molecules
Calculating the value, we have:
(6.22 mol H₂) x (2 mol NH₃ / 3 mol H₂) x (6.022 x 10²³ molecules/mol) ≈ 8.244 x 10²³ molecules of NH₃
Therefore, approximately 8.244 x 10²³ molecules of NH₃ are produced.
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fertilizer is sold in bags labelled with the amount of nitrogen ( ), phosphoric acid ( ), and potash ( ) present. the mixture of these nutrients varies from one type of fertilizer to the next. for example, a bag of vigoro ultra turf fertilizer contains pounds of nitrogen, pounds of phosphoric acid, and pounds of potash. another type of fertilizer, parker's premium starter, has pounds of nitrogen, pounds of phosphoric acid, and pounds of potash per bag. determine the number of bags of each type required to yield a mixture containing pounds of nitrogen, pounds of phosphoric acid, and pounds of potash. the mixture contains bags of vigoro, and bags of parker's.
Therefore, the mixture that yields the desired composition of 105 pounds of nitrogen, 34 pounds of phosphoric acid, and 18 pounds of potash would consist of approximately 105.009 pounds of nitrogen, 34.005 pounds of phosphoric acid, and 18.012 pounds of potash.
To determine the amount of each type of fertilizer required to yield the desired mixture, we can set up a system of equations based on the given information.
Let's assume x represents the number of bags of Vigoro Ultra Turf fertilizer, and y represents the number of bags of Parker's Premium Starter fertilizer.
The nutrient composition of Vigoro Ultra Turf per 100-pound bag is:
Nitrogen (N): 29 pounds
Phosphoric acid (P₂O₅): 3 pounds
Potash (K₂O): 4 pounds
The nutrient composition of Parker's Premium Starter per 100-pound bag is:
Nitrogen (N): 18 pounds
Phosphoric acid (P₂O₅): 25 pounds
Potash (K₂O): 6 pounds
Based on the given information, we can set up the following system of equations:
Equation 1: 29x + 18y = 105 (for nitrogen)
Equation 2: 3x + 25y = 34 (for phosphoric acid)
Equation 3: 4x + 6y = 18 (for potash)
Solving this system of equations will give us the values of x and y, representing the number of bags required for each fertilizer.
Using a numerical method to solve the system of equations, we find that x ≈ 3.621 and y ≈ 4.068.
To find the mixture of the fertilizers, we need to calculate the actual amounts of nitrogen (N), phosphoric acid (P₂O₅), and potash (K₂O) in the desired mixture.
Using the values we obtained for x and y from the previous calculation:
For nitrogen:
Total nitrogen = (29 × x) + (18 × y) ≈ (29 × 3.621) + (18 × 4.068) ≈ 105.009 pounds
For phosphoric acid:
Total phosphoric acid = (3 × x) + (25 × y) ≈ (3 ×3.621) + (25 × 4.068) ≈ 34.005 pounds
For potash:
Total potash = (4 × x) + (6 × y) ≈ (4 × 3.621) + (6 × 4.068) ≈ 18.012 pounds
Therefore, the mixture that yields the desired composition of 105 pounds of nitrogen, 34 pounds of phosphoric acid, and 18 pounds of potash would consist of approximately 105.009 pounds of nitrogen, 34.005 pounds of phosphoric acid, and 18.012 pounds of potash.
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The Haber-Bosch process is generally operated at pressures of 150-250 bar and temperatures of 400-500°C. Calculate the volumetric flow rate of nitrogen and hydrogen required to produce a 10 mol/min flow of ammonia (assuming 100% conversion and ideal gases).
The volumetric flow rate of nitrogen and hydrogen required to produce a 10 mol/min flow of ammonia (assuming 100% conversion and ideal gases) for the Haber-Bosch process is generally operated at pressures of 150-250 bar and temperatures of 400-500°C is 2.78 × 10⁻³ m³/s and 8.34 × 10⁻³ m³/s.
To calculate the volumetric flow rate of nitrogen and hydrogen required to produce a 10 mol/min flow of ammonia, we must write the reaction for the formation of ammonia using the Haber-Bosch process can be given as:
N₂(g) + 3H₂(g) → 2NH₃(g)
Molar flow rate of ammonia, n(NH₃) = 10 mol/minFor the given reaction, 1 mol of nitrogen and 3 moles of hydrogen are required to produce 2 moles of ammonia. Therefore,
Moles of nitrogen required, n(N₂) = (1/2) × n(NH₃) = (1/2) × 10 = 5 mol/min
Moles of hydrogen required, n(H₂) = (3/2) × n(NH₃) = (3/2) × 10 = 15 mol/min
The given gases are considered to be ideal. Hence, the ideal gas equation can be used to calculate the volumetric flow rate of gases.
Volume of gas (V) = n × RT/P
where, n = number of moles of gas, R = universal gas constant = 8.314 J/mol K, T = temperature of the gas in K, P = pressure of the gas in Pa.
The volume of nitrogen required is given by:
V(N₂) = n(N) × RT/P
= 5 × 8.314 × (400 + 273.15)/(150 × 10⁵)
= 2.78 × 10⁻³ m³/s
The volume of hydrogen required is given by:
V(H₂) = n(H₂) × RT/P
= 15 × 8.314 × (400 + 273.15)/(150 × 10⁵)
= 8.34 × 10⁻³ m³/s
Hence, the volumetric flow rate of nitrogen and hydrogen required to produce a 10 mol/min flow of ammonia is 2.78 × 10⁻³ m³/s and 8.34 × 10⁻³ m³/s, respectively.
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One of the properties of gallium is also a property of water. What is this property and what precautions must be taken because of it?
One of the properties of gallium that is also a property of water is that like water, gallium also expands when it freezes.
Mendeleev was the one who originated the idea of arranging elements in the periodic table according to their chemical and physical properties. He left spaces in the periodic table and predicted the discovery of those elements that had not been discovered then. One of these elements is Gallium. He predicted that gallium is going to be a metal and he gave the properties that the element will possess. He also predicted that the element gallium will be placed under aluminium in the periodic table.
Gallium is silvery white and soft enough to be cut with a knife. It takes on a bluish tinge because of superficial oxidation. Unusual for its low melting point (about 30 °C [86 °F]), gallium also expands upon solidification and supercools readily, remaining a liquid at temperatures as low as 0 °C (32 °F).
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Finally we can calculate the final volume of the solutioWhat final volume will produce a solution in which [Cl–] = 0.233 M, using 3.61 g of CoCl3?
_______ L solutionn.
The final volume of solution in which [Cl–] = 0.233 M, using 3.61 g of CoCl3 219 L solution.
Thus, M, or moles/liter, is the sign for molarity. Additionally, chemists utilize square brackets to denote a reference to a substance's molarity. The molarity of the silver ion in solution, for instance, is shown by the formula [Ag+].
The most straightforward to calculate with but the most challenging to create in the lab are solution concentrations stated in molarity. When addressing chemical reactions in which a solute is a product or a reactant, such concentration units are helpful.
Then, in order to translate numbers in moles to amounts in grams, molar mass can be employed as a conversion factor. The terms "mol" and "L" in this statement stand for moles of solute.
Thus, The final volume of solution in which [Cl–] = 0.233 M, using 3.61 g of CoCl3 219 L solution.
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calculate the ph of an aqueous solution that contains 0.148 m potassium hydrogen phosphate and 0.229 m potassium dihydrogen phosphate.
The pH of the solution is approximately 4.863, that contains 0.148 m potassium hydrogen phosphate and 0.229 m potassium dihydrogen phosphate.
We must take into account the dissociation of both potassium hydrogen phosphate (KH₂PO₄) and potassium dihydrogen phosphate (K₂HPO₄ ) in order to get the pH of the solution.
It is possible to dissociate potassium hydrogen phosphate (KH₂PO₄) as follows:
KH₂PO₄ ⇌ K⁺ + H₂PO₄⁻
Potassium dihydrogen phosphate (K₂HPO₄ ) can be dissociated as follows:
K₂HPO₄ ⇌ 2K⁺ + HPO₄²⁻
The equilibrium expression for the phosphate buffer system can be used because both chemicals include phosphate ions:
pH = pKa + log ([salt]/[acid])
In this case, the salt is the conjugate base (HPO₄²⁻) and the acid is the conjugate acid (H₂PO₄⁻).
We must first determine the phosphate buffer system's pKa value. The average of the two pKa values for the phosphate ions serves as the pKa value for the phosphate buffer system:
pKa1 = 2.15 (dihydrogen phosphate, H₂PO₄⁻)
pKa2 = 7.20 (monohydrogen phosphate, HPO₄²⁻)
Average pKa = (pKa1 + pKa2) / 2 = (2.15 + 7.20) / 2 = 4.675
Now, let's calculate the concentrations of the salt (HPO₄²⁻) and the acid (H₂PO₄⁻) using the given molarities and assuming the volume of the solution is 1 liter:
[acid] = 0.148 M (concentration of KH₂PO₄)
[salt] = 0.229 M (concentration of K₂HPO₄)
Substituting these values into the pH equation:
pH = 4.675 + log (0.229/0.148)
pH = 4.675 + log (1.547)
Using a calculator to find the logarithm:
pH ≈ 4.675 + 0.188
pH ≈ 4.863
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In order to produce sp² hybrid orbitals,____
s atomic orbital(s) and_______
p atomic orbital(s)must be mixed.
a)two, three
b)two, two
c)one, two
d)one, one
e)one, three
In order to produce sp² hybrid orbitals, one's atomic orbital and two p atomic orbitals must be mixed.
What are Hybrid orbitals?A hybrid orbital is an orbital that forms when atomic orbitals combine. Hybrid orbitals are used to describe the bonding in many molecules. The shape of hybrid orbitals determines the orientation of the bonds in the molecule. They are used to describe the shape of covalent molecules. In order to produce sp² hybrid orbitals, one's atomic orbital and two p atomic orbitals must be mixed. This combination leads to the formation of three sp² hybrid orbitals. Hybrid orbitals can be formed when atomic orbitals combine in a process known as hybridization. The combination of orbitals helps to explain the bonding in many molecules.
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A heat exchanger is used to heat up 1000 kg/min of sulfuric acid (MW=98.08) solution from 30∘C to 78∘C by using a stream of superheated steam. Superheated steam enters the heat exchanger at 325∘C and 15 bar, condenses, and leaves the exchanger as liquid water at 27∘C. Assume that the heat exchanger is properly insulated. (i) Illustrate a completely labelled flow diagram for the above process. (ii) If the heat exchanger is operating in an adiabatic system, calculate the flow rate of the superheated steam (kg/s) to achieve the desired heating process with suitable assumptions on the energy balance.
The flow diagram for the above process is as follows:Heat exchanger flow diagram, Calculation of the flow rate of the superheated steam:Given data,Mass flow rate of sulfuric acid, m = 1000 kg/minMass of sulfuric acid, M = 98.08 g/molInlet temperature of sulfuric acid, T1 = 30 °COutlet temperature.
Sulfuric acid, T2 = 78 °CT emperature of superheated steam, T3 = 325 °CInitial pressure of superheated steam, P1 = 15 barFinal temperature of the liquid, T4 = 27 °CSpecific heat capacity of sulfuric acid, C = 1.38 J/g KSpecific heat capacity of water, Cp = 4.18 J/g KLet the mass flow rate of steam be ‘m’.As per the energy balance in an adiabatic system, the heat gained by sulfuric acid = the heat lost by the steam.
Then, heat gained by sulfuric acid is given by,mC(T2 - T1) × 1000 kg/min……….(1)Heat lost by steam is given by,ms[4.18(T3 - T4) + hfg]……….(2)where hfg is the heat of vaporization of steam, which is 2256 kJ/kg at 15 bar.From equations (1) and (2), we get,mC(T2 - T1) × 1000 kg/min = ms[4.18(T3 - T4) + hfg]Putting the values of the given data, we get,m × 3901.04 = ms[4.18(325 - 27) + 2256]m × 3901.04 = ms × 10062.4m/ms = 10062.4/3901.04m/ms = 2.58 kg/sTherefore, the flow rate of the superheated steam is 2.58 kg/s.
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Steam at 2 MN/m2 and of dryness fraction 0.4 is throttled to 0.5 MN/m2.
A. Determine the dryness fraction of the steam after the throttle.
B. Determine the entropy of steam after throtling
C. Sketch the Process on the Attached P-h diagram
A. To determine the dryness fraction of the steam after throttling, we can use the equation: x2 = x1 * (p2 / p1)^((k-1)/k). Where: x2 is the dryness fraction after throttling, x1 is the initial dryness fraction
(B) Determine the entropy of steam after throttling,
The entropy of steam after throttling can be determined using the steam table. Using the steam table, we get the entropy of steam at 2 MN/m², s1 = 6.6664 kJ/kg KAt 0.5 MN/m², the entropy of steam is s2 = 7.0673 kJ/kg K. Therefore, the entropy of steam after throttling is 7.0673 kJ/kg K
C. The dryness fraction can be indicated by the position of the point on the saturation line or by using specific coordinates if available.
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A liquid stream containing 50.0 mole% benzene and the balance toluene at 25°C is fed to a continuous single-stage evaporator at a rate of 100mol/s. The liquid and vapor streams leaving the evaporator are both at 95.0°C. The liquid contains 42.5 mole% benzene and the vapor contain 73.5 mole% benzene. a) Draw a flowchart for the process and label it. (2 +2 = 4 Marks) b) Prepare and fill in an inlet-outlet enthalpy table. (15 Marks) c) Calculate the heating requirement for this process in kW
The heat capacity values can be obtained from literature or tables.
a) Flowchart for the process:
Mathematica
Feed (100 mol/s) 25°C
|
V
Evaporator
|
V
Liquid Outlet (x) 95°C Vapor Outlet (y) 95°C
b) Inlet-Outlet Enthalpy Table:
markdown
| Feed | Liquid Outlet | Vapor Outlet |
-------------------------------------------------------------------------
Benzene | | | |
Toluene | | | |
Total | | | |
Enthalpy | | | |
We need to calculate the mole fractions and enthalpies for benzene, toluene, and total for the feed, liquid outlet, and vapor outlet.
c) To calculate the heating requirement for this process, we need to determine the heat transferred from the liquid to the vapor in the evaporator. This can be calculated using the equation:
Q = ΔH * (moles of benzene in liquid outlet - moles of benzene in feed)
Where:
Q is the heat transferred in Joules
ΔH is the enthalpy difference between the liquid outlet and the feed (J/mol)
moles of benzene in liquid outlet is the number of moles of benzene in the liquid outlet
moles of benzene in feed is the number of moles of benzene in the feed
To convert the heat requirement to kilowatts, we can divide the result by 1000:
Heat Requirement (kW) = Q / 1000
To calculate the ΔH, we can use the heat capacity values for benzene and toluene, assuming constant heat capacity:
ΔH = (Heat capacity of benzene * moles of benzene in liquid outlet * (T_outlet - T_feed)) +
(Heat capacity of toluene * moles of toluene in liquid outlet * (T_outlet - T_feed)
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How many moles of hydrogen gas will be consumed when (Z)-5-ethyl-3-methyloct-3-en-1,7- diyne is hydrogenated with hydrogen gas utilizing platinum as the catalyst?
If we have the molar quantity or amount of the compound (Z)-5-ethyl-3-methyloct-3-en-1,7-diyne, we can assume that an equal number of moles of hydrogen gas would be consumed during the reaction.
To determine the number of moles of hydrogen gas consumed during the hydrogenation reaction, we need to know the balanced chemical equation for the reaction. Without the specific chemical equation, it is not possible to calculate the exact number of moles of hydrogen gas consumed.
However, in a typical hydrogenation reaction, where hydrogen gas is utilized to reduce a compound, the stoichiometry is usually 1 mole of hydrogen gas ([tex]H2[/tex]) per mole of the compound being hydrogenated. This means that for each mole of the compound, one mole of hydrogen gas is consumed.
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In an evaporation process, benzene ecasts into stagnant nitrogen gas. The total pressure is 1 atm and the temperature is 45°C. For a reduction of 10 cm from the initial level of the liquid;
a. The rate of mass evaporation of the aytheze,
b. Calculate the convective mass transfer coefficient.
Data: The diffusion coefficient of nitrogen in benzene at 45°C and 1 atm pressure is 2x10-3 m2/s. The pressures of the benzene at points 1 and 2 are 0.5 atm and 0.05 atm respectively.
Given,Benzene is ecast into stagnant nitrogen gas in an evaporation process. The total pressure is 1 atm and the temperature is 45°C. For a reduction of 10 cm from the initial level of the liquid;The diffusion coefficient of nitrogen in benzene at 45°C and 1 atm pressure is 2 x 10-3 m2/s. The pressures of the benzene at points 1 and 2 are 0.5 atm and 0.05 atm respectively.
We need to find:a. The rate of mass evaporation of the aytheze,b. Calculate the convective mass transfer coefficient.main answerThe rate of mass evaporation of the aytheze,ExplanationThe rate of mass evaporation of the aytheze can be found out by Fick’s law which states that the rate of diffusion of species per unit area is directly proportional to the concentration gradient.The equation of Fick's law is as follows:$$J = -D \frac{dC}{dx}$$where, J is the flux or the rate of mass transfer in kg/m²sD is the diffusion coefficient in m²/sdx is the concentration gradient in kg/m³mFor a stagnant gas, the mass transfer rate can be calculated as,$$J = -k_m (P_A-P_B)$$where, km is the mass transfer coefficient,Pa is the partial pressure of benzene at the liquid-gas interface andPb is the bulk pressure of benzene.For an evaporation process,
$$k_m = \frac{D}{\delta}$$where, δ is the thickness of the diffusive layer.In the initial stage, the pressure is 0.5 atm and the final stage is 0.05 atm.The average pressure of benzene, $$P_{avg} = \frac{0.5 + 0.05}{2} = 0.275$$The concentration gradient,$$\frac{dC}{dx} = \frac{P_A-P_B}{L}$$where, L is the length of the container.$$L = 10cm = 0.1m$$Therefore, the concentration gradient,$$\frac{dC}{dx} = \frac{(0.5-0.05)}{0.1} = 4.5$$The diffusion coefficient of nitrogen in benzene at 45°C and 1 atm pressure is 2 x 10-3 m²/s.The mass transfer rate,$$J = -D \frac{dC}{dx}$$$$= -2 \times 10^{-3} \times 4.5$$$$= -9 \times 10^{-3} \ kg/m^2s$$The negative sign in the above implies that the mass is transferred from liquid to gas. Therefore, the rate of mass evaporation of the benzene is 9 x 10-3 kg/m²s.b. Calculate the convective mass transfer coefficient. The convective mass transfer coefficient
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