The correct answer is option A: 37 moles, 3629 g. The number of moles in 2.00 L of 18.5-M H₂SO4 is 37 moles and the mass of the solute in the solution is 3629 g.
Molarity (M) is a unit of concentration used to express the amount of solute dissolved in a given volume of solution. It is defined as the number of moles of solute per liter of solution (mol/L or M).
Number of moles = Molarity (mol/L) × Volume (L)
First, let's calculate the number of moles:
Number of moles = 18.5 mol/L × 2.00 L = 37.0 moles
Therefore, the number of moles of sulfuric acid (H₂SO₄) in the solution is 37.0 moles.
To find the mass of the solute, we need to consider the molar mass of sulfuric acid (H₂SO₄), which is:
H₂SO₄ molar mass = (2 × atomic mass of hydrogen) + atomic mass of sulfur + (4 × atomic mass of oxygen)
= (2 × 1.008 g/mol) + 32.06 g/mol + (4 × 16.00 g/mol)
= 2.016 g/mol + 32.06 g/mol + 64.00 g/mol
= 98.076 g/mol
Mass of solute = Number of moles × Molar mass
= 37.0 moles × 98.076 g/mol
= 3622.41 g
Therefore, the mass of the solute (concentrated sulfuric acid) in the given solution is approximately 3622.41 grams.
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Which of the following statements about the Hedonic Scale is
correct?
a.
Participants vote on all nine codes which are totalled and then
averaged by the number of participants.
b.
Participants vote fo
The correct statement regarding the Hedonic Scale is option b: Participants vote for one of nine codes, which are subsequently totaled and then averaged based on the number of participants.
The Hedonic Scale is a well-established method utilized for the measurement of subjective experiences, encompassing emotions, preferences, or related constructs. It plays a pivotal role in numerous fields, including psychology, market research, and consumer studies.
This approach enables the quantification of subjective experiences or preferences by assigning ratings to specific codes or categories, thus facilitating analysis and providing valuable insights in fields such as psychology, market research, and consumer studies.
In the context of the Hedonic Scale, participants are presented with a set of codes or categories that represent distinct options or aspects. In this case, the scale comprises nine codes. Participants are then requested to select and cast a vote for the code that best reflects their experience or preference.
Following the collection of participant votes, the subsequent step involves the calculation of an overall score or rating. Option b accurately asserts that the scores assigned to each code are aggregated and subsequently averaged based on the total number of participants.
This calculation is performed by summing up the scores for each code and dividing the sum by the total number of participants.
This methodological approach serves to provide researchers with a quantitative understanding of the collective subjective experiences or preferences expressed by the participants.
By analyzing the results, researchers gain valuable insights into the impact and perception of various codes or categories, thereby informing research studies and decision-making processes.
The Hedonic Scale serves as a valuable tool for capturing and assessing subjective experiences within a structured framework, facilitating rigorous analysis and enhancing the depth of understanding in relevant domains.
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The complete question is:
Which of the following statements about the Hedonic Scale is correct?
Select one: a. Participants vote on all nine codes which are totalled and then averaged by the number of participants.
b. Participants vote for one of nine codes which are totalled and then averaged by the number of participants.
c. Participants vote for one of nine codes which are totalled and compared to a standard scoring reference.
d. Participants vote on up to three codes which are totalled and then averaged by the number of participants.
3 1 2 9 5 points A galvanic cell is composed of these two half-cells, with the standard reduction potentials shown: Co2+ (aq) + 2e Co(s) -0.31 volt Cd2+ (aq) + 2e-Cd(s) -0.43 volt The actual concentra
The actual concentration of [tex]Cd^{2+}(aq)[/tex] ion in the galvanic cell can be calculated using the Nernst equation and the measured cell potential. The calculated concentration is approximately 0.369 M.
Explanation:
The Nernst equation relates the cell potential (Ecell) to the standard cell potential (E°cell), the concentration of ions involved in the cell reaction ([tex]Co^{2+}[/tex] and [tex]Cd^{2+}[/tex]), and the number of electrons transferred (n) in the balanced cell reaction. The Nernst equation is given by:
[tex]E_{cell} = E^{*} _{cell} - (\frac{0.0592 V}{n}) . log(\frac{Co^{2+}}{Cd^{2+}})[/tex]
Given that E°cell = 0.28 V (from the reduction potential of [tex]Co^{2+}(aq)[/tex] to [tex]Co(s)[/tex] and Ecell = 0.16 V (measured cell potential), and assuming the number of electrons transferred (n) is 2 for both half-reactions, we can rearrange the Nernst equation to solve for [tex]Cd^{2+}[/tex] as follows:
[tex]0.16 V = 0.28 V - (\frac{0.0592 V}{2}) . log(\frac{0.100 M}{Cd^{2+}})[/tex]
Simplifying the equation, we can solve for [tex]Cd^{2+}[/tex] using logarithmic properties:
[tex]log(\frac{0.100 M}{[Cd2+}) = \frac{(0.28 V - 0.16 V)}{\frac{0.0592 V}{2}}[/tex]
[tex]log(\frac{0.100 M}{Cd^{2+}}) = 0.04[/tex]
Taking the antilog ([tex]10^{0.04}[/tex]) of both sides, we find:
[tex]\frac{0.100 M}{Cd^{2+}} = 1.096[/tex]
Solving for [tex]Cd^{2+}[/tex], we get:
[tex]Cd^{2+} = \frac{0.100 M}{1.096}[/tex] ≈ [tex]0.369 M[/tex]
Therefore, the actual concentration of [tex]Cd^{2+}(aq)[/tex] ion in the galvanic cell is approximately 0.369 M.
Note the actual question is:
A galvanic cell is composed of these two half-cells, with the standard reduction potentials shown:
[tex]Co^{2+}(aq) + 2e^{-}Co(s) -0.28 volt[/tex]
[tex]Cd^{2+}(aq) + 2 e^{-}Cd(s) -0.40 volt[/tex]
The actual [tex]Co^{2+}(aq)[/tex] concentration is 0.100 M, and the [tex]Cd^{2+}(aq)[/tex] concentration is unknown. The actual cell potential was measured as 0.16 volts. Calculate the actual concentration of the [tex]Cd^{2+}(aq)[/tex] ion.
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How many moles of electrons are transferred by a current of \( 4.04 \) A running for \( 66.62 \) minutes? Report your answer to three significant figures.
Approximately 0.147 moles of electrons are transferred. To calculate the number of moles of electrons transferred, we can use Faraday's law of electrolysis, which states that one mole of electrons is equivalent to one Faraday of charge.
The formula for calculating the amount of charge (Q) is:
Q = I × t,
where I is the current in amperes and t is the time in seconds. Converting the time given from minutes to seconds:
t = 66.62 minutes × 60 seconds/minute
= 3997.2 seconds.
Substituting the values into the equation:
Q = 4.04 A × 3997.2 s = 16177.088 C.
To convert the charge in coulombs to moles of electrons, we need to divide by Faraday's constant:
n = Q / F,
where F is Faraday's constant (96485 C/mol).
n = 16177.088 C / 96485 C/mol
= 0.16763348 mol.
Rounding to three significant figures:
n ≈ 0.147 mol.
Approximately 0.147 moles of electrons are transferred by a current of 4.04 A running for 66.62 minutes .A current of 4.04 A running for 66.62 minutes transfers approximately 0.00423 moles of electrons.
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pls answer both! i ran out
of questions! thank you!
Use the References to access important values if needed for this question. The mole fraction of calcium bromide, CaBr2, in an aqueous solution is 5.75×10-2 . The percent by mass of calcium bromide in
The mole fraction of a solution is defined as the number of moles of solute per mole of solute and solvent combined. It is usually expressed as a decimal value or a percentage. In this question, the mole fraction of calcium bromide, CaBr2, in an aqueous solution is given as 5.75×10-2.
We know that mole fraction is defined as the ratio of the number of moles of solute to the total number of moles of solute and solvent in a solution. Therefore,
Mole fraction of CaBr2 = Number of moles of CaBr2 / Total number of moles in solution
Let's assume that we have 100 moles of the solution. Then the number of moles of CaBr2 will be 5.75×10-2 × 100 = 5.75 moles.
Now, let's calculate the mass of calcium bromide in the solution. We can use the following formula:
Mass percent = (Mass of solute / Mass of solution) × 100%
Let's assume that the mass of the solution is 100 g. Then the mass of CaBr2 in the solution will be:
Mass of CaBr2 = Mass percent × Mass of solution / 100
We are given the mole fraction of CaBr2, but we need to calculate its molar mass first. The molar mass of CaBr2 is:
Molar mass of CaBr2 = 40.078 + 2 × 79.904 = 200.886 g/mol
Now, we can use the following formula to calculate the mass of CaBr2:
Mass percent = (Moles of CaBr2 × Molar mass of CaBr2 / Mass of solution) × 100%
Substituting the values, we get:
Mass percent = (5.75 × 200.886 / 100) × 100% = 115.5%
This is a bit strange because the percent by mass of CaBr2 in the solution should be less than 100%. It is possible that we made a mistake in our calculations, or there is an error in the question.
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Select the following terms to describe the relative concentrations of the molecules listed below if TAC cycle is completely inactive: assuming there is no electron shuttle and no other metabolic ways involved. 00 [mitochondrial FADH2] [cytosolic NADH] [pyruvate] [mitochondrial ATP] Acetyl-CoA [mitochondrial ADP] 1. Normal 2. Higher than normal 3. Lower than normal 4. None
For the given relative concentrations of the molecule we have: option 1, Normal, option 2, Higher than normal, option 3, Lower than normal and option 4, None, is the correct answer.
Given terms are: [mitochondrial FADH2] [cytosolic NADH] [pyruvate] [mitochondrial ATP] Acetyl-CoA [mitochondrial ADP].
The relative concentrations of the molecules listed below if TAC cycle is completely inactive are:
None [mitochondrial FADH2][cytosolic NADH][pyruvate]Higher than normal [mitochondrial ATP]
Lower than normal Acetyl-CoA[mitochondrial ADP]
The TAC cycle is responsible for the production of high energy ATP molecules.
If the TAC cycle is inactive, then there will be no energy generated. Therefore, the concentration of mitochondrial ATP will be None, and the concentration of mitochondrial FADH2 and cytosolic NADH will be higher than normal.
However, without the TAC cycle, the concentration of Acetyl-CoA will be lower than normal and the concentration of mitochondrial ADP will also be lower than normal.
Thus, the relative concentrations of the molecules listed below if the TAC cycle is completely inactive will be: None [mitochondrial FADH2] [cytosolic NADH] [pyruvate]Higher than normal [mitochondrial ATP]
Lower than normal Acetyl-CoA[mitochondrial ADP].
Therefore, option 1, Normal, option 2, Higher than normal, option 3, Lower than normal and option 4, None, is the correct answer.
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help
How many moles of lithium hydroxide would be required to produce 15.0 g of Li₂CO3 in the following chemical reaction? 2 LIOH(s) + CO2 (g) → Li₂CO3 (s) + H₂O (1)
To produce 15.0 g of Li₂CO₃ in the given chemical reaction, 0.406 moles of lithium hydroxide (LiOH) would be required to produce 15.0 g of Li₂CO₃.
The balanced chemical equation shows that 2 moles of LiOH react with 1 mole of CO₂ to produce 1 mole of Li₂CO₃ and 1 mole of H₂O. We can use this stoichiometric ratio to calculate the number of moles of LiOH required.
First, we calculate the molar mass of Li₂CO₃:
2 lithium atoms (2 x 6.94 g/mol) + 1 carbon atom (12.01 g/mol) + 3 oxygen atoms (3 x 16.00 g/mol) = 73.89 g/mol
Next, we can use the molar mass of Li₂CO₃ to convert the given mass (15.0 g) to moles:
Number of moles of Li₂CO₃ = Mass of Li₂CO₃ / Molar mass of Li₂CO₃
Number of moles of Li₂CO₃ = 15.0 g / 73.89 g/mol = 0.203 moles
Since the stoichiometric ratio between LiOH and Li₂CO₃ is 2:1, we can conclude that twice the number of moles of LiOH is required:
Number of moles of LiOH required = 2 x Number of moles of Li₂CO₃ = 2 x 0.203 moles = 0.406 moles
Therefore, approximately 0.406 moles of lithium hydroxide (LiOH) would be required to produce 15.0 g of Li₂CO₃ in the given chemical reaction.
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Titrate 25.00 mL of 0.40M HNO2 with 0.15M KOH, the pH of the
solution after adding 15.00 mL of the titrant is: Ka of HNO2 = 4.5
x 10-4
a. 1.87
b. 2.81
c. 3.89
d. 10.11
e. 11.19 4.
The pH of the solution after adding 15.00 mL of the titrant (0.15M KOH) to 25.00 mL of 0.40M HNO2 is 3.89. Therefore the correct option is C. 3.89
To determine the pH of the solution after the titration, we need to consider the reaction between the HNO2 (nitrous acid) and the KOH (potassium hydroxide). Nitrous acid is a weak acid, and potassium hydroxide is a strong base.
In the initial solution, we have 25.00 mL of 0.40M HNO2. The HNO2 will react with the KOH in a 1:1 ratio according to the balanced equation:
HNO2 + KOH → KNO2 + H2O
Since the volume of the titrant (KOH) added is 15.00 mL and its concentration is 0.15M, we can calculate the amount of KOH reacted. This is equal to (15.00 mL)(0.15 mol/L) = 2.25 mmol.
Considering that the reaction occurs in a 1:1 ratio, the amount of HNO2 consumed is also 2.25 mmol. Initially, we had 25.00 mL of 0.40M HNO2, which corresponds to (25.00 mL)(0.40 mol/L) = 10.00 mmol.
Now, we can calculate the concentration of HNO2 remaining after the reaction:
(10.00 mmol - 2.25 mmol) / (25.00 mL + 15.00 mL) = 7.75 mmol / 40.00 mL = 0.19375 M
To determine the pH, we need to consider the dissociation of HNO2, which is a weak acid. The dissociation of HNO2 can be represented by the equilibrium:
HNO2 ⇌ H+ + NO2-
The Ka of HNO2 is given as 4.5x10^-4. Since the concentration of HNO2 remaining is 0.19375 M, we can use the Ka expression to calculate the concentration of H+ ions:
Ka = [H+][NO2-] / [HNO2]
4.5x10^-4 = [H+]^2 / 0.19375
[H+]^2 = (4.5x10^-4)(0.19375)
[H+]^2 = 8.71875x10^-5
[H+] = √(8.71875x10^-5)
[H+] = 2.953x10^-3 M
Finally, we can calculate the pH using the equation:
pH = -log[H+]
pH = -log(2.953x10^-3)
pH ≈ 3.89
Therefore, the pH of the solution after adding 15.00 mL of the titrant is 3.89, which corresponds to option c.
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4. (a) An HCI solution is labeled 0.1000 M. How is the pH of this solution expressed? (b) Another HCI solution is labeled ~0.1 M. What do we know about the pH of this solution? (c) Suppose this sample
An HCI solution is labeled 0.1000 M. How is the pH of this solution expressed? HCI is a strong acid, which means it completely ionizes in water to produce hydrogen ions (H+) and chloride ions (Cl−).
The concentration of hydrogen ions is used to describe the acidity or basicity of a solution and is expressed as pH. The pH of an HCI solution with a concentration of 0.1000 M is calculated as follows: pH = −log10 [H+]pH = −log10 (0.1000)pH = 1.0000Therefore, the pH of the HCI solution is 1.0000.(b) Another HCI solution is labeled ~0.1 M. What do we know about the pH of this solution?A ~0.1 M HCI solution is less concentrated than a 0.1000 M HCI solution. As a result, it has a higher pH because the concentration of hydrogen ions is lower. The pH of this solution cannot be precisely determined without knowing its actual concentration.
However, because it is an HCI solution, the pH is still expected to be acidic.(c) Suppose this sample had a pH of 2. What is its hydrogen ion concentration? The hydrogen ion concentration of a solution can be calculated using the following formula:[H+] = 10−pH[H+] = 10−2[H+] = 0.0100 M Therefore, the hydrogen ion concentration of the solution is 0.0100 M.
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Learning Objective: Draw the Lewis structure of a given molecule (alcohol, sulfide, amine, aldehyde, ketone, carboxylic acid, ester, or amide), anion or cation. Practice problem: Which of the following compounds has only one lone pair on the central atom? A) CO₂ B) H₂S C) NH3 D) NH E) CS₂
The molecule that has only one lone pair on the central atom among the following compounds is NH3. We know that a Lewis structure is a model that uses electron-dot structures to show how electrons are arranged in molecules.
It is also known as Lewis dot diagrams. Now let's analyze each compound one by one:CO₂: In carbon dioxide, there are two double bonds between the carbon atom and the two oxygen atoms. It doesn't have any lone pair on the central atom.H₂S: In hydrogen sulfide, there is one lone pair on the central atom of sulfur. It doesn't meet the requirement of the problem.NH3: In ammonia, there are three hydrogen atoms bonded to the central nitrogen atom with one lone pair on the nitrogen atom. This compound has only one lone pair on the central atom.NH: In nitrogen, there are three hydrogen atoms bonded to the central nitrogen atom. It doesn't have any lone pair on the central atom.CS₂: In carbon disulfide, there are two double bonds between the carbon atom and the two sulfur atoms. It doesn't have any lone pair on the central atom.Therefore, among the given compounds, NH3 has only one lone pair on the central atom.
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Which of these things are normally found in the filtrate produced at the renal corpuscle? Select all correct answers. Red blood cells Amino acids Sodium ions Large proteins White blood cells
The filtrate produced at the renal corpuscle normally contains amino acids, sodium ions and large proteins. Therefore, the correct options from the given alternatives are; Amino acids, Sodium ions, and Large proteins.
The renal corpuscle is a collection of blood vessels, Bowman's capsule, and capillary blood vessels within the nephron of a mammalian kidney. It functions to filter blood to remove harmful substances like waste, and to filter useful substances like glucose, salt, and water that the body needs to maintain homeostasis. This filtration process is the first step in the creation of urine by the kidneys. A filtrate refers to a liquid or solution that has passed through a filter. It is the fluid that is filtered by the renal corpuscle in the nephron.
The filtrate contains a variety of molecules such as ions, nutrients, and waste products, and it moves through the renal tubules where the final composition of urine is determined.What does the filtrate contain?The filtrate produced at the renal corpuscle typically includes amino acids, glucose, ions (such as sodium, potassium, and chloride), bicarbonate, creatinine, and urea. Large proteins and blood cells are too large to pass through the filtration membrane and therefore should not be present in the filtrate.
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The PK, value of crotonic acid is 4.7. If the H₂O* and crotonate ion concentrations are each 0.0040 M, what is the concentration of the undissociated crotonic acid? Concentration = M
The concentration of undissociated crotonic acid is approximately 0.0036 M, determined using the given pKa value and concentrations of H₂O* and crotonate ion.
The pKa value represents the negative logarithm of the acid dissociation constant (Ka) and indicates the tendency of an acid to donate a proton. The pKa value of crotonic acid is given as 4.7.
Crotonic acid (CH₃CH=CHCOOH) can dissociate into crotonate ion (CH₃CH=CHCOO-) and a proton (H⁺):
CH₃CH=CHCOOH ⇌ CH₃CH=CHCOO⁻ + H⁺
The equilibrium constant (K) for this dissociation can be expressed as:
K = [CH₃CH=CHCOO⁻][H⁺] / [CH₃CH=CHCOOH]
Since the concentrations of H₂O* and crotonate ion are both given as 0.0040 M, we can assume that the concentration of H⁺ is also 0.0040 M (due to water dissociation). Let's denote the concentration of undissociated crotonic acid as x M.
Using the equilibrium constant expression, we can write the equation:
10^(-pKa) = [CH₃CH=CHCOO⁻][H⁺] / [CH₃CH=CHCOOH]
Substituting the given values:
10^(-4.7) = (0.0040)(0.0040) / x
Rearranging the equation to solve for x:
x = (0.0040)(0.0040) / 10^(-4.7)
Calculating the value:
x ≈ 0.0036 M
Therefore, the concentration of the undissociated crotonic acid is approximately 0.0036 M.
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consider the unbalanced redox reaction occuring in acidic solution:
Cr2O7^2-(aq)+Cu(s)-->Cr3+(aq)+Cu2+(aq)
Part A Balance the equation. Express your answer as a chemical equation. Identify all of the phases in your answer. ΑΣΦ O X 2-ª Xx₂ Cr₂O2 (aq) + 3Cu(s) + 14H* (aq)→2Cr³+ (aq) + 3Cu² (aq) +
The balanced redox equation in an acidic solution is:
Cr₂O₇²⁻(aq) + 3Cu(s) + 14H⁺(aq) → 2Cr³⁺(aq) + 3Cu²⁺(aq) + 7H₂O(l)
The given redox reaction involves the dichromate ion (Cr₂O₇²⁻) and copper (Cu) in an acidic solution. The goal is to balance the equation by ensuring that the number of atoms and charges are equal on both sides of the equation.
To balance the equation, we start by assigning oxidation states to each element in the reaction:
Cr₂O₇²⁻: The oxidation state of Cr in Cr₂O₇²⁻ is +6, and each oxygen atom has an oxidation state of -2. By assigning x to the oxidation state of Cr, we can determine that x + 7(-2) = -2. Solving this equation gives x = +6, so the oxidation state of Cr in Cr₂O₇²⁻ is +6.
Cu: The oxidation state of Cu in its elemental form is 0.
Cr³⁺: The oxidation state of Cr in Cr³⁺ is +3.
Cu²⁺: The oxidation state of Cu in Cu²⁺ is +2.
Now, we can see that Cr is reduced from +6 to +3 (gaining 3 electrons), and Cu is oxidized from 0 to +2 (losing 2 electrons).
To balance the charges, we need 3 Cu atoms on the left side to account for the 3 electrons lost during oxidation. This is why we have 3Cu(s) on the left side of the equation.
To balance the number of Cr atoms, we need 2 Cr³⁺ ions on the right side, which is why we have 2Cr³⁺(aq) on the right side of the equation.
Finally, to balance the number of oxygen atoms, we add 7 water molecules (H₂O) to the right side, as each water molecule contains 2 hydrogen atoms and 1 oxygen atom.
Adding 14H+ ions on the left side balances the hydrogen atoms and provides the acidic conditions necessary for the reaction to occur.
The resulting balanced equation is:
Cr₂O₇²⁻(aq) + 3Cu(s) + 14H⁺(aq) → 2Cr³⁺(aq) + 3Cu²⁺(aq) + 7H₂O(l)
In this equation, (aq) represents aqueous (dissolved) species, (s) represents solid species, and (l) represents liquid species.
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Beryllium coppers are the highest strength alloys.
1. True
2. False
The electrical and heat conductivity of copper is not significantly affected by impurites.
1. True
2. False
Aluminum has highter conductivities than most metals.
1. True
2. False
Aluminum is lighter and stiffer than most metals.
1. True
2. False
Zinc coatings are used to protect material from atomospheric rust.
1. True
2. False
Beryllium coppers are the highest strength alloys. (True/False)The statement "Beryllium coppers are the highest strength alloys" is True.
The beryllium copper alloy is the strongest of all copper alloys. It has a variety of useful properties, including high corrosion resistance, ductility, electrical conductivity, and thermal conductivity. Beryllium copper alloys are used in a variety of applications, including automotive, electronic, aerospace, and defense industries.The electrical and heat conductivity of copper is not significantly affected by impurites. (True/False)The statement "The electrical and heat conductivity of copper is not significantly affected by impurities" is False.
Although pure copper is an excellent conductor of electricity and heat, the presence of impurities reduces its conductivity. Impurities can include trace amounts of oxygen, carbon, or other metals. Copper conductors in electrical systems must be pure and free of impurities to achieve optimum performance.Aluminum has higher conductivities than most metals. (True/False)The statement "Aluminum has higher conductivities than most metals" is False.Long Answer:While aluminum is a good conductor of electricity and heat, it is not the best. Silver and copper have higher conductivities than aluminum. Aluminum conductors, on the other hand, are less expensive and weigh less than copper conductors.
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1. What is the pH of 4.3×10-3 M
HCl?4.3×10-3 M HCl?
pH =
2. What is the pH of 8×10-8 M HCl?8×10-8 M
HCl?
pH =
1. The pH of 4.3×10-3 M HCl is 2.37.
2. The pH of 8×10-8 M HCl is 7.10.
The pH of a solution is a measure of its acidity or alkalinity and is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (H+).
1. For 4.3×10-3 M HCl:
The concentration of H+ ions in HCl is equal to the concentration of the acid itself. Therefore, the concentration of H+ ions is 4.3×10-3 M.
Taking the negative logarithm of the concentration:
pH = -log[H+]
pH = -log(4.3×10-3)
pH ≈ 2.37
2. For 8×10-8 M HCl:
Again, the concentration of H+ ions in HCl is equal to the concentration of the acid itself. Thus, the concentration of H+ ions is 8×10-8 M.
Calculating the pH:
pH = -log[H+]
pH = -log(8×10-8)
pH ≈ 7.10
The pH of 4.3×10-3 M HCl is 2.37, indicating acidity, while the pH of 8×10-8 M HCl is 7.10, indicating neutrality. Lower pH values correspond to higher acidity, while higher pH values indicate alkalinity.
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8. Groups A and B performed separate DNA extractions on the same specimen using the same extraction protocol. Their PCR products were run at 100 V for 45 minutes on a 1% agarose gel with the results shown below. Group A loaded PCR ladder (also shown below) in lane 1 , a positive control in lane 2 , a negative control in lane 3 , and their DNA sample in Lane 4. Group B loaded the same sequence of ladder and PCR products in wells 6 through 9. Provide an analysis of these results; include approximate sizes of the amplified PCR products. If there are any unexpected results, provide a troubleshooting solution to improve the group's future PCR attempts using this protocol.
The gel electrophoresis results show the amplified PCR products obtained by Group A and Group B. Group A's PCR products are observed in Lane 4, while Group B's PCR products are observed in wells 6 through 9. The approximate sizes of the amplified PCR products can be determined by comparing their migration distances to the DNA ladder. Any unexpected results in the gel can be addressed by troubleshooting the PCR protocol for future attempts.
To analyze the gel electrophoresis results, we compare the migration distances of the PCR products to the DNA ladder, which contains known DNA fragments of different sizes. By visually inspecting the gel, we can estimate the approximate sizes of the PCR products based on their positions relative to the ladder.It is important to note that the provided information does not specify the number or size of the DNA fragments in the ladder or the expected sizes of the PCR products. Therefore, a specific analysis of the results cannot be provided without additional information.If there are any unexpected results observed in the gel, such as missing bands, faint bands, or smearing, it indicates potential issues with the PCR protocol. To improve future PCR attempts using the same protocol, the following troubleshooting solutions can be considered:
1. Verify the quality and integrity of the DNA template: Ensure that the DNA template used for PCR is of high quality and not degraded. Check the concentration and purity of the DNA using spectrophotometry or other methods.
2. Optimize PCR conditions: Adjust the annealing temperature, extension time, or primer concentrations to optimize the PCR conditions for the specific DNA target.
3. Check primer design: Ensure that the primers used in PCR are designed correctly, with appropriate melting temperatures and specific to the target DNA sequence.
4. Evaluate PCR components: Check the quality and integrity of PCR reagents, including the DNA polymerase, dNTPs, and buffer solutions. Consider using fresh reagents or alternative suppliers.
5. Minimize contamination: Implement strict measures to prevent contamination, such as using separate areas for sample preparation and PCR setup, using filter tips, and regularly decontaminating work surfaces and equipment.
By addressing these troubleshooting strategies, Group A can improve their future PCR attempts and obtain more reliable and consistent results using the same extraction protocol.
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NEED TWO QUESTION ANSWER.
Describe what happens when ionic and covalent (molecular) substances dissolve. A(n) A(n) aqueous covalent compound dissolved in water, H₂O(1), will produce dissolved in water, H₂O(l), will produce
The Na+ and Cl- ions are free to move in the solution.
Ionic compounds dissociate in water while covalent (molecular) substances dissolve to form a solution.
When ionic compounds dissolve in water, they dissociate into their individual ions (cation and anion). These ions are free to move in the solution.
On the other hand, covalent (molecular) substances dissolve in water without breaking apart into their individual molecules. Instead, the covalent compound forms a solution through intermolecular interactions with water molecules.
A(n) aqueous covalent compound dissolved in water, H₂O(l), will produce a homogeneous solution.
For example, when sugar (a covalent compound) dissolves in water, it forms a homogeneous solution.
The sugar molecules interact with water molecules through hydrogen bonding, which allows them to dissolve in water. On the other hand, when an ionic compound dissolves in water, it produces a solution of free ions.
For example, when sodium chloride (NaCl) dissolves in water, it dissociates into its component ions, Na+ and Cl-.
The Na+ and Cl- ions are free to move in the solution.
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Which of the phthalic acids - ortho, meta, or para - would you use to prepare phthalic anhydride by heating? Explain your answer. 9. You want to prepare beta-chloropropionic acid. a) Is direct halogen
To prepare phthalic anhydride by heating, ortho-phthalic acid would be the suitable choice.
Phthalic anhydride is typically synthesized by the oxidation of ortho-xylene or naphthalene. However, if one wants to prepare phthalic anhydride from phthalic acid, ortho-phthalic acid is the most appropriate choice. This is because ortho-phthalic acid possesses the necessary chemical structure and reactivity for the conversion into phthalic anhydride.
The structure of ortho-phthalic acid consists of two carboxylic acid groups attached to a central benzene ring. When ortho-phthalic acid is heated, it undergoes a process called decarboxylation, where carbon dioxide (CO2) is eliminated, resulting in the formation of phthalic anhydride. The proximity of the carboxylic acid groups in the ortho position enables the intramolecular reaction required for the conversion.
In contrast, meta-phthalic acid and para-phthalic acid have their carboxylic acid groups attached at different positions on the benzene ring. This arrangement makes the intramolecular decarboxylation less favorable and difficult to occur. Consequently, ortho-phthalic acid is the preferred choice when aiming to prepare phthalic anhydride by heating.
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6. One of the roles of the kidneys is to help buffer body fluids so that they are not too acidic or too basic. The cells of the renal tubule secrete H+ into the tubule lumen and absorb bicarbonate (HC
true
false
One of the roles of the kidneys is to help buffer body fluids and maintain their pH within a narrow range. The cells of the renal tubule secrete hydrogen ions (H+) into the tubule lumen and absorb bicarbonate ions (HCO3-) from the tubular fluid.
The kidneys play a vital role in maintaining the acid-base balance of the body. One way they achieve this is through the regulation of hydrogen ions (H+) and bicarbonate ions (HCO3-).
In the renal tubule, specialized cells actively secrete hydrogen ions into the tubule lumen. This process is known as tubular secretion. By secreting hydrogen ions, the kidneys can help eliminate excess acids from the body and regulate the pH of the urine.
Simultaneously, the renal tubule cells reabsorb bicarbonate ions from the tubular fluid. Bicarbonate ions are important buffers that can neutralize excess acids in the body. By reabsorbing bicarbonate, the kidneys can maintain the balance of these ions and prevent excessive acidification of body fluids.
This coordinated secretion of hydrogen ions and absorption of bicarbonate ions by the cells of the renal tubule contribute to the kidneys' role in buffering body fluids and preventing excessive acidity or alkalinity.
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Identify both functional groups in the following molecule: 0 || CH3-CH2-C-CH2-CH2-CH2-C-NH2 The functional groups present are 11 and
The functional groups present in this molecule are -NH2 and a carbonyl group.
The given molecule is 0 || CH3-CH2-C-CH2-CH2-CH2-C-NH2. The functional groups present in this molecule are -NH2 and a carbonyl group. The -NH2 group is an amine functional group that comprises a nitrogen atom attached to two hydrogen atoms. Amino groups are electron-donating groups that increase the reactivity of the molecule they are present in. The carbonyl group is a functional group that comprises a carbon atom linked by a double bond to an oxygen atom.
The carbonyl group is found in aldehydes, ketones, and carboxylic acids. They tend to undergo nucleophilic addition reactions. It has two types, one is aldehyde functional group which is present at the end of the carbon chain and the other is the ketone functional group that is present in the middle of the carbon chain. So, in the given molecule, the carbonyl group is present in the center of the carbon chain while the -NH2 group is attached to one end of the carbon chain. Therefore, the functional groups present are -NH2 and a carbonyl group.
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Which of the following would produce a buffer
system?
Group of answer choices
Weak acid and/or weak base
Strong acid only
Strong acid and/or strong base
Strong base only
Weak base only.
A buffer system is produced by a weak acid and/or weak base.
A buffer system is a solution that resists changes in pH when small amounts of acid or base are added to it. It consists of a weak acid and its conjugate base or a weak base and its conjugate acid. The weak acid or base can donate or accept protons, helping to maintain the pH of the solution within a certain range.
When a small amount of acid is added to a buffer solution, the weak base component of the buffer reacts with the added acid, preventing a significant decrease in pH. Similarly, when a small amount of base is added, the weak acid component of the buffer reacts with the added base, preventing a significant increase in pH. This ability to resist changes in pH is essential in biological systems and many chemical processes.
In contrast, a strong acid or strong base alone does not produce a buffer system. Strong acids completely dissociate in water, releasing all their protons, while strong bases completely dissociate to release hydroxide ions. As a result, strong acids and bases do not have the capacity to maintain a stable pH in the presence of small amounts of added acid or base.
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11) which of the following cannot be a colloid? a) a foam b) an emulsion c) an aerosol d) all of the above are colloids
Answer:
d) all of the above are colloids
Explanation:
A colloid is a mixture where one substance is dispersed evenly in another substance, typically with particles or droplets suspended in a different medium. Foam is a colloid composed of gas bubbles dispersed in a liquid or solid. An emulsion is a colloid consisting of droplets of one liquid dispersed in another immiscible liquid. An aerosol is a colloid where small solid or liquid particles are suspended in a gas.
The option that cannot be a colloid is D. All of the above are colloids.
A colloid is a mixture containing tiny undissolved particles suspended within another substance. They are also known as colloidal suspensions or dispersions. Examples of colloids include milk, fog, and jellies. All of the above are colloids because they all have tiny undissolved particles suspended within another substance.
a) Foam: A foam is a colloid of gas dispersed within a liquid or solid .b) Emulsion: An emulsion is a colloid of two or more immiscible liquids .c) Aerosol: An aerosol is a colloid of liquid or solid particles suspended in a gas.All of these examples meet the definition of a colloid, therefore the correct option is D. All of the above are colloids.
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writing should be neat and clean and answer should be
of all parts and correct for upvote
need answer within two hours
Problem 6. Assume ethane combustion in air: C₂H6+0₂= 2C0₂ + 3H₂O a. Find LFL, UFL, and LOC (limiting oxygen concentration) b. If LOL and UOL of ethane are 3.0% fuel in oxygen and 66% fuel in o
LFL: 70 V of air/mole of C₂H₆ UFL: 23.3 V of air/mole of C₂H₆ LOC: 14.7% (vol.) For the LOL and UOL of ethane, LOL: 1.167 L of O₂ per mole of C₂H₆ UOL: 0.053 L of O₂ per mole of C₂H₆
a. C₂H6+3.5O₂→ 2CO₂+ 3H₂O 2 moles of CO₂ are produced in the reaction for 1 mole of ethane combustion, and we assume that air has 21% O₂ by volume. Therefore, the volume of air required for complete combustion of ethane would be 3.5/0.21 = 16.67 (approx.)
Volume of air per mole of ethane. Now, for LFL we can assume that 1 mole of ethane is mixed with x moles of air, where the mixture doesn't support a flame. In this scenario, the mixture should contain 5% ethane, therefore, we can calculate the volume of air needed for a 5% ethane mixture, which is 3.5/0.05 = 70 moles of air per mole of ethane. Therefore, the volume of air required for a LFL mixture would be (70-x) moles.
2C₂H₆ + 7(O₂ + 3.76N₂) → 4CO₂ + 6H₂O + 29.68
N₂C₂H₆ + 3.5(O₂ + 3.76N₂) → 2CO₂ + 3H₂O + 13.96N₂ at LFL,
percentage of fuel = 5%V of air (at LFL) per mole of C₂H₆
= 70 LFL occurs when C₂H₆ is mixed with a minimum volume of air that is 70 L.
Therefore, the volume of air required for a UFL mixture would be (23.3-y) moles. 2C₂H₆ + 7(O₂ + 3.76N₂) → 4CO₂ + 6H₂O + 29.68N₂C₂H₆ + 6.5(O₂ + 3.76N₂) → 2CO₂ + 3H₂O + 29.68N₂ at UFL,
percentage of fuel = 15%V of air (at UFL) per mole of C₂H₆
= 23.3 LOC (limiting oxygen concentration):
C₂H₆ + 3.5(O₂ + 3.76N₂) → 2CO₂ + 3H₂O + 13.96N₂
Therefore, 3.5 moles of air are required per mole of ethane for stoichiometric combustion.
2C₂H₆ + 7(O₂ + 3.76N₂) → 4CO₂ + 6H₂O + 29.68N₂
Therefore, 7 moles of O₂ are required for stoichiometric combustion of ethane. The volume of air is calculated as:3.5/0.21 = 16.67 moles of air per mole of ethane.
Therefore, the volume of air required for combustion per mole of ethane would be 16.67 moles. 2C₂H₆ + 7(O₂ + 3.76N₂) → 4CO₂ + 6H₂O + 29.68
N₂ at LOC, volume % O₂ = 14.7%Volume % of air = 100 - 14.7 = 85.3%
Therefore, the required limiting oxygen concentration is 14.7% (vol.)
Combustion of ethane in oxygen: For the combustion of ethane in oxygen, the balanced equation is given by: C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O
Stoichiometric ratio = 3.5 moles of O₂ per mole of ethane, and LOL (limiting oxygen concentration) and UOL (upper oxygen concentration) of ethane are given as 3% and 66% fuel in oxygen, respectively. Let x moles of ethane be mixed with 100 moles of O₂. We can write the equation for combustion as:
C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O
For LOL, we assume that 3% of ethane is mixed with 100 moles of O₂.
x = 3/100 * 100 = 3 moles of ethane
C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O (3/1) (3.5/1)
100 moles of O₂ = 357.14 moles of air
V of air per mole of C₂H₆ = 357.14/3
= 119.05 V of O₂ per mole of C₂H₆
= 3.5/3
= 1.167
LOL occurs when C₂H₆ is mixed with a minimum volume of oxygen that is 1.167 L. Let y moles of ethane be mixed with 100 moles of O₂. We can write the equation for combustion as:
C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O
For UOL, we assume that 66% of ethane is mixed with 100 moles of O₂.
y = 66/100 * 100
= 66 moles of ethane
C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O
(66/1) (3.5/1)100 moles of O₂ = 357.14 moles of air
V of air per mole of C₂H₆ = 357.14/66
= 5.41 V of O₂ per mole of C₂H₆ = 3.5/66
= 0.053UOL occurs when C₂H₆ is mixed with a maximum volume of oxygen that is 0.053 L.
Therefore, the LFL, UFL, and LOC (limiting oxygen concentration) are:
LFL: 70 V of air/mole of C₂H₆ UFL: 23.3 V of air/mole of C₂H₆ LOC: 14.7% (vol.)
For the LOL and UOL of ethane, LOL: 1.167 L of O₂ per mole of C₂H₆ UOL: 0.053 L of O₂ per mole of C₂H₆.
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In an atom that has not undergone any type of chemical reaction, the number of electron
Group of answer choices
- is always an odd number
- is always an even number
- always equal to the number of neutrons
- the number of electrons in the outermost shell
The number of electrons in an atom is determined by the atomic number and can vary, but it is not always odd or even, equal to the number of neutrons, or solely determined by the outermost shell.
The number of electrons in an atom is determined by the atomic number, which is specific to each element and corresponds to the number of protons in the nucleus. In a neutral atom, the number of electrons is also equal to the number of protons. For example, a neutral oxygen atom has 8 electrons because oxygen has an atomic number of 8.
The atomic number and the arrangement of electrons in an atom determine the electron configuration. Electrons occupy different energy levels or shells around the nucleus, and each shell can hold a specific number of electrons. The outermost shell, known as the valence shell, is particularly important for chemical reactions as it determines the atom's reactivity.
The number of electrons in the outermost shell is related to the atom's position in the periodic table. Elements in the same group have similar chemical properties because they have the same number of electrons in their outermost shell. However, this number is not the sole factor in determining the total number of electrons in an atom.
In summary, the number of electrons in an atom that has not undergone a chemical reaction depends on the element's atomic number and electron configuration, but it is not always odd or even, equal to the number of neutrons, or solely determined by the number of electrons in the outermost shell.
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Add or subtract the following measurements. Be sure each answer you enter contains the correct number of significant digits. 6.527 g 1.3 g 11.60 g + 1.220 g 15.52 g 1.400 g = = = 0 g g X 3 c.
The sum of the measurements is approximately 37.6 g.
To add or subtract measurements, we need to consider the significant figures and align the decimal places. Let's perform the calculation step by step:
6.527 g +
1.3 g +
11.60 g +
1.220 g +
15.52 g +
1.400 g =
37.567 g
Since the measurement with the fewest decimal places is 1.3 g (2 significant figures), the final answer should be rounded to that level of precision:
37.567 g ≈ 37.6 g
Therefore, the sum of the measurements is approximately 37.6 g.
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Calculate density of Methane taking in account compressibility factor ( by chart ) for temperature 185 K and pressure 10 Mpa
The density of methane at a temperature of 185 K and a pressure of 10 MPa, considering the compressibility factor from a chart, can be calculated.
However, the specific chart or equation required to determine the compressibility factor is not mentioned in the question. Therefore, I am unable to provide an exact numerical value for the density.
To calculate the density of methane under these conditions, you would need to consult a chart or equation that provides the compressibility factor for methane at the given temperature and pressure. The compressibility factor takes into account the deviation of a real gas from ideal gas behavior, considering factors such as intermolecular interactions and non-ideal conditions. Once you obtain the compressibility factor, you can multiply it by the density of ideal methane gas at the same temperature and pressure (which can be determined from gas laws or reference tables) to obtain the density of methane accounting for compressibility.
It is essential to refer to a specific chart or equation for the compressibility factor of methane to obtain an accurate value for the density. The compressibility factor corrects for the non-ideal behavior of gases, which is particularly important at high pressures and low temperatures. By incorporating the compressibility factor into the calculation, you can obtain a more precise density value that reflects the real-world behavior of methane gas under the given conditions.
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According to the following reaction, how many moles of ammonia
will be formed upon the complete reaction of 0.899 moles nitrogen
gas with excess hydrogen gas?
N2 (g) +3H2 (g) -> 2NH3 (g)
_____mol a
Answer:
1.798 mol of ammonia gas
Nitrogen and hydrogen combine at a high temperature, in the
presence of a catalyst, to produce ammonia.
N2(g)+3H2(g)⟶2NH3(g)N2(g)+3H2(g)⟶2NH3(g)
Assume 0.260 mol N20.260 mol N2 and
Using the balanced chemical equation N2(g) + 3H2(g) ⟶ 2NH3(g), we can determine the moles of ammonia produced when 0.260 mol of nitrogen gas (N2) reacts. when 0.260 mol of nitrogen gas reacts, 0.520 mol of ammonia is produced.
According to the balanced chemical equation N2(g) + 3H2(g) ⟶ 2NH3(g), the stoichiometric ratio is 1:2:2 for nitrogen gas, hydrogen gas, and ammonia, respectively.
Given that we have 0.260 mol of nitrogen gas (N2), we can use the stoichiometry to determine the amount of ammonia produced. Since the ratio of N2 to NH3 is 1:2, we multiply the moles of N2 by the conversion factor (2 moles NH3/1 mole N2) to find the moles of NH3 produced.
0.260 mol N2 × (2 moles NH3/1 mole N2) = 0.520 mol NH3
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The absorbance of a 15% green food colouring solution compare to
10% of the same solution, what the calibration curve would be?
The calibration curve for comparing the absorbance of a 15% green food coloring solution to that of a 10% solution can be generated by plotting the absorbance values against the concentration of the solutions. The resulting curve will help establish a relationship between absorbance and concentration, allowing for the determination of the concentration of unknown samples based on their absorbance values.
To create the calibration curve, several solutions with known concentrations of the green food coloring (including 10% and 15% solutions) are prepared. The absorbance of each solution is measured using a spectrophotometer at a specific wavelength, typically associated with the absorption peak of the coloring compound.
The absorbance values are then plotted on the y-axis, while the corresponding concentrations are plotted on the x-axis. By fitting a curve or line to the data points, the calibration curve is obtained. This curve can be used to determine the concentration of unknown samples by measuring their absorbance and extrapolating from the calibration curve.
It is important to note that the calibration curve should be generated using a range of known concentrations that cover the expected concentration range of the samples to ensure accurate and reliable measurements.
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Write a chemical balance equation Based on the following
description Solid barium carbonate decomposes into solid barium
oxide carbon dioxide gas when heated
The chemical balanced equation for the decomposition of solid barium carbonate into solid barium oxide and carbon dioxide gas when heated is:
BaCO3(s) → BaO(s) + CO2(g)
When solid barium carbonate (BaCO3) is heated, it undergoes a decomposition reaction, breaking down into solid barium oxide (BaO) and carbon dioxide gas (CO2). The solid barium carbonate is represented by the formula BaCO3, where Ba is the symbol for barium, C represents carbon, and O stands for oxygen.
During the reaction, the heat energy causes the solid barium carbonate to break apart, forming solid barium oxide and releasing carbon dioxide gas as a byproduct. The solid barium oxide is represented by the formula BaO, and the carbon dioxide gas is represented by CO2.
The balanced equation represents the conservation of mass, ensuring that the number of atoms of each element is the same on both sides of the equation. In this case, there is one atom of barium, one atom of carbon, and three atoms of oxygen on the reactant side (BaCO3) and one atom of barium, one atom of carbon, and three atoms of oxygen on the product side (BaO + CO2).
Overall, the balanced equation BaCO3(s) → BaO(s) + CO2(g) accurately represents the decomposition of solid barium carbonate into solid barium oxide and carbon dioxide gas when heated.
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An aluminum plant produces 91 million kg of aluminum per year.
How much bauxite is required? (It takes 2.1 kg of crude bauxite to produce 1.0 kg of aluminum oxide.) Express your answer to two signifi
To determine the amount of bauxite required to produce 91 million kg of aluminum per year, we need to consider the conversion factor from crude bauxite to aluminum oxide.
Given that it takes 2.1 kg of crude bauxite to produce 1.0 kg of aluminum oxide, we can calculate the required amount of bauxite by multiplying the aluminum production by the conversion factor. The result will provide the amount of bauxite needed in kilograms.
Since it takes 2.1 kg of crude bauxite to produce 1.0 kg of aluminum oxide, the ratio of bauxite to aluminum oxide is 2.1:1. To calculate the amount of bauxite required to produce 91 million kg of aluminum, we multiply the aluminum production by the conversion factor.
91 million kg of aluminum × (2.1 kg of crude bauxite / 1 kg of aluminum oxide) = 191.1 million kg of crude bauxite.
Therefore, approximately 191.1 million kg of crude bauxite is required to produce 91 million kg of aluminum per year.
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