Aluminum and iron oxide (Fe₂O₃) react together to produce aluminum oxide, 20 g of Fe₂O3 will produce 14 g of Fe.
What is meant by oxide?Oxide is a category of chemical compound that has one or more oxygen atoms and also another element in its composition.
2Al + Fe₂O₃ -> Al₂O₃ + 2Fe
Molar mass of Fe₂O₃ is: (2 x 56) + (3 x 16) = 160 g/mol
moles of Fe2O3 = 20 g / 160 g/mol = 0.125 mol
As the reaction is carried out with an excess of Al, we can assume that all of the Fe₂O₃ is consumed and that 2 moles of Fe are produced for every mole of Fe₂O₃
moles of Fe = 2 x moles of Fe₂O₃ = 0.25 mol
Molar mass of Fe is : 56 g/mol
So the mass of Fe produced is:
mass of Fe = moles of Fe x molar mass of Fe
mass of Fe = 0.25 mol x 56 g/mol = 14 g
Therefore, 20 g of Fe₂O₃ will produce 14 g of Fe.
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Suppose 2,560 grams of low-level radioactive waste is buried at a waste disposal site. Assume that 10 grams of radioactive material gives off an acceptable level of radiation and that one half-life is 5.26 years. Write a paragraph in which you explain to townspeople how much time must pass before there is an acceptable ratiation level at the site.
However, keep in mind that 20 mSv per year is the recommended amount for any radiation worker and is still regarded quite safe. This is the most radiation most of us will ever be exposed to.
after 1st half life , remaining sample would be 100/2=50 g
after 2nd half life , remaining sample would be 50/2=25 g
after 3rd half life , remaining sample would be 25/2= 12.5 g
What is a radioactive material's half-life?The half-life of a radionuclide is the amount of time it takes for half of its radioactive atoms to decay. A decent rule of thumb is that you will have less than 1% of the initial quantity of radiation after seven half-lives. Click here to learn more about half life.
A short-term and whole-body dosage would result in rapid sickness, such as nausea and a reduction in white blood cell count, followed by death.
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42.08 years of time must pass before there is an acceptable radiation level at the site.
What is the half-life of a radioactive material?
The half life of a radioactive substance is the period of time during which its mass or number of atoms is decreased to half of what it was initially. The time it takes for a radioactive substance (or half of its atoms) to break down or transform into another substance is commonly used to define half-life.
Radioactivity, as its name suggests, is the act of generating radiation without any external cause. This is accomplished by an atomic nucleus that is unstable for whatever reason and "wants" to surrender some energy in order to change its configuration to one that is more stable.
After first half-life will remain 2560/2 i.e. 1280g of radioactive substance.
After second half-life will remain 1280/2 i.e. 640g
After 3rd half-life will remain 640/2 i.e. 320g
After 4th half-life will remain 160g
After 5th half-life will remain 80g
After 6th, 7th and 8th half-life will remain 40g, 20g and 10g respectively
It takes 8 half-lives to reach acceptable level of radiation i.e. 8*5.26 years
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an acid and base react to form a salt and water in a(n) reaction. group of answer choices oxidation ionization neutralization reduction dissociation
An acid and a base react, to form salt and water in a neutralization reaction. The correct answer is neutralization.
Neutralization reaction is a chemical reaction that takes place when an acid and a base are mixed, resulting in the creation of a neutral solution, a salt, and water. The acid and base cancel each other out, producing a solution that is neither acidic nor basic.
An example of a neutralization reaction is the reaction between hydrochloric acid and sodium hydroxide, which creates salt and water as the only products.
The reaction can be represented as follows:
[tex]HCl[/tex] + [tex]NaOH[/tex] →[tex]NaCl[/tex] + [tex]H_2O[/tex]
An acid is a compound that contains hydrogen ions[tex](H^+)[/tex] and can dissolve in water to produce a sour taste.
Bases are chemical substances that generate hydroxide ions [tex](OH^-)[/tex] when dissolved in water.
Salt is a compound that can be formed by combining an acid and a base. Therefore, "neutralization" is the correct answer.
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in order for the reaction shown to occur, the procedure calls for the addition of sulfuric acid to nitric acid. what might be the consequence of adding nitric acid to the sulfuric acid instead? the nitration would occur exclusively at the meta position. the nitration would occur exclusively at the ortho position. there are no regioselectivity or safety consequences. the exothermic reaction could cause the concentrated acid to boil.
Adding nitric acid to sulfuric acid could cause the concentrated acid to boil, potentially leading to safety hazards.
The result of adding nitric corrosive to sulfuric corrosive rather than the opposite way around is that the exothermic response could make the concentrated corrosive bubble, possibly prompting security risks. Nitration responses regularly require a combination of sulfuric corrosive and nitric corrosive as the nitrating specialist, as the sulfuric corrosive goes about as an impetus to produce the electrophilic nitrating species. Assuming that nitric corrosive were added first, the concentrated sulfuric corrosive might actually bubble, which could bring about the deficiency of reactants, decline in yield, and posture wellbeing perils because of the potential for arrival of harmful exhaust and additionally blast.
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why does hydrogen bonding lead to alpha helices
In terms of the specific question asked, hydrogen bonding leads to alpha helices because it stabilizes the structure of proteins.
Hydrogen bonds can form between the carbonyl and amino groups in a protein backbone, which results in the formation of a helix structure.
This helix structure is the alpha helix, and it is a common protein secondary structure.
The hydrogen bonds that form between the carbonyl and amino groups are responsible for the stability of the alpha helix, and without these hydrogen bonds, the structure would be unstable and could not exist.
Therefore, hydrogen bonding is essential to the formation of alpha helices in proteins.
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how many electrons (including bonding electrons) are around the sulfur atom in h2so4 ? in other words, how many valence electrons are in the orbitals of the s atom in this molecule?
According to the Lewis structure of [tex]H_{2} SO_{4}[/tex] each hydrogen has one, and sulfur and oxygen each have 6 valence electrons.
The Lewis Structure can be representation of the valence shell electrons in a molecule. The Lewis representation is used to show how the electrons are arranged around individual atoms of a molecule. Electrons in the lewis structure are represented as "dots". This is also known as Lewis dot formulas, Lewis dot structure and electron dot structures or Lewis electron dot structures.
Valence electrons are defined as the electrons present in the outermost shell or in the energy level of an atom. It is an electron in the outer shell associated with an atom which can participate in the formation of a chemical bond of the molecule.
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a litre container holds hydrogen gas. an electric discharge is applied to the gas, resulting in the dissociation of some of the h2 to h. after the discharge, the system is at 1.1 atm, 298 k, and the mole fraction of h is 5%. assuming that the only final product in the system is h2, what is the final system temperature and pressure? you may assume constant specific heat. properties at 298 k and 1 atm:
Q is the heat transferred, m is the mass, C is the specific
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since we are given the pressure and temperature of the system, we need to find the volume and number of moles of gas.
First, we can find the number of moles of hydrogen that dissociates from the H₂ gas. If the mole fraction of H is 5%, then the mole fraction of H₂ is 95%. Let's assume we start with 1 mole of H₂ . Then, we have:
0.05 moles of H
0.95 moles of H2
Some of the H₂ gas dissociates to H, so let's say x moles of H₂ dissociate. Then, we have:
(0.05 + x) moles of H
(0.95 - x) moles of H₂
Since we are assuming that the only final product is H₂, we know that all of the H atoms will recombine to form H₂:
(0.05 + x - 2x) moles of H₂
(0.95 - x) moles of H2
Simplifying this, we get:
(0.05 - x) moles of H₂
(0.95 - x) moles of H₂
Since we started with 1 mole of H₂, we know that the total number of moles is:
(0.05 - x) + (0.95 - x) = 1 - 2x
Now, we can use the ideal gas law to find the volume of gas at the final pressure and temperature. Since we know the number of moles of gas and the temperature, we just need to find the volume. We can rearrange the ideal gas law to get:
V = (nRT)/P
At 298 K and 1 atm, the gas constant R is 0.08206 L·atm/(mol·K). So, for the initial conditions, we have:
V_initial = (1 mol * 0.08206 L·atm/(mol·K) * 298 K) / 1 atm = 24.4658 L
Now, we can use the mole fractions and the number of moles to find the final volume of gas. We have:
(0.05 - x) moles of H₂
(0.95 - x) moles of H₂
The total number of moles is 1 - 2x, so we have:
(0.05 - x)/(1 - 2x) moles of H₂
(0.95 - x)/(1 - 2x) moles of H₂
Using the ideal gas law again, we can find the final volume:
V final = [(0.05 - x)/(1 - 2x) + (0.95 - x)/(1 - 2x)] * (0.08206 L·atm/(mol·K) * T) / 1.1 atm
Simplifying this, we get:
V final = [1 - 2x] * (0.08206 L·atm/(mol·K) * T) / (1.1 atm)
Now, we need to use the fact that the specific heat is constant to find the final temperature. We can use the formula:
Q = mCΔT
where Q is the heat transferred, m is the mass, C is the specific
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15 grams of nh3 are placed in a 1.4 liter container with no reactant present. at equilibrium 7.91 g of nh3 are present. find the equilibrium constant, kc.
The equilibrium constant, Kc, is 2.93 x 10^-4.
The balanced equation for the reaction is;
NH₃ (g) ⇌ ½ N₂ (g) + 3/2 H₂ (g)
The equilibrium expression for the reaction is;
Kc = ([N₂[tex]]^{(1/2)}[/tex] [H₂[tex]]^{(3/2)}[/tex]) / [NH₃]
We are given the initial and equilibrium concentrations of NH₃. The change in NH₃ concentration is;
Δ[NH₃] = [NH₃]equilibrium - [NH₃]initial = 7.91 g/L - 15 g/L = -7.09 g/L
Since NH₃ is a reactant, its concentration decreases at equilibrium. Therefore, we can assume that the reaction proceeds in the forward direction, and that the concentration of N₂ and H₂ at equilibrium are both equal to x.
Using the ideal gas law, we can find the concentration of NH3 at equilibrium; PV = nRT
n/V = P/RT
[n(NH₃)]equilibrium = (P/RT) x V
[n(NH₃)]initial = (P/RT) x V
Substituting the values, we get;
[n(NH₃)]equilibrium = (1 atm / (0.0821 L·atm/mol·K x 298 K)) x 1.4 L = 0.0652 mol/L
[n(NH₃)]initial = (1 atm / (0.0821 L·atm/mol·K x 298 K)) x 1.4 L = 0.0652 mol/L
Using the balanced equation, we can relate the concentration of NH3 to the concentrations of N₂ and H₂.
[NH₃]initial = 0.0652 mol/L
[N₂]equilibrium = [H₂]equilibrium = x
[N₂]initial = [H₂]initial = 0 mol/L
Substituting into the equilibrium expression, we get;
Kc = ([N₂[tex]]^{(1/2)}[/tex] [H₂[tex]]^{(3/2)}[/tex]) / [NH₃]
Kc = ([tex]X^{(1/2)}[/tex] [tex]X^{(3/2)}[/tex]) / (0.0652 - 7.09x10⁻³)
Kc = 2.93 x 10⁻⁴
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why does starch test negative in benedict's assay, even though starch contains a reducing group
The starch tests negative in Benedict's assay even though it contains a reducing group, the answer is as follows:Starch tests negative in Benedict's assay even though it contains a reducing group because the reducing group is not exposed due to the molecule's highly branched structure.
Benedict's assay is used to test for reducing sugars, which are monosaccharides or disaccharides that have a free carbonyl group that can reduce copper ions in the reagent to form a red precipitate.
However, the glucose units in starch are joined together by alpha glycosidic bonds, which link the anomeric carbon of one glucose molecule to the hydroxyl group of another glucose molecule, forming a highly branched structure.
This structure makes it difficult for Benedict's reagent to react with the reducing groups in the glucose units, resulting in a negative test.
Therefore, even though starch contains a reducing group, it will not test positive in Benedict's assay due to its highly branched structure that hides the reducing group.
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A compound contains 79.3% tungsten (W) and 20.7 % oxygen. What is the empirical formula?
Answer:
To find the empirical formula of a compound, we need to determine the simplest whole number ratio of the atoms present in the compound. Let's assume we have 100 g of this compound. Then we have 79.3 g of tungsten and 20.7 g of oxygen. We can convert these masses to moles by dividing by their respective atomic masses: Number of moles of W = 79.3 g / 183.84 g/mol = 0.431 mol Number of moles of O = 20.7 g / 16.00 g/mol = 1.294 mol To get the simplest whole number ratio of the atoms, we need to divide both of these values by the smallest value, which is 0.431 mol: 0.431 mol W / 0.431 mol = 1 1.
What pressure is exerted by 0.750 mol of a gas at a temperature of 0.00ºC and a volume of 5.00 L?
format: pv= nrt
The pressure exerted by 0.750 mol of gas at a temperature of 0.00ºC and a volume of 5.00 L is 101325 Pa.
To calculate the pressure exerted by a gas, we can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in kelvin.
First, we need to convert the temperature from Celsius to Kelvin:
T = 0.00ºC + 273.15 = 273.15 K
Next, we can plug in the given values and solve for the pressure:
P = nRT/V
P = (0.750 mol)(8.314 J/mol·K)(273.15 K)/(5.00 L)
P = 101325 Pa
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Calculate n−factor of Cl 2 in Cl 2 →ClO 3 − +Cl −
The n-factor of [tex]Cl_2[/tex] is 2.5 in this reaction.
To calculate the n-factor of [tex]Cl_2[/tex] in the given reaction:
[tex]Cl_2[/tex] → [tex]ClO^{3-} + Cl^-[/tex]
We need to identify the oxidation states of Cl in both reactants and products.
In [tex]Cl_2[/tex], since it is a diatomic molecule, both Cl atoms have the same oxidation state, which we can represent as 0.
In [tex]ClO^{3-}[/tex], the oxidation state of Cl is +5, since the oxidation state of O is -2 and there are three O atoms bonded to Cl.
In [tex]Cl^-[/tex], the oxidation state of Cl is -1.
Now, we can calculate the n-factor of [tex]Cl_2[/tex] using the following formula:
n-factor = change in oxidation state / number of electrons transferred
In the given reaction, [tex]Cl_2[/tex] is oxidized to [tex]ClO^{3-}[/tex] and reduced to [tex]Cl^-[/tex]. The change in oxidation state of Cl is:
+5 (in [tex]ClO^{3-}[/tex]) - 0 (in [tex]Cl_2[/tex]) = +5
-1 (in [tex]Cl^-[/tex]) - 0 (in [tex]Cl_2[/tex]) = -1
The maximum of these values is +5, so we will use this value for the n-factor.
The number of electrons transferred is equal to the number of Cl atoms in the balanced equation, which is 2.
The n-factor of [tex]Cl_2[/tex] in the given reaction is:
n-factor = 5/2 = 2.5
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What is the frequency of electromagnetic radiation with wavelength 532 nm. C = 3.00 * 10^17 nm/s a. 5.64 x 10^14 s^-1 b. 6.48 x 10^12 s^-1 c. 4.18 x 10^18 s^-1 d. 6.23 x 10^14 s^-1 e. 3.75 x 10^15 s^-1
The frequency of electromagnetic radiation with wavelength 532 nm is [tex]5.64 x 10^14 s^-1[/tex]. The correct option is (a).
According to the equation c = λν,
where c is the speed of light,
λ is the wavelength,
and ν is the frequency of the wave.
We have been given,
λ = 532 nm = [tex]532 x 10^-9 m[/tex] (since, 1 nm = 10^-9 m)
[tex]c = 3.00 x 10^8 m/s[/tex]
By substituting the values in the equation,
[tex]c = λν3.00 x 10^17 s^-1 = (532 x 10^-9 m)[/tex]
[tex]νν = (3.00 x 10^17 s^-1)/(532 x 10^-9 m)ν = 5.64 x 10^14 s^-1[/tex]
Therefore, the frequency of electromagnetic radiation with wavelength 532 nm is [tex]5.64 x 10^14 s^-1[/tex]. The correct option is (a).
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identify the components of the ionic formula based on the name nickel(iii) sulfide. what is the symbol for the element that forms the positive cation? what is the symbol for the element that forms the negative anion? what is the subscript on the cation in the neutral formula? what is the subscript on the anion in the neutral formula?
The components of the ionic formula for nickel(III) sulfide are as follows: 1. The symbol for the element that forms the positive cation is Ni (nickel). 2. The symbol for the element that forms the negative anion is S (sulfur). 3. The subscript on the cation in the neutral formula is 2, which represents the number of nickel atoms in the formula. 4. The subscript on the anion in the neutral formula is 3, which represents the number of sulfur atoms in the formula.
To find the correct formula, we need to balance the charges of the ions. Nickel(III) indicates that the nickel cation has a charge of +3 (Ni3+), and sulfide has a charge of -2 (S2-). To create a neutral compound, we need two nickel ions (2 x +3 = +6) and three sulfide ions (3 x -2 = -6). Therefore, the balanced ionic formula for nickel(III) sulfide is Ni2S3.
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what is the molartiy of ch3cooh in vinegar containing 4.0% ch3cooh by mass and having a desntiy of 1.0s g/ml
The molarity of acetic acid in the vinegar solution is 0.666 M.
to calculate the molarity of acetic acid ch3cooh in vinegar we need to understand the molecular weight of acetic acid which is 6005 gmol first we want to determine the mass of acetic acid present in 1000 grams 1 liter of vinegar we recognize that the density of vinegar is 10 gml so one thousand ml of vinegar would weigh a thousand g since the vinegar contains 40 acetic acid by using mass we can calculate the mass of acetic acid in one thousand g of vinegar as follows
Mass of acetic acid = 4.0% × 1000 g = 40 g
Now we can calculate the number of moles of acetic acid in 1000 g of vinegar:
Number of moles = mass / molecular weight
Number of moles = 40 g / 60.05 g/mol = 0.666 mol
Finally, we can calculate the molarity of acetic acid in the vinegar solution:
Molarity = number of moles / volume in liters
Molarity = 0.666 mol / 1 L = 0.666 M
Therefore, the molarity of acetic acid in the vinegar solution is 0.666 M.
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9. part e. 1. instead of 6 m nh, being added to the solution, 6 m naoh is added (both are bases) before the addition of the k2c204. what would be the appearance of the solution? explain.
The appearance of the solution may change slightly, depending on the initial color of the solution and the solubility of the products formed.
If 6 M NaOH is added instead of 6 M NH₄OH, the solution would become more basic. NaOH is a stronger base than NH₄OH and therefore would react more strongly with the K₂C₂O₄.
Balanced chemical equation for the reaction between K₂C₂O₄ and NaOH will be;
K₂C₂O₄ + 2NaOH → 2KOH + Na₂C₂O₄ + H₂O
As NaOH is added to the solution, it will react with K₂C₂O₄ forming Na₂C₂O₄, KOH and water. This will result in the production of more hydroxide ions, which will increase the pH of the solution.
However, the increase in pH will be noticeable and the solution may become more cloudy due to the formation of a precipitate of Na₂C₂O₄.
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1.2/0.35an athletic trainer is attempting to produce a carbohydrate-electrolyte solution that is at 27% carbohydrates by mass, which is the maximum amount of saturation allowed by her league. a supply company provides solutions that are at 15% and 35% carbohydrates by mass, respectively. based on the equation above, if the trainer uses 10 quarts of the 15% solution, how many quarts of the 35% solution will she need?
The answer is the athletic trainer will need 15 quarts of the 35% carbohydrate solution to achieve a final 27% carbohydrate-electrolyte solution when mixed with 10 quarts of the 15% solution.
To find out how many quarts of the 35% solution the athletic trainer needs, we can set up a weighted average equation using the given information:
27% = (15% * 10 quarts + 35% * x quarts) / (10 quarts + x quarts)
Step 1: Convert the percentages to decimals by dividing by 100.
0.27 = (0.15 * 10 + 0.35 * x) / (10 + x)
Step 2: Distribute the denominator to both sides of the equation.
0.27(10 + x) = 0.15 * 10 + 0.35 * x
Step 3: Expand the equation.
2.7 + 0.27x = 1.5 + 0.35x
Step 4: Subtract 0.27x from both sides.
2.7 - 1.5 = 0.35x - 0.27x
Step 5: Simplify the equation.
1.2 = 0.08x
Step 6: Divide by 0.08 to find x.
x = 1.2 / 0.08
x = 15
The athletic trainer will need 15 quarts of the 35% carbohydrate solution to achieve a final 27% carbohydrate-electrolyte solution when mixed with 10 quarts of the 15% solution.
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use the equation of the dissociation of the weak acid and a le schatelier principle to answer the following question: what change will be caused by the addition of a small amount of the weak base b- to a solution containing weak acid hb? group of answer choices ph slightly increase; poh slightly decrease ph slightly decrease; poh slightly decrease ph increase; poh increase ph decrease; poh decrease no change at all
The dissociation equation for weak acid can be written as HB + H2O ⇌ H3O+ + B- and there will be no change in Poh at all.
The correct answer is option D.
When a stress is given to an equilibrium system, the system will move to offset the stress and reestablish equilibrium, according to Le Chatelier's principle. In this situation, adding a bitsy volume of the weak base B- to the result will raise the attention of B- in the result.
This indicates that the equilibrium will move to the left to neutralize the rise in B- attention, performing in a drop in H3O ion attention. Because pH measures the attention of H3O ions in a result, a reduction in H3O attention will affect in a rise in pH. As a result, the answer is pH slightly rises; pOH slightly falls.
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Think about other kinds of animals living with the zebra on the savanna. How might changes in the availability of resources affect their survival?
Zebras are social animals that live together in groups. They help the local ecosystem by grazing on dry, hardened grass that is too tough for other species (such as wildebeest, ostriches, and antelopes) to digest. Every zebra is important, But habitat loss due to human encroachment, agricultural practices, and livestock grazing remains an issue in the ongoing conservation of this species. These problems seem to be especially prevalent in the southern half of their range and account for much of the recent population decline.
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a mixture of oxygen and hydrogen gases at a total pressure of 660 mm hg contains oxygen at a partial pressure of 425 mm hg. if the gas mixture contains 3.75 grams of oxygen, how many grams of hydrogen are present?]
There are around 0.118 grammes of hydrogen in the gas mixture. We can use the ideal gas law to solve this problem.
The ideal gas law relates the pressure, volume, number of moles, and temperature of a gas:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
First, we need to calculate the partial pressure of hydrogen in the gas mixture. The total pressure of the mixture is given as 660 mmHg, and the partial pressure of oxygen is given as 425 mmHg. Therefore, the partial pressure of hydrogen is:
P(H2) = P(total) - P(O2) = 660 mmHg - 425 mmHg = 235 mmHg
Next, we need to calculate the number of moles of oxygen in the mixture. We are given that the mixture contains 3.75 grams of oxygen. The molar mass of oxygen is 16.00 g/mol. Therefore, the number of moles of oxygen is:
n(O2) = mass/molar mass = 3.75 g/16.00 g/mol = 0.2344 mol
Since hydrogen and oxygen react in a 2:1 mole ratio to form water, the number of moles of hydrogen in the mixture is:
n(H2) = 0.5 x n(O2) = 0.5 x 0.2344 mol = 0.1172 mol
Finally, we can calculate the mass of hydrogen in the mixture using the number of moles and the molar mass of hydrogen, which is 1.008 g/mol:
mass(H2) = n(H2) x molar mass(H2) = 0.1172 mol x 1.008 g/mol = 0.118 g
Therefore, the gas mixture contains approximately 0.118 grams of hydrogen.
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What group are the alkali metals on the periodic table?
Answer: group 1A (or IA)
Explanation:
Many plants can reproduce both sexually and asexually. For example, rose bushes reproduce asexually when cuttings planted in the ground grow roots and become a new rose bush. They can also reproduce sexually when their flowers are pollinated. What is the primary advantage of sexual reproduction in roses?
The primary advantage of sexual reproduction in roses, and in plants in general, is genetic diversity.
Sexual reproduction involves the fusion of gametes (sperm and egg cells) from different individuals, which results in offspring that have unique combinations of genetic traits from both parents.
This genetic diversity can provide several advantages to the offspring, such as increased resistance to diseases and pests, adaptability to changing environmental conditions, and the ability to colonize new habitats. Additionally, genetic diversity can lead to the development of new and desirable traits, which can be selected and propagated by breeders to improve the quality of roses and other plants.
While asexual reproduction can provide quick and easy propagation of plants, it results in genetically identical offspring, which can be disadvantageous in terms of their adaptability and resistance to environmental stresses. Therefore, sexual reproduction is an essential process for maintaining the genetic diversity and evolution of plant species, including roses.
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GEN CHEM 2 PLEASE HELP
When Q < K (the reaction quotient is less than the equilibrium constant), the reaction will proceed in the forward direction to reach equilibrium. The correct response is option C
What happens when Q < K or Q = K in a reaction?When Q = K (the reaction quotient is equal to the equilibrium constant), the reaction is at equilibrium and the concentrations of the reactants and products will remain constant over time.
At equilibrium, the rates of the forward and reverse reactions are equal and there is no net change in the concentrations of the reactants and products.
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if 9.97 g of nacl reacts with excesses of the other reactants and 3.77 g of nahco3 is isolated, what is the percent yield of the reaction?
The percent yield of the reaction is 52.6%. The balanced chemical equation for the reaction between NaCl and NaHCO3 is:
2 NaCl + NaHCO3 → Na3CO3 + H2O + CO2
The molar mass of NaCl is 58.44 g/mol, and 9.97 g of NaCl corresponds to 9.97 g / 58.44 g/mol = 0.1705 mol of NaCl.
According to the equation, 1 mole of NaCl reacts with 1/2 mole of NaHCO3. So, the amount of NaHCO3 needed to react completely with 0.1705 mol of NaCl is:
0.1705 mol NaCl x (1/2) mol NaHCO3 = 0.08525 mol NaHCO3
The actual amount of NaHCO3 isolated is 3.77 g / 84.01 g/mol = 0.0449 mol NaHCO3.
The percent yield of the reaction is:
(actual yield / theoretical yield) x 100%
= (0.0449 mol / 0.08525 mol) x 100%
= 52.6%
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why would the presence of the side product benzalacetone be minimized? b) why would the presence of the side product 4-hydroxy-4-methyl-2-pentanone be minimized?
a) The formation of benzalacetone as a side product may be undesirable because it reduces the yield of the desired product (e.g., if the reaction is intended to produce acetophenone)
Additionally, the presence of benzalacetone may complicate the purification of the desired product, as the two compounds may have similar physical and chemical properties.
b) The formation of 4-hydroxy-4-methyl-2-pentanone as a side product may be undesirable because it may affect the properties and functionality of the desired product (e.g., if the reaction is intended to produce a specific ketone).
Additionally, the presence of this side product may complicate the purification of the desired product, as the two compounds may have similar physical and chemical properties. In some cases, the formation of this side product may also indicate an undesired competing reaction pathway that reduces the yield of the desired product.
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1. Which statement is correct about intermolecular versus intramolecular forces?
Intramolecular forces exist within a single molecule. Covalent bonding is an example.
Intermolecular forces exist within a single molecule. Hydrogen bonding is an example.
Intermolecular forces exist between separate molecules. Covalent bonding is an example.
Intramolecular forces exist between separate molecules. Hydrogen bonding is an example.
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which of the following substances is a strong electrolyte in aqueous solution? a. ki b. c2h6 (ethane) c. c6h6 (benzene) d. c8h18 (octane) e. ccl4 (carbon tetrachloride)
The substance that is a strong electrolyte in an aqueous solution is KI.
Strong electrolytes are substances that completely dissociate into ions when dissolved in water, resulting in a solution that can conduct electricity well.
KI is an ionic compound, meaning it is made up of a metal (potassium) and a nonmetal (iodide) that are held together by ionic bonds. When KI is dissolved in water, it dissociates into its ions (K+ and I-), which allows the solution to conduct electricity effectively.
The other options are, b. C2H6 (ethane), c. C6H6 (benzene), d. C8H18 (octane), and e. CCl4 (carbon tetrachloride), are all non-electrolytes or weak electrolytes, as they do not dissociate into ions when dissolved in water, and their solutions cannot conduct electricity well.
These substances are composed of nonmetal atoms and are held together by covalent bonds.
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consider the cannizzaro reaction of a base with the pictured structure. aldehyde with a trichloromethyl group attached to the carbonyl what products do you expect from the reaction, after acidifying the reaction mixture? select one or more: two carbon chain with one carbon attached to a hydroxyl group and the other attached to three carbons
The Cannizzaro reaction is a redox reaction in which an aldehyde (or a ketone) is simultaneously oxidized and reduced. The reaction requires a strong base such as a hydroxide ion (OH-) to deprotonate one aldehyde molecule and a second aldehyde molecule is reduced by the resulting hydride ion (H-).
The given aldehyde with a trichloromethyl group attached to the carbonyl group will undergo the Cannizzaro reaction with OH- as the strong base. After the reaction, the mixture is acidified to protonate the products.
The trichloromethyl group (-CCl3) is a strongly electron-withdrawing group that deactivates the carbonyl group toward nucleophilic attack. Therefore, the expected products from the Cannizzaro reaction of the given aldehyde are:
A carboxylic acid, is the oxidized form of the aldehyde.
Alcohol, is the reduced form of the aldehyde.
The final products after acidification will be:
A carboxylic acid with a trichloromethyl group attached to the carbon.
Alcohol with a two-carbon chain attached to a hydroxyl group.
Therefore, the expected product(s) from the reaction, after acidifying the reaction mixture are:
A carboxylic acid with a trichloromethyl group attached to the carbon, and
Alcohol with a two-carbon chain attached to a hydroxyl group.
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a scientist wishes to measure the concentration of methyl benzoate in a plant stream by gas chromatography. he prepares a sample of butyl benzoate to use as an internal standard. the results of a preliminary run, which used a solution known to contain 1.83 mg/ml of methyl benzoate (peak a) and 2.01 mg/ml of butyl benzoate (peak b), are shown. the area of peak a is determined to be 328 and the area of peak b is determined to be 384 measured in arbitrary units by the computer. to measure the sample, 1.00 ml of a standard sample of butyl benzoate containing 2.71 mg/ml is mixed with 1.00 ml of the plant stream material. analysis of the mixture gave a peak area of 477 for peak a and 411 for peak b. what is the concentration of methyl benzoate in the plant stream?
The concentration of methyl benzoate in the plant stream is 1160 µg/mL.
When measuring the concentration of methyl benzoate in a plant stream by gas chromatography, a scientist prepared a sample of butyl benzoate to use as an internal standard. The results of a preliminary run showed that the solution contains 1.83 mg/mL of methyl benzoate and 2.01 mg/mL of butyl benzoate, with peak a area being 328 and peak b area being 384.
To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.71 mg/mL is mixed with 1.00 mL of the plant stream material. The analysis of the mixture gave a peak area of 477 for peak a and 411 for peak b.
Now, let's calculate the concentration of methyl benzoate in the plant stream. We'll start with the ratio of peak areas of peak a to peak b. Peak a's area is to peak b's area as the concentration of methyl benzoate is to the concentration of butyl benzoate.
(328/384) = (x/2.71)
= 0.8542x x
= 2.32 mg/mL.
Finally, we'll find the concentration of methyl benzoate in the plant stream:
2.32 mg/mL x 1 mL/2 mL
= 1.16 mg/mL or 1160 µg/mL.
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Is the following sentence true or false? The theory of plate tectonics explains the formation, movement, and subduction of Earth’s plates.
Answer:
true.
Explanation:
Earth's lithosphere is broken into plates
Plate tectonics explains how plates move and interact with each other
Plate tectonics theory helps to understand earthquakes, volcanic eruptions, and the formation of mountain ranges.
Plate tectonics explains the process of subduction,
Subduction is where one plate is forced under another
One of the plate eventually melts into the mantle
Plate melting recycles Earth's crustal material
carboxylic acids are typically prepared using oxidation reactions. which of the functional groups below can be oxidized to give a carboxylic acid product? select all that apply.
The functional groups that can be oxidized to give a carboxylic acid product are:
Primary alcoholsAlkynes. Options A and D are correct.Carboxylic acids can be prepared by oxidizing a variety of functional groups. However, not all functional groups can be oxidized to give a carboxylic acid product. In general, primary alcohols and alkynes are two types of functional groups that can be oxidized to give a carboxylic acid.
Primary alcohols are oxidized to carboxylic acids via an intermediate aldehyde using a strong oxidizing agent such as potassium permanganate (KMnO₄) or chromium trioxide (CrO₃). The aldehyde is then further oxidized to the corresponding carboxylic acid.
Alkynes, on the other hand, can be oxidized directly to carboxylic acids using a strong oxidizing agent such as ozone (O₃) or potassium permanganate (KMnO₄). Phenols and secondary alcohols, however, cannot be directly oxidized to carboxylic acids. Phenols can be oxidized to quinones, while secondary alcohols can be oxidized to ketones, but these products are different from carboxylic acids.
The complete question is
Carboxylic acids are typically prepared using oxidation reactions. Which of the functional groups below can be oxidized to give a carboxylic acid product? Select all that apply.
A) Primary alcohols
B) Phenols
C) Secondary alcohols
D) Alkynes
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