Answer:
The reduction reaction is: Cl₂ + 2 e⁻ ⇄ 2 Cl⁻
The oxidation reaction is: Cs ⇄ Cs⁺ + 1 e⁻
Explanation:
In a redox reaction, two half-reactions happen simultaneously. The oxidation refers to a species losing electrons. The reduction refers to a species gaining electrons.
Let's consider the following reaction.
Cl₂ + 2 Cs ⇄ 2CsCl
We can write the ionic equation.
Cl₂ + 2 Cs ⇄ 2 Cs⁺ + 2 Cl⁻
The reduction reaction is: Cl₂ + 2 e⁻ ⇄ 2 Cl⁻
We added 2 electrons to the left (electrons gained) to balance the half-reaction electrically.
The oxidation reaction is: Cs ⇄ Cs⁺ + 1 e⁻
We added 1 electron to the right (electrons lost) to balance the half-reaction electrically.
The redox reaction involves the electron transfer between the species. In reduction, the electrons are gained, while in oxidation the electrons are lost.
What is reduction-oxidation?In reduction reactions, the chemical species gains electrons, whereas, in oxidation reactions, the electrons are donated or lost by the chemical species.
The overall balanced reaction is given as,
[tex]\rm Cl_{2} + 2 Cs \leftrightarrow 2CsCl[/tex]
The ionic reaction for the above reaction is given as,
[tex]\rm Cl_{2} + 2 Cs \leftrightarrow 2 Cs^{+} + 2 Cl^{-}[/tex]
The reduction half of the reaction is given by adding electrons on the left side of the reaction as:
[tex]\rm Cl_{2} + 2 e^{-} \leftrightarrow 2 Cl^{-}[/tex]
The oxidation half of the reaction is given by adding electrons on the right side of the reaction as:
[tex]\rm Cs \leftrightarrow Cs^{+} + 1 e^{-}[/tex]
Therefore, electrons are lost in the oxidation reaction and are gained in the reduction reaction.
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In this experiment, 1-2 mL of saturated sodium chloride is used to transfer the crude product after the initial distillation. Why is saturated sodium chloride, rather than pure water, used for this procedure?
Answer:
Because it helps to remove water from the system.
Explanation:
The saturated sodium chloride solution has a strong affinity for water molecules and there is the possibility of changing the saturated solution to a dilute solution in the presence of pure water. Because of these reasons, the saturated sodium chloride solution removes water molecules from the system to become a diluted solution. That is the reason why the saturated solution was used instead of pure water.
Final answer:
Saturated sodium chloride solution is used over pure water to transfer crude product in an experiment due to its higher density and the ability to maintain phase separation, aiding in a more efficient and precise transfer process.
Explanation:
In the context of a chemical experiment, saturated sodium chloride solution is used instead of pure water to transfer the crude product after an initial distillation due to its unique properties. A saturated NaCl solution has a higher density compared to water, which is around 1.2 g/mL. This increased density is significant because it helps to separate the crude product from solvents that may have similar densities to water.
When dealing with a sparing soluble hydrocarbon (HC), phase separation occurs after the saturation point. The presence of a high concentration of NaCl in the transfer medium helps ensure that the HC and water remain separate, thus making the transfer of the HC more efficient. In essence, the use of saturated sodium chloride creates a denser medium which can assist in the separation due to the difference in solubility and density between the aqueous layer and organic compounds.
Furthermore, the saturated solution is at a state of solution equilibrium, where the rate of dissolution equals the rate of recrystallization. This ensures that adding the hydrocarbon or further NaCl will not significantly change the composition of the transfer medium.
how many particles are there in 1.43 g of a molecular compound with a gram molecular mass of 233g
The number of particles in 1.43 g or a molecular compound is 3.69 × 10²¹.
• The mass of the compound given is 1.43 grams and the molecular mass of the compound is 233 g/mol.
• The mole of a compound can be determined by using the formula,
n(number of moles) = Weight/Molecular mass = 1.43/233 = 0.00613 moles
• The number of particles in 1 mole is 6.022 × 10²³ particles
The number of particles in 0.00613 moles is,
= 0.0063 × 6.022 × 10²³
= 3.69 × 10²¹ particles
Thus, the number of particles in the given case is 3.69 × 10²¹ particles.
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a compound with the empirical formula CH2 has a molar mass of 112 g/mol. What is the molecular formula for this compound?
The molecular formula for [tex]CH_{2}[/tex] carrying a molar mass of 112 g/mol would be:
- [tex]C_{8} H_{16}[/tex]
Given that,
Empirical Formula of the compound [tex]=[/tex] [tex]CH_{2}[/tex]
The molar mass of the compound [tex]= 112 g/mol[/tex]
As we know,
Empirical Mass = [tex]12 + 2[/tex]
[tex]= 14[/tex]
It is known that,
Molar mass [tex]= n[/tex] × [tex]empirical mass[/tex]
∵ [tex]n = molar mass/empirical mass[/tex]
[tex]= 112/14[/tex]
[tex]= 8[/tex]
∵ The Molecular Formula of the compound would be determined by multiplying n into the empirical formula,
so,
Molecular Formula = [tex]n[/tex] × [tex]emprical formula[/tex]
[tex]=[/tex] ([tex]CH_{2}[/tex]) × [tex]8[/tex]
= [tex]C_{8} H_{16}[/tex]
Thus, [tex]C_{8} H_{16}[/tex] is the correct answer.
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how many formula units in MgCl2
Arrange the following alkyl bromides in order from most reactive to least reactive in an SN2 reaction: 1-bromo-2-methylbutane, 1-bromo-3-methylbutane, 2-bromo-2-methylbutane, and 1-bromopentane.
Rank the alkyl bromides from most reactive to least reactive. To rank items as equivalent, overlap them.
And what makes an alkyl bromide more reactive in this case?
Answer:
See explanation
Explanation:
First, in order to know this it's neccesary to remember how a SN2 reaction takes place. A Sn2 reaction is a bimolecular concerted reaction where all bonds are broken and making in only one step.
For this to occur, we need a strong nucleophyle (such a strong base) and a substract with a great outgoing group (The halides are great leaving groups).
The nucleophyle attacks on the back side of the molecule with the bromine, and the result is a molecule with inverted configuration and the bromine is replaced by the nucleophyle.
However, this step is fast and concerted, and in order to do this faster, the reactant must be (Ideally) with no substituent, because if the molecule is bulky, the nucleophyle's attack to the back side is hard. This doesn't mean that it will not undergo, but it will be harder and slower.
Because of this reason, we can see that from all the alkyl bromides there, the least bulky is the 1-bromopentane, so this will be the more reactive in Sn2, followed by 1-bromo-methylbutane, then the 1-bromo-2-methylbutane and finally the 2 - bromo - 2 - methylpentane.
In the picture, you have the structures of these molecules, so you can see how the steric hindrance affects this.
A 70.0 mL sample of water is heated to its boiling point. How much heat is required to vaporize it? (Assume a density of 1.00 g/mL.)
Explanation:
As density is the amount of mass divided by volume of the substance.
Mathematically, Density = [tex]\frac{mass}{volume}[/tex]
It is given that volume is 70.0 ml and density is 1.00 g/ml. Therefore, mass of the given substance will be as follows.
Density = [tex]\frac{mass}{volume}[/tex]
1.00 g/ml = [tex]\frac{mass}{70.0 ml}[/tex]
mass = 70.0 g
As we known that heat of vaporization of water is 2260 J/g for 1 g of a substance. Therefore, heat of vaporization of water for 70.0 g will be as follows.
[tex]70.0 g \times 2260 J/g[/tex]
= 158200 J
or, = 158.2 kJ (as 1 kJ = 1000 J)
Thus, we can conclude that 158.2 kJ heat is required to vaporize given sample of water.
What volume of H2O(g) is produced when 8.00 mol of C2H4(g) reacts at STP?
C2H4 + 3O2 >>>>> 2CO2 + 2H2O
In chemistry a formula is used to represent
Organic compounds are produced only by living systems, and cannot be synthesized in the laboratory true or false
A sample of carbon dioxide is contained in a 125.0 mL flask at 0.984 atm and 16.4 °C. How many molecules of gas are in the sample?
A chemist wants to produce 12.00 grams of barium sulfate by reacting a .6000 M BaCl2 solution with excess H2SO as show in the reaction below. What volume of the barium chloride should be used?
BaCl2 + H2SO4 --> BaSO4 +2HCl
Answer: 0.0857 L [tex]BaCl_2[/tex]
Explanation: It's a stoichiometry problem. Balanced equation is given from which there is 1:1 mol ratio between [tex]BaCl_2[/tex] and [tex]BaSO_4[/tex] .
Chemist wants to produce 12.00 grams of barium sulfate by reacting a 0.6000 M barium chloride solution with excess sulfuric acid.
We know that molarity is moles of solute per liter of solution. Molarity of barium chloride is given. If we know its moles then its volume could easily be calculated.
From given grams of barium sulfate, we calculate its moles and then using mol ratio we calculate the moles of barium chloride.
Molar mass of barium sulfate is 233.38 gram per mol.
The complete set up is shown below using dimensional analysis:
[tex]12.00gBaSO_4(\frac{1molBaSO_4}{233.38gBaSO_4})(\frac{1molBaCl_2}{1molBaSO_4})(\frac{1LBaCl_2}{0.6000molBaCl_2})[/tex]
= 0.0857 L [tex]BaCl_2[/tex]
So, 0.0857 L or 85.7 mL of [tex]BaCl_2[/tex] should be used.
Why does saponification produce sodium lactate (and not lactic acid) as the final product??
Answer:
Explanation:
Hello,
Saponification chemical reactions are defined as such reactions between either potassium hydroxide or sodium hydroxide and a fatty acid to break hydroxiles in order to attach sodium or potassium to the organic chain. They are a type of esterification chemical reaction.
In this case, lactic acid reacts with sodium hydroxide as shown below:
[tex]CH_3CHOHCOOH+NaOH-->CH_3CHOHCOONa+H_2O[/tex]
As it is seen, sodium lactate is produced rather than lactic acid due to the easiness that the hydroxile has to break and to subsequently attract the ionized sodium to form the lactate, even do, free sodium cations easily break the hydroxile and form sodium lactate.
Best regards.
Which of these structures represent a soap
The given structures represents a soap is E. more than one of the compounds is a soap.
To understand Soaps are typically the sodium or potassium salts of long-chain fatty acids, also known as carboxylic acids. These long-chain fatty acids contain a hydrophobic (nonpolar) hydrocarbon tail and a hydrophilic (polar) carboxylate head, which allows them to interact with both water and oils.
Now, let’s analyze the options provided:
a. [tex]CH_3CO_2^- K^+[/tex]- This compound does not have a long hydrocarbon chain. It is the acetate ion, which does not classify as a soap.
b. [tex]CH_3(CH_2)_{14}CO_2^- Na^+[/tex] - This is sodium hexadecanoate, which is indeed a soap, as it features a long hydrocarbon chain and a carboxylate group.
c. [tex]CH_3(CH_2)_{12}COOH[/tex] - This is lauric acid, which is not a salt and thus is not a soap.
d. [tex]CH_3(CH_2)_7CO_2(CH_2)_7 Na[/tex] - This represents a soap because it consists of a long hydrocarbon chain (from both ends) and includes a sodium carboxylate.
e. More than one of the compounds is a soap. - From our analysis, options b and d are soaps.
Complete question
Which of these structures represents a soap?
a. [tex]CH_3CO_2^- K^+[/tex]
b.[tex]CH_3(CH_2)_{14}CO_2^- Na^+[/tex]
c.[tex]CH_3(CH_2)_{12}COOH[/tex]
d. [tex]CH_3(CH_2)_7CO_2(CH_2)_7 Na[/tex]
e. More than one of the compounds is a soap.
If 71.5 moles of an ideal gas is at 5.03 atm at 6.80 °C, what is the volume of the gas?
Which energy source causes the greatest air pollution?
Meteorologists use weather balloons to carry weather instruments high into the atmosphere. When it is first released at Earth’s surface, a weather balloon typically appears underinflated. What happens to the balloon’s volume as it rises into the atmosphere? Explain your answer in terms of pressure.
As the balloon’s altitude increases, its volume also increases.
At high altitude, the atmospheric pressure on the outside of the balloon is less that it is at Earth’s surface.
As the pressure on the outside of the balloon decreases, the balloon’s volume increases because the pressure inside the balloon pushes the balloon outward.
As the balloon’s volume increases, the pressure inside the balloon decreases until it is equal to the pressure on the outside of the balloon.
are answers for edg
A weather balloon's volume increases as it rises due to decreased atmospheric pressure in accordance to Boyle's Law. The ideal gas law better explains this behavior considering the changing temperatures. Eventually, the balloon will burst when its material's limit is reached.
As a weather balloon rises into the atmosphere, the atmospheric pressure decreases. This occurs because the atmosphere becomes thinner with elevation, resulting in fewer air molecules to exert force on any given surface area. According to Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when temperature is held constant, the volume of the gas inside the balloon will increase as the balloon ascends and the outside pressure decreases. However, since the temperature is not constant during ascent and typically drops, the ideal gas law, which combines Boyle's Law and Charles's Law, is a better descriptor of the balloon's behavior. This law indicates that the relationship between the temperature, volume, and pressure of the gas will dictate the expansion of the balloon. Notably, as the volume expands and the balloon material reaches its stretching limit, it will eventually burst.
the half-life of iodine is 8 days. If 700.00 g sample decays to 43.75, how much time has passed
To calculate the time passed for a 700 g sample of iodine to decay to 43.75 g, we determine the number of half-lives that have elapsed, which is 4, as 43.75 g is approximately 1/16th of 700 g. Since the half-life for iodine is 8 days, the total time passed is 4 half-lives times 8 days, equalling 32 days.
Explanation:The half-life of iodine is 8 days. If a 700.00 g sample decays to 43.75 g, to find out how much time has passed, we use the concept of half-lives. A half-life is the time taken for half of a radioactive substance to decay into another element. Knowing the original amount and the final amount of the substance, we can calculate the number of half-lives that have elapsed.
To determine the number of half-lives, we divide the final amount by the initial amount and keep halving until we reach a value less than or equal to 1. In this case, we divide 43.75 g by 700 g and get approximately 0.0625. This corresponds to 4 half-lives since (1/2)^4 = 1/16, and when multiplied by the initial amount (700 g), gives us approximately 43.75 g. Since each half-life is 8 days, 4 half-lives would be 32 days.
Therefore, the time that has passed since the 700 g of iodine began to decay is 32 days.
In which case would you revise a hypothesis?
A.When you have no prior knowledge of a subject
B.When you do not have enough data to make a graph
C.When your experiment design does not work
D.When experimental results do not support the original hypothesis
You should revise a hypothesis when the experimental results do not support the original hypothesis (Option D), based on the process of the scientific method.
Explanation:In the process of scientific investigation, a hypothesis is essentially a predicted answer to a research question, which is then tested through experiments. You would revise a hypothesis in the instance where experimental results do not support the original hypothesis (Option D). This is based on the scientific method, where the hypothesis is accepted or revised based on the evidence collected.
For example, if you hypothesize that plants grow faster when exposed to classical music and you conduct an experiment that shows no significant difference in growth rates between plants exposed to classical music and those that weren't, you would then revise your hypothesis, potentially considering variables you hadn't initially accounted for or a different predicted outcome.
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which of the following radioactive emissions is the least penetrating, aloha particles , beta particles or gamma rays
Answer: The least penetrating of the given radioactive emissions will be alpha particles.
Explanation:
There are 3 radioactive particles which are emitted during radioactive processes:
1. Alpha particles: These particles are emitted when a nuclei undergoes alpha decay process. These particles have low energy associated with them.
2. Beta particles: These particles are emitted when a nuclei undergoes beta decay process. These particles have higher energy than alpha particles.
3. Gamma radiations: These radiations are emitted when an unstable nuclei undergoes gamma ray emission process and gives an excess energy through a spontaneous electromagnetic process. These radiations have the highest energy of all the radioactive particles.
Penetrating power of the particles is directly proportional to the energy of the particles, therefore:
[tex]\text{Penetrating power}\propto \text{Energy of the particles}[/tex]
More the energy of the particles, more will be the penetrating power and vice-versa.
Increasing order of penetrating power will be:
[tex]\text{Alpha particles}<\text{Beta particles}<\text{Gamma radiations}[/tex]
Hence, alpha particles is the least penetrating among the following radioactive particles.
Where, approximately, is the negative pole on each of these molecules?
Orbital shell notation of fluorine is 2. 7 while that of oxygen s 2. 6. This means that these elements (that follow each other in the periodic table) will have high electronegativity in molecules due to their high atomic number (which causes them to strongly attract electron orbital shell closer to their nucleus). NB: Atomic number of a peroid increased from left to right of the periodic table.
Therefore, in the first molecule, the negative dipole would most likely be located between the F atoms In the second molecule the negative molecule would be most likely located in the between the O and F atoms.
In the given question, the oxygen atom serves as the negative pole in both [tex]\rm CF_2O \ and\ CHFO[/tex].
Molecules are the smallest units of a compound that retain all of the chemical properties of that compound. They consist of two or more atoms that are bonded together chemically.
In the molecule [tex]\rm CF_2O[/tex], the negative pole is located on the oxygen atom. This is because oxygen is more electronegative than carbon and fluorine, and therefore attracts the bonding electrons closer to itself. As a result, the oxygen atom has a partial negative charge (δ-) and the fluorine atoms have partial positive charges (δ+).In the molecule CHFO, the negative pole is also located on the oxygen atom. This is because the oxygen atom is more electronegative than the carbon and hydrogen atoms, and attracts the bonding electrons closer to itself. As a result, the oxygen atom has a partial negative charge (δ-) and the hydrogen atoms have partial positive charges (δ+).Therefore, the negative pole in both [tex]\rm CF_2O \ and\ CHFO[/tex] is located on the oxygen atom.
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The given question is not complete. The complete question is:
Where, approximately, is the negative pole on each of these molecules [tex]\rm CF_2O \ and\ CHFO[/tex] ?
If there are 0.064g of theobromine in a 43g chocolate bar, how many moles of theobromine are there?
Answer: [tex]3.5\times 10^{-4}[/tex]
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
Given mass of theobromine [tex](C_7H_8N_4O_2)[/tex] = 0.064 g
Molar mass of theobromine [tex](C_7H_8N_4O_2)=12\times 7+1\times 8+14\times 4+16\times 2=180 g[/tex]
[tex]\text{Number of moles}=\frac{0.064g}{180g/mol}=3.5\times 10^{-4}moles[/tex]
Final answer:
To calculate the moles of theobromine in a chocolate bar, divide the mass of theobromine by its molar mass. There are approximately 0.000355 moles of theobromine in a 43g chocolate bar containing 0.064g of theobromine.
Explanation:
To find out how many moles of theobromine are in a 43g chocolate bar that contains 0.064g of theobromine, we first need to know the molar mass of theobromine.
The molar mass of theobromine (C₇H₈N₄O₂) is 180.16 g/mol. Using the formula:
Moles = mass (g) / Molar Mass (g/mol)
We can calculate the number of moles:
Moles of theobromine = 0.064g / 180.16 g/mol = 0.000355 moles (approximately).
Therefore, there are approximately 0.000355 moles of theobromine in a 43g chocolate bar.
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Chloric acid, HClO3, contains what percent hydrogen by mass?
Chloric acid, HClO3, contains 1.19% percent hydrogen by mass
What is chloric acid used for?Chloric acid is a colorless liquid. It will accelerate the burning of combustible materials and can ignite most on contact. It is corrosive to metals and tissue. It is used as a reagent in chemical analysis and to make other chemicals.
Where is chloric acid found?Chloric acid is not found in nature. It is prepared. Physical properties: Chloric acid is a colorless liquid. Its density is 1 g mL-1.
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A gas exerts a pressure of 450 atmospheres (atm). What is this pressure in mmHg?
To convert 450 atmospheres to millimeters of mercury, multiply 450 atm by the conversion factor of 760 mmHg per 1 atm, resulting in 342,000 mmHg.
To convert the pressure of a gas from atmospheres to millimeters of mercury (mmHg), we use the standard conversion factor between these two units of pressure. The relationship is 1 atmosphere (atm) is equivalent to 760 millimeters of mercury (mmHg).
Given that the gas pressure is 450 atmospheres (atm), the conversion to mmHg would be as follows:
(450 atm)
(760 mmHg / 1 atm) = 342,000 mmHg
Step-by-Step Calculation:
Start with the given pressure in atmospheres (450 atm).Multiply by the conversion factor 760 mmHg / 1 atm.Calculate the product to find the pressure in mmHg.Therefore, a pressure of 450 atm is equal to 342,000 mmHg.
Gold has a density of 0.01932 kg/cm3. What volume (in cm3) would be occupied by a 33.3 get sample of gold?
The volume would be occupied by a 33.3 g sample of gold is 1723.6cm³. Density is the mass of a specific material per unit volume.
What is density?Density is the mass of a specific material per unit volume. d = M/V, in which d is density, M is weight, and V is volume, is the formula for density. Grams per cubic centimeter are a typical unit of measurement for density. For instance, whereas Earth has a density of 5.51 grams, water has a density of 1 grams.
Another way to state density is in kilograms per cubic meter (in metre-kilogram-second or SI units). For instance, air weighs 1.2 pounds per cubic metre. In textbooks and manuals, the densities of typical solids, liquids, as well as gases are stated.
Density = mass / volume
0.01932 kg/cm³= 33.3 g/ volume
volume =33.3/ 0.01932=1723.6cm³
Therefore, the volume would be occupied by a 33.3 g sample of gold is 1723.6cm³.
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How many kg of carbon dioxide are used per year to produce their arm and hammer baking soda?
Final answer:
The question about how many kilograms of CO₂ are used per year to produce Arm and Hammer baking soda requires specific data from the company, but chemical principles and reactions illustrate how CO₂ is used in similar processes. Examples like the reaction of KOH with CO₂ to produce potassium carbonate, and the CO₂ emissions from gasoline, offer an educational understanding of CO₂ use and release in the environment.
Explanation:
Quantifying CO₂ Emissions in Industry and Chemical Reactions
The original question regarding the amount of carbon dioxide (CO₂) used per year to produce Arm and Hammer baking soda cannot be directly answered without specific industrial data from the company. However, we can discuss the chemical processes and general principles that may be involved in the production and the use of CO₂ in related reactions.
For example, in the production of sodium bicarbonate (baking soda), a reaction occurs between sodium carbonate and carbon dioxide to form sodium bicarbonate.
Similarly, other reactions outlined for educational purposes, like the reaction of potassium hydroxide with carbon dioxide producing potassium carbonate and water, demonstrate how CO₂ is utilized in chemical processes.
Specifically, in the reaction between 224.4 grams of KOH and 88.0 grams of CO₂, 138.4 grams of potassium carbonate and 36.0 grams of water are formed.
When comparing CO₂ emissions from gasoline consumption, it's noted that a 40-liter tank of gasoline, with a density of 0.75 kg/L, would release a certain mass of CO₂ upon combustion.
This amount of CO₂ can be compared to typical human mass for perspective. For instance, the combusted gasoline might release more CO₂ than the average weight of a person.
Furthermore, general environmental data is presented to understand the mass concentration of CO₂ in the atmosphere resulting from oil combustion, leading to an increase in atmospheric CO₂ concentration measured in parts per million by mass (ppmm) and volume (ppmv).
What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drops of equal size? The surface tension, γ, of water at room temperature is 72.0 mJ/m2.
The energy required to split a spherical drop of water into three smaller ones of the same size can be calculated considering the change in surface area of the drops and the surface tension of water. The energy is equivalent to the work done against the surface tension.
Explanation:To solve this question, we need to understand that when a spherical drop of water splits into multiple smaller drops, energy is required. This energy is equivalent to the work done against surface tension. The energy required, also known as the surface energy, is derived from the change in the surface area of the water drops. Surface tension (γ) is the energy required per unit increase in area.
Let's consider the initial drop of water has a diameter D and the smaller drops each having diameter d. The initial surface area of the spherical drop, using the formula for the surface area of a sphere 4πr² (where r is the radius), is 4π(D/2)² and the total surface area of the 3 smaller drops is 3 * 4π(d/2)². The change in surface area ΔA = 3 * 4π(d/2)² - 4π(D/2)².
The energy required which is derived from the surface tension formula is ΔE = γ ΔA. By substituting the ΔA in this equation, we can calculate the energy required. This problem might require additional details like the ratio between the diameters D and d.
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What mass of ether will be produced is 221 g of ethanol reacts with a 66 percent yield?
what is solution in chemistry
A solution in chemistry is a homogeneous mixture of two or more substances, where one substance (the solute) is dissolved in another substance (the solvent). It might involve physical or chemical changes.
Explanation:In chemistry, a solution is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as the solvent. The process of dissolving can involve physical changes, like if you dissolve granulated sugar in water, you see it disappear since it changes to the microscopic level. However, it can also involve chemical changes, for instance, when you dissolve baking soda in vinegar. You observe a chemical reaction producing carbon dioxide gas, among other things, which indicates a new substance is created.
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When the reaction
CH3Cl(g) + H2O(g) →CH3OH(g) + HCl(g)
Was studied, the tabulated data were obtained. Based on thesedata, what are the reaction orders?
Initial Concentrations, M; Initial Rates,Ms-1
Exp CH3Cl H2O Rate, Ms-1
1 0.100 0.100 0.182
2 0.200 0.200 1.45
3 0.200 0.400 5.81
A. CH3Cl: firstorder H2O:first order
B. CH3Cl: second order H2O: first order
C. CH3Cl: firstorder H2O: second order
D. CH3Cl: second order H2O: second order
Please EXPLAIN your answer.
The reaction orders for CH3Cl and H2O are determined to be second order for both reactants as evidenced by the quadrupled rate increase when their concentrations are doubled. So the correcct option is D.
Explanation:To determine the reaction orders for CH3Cl and H2O in the given reaction, we analyze the provided experimental data by comparing initial concentrations and initial rates.
Comparing Experiment 1 and Experiment 2: The concentration of both CH3Cl and H2O is doubled, and the rate increases from 0.182 to 1.45, an 8-fold increase. This suggests that both are of second order because doubling the concentration of a second-order reactant increases the rate by a factor of 22, which is 4, and since both reactants are doubled, 4*4=16, accounting for the 8-fold increase.Comparing Experiment 2 and Experiment 3: Only the concentration of H2O is doubled, and the rate increases from 1.45 to 5.81, a 4-fold increase. This change indicates that H2O is second-order because doubling the concentration of a second-order reactant increases the rate by a factor of 22 or 4.Based on these observations, the answer is D. CH3Cl: second order and H2O: second order. Both the concentrations of CH3Cl and H2O have a squared relationship to the rate, characteristic of second-order reactions.