Hello! I am taking chemistry in Acellus and I'm looking for someone to sign in and help me catch up. I'm willing to give credits, money etc. I just need to pass and be done with it SOON.

Centrifugation, chromatography, and long standing are the procedures that would work well for separating a colloid mixture but would not be effective with solutions.

Stratification that is "inverted" is created when the larger colloids are forced to the bottom. The reason for this qualitative segregation is that evaporation causes a local rise in colloid concentration close to the film-air contact, which results in a chemical potential gradient for both colloid species.

Dialysis: The removal of ions from a solution through the diffusion process through a permeable membrane is known as dialysis. This procedure involves filling a permeable membrane bag with a sol made up of ions or molecules and submerging it in water.

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The volume of the solution is 2.12 L.

Molarity (M) is defined as the number of moles of solute per liter of solution. So, if we know the molarity of a solution and the amount of solute present, we can calculate the volume of the solution.

To find the volume (in L) of a 1.98 M solution containing 4.2 moles of solute, we can use the following formula:

moles of solute = molarity x volume (in L)

We can rearrange this formula to solve for volume:

volume (in L) = moles of solute / molarity

Substituting the given values, we get:

volume (in L) = 4.2 moles / 1.98 M

volume (in L) = 2.12 L

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--The complete question is, What is the volume (in L) of a 1.98 M solution containing 4.2 moles of solute?--

Yes, half the C2H2 (acetylene) can react with the given amount of O2 (oxygen) in the reactor and 7.0 mol of CO2 would be produced after half the C2H2 is used up.

To determine how many moles of CO2 (carbon dioxide) would be produced after half the C2H2 is used up, we need to first write the balanced chemical equation:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

Given 7.0 mol of C2H2 and 1.0 mol of O2, let's find out how much C2H2 can react:

(1.0 mol O2) * (2 mol C2H2 / 5 mol O2) = 0.4 mol C2H2

Since 0.4 mol is less than half of the initial 7.0 mol of C2H2 (which is 3.5 mol), half the C2H2 can react.

Now, we'll calculate the moles of CO2 produced after half the C2H2 is used up:

(3.5 mol C2H2) * (4 mol CO2 / 2 mol C2H2) = 7.0 mol CO2

Thus, 7.0 mol of CO2 would be produced after half the C2H2 is used up, rounded to the nearest 0.1 mol.

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after half of the C₂H₂ is used up, 7.0 mol of CO₂ would be produced.

Yes, half of the 7.0 mol of C₂H₂ can react, leaving 3.5 mol of C₂H₂ unreacted. The balanced chemical equation for the reaction is

C₂H₂ + 2.5 O₂ -> 2 CO₂ + H₂O

From the equation, we can see that 1 mol of C₂H₂ reacts with 2.5 mol of O₂ to produce 2 mol of CO₂ So, to determine how many moles of CO₂are produced when half of the C₂H₂ is used up, we need to calculate how many moles of O2 are required to react with 3.5 mol of C₂H₂.

3.5 mol C₂H₂ × (2.5 mol O / 1 mol C₂H₂) = 8.75 mol O₂

So, to react with 3.5 mol of C2H2, we need 8.75 mol of O2. Since we only have 1.0 mol of O2, we can only react with 0.4 mol of C2H2. Therefore, half of the 7.0 mol of C₂H₂, which is 3.5 mol, can react with 1.0 mol of O₂.

When 3.5 mol of CH₂ reacts, it produces 2/1 x 3.5 = 7.0 mol of CO₂(rounded to the nearest 0.1 mol).

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The mass percentage of C in codeine, C₁₈H₂₁NO₃, is 63.16%.

To calculate the mass percentage of C in codeine, we need to find the molar mass of the compound first.

Molar mass of C₁₈H₂₁NO₃ = (18 x 12.011) + (21 x 1.008) + (1 x 14.007) + (3 x 15.999) = 299.37 g/mol

Next, we need to find the mass of the carbon atoms in one mole of codeine. Since there are 18 carbon atoms in one mole of codeine, we can multiply the molar mass by the number of carbon atoms and divide by the total molar mass of the compound:

Mass of carbon atoms = 18 x 12.011 g/mol = 216.198 g/mol

Mass percentage of C = (mass of carbon atoms / molar mass of codeine) x 100% = (216.198 g/mol / 299.37 g/mol) x 100% = 63.16%

As a result, codeine has a mass proportion of C of 63.16%.

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The best describes her speed and velocity is; Her speed is 4. 4 m/s, and her velocity will be 0 m/s. Option A is correct.

Lila's speed will be calculated by dividing the total distance she covered by  time it took her to complete the race;

Speed = Total distance/Time

In this case, Lila will covered 4 laps, which is a total distance of 4 x 400 = 1600 meters. She completed the race in 6 minutes, which is 6 x 60 = 360 seconds. Therefore, her speed is;

Speed = 1600 meters / 360 seconds

Speed = 4.44 m/s (rounded to two decimal places)

Velocity, on the other hand, is a vector quantity that takes into account both speed and direction. Since Lila ran four laps around a circular track, she ended up at the same position where she started. However, her displacement (change in position) is zero, which means her velocity is also zero.

Hence, A. is the correct option.

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--The given question is incomplete, the complete question is

"Lila is a track and field athlete. She must complete four laps around a circular track. The track itself measures 400 meters from start to finish and the race took her 6 minutes to complete. Which best describes her speed and velocity? A) Her speed is 4. 4 m/s, and her velocity is 0 m/s. B) Her speed is 1. 1 m/s, and her velocity is 0 m/s. C) Her speed is 0 m/s, and her velocity is 2400 m/s. D) Her speed is 4. 4 m/s, and her velocity is 4. 4 m/s"--

pH of the solution hat is composed of 175 mL of 0.5 M ammonia, NH₃, with 0.125 mL of 0.5 M HCl is calculated as 9.52.

Molarity (M) is a unit of concentration that represents the number of moles of solute per liter of solution.

Kb = [NH₄⁺][OH⁻]/[NH₃]

Kb = 1.0 x 10⁻¹⁴ / Kb for ammonia, [NH₄⁺] is the concentration of ammonium ion, [OH⁻] is concentration of hydroxide ion, and [NH₃] is  concentration of ammonia.

moles of NH₃ = (175. mL) x (0.5 mol/L) = 87.5 mmol

moles of HCl = (0.125 mL) x (0.5 mol/L) = 0.0625 mmol

moles of NH₃ that react = 0.0625 mmol

moles of [NH₄⁺] produced = 0.0625 mmol

Total moles of ammonia are now: moles of NH₃ = 87.5 mmol - 0.0625 mmol = 87.4375 mmol

[NH₃] = 87.4375 mmol / 0.175 L = 0.4996 M

The concentration of ammonium ion is the same as the moles produced by the reaction and can be calculated as follows: [NH₄⁺] = 0.0625 mmol / 0.175 L = 0.357 M

Kb = [NH₄⁺][OH⁻]/[NH₃]

4.75 = (0.357 M)([OH⁻])/0.4996 M

[OH⁻] = 3.359 x 10⁻⁵M

Kw = [H⁺][OH⁻] = 1.0 x 10⁻¹⁴

pH = -㏒ [H⁺]

[H⁺] = Kw/[OH⁻] = 1.0 x 10¹⁴ / 3.359 x 10⁻⁵ = 2.98 x 10⁻¹⁰

pH = -㏒ (2.98 x 10⁻¹⁰) = 9.52

Therefore, the pH of the solution is 9.52.

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The statement that is not true is that energy is not transferred to the walls of the container when particles collide with the container walls. That is option C.

The kinetic theory of gases states that gases are made up of large number of atoms or molecules which are in constant motion.

It also states that the molecules constantly collide with each other and with the walls of their container.

The collision between gas particles or collisions with the walls of the container are perfectly elastic.

None of the energy of a gas particle is lost when it collides with another particle or with the walls of the container. This means that the energy should be transferred to the walls of the container when they collide with it.

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The hammer and the feather are dropped from  same height by the astronaut on the planet without the air. The feather will fell at the same rate as the the hammer.

The hammer and the feather are dropped from equal  height by the astronaut on the planet without the air. They were the essentially in the vacuum, and there was the no air resistance and because of the feather will fell at the same rate as compared to the hammer, the Galileo had to concluded that the hundreds of the years before.

All the objects that released together will fall at the same rate excluding the factor of the mass.

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Answer:

In order is 2, 3, 4, 1

Explanation:

It starts when you come into contact with a trigger that you inhale, swallow, or get on your skin. In response, your body starts to make a protein called IgE, which grabs onto the allergen. Then histamine and other chemicals get released into the blood.

Hydroboration of terminal alkynes yields aldehydes.

Here's a step-by-step explanation:

1. Terminal alkynes are reacted with a borane reagent (such as BH₃) in a process called hydroboration.
2. During hydroboration, the borane adds across the triple bond of the terminal alkyne in a syn addition, resulting in the formation of an organoborane intermediate.
3. The organoborane intermediate is then oxidized with hydrogen peroxide (H₂O₂) and a hydroxide ion (OH⁻) in a process called oxidation.
4. This oxidation step converts the organoborane intermediate into an aldehyde functional group.

So, the correct answer is (c) aldehyde.

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The hydroboration of terminal alkynes yields an aldehyde. During the hydroboration process

boron and hydrogen are added across the carbon-carbon triple bond of the alkyne, which yields an unstable intermediate. This intermediate quickly reacts with water to form an aldehyde.

It is important to note that the product of the hydroboration of an alkyne can be influenced by the reaction conditions and the substituents present on the alkyne. However, for a terminal alkyne, the product is typically an aldehyde.

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A reaction is spontaneous under a certain set of conditions are Q < K , ΔGrxn < 0. The correct options are 1) and 14).

The spontaneous reaction is the reaction which will favors the formation of the products under the conditions and under that the reaction is occurring. The spontaneous process is defines as that it occurs by its own, without the external energy input from the outside.

If ΔG°rxn < 0, the reaction will be the spontaneous in the forward direction. The Q < K is when the reaction is the spontaneous reaction or  the reaction is spontaneous for the forward.

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The equilibrium equation for the reaction:

NH₄NO₃(s) ⇌  N₂O(g) + 2H₂O(g) is Kp =  [ N₂O][H₂O]².

In the equilibrium expressions we only considered the gases and the aqueous compounds.  When the one or the more of the substances in the system that will exists in the gaseous phase, and the partial pressure of the species which can be used for the equilibrium expression.

The chemical reaction is as :

NH₄NO₃(s) ⇌  N₂O(g) + 2H₂O(g)

The equilibrium expression is as :

Kp =  [ N₂O][H₂O]².

The ratio of the concentrations for the reactants and the products is called as the equilibrium constant expression.

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A sample of [tex]Br_2[/tex](g) takes 14.0 min to diffuse out through a membrane. It would take 7 min for the same number of moles of Ar(g) to effuse through the same membrane.

According to Graham's Law of Diffusion, it is known that the rate of diffusion of a gas is proportional to the reciprocal of the square root of the molar mass of the gas. The rate of diffusion is recorded under the same pressure and temperature conditions.

It can be written as [tex]\frac{r_1}{r_2}[/tex] ∝ [tex]\sqrt\frac{m_2}{m_1}[/tex]

where [tex]r_1[/tex] is the rate of diffusion of one of the gas

[tex]r_2[/tex]  is the rate of diffusion of the second gas

[tex]r_1[/tex] is the molar mass of one of the gas

[tex]m_2[/tex]  is the molar mass of the second gas

According to the question,

[tex]\frac{14}{r_2}=\sqrt\frac{160}{40}\\\frac{14}{r_2}=\sqrt\frac{4}{1} \\\\r_2 = 14\sqrt{\frac{1}{4} } = 7[/tex]

Therefore the time taken for the diffusion of Ar (g) is 7 min

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It would take approximately 7.0 minutes for the same number of moles of Ar(g) to effuse through the same membrane.

Using Graham's law of effusion, we can compare the rates of effusion for Br2(g) and Ar(g). The formula for Graham's law is:

Rate₁ / Rate₂ = √(M₂ / M₁)

Here, Rate₁ and Rate₂ are the effusion rates of the two gases, and M₁ and M₂ are their molar masses. In this case, Br2(g) is gas 1 and Ar(g) is gas 2. The molar mass of Br2 is 159.8 g/mol, and the molar mass of Ar is 39.95 g/mol.

Since we know the time it takes for Br2 to effuse, we can write:

Time₁ / Time₂ = Rate₂ / Rate₁ = √(M₁ / M₂)

Plugging in the given time and molar masses:

14.0 min / Time₂ = √(159.8 g/mol / 39.95 g/mol)

Solving for Time₂:

Time₂ = 14.0 min * √(39.95 g/mol / 159.8 g/mol) ≈ 7.0 min

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The total of 12 electrons are removed from glucose during cellular respiration to produce 6CO2 and water.

During cellular respiration, glucose (C6H12O6) is broken down into carbon dioxide (6CO2) and water. This process involves the removal of electrons from glucose molecules, which are then used to create ATP, the energy currency of cells.

Specifically, in glycolysis, two electrons are removed from glucose to form NADH, which carries these electrons to the electron transport chain (ETC) for further energy production. In the ETC, the electrons are transferred between different electron carriers and ultimately used to produce ATP.  

It is important to note that cellular respiration is a complex process that involves multiple steps and different electron carriers, and the removal of electrons from glucose is not a single event but a continuous process that generates energy for cells.

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Answer:

A

Explanation:

M.G has no cure. it's an autoimmune disease that affects communication between nervous and the muscles.

The amount of copper obtained by the reaction depends on the amount of copper chloride reacted and the limiting reagent in the reaction.

Without information on the amounts of reactants used, it is not possible to calculate the amount of copper obtained by the reaction or compare it to the expected amount. However, if the reaction was carried out under controlled conditions and all reactants were used in the stoichiometric ratio, the amount of copper obtained should be the expected amount calculated based on the balanced chemical equation.

In order to calculate the expected amount of copper obtained, the amount of copper chloride reacted should be known, and the limiting reagent should be identified. The amount of copper obtained will be determined by the limiting reagent.

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The complete question is:

Calculate the amount of copper obtained by the reaction. Was it the amount you expected? Explain. ​

3CuCl2(aq) +2 Al (s) ----> 2AlCl3(aq) + 3Cu(s)

The volume of the gas, given that the gas was compressed at a constant temperature until the pressure becomes 1.43 atm is 514.69 mL

We'll begin by listing out the various parameters given from the question. This is given below:

The final volume of the gas can be obtained by using the Boyle's law equation as shown below:

P₁V₁ = P₂V₂

1.15 × 640  = 1.43 × V₂

736  = 1.43 × V₂

Divide both side by 1.43

V₂ = 736 / 1.43

V₂ = 514.69 mL

Thus, we can conclude from the above calculation that the volume of the gas is 514.69 mL

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If the temperature of water is increased, then it will change from a solid state (ice) to a liquid state, and then to a gaseous state.

A solid is one of the three basic states of matter, alongside liquid and gas. In a solid, the particles are tightly packed together and held in fixed positions by intermolecular forces. This gives solids a definite shape and volume that is maintained unless acted upon by an external force.

In a solid, the particles vibrate in place but do not move past one another. This is in contrast to a liquid, where the particles are more loosely packed and can move past one another, and a gas, where the particles are widely spaced and move freely in all directions.

Solids can be classified based on their physical properties, such as their hardness, density, and melting point. Some common examples of solids include metals, minerals, and various types of rock. Solids also play an important role in many areas of science and technology, from materials science and engineering to biology and medicine.

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The solutes in the new solution are alcohol and salt. Therefore, option D is correct.

A solution is a homogeneous mixture composed of two or more substances. A solution consists of a solute and a solvent. The solute is the substance that is dissolved in the solvent, and the solvent is the substance that dissolves the solute.

A solution is considered homogeneous because the solute particles are evenly distributed throughout the solvent. This means that the composition of the solution is the same throughout, and there are no visible differences or separations of the components.

Solutions can be in different states of matter, such as a gas in a gas, a gas in a liquid, a liquid in a liquid, or a solid in a liquid. Therefore, option D is correct.

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The normality of acetic acid in the vinegar is 0.2919 N.

To determine the normality of acetic acid in the vinegar, you can follow these steps:

1. Write the balanced equation for the reaction:
  CH₃COOH (acetic acid) + NaOH → CH₃COONa + H₂O

2. Calculate the milliequivalents of NaOH used in the titration:
  Milliequivalents of NaOH = volume (mL) × normality
  Milliequivalents of NaOH = 34.80 mL × 0.4164 N
  Milliequivalents of NaOH = 14.49 meq (rounded to two decimal places)

3. Since the reaction is a 1:1 ratio, the milliequivalents of acetic acid will be equal to the milliequivalents of NaOH:
  Milliequivalents of acetic acid = 14.49 meq

4. Calculate the normality of acetic acid in the vinegar:
  Normality of acetic acid = milliequivalents of acetic acid/volume of vinegar (mL)
  Normality of acetic acid = 14.49 meq / 49.6 mL
  Normality of acetic acid = 0.2919 N (rounded to four decimal places)

So, the normality of acetic acid in the vinegar is 0.2919 N.

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Vinegar is an aqueous solution of acetic acid, ch3cooh. suppose you titrate a 49.6 ml sample of vinegar with 34.80 ml of a standardized 0.4164 n solution of NaOH. The normality of acetic acid in this vinegar is 0.292 N.

To find the normality of acetic acid in the vinegar, we can use the titration information provided. Here's a step-by-step explanation:
1. Write down the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
  CH3COOH + NaOH → CH3COONa + H2O
2. Calculate the number of equivalents of NaOH used in the titration:
  Number of equivalents of NaOH = Volume of NaOH solution × Normality of NaOH solution
  Number of equivalents of NaOH = 34.80 mL × 0.4164 N
  (Note: Convert mL to L by dividing by 1000)
  Number of equivalents of NaOH = 0.03480 L × 0.4164 N = 0.0145 equivalents
3. Since the balanced equation shows a 1:1 ratio between acetic acid and NaOH, the number of equivalents of acetic acid in the vinegar is equal to the number of equivalents of NaOH used in the titration.
4. Calculate the normality of acetic acid in the vinegar:
  Normality of acetic acid = Number of equivalents of acetic acid / Volume of vinegar sample
  Normality of acetic acid = 0.0145 equivalents / 0.0496 L = 0.292 N
So, the normality of acetic acid in the vinegar is 0.292 N.

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When acid is added to the slime, the crosslinking is broken, resulting in a liquid with lesser viscosity. When alkali is added, the process is reversed, and the slime needs to be regenerated.

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Here are some general steps you can follow to write a lab report:

Understand the purpose of the lab report: Before you begin writing, make sure you understand the purpose of the lab report. What are the objectives of the experiment? What are the research questions being investigated? What hypothesis is being tested?

Gather your data: Make sure you have all the data you need to write your report. This includes raw data, observations, and any notes you took during the experiment. Organize your data in a clear and logical manner so that you can easily refer to it when writing your report.

Write an outline: Create an outline for your report that includes the main sections you need to cover. These typically include an introduction, methods, results, discussion, and conclusion.

Write the introduction: The introduction should provide an overview of the experiment and explain its significance. You should also provide some background information to help the reader understand the context of the experiment.

Write the methods: In the methods section, describe the experimental design, materials used, and procedures followed. Be sure to include enough detail so that someone else could repeat the experiment.

Write the results: In the results section, present your data in a clear and organized manner. Use tables, graphs, and figures to help illustrate your findings. Make sure to include any statistical analyses you performed.

Write the discussion: In the discussion section, interpret your results and explain what they mean. Discuss any patterns or trends you observed and explain how they relate to the research question. Compare your results to previous research in the field, and discuss any limitations or potential sources of error.

Write the conclusion: The conclusion should summarize the main findings of the experiment and explain their significance. You should also discuss any future directions for research in the field.

Proofread and revise: Once you have completed your first draft, proofread your report carefully to check for errors and inconsistencies. Revise your report as necessary to make sure it is clear, concise, and well-organized.

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Glyphosate was initially thought to have no effect on human health because it primarily targets enzymes found only in plants and bacteria, not humans.

Additionally, the compound was believed to have a low toxicity level and was considered to be safe when used according to the labeled instructions.

However, recent scientific studies have suggested potential health risks associated with glyphosate exposure, including links to cancer and other health issues.

These studies have prompted further investigation and controversy surrounding the safety of glyphosate in herbicides.

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The one-step Wittig reaction is preferred over the two-step Grignard reaction for the conversion of aldehydes to alkenes due to its simplicity, higher yields, and fewer side reactions.

The change includes switching an aldehyde over completely to an alkene, which can be accomplished through a one-step Wittig response or a two-step Grignard response.

The one-step Wittig response includes responding the aldehyde with a Wittig reagent, which is a phosphonium ylide, within the sight of a base. The response delivers an alkene and a phosphine oxide as a result.

Then again, the two-step Grignard response includes the development of a Grignard reagent from an alkyl or aryl halide, trailed by the expansion of the subsequent Grignard reagent to the aldehyde to shape a liquor. The liquor is then got dried out to shape the ideal alkene.

As a general rule, the one-step Wittig response is liked over the two-step Grignard response for the transformation of aldehydes to alkenes since it is a less difficult and more straightforward interaction. The Wittig response additionally has better returns and less side responses contrasted with the Grignard response. In any case, the decision of response might rely upon the particular beginning material and the ideal item.

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The transformation in question involves the conversion of a carbonyl compound (such as an aldehyde or ketone) to an alkene.

One method to achieve this transformation is through a one-step reaction called the Wittig reaction. In this method, a phosphorus ylide is used to transfer a carbonyl group to the ylide, forming an intermediate called an oxaphosphetane. This intermediate then undergoes an elimination reaction to give the desired alkene.

Another method to achieve this transformation is through a two-step reaction called the Grignard reaction. In this method, a Grignard reagent is first prepared by reacting an alkyl or aryl halide with magnesium in the presence of an ether solvent. This Grignard reagent can then react with a carbonyl compound to form an intermediate called an alkoxide. This intermediate can then undergo an elimination reaction to give the desired alkene.

The preferred route depends on several factors, such as the nature of the starting material, the desired product, and the availability of reagents. In general, the Wittig reaction is preferred for the conversion of aldehydes to alkenes, while the Grignard reaction is preferred for the conversion of ketones to alkenes. However, both methods can be useful in different situations and should be considered based on the specific requirements of the synthesis.

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We can see that the hydrobromic acid is the limiting reagent because it is completely consumed when 0.599 mol of it reacts with 0.599 mol of ethanol. After the reaction is complete, there will be some excess ethanol left over.

To determine the limiting reagent, we need to compare the amount of moles of each reactant used in the reaction. We can calculate the number of moles of each reactant by dividing their mass by their molar mass. Let's assume the starting alcohol is ethanol and has a molar mass of 46.07 g/mol, and hydrobromic acid has a molar mass of 80.91 g/mol. Then we have:

Moles of ethanol = 81.3 g / 46.07 g/mol = 1.765 mol

Moles of hydrobromic acid = 48.5 g / 80.91 g/mol = 0.599 mol

According to the balanced chemical equation, the stoichiometric ratio of ethanol to hydrobromic acid is 1:1.

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The new volume of the gas with 1.90 L of gas are heated from 25.00°C to 95.00°C. If the pressure remains constant, is 72.2L.

According to Charles Law the formula given here is V1/T1=V2/T2

Substituting the values in the equation we get

1.9/25=V2/ 95

V2= 1.9× 95/25

V2= 72.2 L

The experimental gas law known as Charles's law outlines how gases have a tendency to expand when heated. Charles's law is now expressed as follows: The volume and Kelvin temperature of a sample of a dry gas will be directly proportional to the pressure, when the pressure is held constant.

The volume of a gas is equal to a constant amount multiplied by the temperature of the gas as determined by the Kelvin scale (zero Kelvin is equivalent to -273.15 degrees Celsius), according to the physical law known as Charles' law.

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Answer: flowing water

Explanation:

The standard enthalpy of reaction for the combustion of propane is approximately -2044 kJ/mol.

The balanced equation for the combustion of propane is:

To calculate the standard enthalpy of reaction (ΔH°) for this reaction, we need to use the heats of formation (ΔHf°) for the compounds involved. Using the given values:

We can calculate the ΔH° for the reaction by subtracting the sum of the heats of formation of the reactants from the sum of the heats of formation of the products:

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The steps that compose a rationale for the cation Pb+2 being absent in an unknown (but Ag+ is present) include:

1. No white precipitate formed when 6 M HCl was added to the unknown solution in step 1-A. This indicates that the unknown does not contain chloride ions, which could form a white precipitate with Pb+2.

2. A yellow precipitate did not form when K2CrO4 was added in step 1-C. This indicates that the unknown does not contain chromate ions, which could form a yellow precipitate with Pb+2.

3. The presence of a light blue decantate in step 1-A. This indicates the presence of Ag+ ions, which could form a light blue precipitate with NH3 in step 2-A.

4. A white precipitate did not form in step 2-B. This indicates that the unknown does not contain sulfate ions, which could form a white precipitate with Pb+2.

5. The white solid did not turn black upon addition of NaOH and SnCl2. This indicates that the unknown does not contain lead(II) sulfide, which could react with NaOH and SnCl2 to form black lead(II) oxide.

6. A reddish brown precipitate did not form after adding K4Fe(CN)6. This indicates that the unknown does not contain lead(II) ions, which could form a reddish brown precipitate with K4Fe(CN)6.

7. A lack of dark blue colored solution after addition of 15 M NH3. This indicates that the unknown does not contain copper(II) ions, which could form a dark blue solution with NH3.

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The step(s) that compose a rationale for the cation Pb+2 being absent in an unknown (but Ag+ is present) include:
- No white precipitate formed when 6 M HCl was added to the unknown solution in step 1-A, indicating that PbCl2 is not present.
- All of the white precipitate from step 1-A dissolved in hot water, further indicating that PbCl2 is not present.
- A yellow precipitate did not form when K2CrO4 was added in step 1-C, ruling out the presence of PbCrO4.
- The white precipitate from step 1-B dissolved in 6 M NH3 and then reformed when 6 M HNO3 was added, indicating the presence of AgCl.
- The white solid did not turn black upon addition of NaOH and SnCl2, ruling out the presence of Hg2+.
- A reddish brown precipitate did not form after adding K4Fe(CN)6, ruling out the presence of Fe2+ or Fe3+.
- A lack of dark blue colored solution after addition of 15 M NH3, ruling out the presence of Cu2+.

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