calculate the standard enthalpy of reaction for the combustion of propane. note: this equation is not balanced. round to the nearest whole number. c3h8(g) o2 --> co2(g) h2o(l) kj/mol compound hf (kj/mole) c3h8(g) -105 co2(g) -394 h2o(l) -284

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"--

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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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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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:

A

Explanation:

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

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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Changes in the chemical environment of a basic buffer solution can cause shifts in the equilibrium between ammonia, hydroxide ion, and ammonium ion, leading to changes in their respective concentrations.

When a basic buffer solution of equal concentrations of ammonia and ammonium nitrate is subjected to changes in its chemical environment, the concentrations of ammonia, hydroxide ion, and ammonium ion will be affected as follows:

1) Addition of an acid: The acid reacts with the hydroxide ion present in the buffer solution to form water. This decreases the concentration of hydroxide ion and shifts the equilibrium towards the formation of more ammonia and ammonium ion, thus increasing their concentrations.

2)Addition of a base: The base reacts with the ammonium ion present in the buffer solution to form ammonia and water. This decreases the concentration of ammonium ion and shifts the equilibrium towards the formation of more hydroxide ion, thus increasing its concentration.

3)Dilution: Diluting the buffer solution with water decreases the concentrations of both ammonia and ammonium ion, but does not affect the concentration of hydroxide ion.

4)Addition of ammonium chloride: The ammonium chloride dissociates into ammonium ion and chloride ion in the buffer solution. The increase in ammonium ion concentration causes the equilibrium to shift towards the formation of more ammonia and hydroxide ion, thus increasing their concentrations.

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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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Answer: flowing water

Explanation:

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.

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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1) 5.1167 x 10²⁴atoms of chlorine. 2) 248.00 g. 3) 4.8825 x 10⁷ moles of helium. 4) 2.7757 x 10²⁴ atoms of sulfur.  5) Molar mass of Co (cobalt) is 58.93 g/mol.  6) Formula mass =  310.18 g/mol.

Sum of the atomic masses of all the atoms in chemical formula is called formula mass

1.)  Number of atoms = 8.500 moles x 6.022 x 10²³ atoms/mole = 5.1167 x 10²⁴ atoms of chlorine.

2.) Molar mass of oxygen is 16.00 g/mol. Therefore:

Mass of 15.50 moles of oxygen = 15.50 moles x 16.00 g/mol = 248.00 g.

3.) Molar mass of helium is 4.00 g/mol. Therefore, the number of moles of helium in 1.953 x 10⁸ g is:

Number of moles = 1.953 x 10⁸ g / 4.00 g/mol = 4.8825 x 10⁷ moles of helium.

4.) Molar mass of sulfur is 32.06 g/mol. Therefore, the number of moles of sulfur in 147.82 g is:

Number of moles = 147.82 g / 32.06 g/mol = 4.6084 moles of sulfur.

To find the number of atoms, we can use Avogadro's number again:

Number of atoms = 4.6084 moles x 6.022 x 10²³ atoms/mole = 2.7757 x 10²⁴ atoms of sulfur.

5.) Molar mass of Co (cobalt) is 58.93 g/mol.

6.) Ca₃(PO₄)₂ contains 3 calcium atoms, 2 phosphorus atoms, and 8 oxygen atoms.

Atomic masses of these elements are:

Calcium (Ca) = 40.08 g/mol

Phosphorus (P) = 30.97 g/mol

Oxygen (O) = 16.00 g/mol

Therefore, formula mass of Ca₃(PO₄)₂ is:

Formula mass = (3 x 40.08 g/mol) + (2 x 30.97 g/mol) + (8 x 16.00 g/mol)

= 120.24 g/mol + 61.94 g/mol + 128.00 g/mol

= 310.18 g/mol.

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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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If you had 6 M acetic acid rather than glacial acetic acid in the first step of the Friedel-Crafts reaction, less product would likely have formed.

The Friedel-Crafts reaction requires a strong Lewis acid catalyst, such as aluminum chloride (AlCl3), which reacts with the acylating agent to form a reactive electrophile. In this case, the acylating agent is acetic anhydride, which reacts with AlCl3 to form an acylium ion that can then react with the aromatic ring. However, the reaction is sensitive to the amount of water present, and the presence of excess water (which would be more likely in the case of dilute acetic acid) can lead to hydrolysis of the acylium ion and a decrease in the yield of the desired product.

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The sour mash used to make distilled spirits is blank is the liquid fermented grain that is collected from the previous batch.

The Sour mash or we can say that the  sourmash is the process that is used in the industry of the distilling which uses the material from the older batch of the mash and to start the fermentation for the new batch, the analogous to the make of the sourdough bread with the starter.

Therefore, the liquid fermented grain that is collected from the previous batch is the sour mash used to make the distilled spirits is blank.

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When an ionic compound was dissolved in 250 mL of water and the temperature of the solution increased from 25°C to 42°C, the value of ΔH is lesser than zero and has a small magnitude and the value of ΔG is also lesser than zero.

When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation.

In this there is increase in temperature which means that heat energy is released which makes the dissolution exothermic and thus it has ΔH negative. Since the reaction is feasible, the ΔG will be less than zero.

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True. The standard conditions for measuring thermodynamic properties such as enthalpy, entropy, and Gibbs free energy are well-defined and standardized.

These conditions are used to compare and evaluate the relative stability and reactivity of different chemical species. The standard conditions for measuring these properties include a concentration of 1.0 m for soluble aqueous species, pure solids and liquids, and

a partial pressure of 1 atm for gaseous species. This means that the molar concentration of soluble aqueous species is set at 1.0 mol/L, and the pressure of gaseous species is set at 1 atm.

Pure solids and liquids are considered to have an activity of 1, which means that they do not affect the thermodynamic properties.

These conditions are used to determine the standard thermodynamic properties of chemical reactions, which are used to predict the direction and extent of chemical reactions.

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The rate of the reaction is directly proportional to the stability of the carbocation intermediate, and any changes in the solvent will affect the rate of the reaction.

In an SN1 reaction, the rate-determining step is the formation of a carbocation intermediate. The stability of the carbocation intermediate affects the rate of the reaction.

In this case, if 2-propanol (isopropanol) was used instead of 2-methyl-2-propanol (t-butanol), the rate of the reaction would decrease. This is because the carbocation intermediate formed in 2-propanol is less stable compared to the one formed in t-butanol.

The carbocation intermediate formed in t-butanol is tertiary, which is more stable than the one formed in isopropanol, which is secondary. This means that the reaction will be slower in isopropanol due to the less stable carbocation intermediate.

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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?--

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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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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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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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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Answer:

the answer is D. the molecules of a gas are small compared to the distance between them

Explanation:

The postulates of Kinetic-Molecular Theory are as follows:

The molecules of a gas move rapidly, constantly, and in straight lines.

The average kinetic energy of the molecules of a gas is directly proportional to the Kelvin temperature.

A gas exerts pressure because the gas molecules repel each other.

The molecules of a gas are small compared to the distance between them1.

Therefore, “the molecules of a gas are small compared to the distance between them” is not a postulate of Kinetic-Molecular Theory

In changing the size of the zinc electrode would not impact the voltage of the cell.

If the size of the zinc electrode were doubled, the cell voltage would stay the same. This is because the voltage of a cell is dependent on the difference in potential between the two electrodes, not their size.

Doubling the size of the zinc electrode would not change the potential difference between the zinc and copper electrodes, therefore the cell voltage would remain constant.

The only factor that would affect the cell voltage would be a change in the concentration or temperature of the electrolyte solution or a change in the material or surface area of the copper electrode.

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We separate porridge from cooked rice using a strainer or a sieve.

Porridge and cooked rice are both similar in texture and appearance, making it difficult to separate them. One way to do it is to use a strainer or a colander with small holes. Pour the mixture of porridge and rice into the strainer or colander and let the liquid portion drain out. You can also use a cheesecloth or a muslin cloth to squeeze out the liquid while retaining the rice grains.

Another method is to use a spoon to scoop out the rice from the top, leaving the porridge at the bottom. However, this method may not be as effective as the others. Regardless of the method you choose, it is important to be gentle and patient to avoid breaking the rice grains and mixing them with the porridge.

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The concentration of stock solution is 15.2 M if You have 450.80 mL of 0.86 M dilute solution that was prepared with 25.50 of stock solution.

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When lactic acid accumulates, the compensatory mechanism initially triggered to maintain acid-base balance is the respiratory system. It increases the rate and depth of breathing to eliminate more carbon dioxide, which helps reduce acidity and restore pH balance in the body.

As lactic acid accumulates, the compensatory mechanism that is initially triggered to maintain acid-base balance is increased ventilation or hyperventilation. This is because the increased ventilation leads to a decrease in carbon dioxide (CO2) levels in the blood, which helps to offset the acidic effects of lactic acid accumulation. This is a temporary compensatory mechanism, however, as the body will eventually need to eliminate the excess lactic acid through other means such as metabolism and excretion.

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