how would the value of the atomic mass of the metal calculated be affected if the hot metal sample cooled off before it was transferred to the water in the calorimeter? would it be too high or too low?

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Answer 1

The value of the atomic mass of the metal calculated would be too high if the hot metal sample cooled off before it was transferred to the water in the calorimeter.

How is atomic mass calculated?

The atomic mass of an element is defined as the mass of an atom of an element in atomic mass units (amu). One atomic mass unit is defined as 1/12th of the mass of an atom of carbon-12.

The atomic mass of an element can be calculated using the following formula:

Atomic mass = (mass of isotope 1 × % abundance of isotope 1) + (mass of isotope 2 × % abundance of isotope 2) + (mass of isotope 3 × % abundance of isotope 3) + ...

If the hot metal sample cooled off before it was transferred to the water in the calorimeter, the temperature of the sample would have decreased. The decrease in temperature would result in a decrease in the thermal energy of the sample. Consequently, the amount of heat absorbed by the water in the calorimeter would decrease, leading to a lower value of the heat capacity of the metal.

Since the heat capacity is directly proportional to the mass of the sample, a lower value of the heat capacity would lead to a higher value of the atomic mass of the metal calculated. Therefore, the value of the atomic mass of the metal calculated would be too high if the hot metal sample cooled off before it was transferred to the water in the calorimeter.

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Related Questions

determine the enthalpy change when 18.6 g of carbon is reacted with oxygen according to the reaction: c(s) o2 (g) --> co2 (g) the change in enthalpy for this reaction is -349 kj/mol.

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The enthalpy change when 18.6 g of carbon is reacted with oxygen according to the reaction: c(s) + O2 (g) --> CO2 (g) is -349 kJ/mol. This enthalpy change is referred to as the heat of reaction, or enthalpy of reaction, and can be calculated using the enthalpy of formation of each reactant and product in the reaction.

The enthalpy of formation for carbon is given as +716 kJ/mol and for oxygen it is given as 0 kJ/mol. The enthalpy of formation for CO2 is given as -393.5 kJ/mol. Using Hess’s law, we can calculate the enthalpy of reaction using the following equation:  ΔHreaction = (ΔHformation CO2) - (ΔHformation C + ΔHformation O2)

Using the values for the enthalpies of formation for the reactants and products, the enthalpy of reaction can be calculated as follows: ΔHreaction = (-393.5) - (716 + 0) = -349 kJ/mol.This is the same enthalpy change as given in the question.

In conclusion, the enthalpy change when 18.6 g of carbon is reacted with oxygen according to the reaction: c(s) + O2 (g) --> CO2 (g) is -349 kJ/mol.

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How many atoms are in 0.75mol of H2O

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There are approximately 4.5 x 10^23 atoms in 0.75 mol of H2O.

Or 4,500,000,000,000,000,000,000.

In the illustration, which solute will dissolve first? A) solute in tank B will dissolve first B) solute in tanks A and B will dissolve at equal rates C) solute in tank A will dissolve first

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A) The solute in tank B will dissolve first, is the key response.Temperature, pressure, and concentration are only a few examples of the variables that affect a solute's solubility in a solvent. As the water in both tanks A and B is originally pure.

in this instance the solute in tank B will dissolve first due to its larger concentration than in tank A. The concentration gradient between the solute and the water narrows as the solute in tank B dissolves and diffuses into the surrounding water, slowing the rate of dissolution. The solute in tank A will also eventually dissolve, but because of its lower initial concentration, it will do so more gradually.I am unable to tell which solute will dissolve first because the relevant illustration is not given. However, a number of variables, including temperature, pressure, and the chemical makeup of the solute and solvent, affect how soluble a solute is in a solvent. The solute that is more soluble in the given solvent will often dissolve first. It is impossible to predict which solute will dissolve first without more details or context.

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Select an INCORRECT expression about a first-order reaction of A→2P. Note: k is rate constant, [A]0​ is initial reactant concentration. A) Rate Law for this reaction: Rate=k[ A] B) For rates of concentration changes: −2Δ[A]/t=Δ[P]/t C) Reactant half-life (t1/2​)=0.693/k D) For Product Concentration: [P]=[A]0​−[A]0​exp(−kt) A B C D

Answers

Option C is incorrect as it does not use the correct equation to calculate the half-life of a first-order reaction.

The correct answer is C. The reactant half-life of a first-order reaction is not equal to 0.693/k, as expressed in option C.

The equation for the half-life of a first-order reaction is: t1/2 = 0.693/k[A]0, where k is the rate constant and [A]0 is the initial reactant concentration.

To understand why this equation is correct, we need to understand how the half-life of a reaction is calculated. The half-life of a reaction is defined as the time taken for the concentration of a reactant to be halved.

This means that after a period of time, the concentration of the reactant will be equal to half of its initial concentration.

We can calculate this time using the equation for the reaction rate law: rate = k[A]0. By rearranging this equation and solving for t, we get t = 0.693/k[A]0.

This equation is known as the integrated rate law and is used to calculate the half-life of a first-order reaction.

Therefore, option C is incorrect as it does not use the correct equation to calculate the half-life of a first-order reaction. The correct options are A, B, and D.

Option A states that the rate law for this reaction is rate = k[A]0. Option B states that for rates of concentration changes, the equation is -2Δ[A]/t = Δ[P]/t,

where Δ[A] and Δ[P] are changes in the concentrations of the reactant and product, respectively.

Option D states that for product concentration, the equation is [P] = [A]0 - [A]0exp(-kt), which is correct.

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describe the chemistry of biurets reagent, explaining how it works and, specifically, why you used absorbance of 550 nm to quantify protein concentration.

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Biurets reagent is a solution of potassium hydroxide and copper sulfate used to measure the concentration of proteins. The reagent works by breaking down peptide bonds and creating a pink or purple solution when proteins are present. The absorbance of 550 nm is used to quantify the protein concentration because it is the wavelength that best corresponds to the color change of the solution.


Biurets reagent is a solution containing copper sulfate, sodium hydroxide, and potassium sodium tartrate. The copper ions in the biuret reagent combine with the peptide bonds present in proteins, forming a violet-colored complex. The intensity of the violet coloration is proportional to the concentration of proteins in the sample being analyzed. Absorbance at 550 nm is used to quantify protein concentration because this is the wavelength at which the violet color produced by the copper ion-peptide bond complex has maximum absorbance. By measuring the absorbance at this wavelength, the concentration of the protein in the sample can be determined through a standard curve that relates the absorbance values to known protein concentrations. The biuret test is commonly used to determine protein concentration in a variety of biological and chemical samples. The test is widely used because it is relatively simple and can be performed quickly. The biuret test is often used in combination with other analytical techniques to obtain more detailed information about protein samples.

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Activity 2: Who's My Family? A fire has occurred in a nearby maternity clinic. The assigned nurse quickly rushed out of the place to secure the newly born babies. Unfortunately, there were some babies without their identification bracelets. Using your knowledge about codominance inheritance will help bring these babies back to their correct parents. ​

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Codominance is a type of inheritance pattern in which both alleles of a gene are expressed equally in the phenotype of the individual. This means that if a baby inherits two different alleles for a particular trait, both will be expressed in the baby's physical appearance.

In the case of the missing identification bracelets, the nurse could use the principle of codominance to help identify the babies and return them to their correct parents. For example, if one baby has a parent with blood type A and the other has a parent with blood type B, and both babies have blood type AB due to codominance, then the nurse could match the babies with their correct parents based on their blood type.

Similarly, if there are other observable traits that exhibit codominance, such as eye color or skin tone, the nurse could use these to help identify the babies and return them to their correct parents. By understanding and applying the principles of codominance inheritance, the nurse could help ensure that each baby is reunited with their rightful family.

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write the rate law for each of the following elementary steps and tell whether the reaction unimolecular, bimolecular or termolecular a) o3 cl --> o2 clo b) no2 no2 --> no3 no c) 2no h2 --> h2o2 n2

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a. The rate law for the elementary step [tex]O_{3} + Cl[/tex] --> [tex]O_{2} + ClO[/tex] is k[[tex]O_{3}[/tex]][Cl], indicating that the reaction is bimolecular.

b. The rate law for the elementary step [tex]NO_{2}[/tex] + [tex]NO_{2}[/tex] --> [tex]NO_{3}[/tex] + NO is k[[tex]NO_{2}[/tex]]2, indicating that the reaction is termolecular.

c. The rate law for the elementary step 2NO + [tex]H_{2}[/tex] --> [tex]H_{2}O_{2}[/tex] + [tex]N_{2}[/tex] is k[NO][[tex]H_{2}[/tex]], indicating that the reaction is bimolecular.

The moleculаrity of а reаction refers to the number of reаctаnt pаrticles involved in the reаction. Becаuse there cаn only be discrete numbers of pаrticles, the moleculаrity must tаke аn integer vаlue. Moleculаrity cаn be described аs unimoleculаr, bimoleculаr, or termoleculаr. А unimoleculаr reаction occurs when а molecule reаrrаnges itself to produce one or more products. Аn exаmple of this is rаdioаctive decаy, in which pаrticles аre emitted from аn аtom.

А bimoleculаr reаction involves the collision of two pаrticles. Bimoleculаr reаctions аre common in orgаnic reаctions such аs nucleophilic substitution.  А termoleculаr reаction requires the collision of three pаrticles аt the sаme plаce аnd time. This type of reаction is very uncommon becаuse аll three reаctаnts must simultаneously collide with eаch other, with sufficient energy аnd correct orientаtion, to produce а reаction.

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Will the following reaction result in a precipitate? If so, identify the precipitate.K3PO4 + Cr(NO3)+ 3 KNO3 + CrPO4A. No, a precipitate will not formB. Yes, CrPO4 will precipitateC. Yes, KNO3 will precipitate

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Answer: B. Yes, CrPO4 will precipitate. In the given reaction: K3PO4 + Cr(NO3)3 → 3 KNO3 + CrPO4A precipitate is formed when two aqueous solutions are mixed that resulting in the formation of an insoluble compound.

The insoluble compound is called a precipitate. In the given reaction, K3PO4 and Cr(NO3)3 are the reactants. On mixing the two reactants, we can see that there are no common ions present in the reactants that could result in the formation of an insoluble compound. So, no precipitate is formed.

Based on solubility rules, CrPO4 is an insoluble compound. When K3PO4 reacts with Cr(NO3)3, it forms CrPO4. So, the precipitate that is formed is CrPO4. Hence, the correct option is B. Yes, CrPO4 will precipitate.

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identify the beaker that best represents a saturated solution. assume all the solutions are at the same temperature.

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The beaker that best represents a saturated solution is the one in which the solution is at its maximum level of solubility, meaning it cannot dissolve any more solute at the same temperature.

Saturated solutions are solutions in which no more solute can dissolve in the solvent at the same temperature. A solution is a homogeneous mixture composed of a solvent and a solute.

The solvent is the major component of the solution, and the solute is the minor component. The solute dissolves in the solvent to create a homogeneous solution.

A solution is said to be saturated when it has the maximum amount of solute that can dissolve in it at the same temperature. If the temperature changes, the solubility of the solute will also change, and the solution will become unsaturated or supersaturated.

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both the cno cycle and the proton-proton chain combine 4 h nuclei to produce 1 he nucleus. would those two processes release the same amount of energy per he nucleus produced? why or why not?

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The CNO cycle and the proton-proton chain don't release the same amount of energy per He nucleus produced.

Let's understand this in detail:

1. The CNO cycle produces more energy than the proton-proton chain per He nucleus produced. The proton-proton chain and CNO cycle produce energy by nuclear fusion in the sun's core.

2. In the core of the Sun, the proton-proton chain occurs. It converts four hydrogen nuclei (protons) into one helium nucleus via a series of nuclear reactions. This reaction liberates a significant amount of energy through gamma rays and neutrinos.

3. The CNO cycle also takes four hydrogen nuclei, producing one helium nucleus. The key difference between these two processes is the method in which helium is produced.

4. In the proton-proton chain, two protons combine to form deuterium. This then combines with another proton to form helium-3, and two helium-3 nuclei combine to form helium-4.

5. In the CNO cycle, hydrogen is fused with carbon, nitrogen, and oxygen isotopes to create helium. The CNO cycle releases more energy than the proton-proton chain per He nucleus produced because it has more intermediate steps.

5. The CNO cycle requires more heat and pressure to function because it involves carbon, nitrogen, and oxygen isotopes, which are heavier elements. The proton-proton chain is simpler because it only involves hydrogen and doesn't require as much energy.

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what mass of silver bromide is formed when 35.5 ml of 0.184 m silver nitrate is treated with an excess of hydrobromic acid?

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The mass of silver bromide formed when 35.5 ml of 0.184 m silver nitrate is treated with an excess of hydrobromic acid is 9.89 g.

When 35.5 mL of 0.184 M silver nitrate is treated with an excess of hydrobromic acid, the reaction forms silver bromide and a salt containing bromide ions. The mass of silver bromide that is formed can be calculated using the following equation:

Mass = Concentration x Volume x Molecular Weight

Where:

Mass = Mass of silver bromideConcentration = Concentration of silver nitrate (0.184 M)Volume = Volume of silver nitrate (35.5 mL)Molecular Weight = 187.81 g/mol

Therefore, the mass of silver bromide formed is:

Mass = 0.184 x 35.5 x 187.81 = 9.89 g

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In which situation are unbalanced forces acting on an object?(1 point)

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An object is said to be acted upon by an unbalanced force only when there is an individual force that is not being balanced by a force of equal magnitude and in the opposite direction.

An unbalanced force refers to a situation where the net force acting on an object is not equal to zero, which causes the object to accelerate in a particular direction. In other words, when the forces acting on an object are unbalanced, the object will either speed up, slow down, or change direction.

According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. Therefore, when an unbalanced force acts on an object, it will experience an acceleration proportional to the force applied. an unbalanced force is a force that causes an object to accelerate in a particular direction due to an imbalance in the forces acting on it.

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What correlates with metallic behavior

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Answer:

large atomic size and low ionization energy.

Explanation:

Metallic behavior correlates with large atomic size and low ionization energy. Thus, metallic behavior increases down a group and decreases from left to right across a period. Elements in Groups 1A(1) and 2A(2) are strong reducing agents; nonmetals in Groups 6A(16) and 7A(17) are strong oxidizing agents.

when combined, solutions of silver nitrate and hydroiodic acid produce a precipitate. what are the spectator ions in this reaction?

Answers

The spectator ions in the reaction between silver nitrate and hydroiodic acid are nitrate ions (NO₃₋) and hydrogen ions (H⁺).

In order to identify the spectator ions in this reaction, we need to first write out the balanced chemical equation for the reaction:

AgNO₃(aq) + HI(aq) → AgI(s) + HNO₃(aq)

In this equation, the silver nitrate (AgNO₃) reacts with hydroiodic acid (HI) to produce a precipitate of silver iodide (AgI) and nitric acid (HNO₃).

The spectator ions are those ions that do not participate in the reaction, but remain in the solution unchanged. In this case, the nitrate ions (NO₃₋) from silver nitrate and the hydrogen ions (H⁺) from hydroiodic acid are the spectator ions, as they are present on both the reactant and product side of the equation.

In other words, the nitrate ions and hydrogen ions are not involved in the formation of the precipitate of silver iodide, and do not undergo any chemical change themselves.

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In the given figure, red litmus paper is inserted in solution and colour remains unchanged then what may be contained in vessel among acid, base and salt solution? How can it be further tested to confirm it?​

Answers

Answer:

Explanation: If the red litmus paper is inserted into the solution and the color remains unchanged, it indicates that the solution is likely a neutral solution or a solution with a pH close to 7. This means that it may contain either water or a salt solution.

To further confirm whether the solution contains a salt or water, we can perform a simple test using blue litmus paper. We can dip a blue litmus paper into the solution, and if it turns red, it indicates that the solution is acidic. If it remains blue, it indicates that the solution is basic.

If the blue litmus paper also does not change its color, it means that the solution is neutral or has a pH close to 7, which supports the possibility that the solution may contain either water or a salt solution.

To further test whether the solution contains a salt or not, we can perform a flame test. We can take a small amount of the solution and place it on a platinum wire loop and hold it in a Bunsen burner flame. If the flame produces a characteristic color, it indicates that the solution contains a salt. The characteristic color of the flame will depend on the metal ion present in the salt.

Overall, based on the initial test with the red litmus paper, the solution is likely neutral or close to neutral, and additional tests with blue litmus paper and flame test can be used to confirm whether the solution contains a salt or water.

Which of the following bonds would be the most polar without being considered ionic?a. F-Hb. Na-Fc. S-Hd. Cl-He. O-H

Answers

The most polar bond without being considered ionic would be O-H.

Ionic bonds are the bond formed by the sharing of electrons between nonmetals to create a molecule that is neutral, while a covalent bond is a bond formed by the sharing of electrons between metals and nonmetals to create a molecule that is neutral.

Polar covalent bonds happen when there is an uneven distribution of electrons between two atoms that are bonded together. This is usually because the electrons are more strongly attracted to one atom over the other.

As a result, one atom will have a partial negative charge, and the other atom will have a partial positive charge.

In the water molecule, the O-H bond is polar because oxygen is more electronegative than hydrogen. Since the difference in electronegativity between hydrogen and oxygen is more significant than between the other atoms in the other bonds, the O-H bond is the most polar.

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Can you tell which one of the four examples corresponds to the making of a hydrocarbon


fuel from CO2 and water?

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A hopeful method for lowering greenhouse gas emissions and creating sustainable energy sources is the process of converting CO2 and water into a hydrocarbon fuel.

The process of making a hydrocarbon fuel from CO2 and water is called "artificial photosynthesis," and it involves using renewable energy sources to convert carbon dioxide and water into a liquid hydrocarbon fuel. This process is similar to photosynthesis in plants, where sunlight is used to convert carbon dioxide and water into glucose and oxygen.

Out of the four examples provided, it is not clear which one corresponds to the making of a hydrocarbon fuel from CO2 and water. However, one possible process involves using solar energy to drive the reaction between carbon dioxide and water, which results in the formation of a liquid hydrocarbon fuel. This process involves capturing carbon dioxide from the air or from industrial processes and combining it with water in the presence of a catalyst to produce a liquid hydrocarbon fuel.

Overall, the process of making a hydrocarbon fuel from CO2 and water is a promising approach to reducing greenhouse gas emissions and producing sustainable energy sources.

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complete the lewis structure for this species: co2 e
nter the total number of valence electrons in the box.
valence electrons:

Answers

The Lewis structure for CO2 is:

O = C = O

The "e" notation typically refers to an electron, so it's unclear what is meant by "CO2 e". However, the total number of valence electrons for CO2 is 16.

In the pictured cell, the side containing zinc is the_________ and the side containing copper is the __________. The purpose of the Na2SO4 is to _________

Answers

In the pictured cell, the side containing zinc is the anode and the side containing copper is the cathode. The purpose of the Na2SO4 is to facilitate the transfer of electrons from the anode to the cathode.

A cell is a unit of life that is the smallest and most simple living organism, it can be classified as a complete organism, with all of the components that make up a living being, including DNA, membranes, and organelles. A voltaic cell is a device that converts chemical energy into electrical energy, it is also known as a galvanic cell or a Daniell cell. It is made up of two different metals that are submerged in an electrolyte solution that enables the transfer of electrons from one electrode to the other. The anode is the electrode that oxidizes and loses electrons during a redox reaction, this electrode is negatively charged, as it is the site of the oxidation reaction that releases electrons and generates an electrical current.

A cathode is an electrode that is reduced and gains electrons in a redox reaction, this electrode is positively charged and acts as a sink for electrons, absorbing them and using them to create a reduction reaction that generates an electrical current. The Na2SO4 in the pictured cell is an electrolyte solution that facilitates the transfer of electrons from the anode to the cathode. The salt dissociates into Na+ and SO42- ions, which then migrate toward the anode and cathode, respectively, where they can participate in redox reactions that generate an electrical current. This flow of ions helps to maintain a balance of charge in the cell and enables the transfer of electrons to occur more efficiently.

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The pH in the intermembrane space of the mitochondria should be_____ compared to the matrix due to the
A. higher; higher concentration of protons in the intermembrane space B. higher; lower concentration of protons in the intermembrane space C. lower; higher concentration of protons in the intermembrane space
D. lower; lower concentration of protons in the intermembrane space

Answers

The pH in the intermembrane space of the mitochondria should be lower compared to the matrix due to the C. higher concentration of protons in the intermembrane space.

What is a Mitochondria?

Mitochondria are organelles found in eukaryotic cells that play a vital role in producing the energy required to sustain cellular activity. Mitochondria produce energy from food and oxygen, which they use to generate ATP, the primary source of cellular energy.

The intermembrane space (IMS) is the region between the mitochondrial inner and outer membranes. The pH of the intermembrane space is significantly lower than that of the matrix due to the higher concentration of protons in the intermembrane space.

The pH gradient of the mitochondria enables the generation of ATP from ADP and Pi by ATP synthase, which pumps protons from the intermembrane space to the matrix, making the pH gradient a source of energy. The proton gradient generated by ATP synthase is used for ATP synthesis. Therefore, the pH in the intermembrane space of mitochondria should be lower compared to the matrix due to the higher concentration of protons in the intermembrane space.

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Oxidation of Alcohols: Practical Methods
1. a) State the reagents & conditions used in the oxidation of alcohols.
b) State the colour change observed for the oxidising agent.
2.a) Explain why oxidation of a primary alcohol under distillation produces an aldehyde whereas oxidation
under reflux produces a carboxylic acid. You could use ethanol oxidation as an example. Include
structural formulae in your explanation.
b) Which experimental set-up below would you use to:
i) oxidise ethanol to ethanoic acid?
ii) oxidise ethanol to ethanal?
2. For ea
drawin
a) ethanc
b) etha
c) pre

Answers

Answer:

1a) The most common reagents used for the oxidation of alcohols are potassium permanganate (KMnO4), chromic acid (H2CrO4), and potassium dichromate (K2Cr2O7). Other oxidizing agents include sodium hypochlorite (NaOCl), pyridinium chlorochromate (PCC), and Jones reagent (CrO3/H2SO4). The conditions vary depending on the reagent used, but generally, the reaction is carried out under acidic or basic conditions and at elevated temperatures.

b) The oxidizing agents generally have a distinctive color, and their color changes during the reaction. For example, potassium permanganate is purple in its initial state, but it turns brown when it is reduced. Similarly, potassium dichromate is orange, but it changes to green when it is reduced.

2a) When a primary alcohol is oxidized, it can produce either an aldehyde or a carboxylic acid, depending on the reaction conditions. When the oxidation is carried out under distillation conditions, the aldehyde is formed as the reaction intermediate, which is then distilled off before it can be further oxidized to a carboxylic acid. On the other hand, when the oxidation is carried out under reflux conditions, the aldehyde is in equilibrium with the carboxylic acid, and the carboxylic acid is formed as the major product. For example, when ethanol is oxidized using potassium dichromate in acidic conditions:

Under distillation conditions:

CH3CH2OH + [O] → CH3CHO + H2O

Under reflux conditions:

CH3CH2OH + 2[O] → CH3COOH + H2O

b) i) The experimental set-up to oxidize ethanol to ethanoic acid would involve refluxing ethanol with an excess of potassium dichromate in acidic conditions.

ii) The experimental set-up to oxidize ethanol to ethanal would involve distilling a mixture of ethanol and a limited amount of oxidizing agent, such as pyridinium chlorochromate or Jones reagent, at a temperature that is lower than the boiling point of ethanal.

See the attached image for the requested drawings of ethane, ethanol, and propanone.

(please could you kindly mark my answer as brainliest)

b) which compound, a or b, was the limiting reagent in this reaction? compound b c) consider the lane that shows the reaction mixture. are the starting materials more or less polar than the reaction product? more polar

Answers

As per the information provided in the question, the compound that is the limiting reagent is "B". And the starting materials were "more polar" than the reaction product.

The limiting reagent is the one that gets consumed completely in the reaction. The other reactant is left behind in excess. The reaction's speed is determined by the amount of the limiting reagent present. In the given reaction, compound B is the limiting reagent. We can prove this by comparing the number of moles of compounds A and B. We can see that compound B has fewer moles. Therefore, it is the limiting reagent. 2 moles of compound A react with 1 mole of compound B. We have 2 moles of A and 1 mole of B in this reaction mixture. Hence, compound B is the limiting reagent. Starting materials are more polar than the reaction product. When a chemical reaction occurs, the reactants combine to form a new compound or product. The product's properties are often different from those of the starting materials. In this reaction, the starting materials are more polar than the reaction product. This can be seen by observing the reaction mixture's lane. We can see that the reaction product has moved ahead of the starting materials on the chromatogram. The starting materials are more polar than the reaction product.

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Which statement BEST describes one of the three main categories of elements?

a. Nonmetals are ductile and malleable.
b. Nonmetals are mostly gas at room temperature.
c. Metals are poor conductors of heat.
d. Metals are dull and brittle.

Answers

The statement that describes one of the three main categories of elements is: b. Nonmetals are mostly gas at room temperature.

What are Nonmetals?

Nonmetals are a group of elements that generally lack metallic properties. They are located on the right-hand side of the periodic table and include elements such as hydrogen, carbon, nitrogen, oxygen, fluorine, and neon, among others.

Nonmetals are typically poor conductors of heat and electricity and tend to have low melting and boiling points. They also tend to be brittle and lack luster, and some are gases at room temperature, while others are solids or liquids.

Nonmetals play important roles in various fields, such as chemistry, biology, and electronics. For example, nonmetals like oxygen, carbon, and nitrogen are essential components of many organic molecules and play critical roles in biological processes. In electronics, nonmetals like silicon and germanium are used to make semiconductors, which are essential components in electronic devices such as computers.

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HOW MANY LITERS OF H2 DO YOU HAVE IF YOU START WITH 1.5 MOLES OF H2?

Answers

If you started with 1.5 moles of H2 at STP, you would have approximately 33.6 liters of volume of hydrogen (H₂) gas.

What is the volume of the hydrogen gas at STP?

To determine the number of liters of H2 you have, we need to consider the conditions under which the gas is being held (i.e. temperature and pressure), as well as the molar volume of H2 at those conditions.

At standard temperature and pressure (STP), which is 0°C (273.15 K) and 1 atm (101.325 kPa), the molar volume of any ideal gas is approximately 22.4 L/mol.

Therefore, at STP, 1.5 moles of H₂ would occupy approximately:

V = n x Vm = 1.5 mol x 22.4 L/mol = 33.6 L

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The complete question is below:

HOW MANY LITERS OF H2 DO YOU HAVE IF YOU START WITH 1.5 MOLES OF H2? (assume STP condition)

Occurs naturally in bedrock and leads to the formation of radon.a.Uranium-238b.Coalc.Natural Gasd.Oile.Solar

Answers

Uranium 238 occurs naturally in bedrock and leads to the formation of radon. So. option (a) is correct.

Uranium-238 is said to be the most common isotope of uranium found in nature having a relative abundance of 99%. Uranium-238 is non-fissile that means it cannot sustain a chain reaction in a thermal-neutron reactor. Depleted uranium that is uranium containing mostly U-238 can be used for radiation shielding or as projectiles in armor-piercing weapons. Uranium-238 occurs naturally in nearly all rock, soil, and water. Uranium-238 is the most abundant form in the environment. Radon is said to be an odorless, invisible, radioactive gas naturally released from rocks, soil, and water. It can get into homes and buildings through small cracks or holes and build up in the air.

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Complete question is,

Occurs naturally in bedrock and leads to the formation of radon.

a. Uranium-238

b. coal

c. natural Gas

d. Oil

e. Solar

For the best system, calculate the ratio of the masses of the buffer components required to make the buffer. Express your answer using two significant figures. NH3/NH4Cl ph=8.95

Answers

Answer : The ratio of the masses of NH3 to NH4Cl required to make the buffer is 1.6 x 10^4 : 1.

The buffer system is one of the most important chemical systems. They are usually composed of a weak acid and a salt of its conjugate base or a weak base and a salt of its conjugate acid. The buffer capacity is important as it helps to resist changes in pH. The Henderson-Hasselbalch equation can be used to calculate the pH of the buffer system.

It's given by: pH = pKa + log [A-] / [HA]Here, NH3 is the weak base and NH4Cl is the salt of its conjugate acid. NH3 + H2O <--> NH4+ + OH- NH4Cl <--> NH4+ + Cl-By combining the above equations, the ratio of the masses of NH3 and NH4Cl can be found as shown below. pH = pKb + log [salt] / [base] pH = 5.09 + log [NH4Cl] / [NH3]pH = 8.95, pKb of NH3 = 4.74Therefore, 8.95 = 4.74 + log [NH4Cl] / [NH3] 4.21 = log [NH4Cl] / [NH3] [NH4Cl] / [NH3] = antilog (4.21) [NH4Cl] / [NH3] = 1.6 x 10^4

Therefore, the ratio of the masses of NH3 to NH4Cl required to make the buffer is 1.6 x 10^4 : 1.

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write a list of rules for recognizing and naming binary molecular compounds from their chemical formulas

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The following are the rules for recognizing and naming binary molecular compounds from their chemical formulas:
1. The first element in the chemical formula will be the name of the first element in the compound.
2. The second element in the chemical formula will be the name of the second element in the compound.
3. If the first element is a metal, the second element will end in “-ide”.
4. If the first element is a nonmetal, the second element will end in “-ate” or “-ite”.
5. The prefixes “mono-, di-, tri-, tetra-, penta-, and hexa-” are used to indicate the number of atoms of each element in the compound.
6. When the prefixes are not used, the number of atoms of each element is implied by the subscript.
7. If the subscript is written as a fraction, the fraction is changed to a whole number when forming the compound name.

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The rules for recognizing and naming binary molecular compounds are written focusing on the lower groups and the higher groups.

The rules for recognizing and naming binary molecular compounds from their chemical formulas are as follows:

1. The element with the lower group number is written first in the formula, and its full name is used.

2. The element with the higher group number is written second in the formula, and its stem name is used along with the suffix -ide.

3. The prefixes mono-, di-, tri-, tetra-, penta-, and so on are used to indicate the number of atoms present for each element in the molecule.

4. The prefix mono- is omitted for the first element in the formula.

5. The ending -a or -o in the prefix is omitted if the element name begins with a vowel, and only the vowel of the prefix is used in the compound name.

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rank the relative rates of the alkyl halides in an sn1 reaction.H3C-1 CH3 CH3 CH₂ H₂C Fastest SN 1 reaction Slowest SN 1 reaction Answer Bank CH3 H3C. CH3 H3C. H₂C₂ CH3 CH3

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The relative rates of alkyl halides from fastest sn 1 to slowest sn1 mechanism is CH3 H3C. CH3 H3C. H₂C₂ CH3 CH3.

Alkyl halides can go through one of two different sorts of significant reactions: substitution or elimination.

Nucleophilic Substitution reaction occurs when the halogen at the alpha-carbon is replaced by a nucleophile after the electrophilic alkyl halide forms a new bond with it.

The SN1 reaction mechanism proceeds step-by-step, starting with the formation of the carbocation through the elimination of the leaving group. The nucleophile then attacks the carbocation. Ultimately, the protonated nucleophile is deprotonated to produce the desired product.

Alkenes are formed by the E1 mechanism while substitution products are produced by the Sn1 process.

The rate law in an SN1 reaction is first order. In other words, the concentration of just one component—the alkyl halide—determines the reaction rate.

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A photon of light has a wavelength of 0. 050 cm. Calculate its energy

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A photon of light has an energy of 3.977 x [tex]10^{-19}[/tex] joules and a wavelength of 0.050 centimetres.

The energy of a photon is related to its wavelength by the formula E = hc/λ, where E is the energy, h is Planck's constant (6.626 x [tex]10^{-34}[/tex] joule seconds), c is the speed of light (2.998 x [tex]10^{8}[/tex] meters per second), and λ is the wavelength of the photon.

To use this formula, we need to convert the wavelength of the photon from centimeters to meters, since c is given in meters per second. We can do this by dividing 0.050 cm by 100, which gives us 5.0 x [tex]10^{-4}[/tex]meters.

Now we can plug in the values we have into the formula: E = (6.626 x [tex]10^{-34}[/tex] joule seconds) x (2.998 x [tex]10^{8}[/tex] meters per second) / (5.0 x [tex]10^{-4}[/tex]meters)

Simplifying the equation, we get:

E = 3.977 x [tex]10^{-19}[/tex] joules

Therefore, a photon of light with a wavelength of 0.050 cm has an energy of 3.977 x [tex]10^{-19}[/tex] joules. It is important to note that photons are the smallest quantifiable packets of electromagnetic energy, and their energy is directly proportional to their frequency and inversely proportional to their wavelength.

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write the full electron configuration for a k− ion.

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A k− ion is a potassium ion that has lost one electron, therefore the full electron configuration is 1s² 2s² 2p² 3s² 3p⁶

How to write an electron configuration?

To write an electron configuration, follow these steps:

Write the symbol of the element or ion you are interested in.Determine the total number of electrons based on the atomic number or ion charge.Write the electron configuration using the Aufbau principle, which states that electrons fill orbitals starting from the lowest energy level.Use the Pauli exclusion principle, which states that each orbital can hold a maximum of two electrons with opposite spins.Use Hund's rule, which states that electrons will occupy orbitals of the same energy level with parallel spins before pairing up in the same orbital.

The electron configuration for a neutral potassium atom is:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

When one electron is removed from the outermost shell, the electron configuration becomes:

1s² 2s² 2p⁶ 3s² 3p⁶

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