In your lab you are studying the kinetics of the degradation of a pain killer in the human liver. You are monitoring the concentration of the pain killer over a period of time. The initial concentration of the pain killer in your experiment was 1.59 M. After 0.80881 hours the concentration was found to be 0.795 M. In another 0.80881 hours the concentration was found to be 0.3975 M (t=27.868 hours overall).

If another experiment were set up where the initial concentration of the pain killer was 0.479 M, how long would it take for the pain killer concentration to reach 0.0161 M?

Answer:

Electrolysis of either the molten or aqueous salt will produce solid iron.

Explanation:

Iron (II) iodide has a relatively low melting point. Electrolysis of the molten salt will probably produce solid metallic iron and liquid I2 as well as iodine vapor. The temperature need not be high enough to cause Fe to melt.

so Electrolysis of either the molten or aqueous salt will produce solid iron.

When The Electrolysis of either the molten or aqueous salt will produce solid iron. So The temperature need not be high enough to cause Fe to melt.

When Iron (II) iodide has a moderately low melting point. Electrolysis of the molten salt will probably construct solid metallic iron and liquid I2 as well as iodine vapor. The temperature requirement not be high enough to cause Fe to melt.

Therefore, The Electrolysis of either the molten or aqueous salt will produce solid iron.

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The question is incomplete; the complete question is:

A chemist must prepare 800.0mL of potassium hydroxide solution with a pH of 13.00 at 25 degree C. He will do this in three steps: Fill a 800.0mL volumetric flask about halfway with distilled water. Weigh out a small amount of solid potassium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to 2 significant digits.

Answer:

4.5g (to 2 significant digits)

Explanation:

Now we must remember that KOH is a strong base, therefore it will practically dissociate completely.

To find the pH we can use the equation pH + pOH = 14.

Firstly to find the pOH:

13.00 + pOH = 14

pOH = 1.00

To find the [OH-]

Since

pOH= -log[OH^-]

[OH^-] = antilog (-pOH)

[OH^-]= antilog (-1)

[OH^-] = 0.1 molL-1

Since we've established that KOH is a strong base, we know that [OH-] = [KOH]

Also, we know that concentration = number of moles/volume

we have the concentration and the volume now so we can calculate the number of number of moles as follows:

The 800mL volume is the same as 0.8L

0.1 molL-1= number of moles/0.8L

0.08 moles = number of moles

now we can calculate the amount of solid KOH required

the molar mass of KOH = 39 + 16 +1 = 56 gmol-1

56 x 0.08 moles = 4.48g

So in 800mL of pH 13.00 KOH there is 4.5g of KOH dissolved.

A current of 620 A for 90 seconds produces 5.209 g of aluminum.

The mass of pure aluminum produced in the Hall-Heroult process can be calculated by using Faraday's laws of electrolysis and the molar mass of aluminum. First, we determine the moles of electrons transferred using the given current and time. Then, using the stoichiometry of the reaction where 4 moles of Al are produced per 12 moles of electrons:

Calculate the charge (Q) passed: Q = current (I) x time (t) = 620 A x 90.0 s = 55800 C

Determine the moles of electrons: Moles of e- = Q / Faraday's constant (F) Assuming the Faraday constant is approximately 96500 C/mol, this gives Moles of e- = 55800 C / 96500 C/mol = 0.5784 mol

Calculate the moles of aluminum: Since 3 moles of electrons yield 2 moles of Al, 0.5784 mol of e- will yield (4/12) x 0.5784 mol of Al = 0.1931 mol of Al.

Compute the mass of aluminum produced: Mass of Al = moles of Al x molar mass of Al = 0.1931 mol x 26.98 g/mol = 5.209 g

The mass of pure aluminum produced is therefore 5.209 g, assuming 100% efficiency and no other side reactions.

To find the mass of pure aluminum produced, we calculate the charge passed, find the moles of electrons transferred, determine the moles of aluminum from the moles of electrons, and use the molar mass of aluminum to find that approximately 5.20 grams of aluminum is produced.

To calculate the mass of pure aluminum produced in the Hall-Heroult process when 620 A of current is passed through a Hall-Heroult cell for 90.0 seconds, we use Faraday's laws of electrolysis. The reduction of aluminum oxide to aluminum involves the transfer of three moles of electrons per mole of aluminum according to the equation:

Al₂O₃ + 3e⁻ → 2Al + 1.5O₂

Firstly, we calculate the amount of charge passed using the formula Q = It, where Q is the charge in coulombs, I is the current in amperes, and t is time in seconds. Then, using Faraday's constant (96485 C/mol), we determine the moles of electrons transferred. Since we need 3 moles of electrons to produce 1 mole of aluminum, we find the moles of aluminum formed by dividing the moles of electrons by 3. Finally, we calculate the mass using the molar mass of aluminum (26.98 g/mol).

Q = 620 A * 90.0 s = 55800 C
Moles of electrons = Q/F = 55800 C / 96485 C/mol ≈ 0.5785 mol
Moles of aluminum = Moles of electrons / 3 ≈ 0.1928 mol
Mass of aluminum = Moles of aluminum * Molar mass of Al
Mass of aluminum ≈ 0.1928 mol * 26.98 g/mol ≈ 5.20 g

The mass of pure aluminum produced is approximately 5.20 grams.

Energy given off by the system during dissolution equals the energy absorbed by the surroundings. Therefore, option (A) is correct.

The law of energy conservation states that energy can only be transmitted or changed. This equation says that the energy produced by the system during exothermic dissolution, where a material dissolves while generating heat, is equal to the energy received by the solvent and container.

Solute molecules spread and interact with the solvent, forming or breaking bonds and releasing heat. The environment absorbs this heat, keeping energy balance. The law of energy conservation emphasises that energy changes in a system must be counterbalanced by energy changes in the surrounding environment, reinforcing the idea that energy is conserved during exothermic dissolution. Therefore, option (A) is correct.

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The law of conservation of energy states that the energy in a system cannot be created or destroyed, only changed in form. In exothermic dissolution processes, the energy given off by the system as heat is equal to the energy absorbed by the surroundings.

According to the law of conservation of energy, energy can neither be created nor destroyed, only changed in form during a chemical or physical change. In the case of exothermic dissolution processes, energy is given off by the system as heat, and this energy is equal to the energy absorbed by the surroundings. This means that the energy released during dissolution is the same as the energy gained by the surroundings.

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Answer:

caffience

Explanation:

Answer:

It opens the can for him

Explanation:

Answer:

Takes the pain off the mans hand.

Explanation:

Saw that you didn't like europa2433, i just kinda wanted to agree he reports to much.

Answer:

A. 145.2 g NH3

B. 0.76 g H2

C. 1.41 x 10^22 molecules NH3

Explanation:

A. 1 mol N2    -> 2 mol NH3

   4.27 mol N2 -> x

x= (4.27 mol N2 * 2 mol NH3)/1 mol N2     x= 8.54 mol NH3

1 mol N2       -> 17 g

8.54 mol N2 -> x        x= 145.2 g NH3

B.

2 g H2  -> 34 g NH3

         x   -> 13.01 g NH3

x= 0.76 g H2

C.

2 g  H2      -> 34 g NH3

0.0235 g H2 -> x

x=   0.39 g NH3

0.39 g NH3 (1 mol NH3/17 g NH3)(6.023 x 10^23 molecules/1 mol NH3) =

1.41 x 10^22 molecules NH3

Answer:

empirical formula: [tex]H_2SO_4[/tex]

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02  :  1.02/1.02  :  4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: [tex]H_2SO_4[/tex].

The three things that should be mentioned are Helium in Balloons, Lithium in Batteries, Beryllium in Emeralds, and Boron in Sports equipment.

Elements or rather compounds seem to be natural pure chemical substances.

The distinction here between element as well as a compound is whether an element is a compound consisting of the same type of atoms as a compound.

Whereas a compound is made up of various elements in specific proportions. Elements include iron, copper, hydrogen, and oxygen.

Metals are on the left side of the staircase, metalloids are on the right, and nonmetals are on the right.

An atom is a component of an element. An element is made up of only one type of atom.

Atoms are further subdivided into subatomic particles known as electrons, protons, and neutrons. Chemical reactions allow elements to combine to form molecules.

Thus, the three examples are helium, lithium, and boron.

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Answer:

solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] can be calculated using the information given.

Let's assume solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

   or, y = [tex]\frac{3.96}{36}[/tex] = 0.11

Hence solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is 0.11 g/mL.

Final answer:

Aluminum, symbol (C) Al, forms a protective oxide layer known as aluminum oxide (Al2O3) when exposed to oxygen, protecting it from further corrosion.

Explanation:

The symbol for the element that forms a protective oxide coating is (C) Al, which stands for Aluminum. When aluminum is exposed to the atmosphere, it reacts with oxygen to form aluminum oxide (Al2O3), a thin, hard layer that helps prevent further oxidation and protects the metal underneath. This property is particularly useful for applications where durability and resistance to corrosion are important. For example, the outside of the aerospace and construction materials are often made from aluminum for this reason.

Answer:

%error = 0.32%

Explanation:

Let's answer both questions, by parts.

1. Percentage error:

In this case, I do not have the video, but I do have the reported melting point of naphtalene which is 80.26 °C.

The expression to calculate the percentage error is the following:

%Error = absolute error / actual percentage. (1)

And the absolute error is:

Abs error = actual value - experimental value  (2)

But the experimental value is a range, so we just have to get a average of that:

Exp value = 77 + 83 / 2 = 80 °C

Now the absolute error:

Abs error = 80.26 - 80 = 0.26 °C

Finally the %error:

%error = (0.26 / 80.26) * 100

2. Meaning of melting point range and %error

The melting point range just means that the sample of naphtalene has impurities, and when a sample of any compound has impurities, melting point tends to be low. However, this decrease of temperature is a wider range. But usually a range of just 5° C means that compound has little traces of impurities but it can still be used for reactions.

The %error means that the impurities of the sample are really low, so the sample is practically pure with little traces of impurities.

Final answer:

The percent error for the melting point of naphthalene is calculated using the formula and considering the literature value of 80.2°C, yielding an approximate percent error of 0.25%. The broad melting range and low percent error suggest the sample is mostly pure with possible minor impurities.

Explanation:

The student's question is regarding the calculation of the percent error for the melting point of a sample of naphthalene and what the results suggest about the sample's purity.

Percent Error Calculation

To find the percent error, we use the formula:

Percent Error = (|Experimental Value - Literature Value| / Literature Value) x 100%

Assuming the literature value of naphthalene's melting point is approximately 80.2°C (the value will need to be verified as it can vary slightly in literature), we calculate percent error using the experimental value's mean (80°C, the midpoint of 77-83°C) as follows:

Percent Error = (|80°C - 80.2°C| / 80.2°C) x 100% = (0.2°C / 80.2°C) x 100% ≈ 0.25%

Suggestion About the Sample

The observed melting range of 77-83°C and the low percent error suggest that the sample is relatively pure but may contain minor impurities since a pure sample would have a narrower melting point range close to the literature value.

Answer:

Laura can look for a transparent and  translucent liquid and hence determine which beaker has water and which has solution

Explanation:

Pure water is a compound that is transparent in color. However, a solution is a liquid mixture comprising of a solvent or a solute. The atoms of solute occupy space between the atoms of solvent and hence are said to dissolve in it. Water can be a solvent.  

Thus, if the beaker has a transparent liquid in it, then it would be pure water while a beaker having a translucent liquid, then it would be a solution  

Final answer:

Laura can differentiate pure water from solutions by using a conductivity test, indicators like pH paper or red cabbage water, measuring boiling and freezing points, or measuring the refractive index with a refractometer.

Explanation:

Laura can determine which beakers contain pure water and which contain solutions by conducting a series of tests that rely on the physical and chemical properties of the substances. Here are potential methods for differentiating between pure water and solutions:

Conducting a conductivity test, as pure water is a poor conductor of electricity while solutions with ions will conduct electricity.

Using indicators such as pH paper or red cabbage water, as solutions might be acidic or basic, changing the color of the indicator while pure water will not.

Examining the boiling and freezing points, as solutions have different boiling and freezing points compared to pure water.

Employing a refractometer or similar devices to measure the refractive index, which would differ between pure water and a solution.

To elaborate, a conductivity test can be set up by inserting electrodes into each beaker and connecting these to a circuit with a bulb or a conductivity meter. A color change when using red cabbage water as an indicator would signify the presence of an acidic or basic solution, since red cabbage juice changes color at different pH levels. Observing the boiling and freezing points would require heating or cooling the liquid and taking note of the temperature at which the change of state occurs. Pure water has specific boiling and freezing points (100°C and 0°C at standard atmospheric pressure), and deviations from these numbers would indicate a solution. Lastly, a refractometer could be used to compare the refraction of light through the liquids against known values for pure substances, revealing which beakers contain pure water and which contain solutions.

Answer:  B) Ag(s) is formed at the cathode, and [tex]Mn^{2+}(aq)[/tex] is formed at the anode.

Explanation:

[tex]Ag^{+}(aq)+e^{-1}\rightarrow Ag(s)[/tex]  E=0.80 V

[tex]Mn^{2+}(aq)+2e^{-1}\rightarrow Mn(s)[/tex]  E=-1.18 V

Reduction takes place easily if the standard reduction potential is higher (positive) and oxidation takes place easily if the standard reduction potential is less(more negative).

Thus as reduction potential of Ag is higher , it undergoes reduction and Manganese with lower reduction potential undergoes oxidation. Here Mn undergoes oxidation by loss of electrons, thus act as anode. Ag undergoes reduction by gain of electrons and thus act as cathode.

Cathode : reduction : [tex]Ag^{+}(aq)+e^{-1}\rightarrow Ag(s)[/tex]

Anode : oxidation : [tex]Mn\rightarrow Mn^{2+}(aq)+2e^{-1}[/tex]

Ag(s) is formed at the cathode, and [tex]Mn^{2+}(aq)[/tex] is formed at the anode.

Answer:

0.382 g

Explanation:

Let's consider the reduction of gallium (III) to gallium that occurs in the electrolysis.

Ga³⁺ + 3 e⁻ → Ga

We can establish the following relations:

We will use this that to determine the mass of gallium deposited from a Ga(III) solution using a current of 0.880 A that flows for 30.0 min

[tex]30.0min \times \frac{60s}{1min} \times \frac{0.880c}{s} \times \frac{1mole^{-} }{96,468c} \times \frac{1molGa}{3 mole^{-}} \times \frac{69.72g}{1molGa} = 0.382 g[/tex]

Answer:

Initial rate of reaction is [tex]2.07\times 10^{-3}M.s^{-1}[/tex].

Explanation:

It is a second order reaction.

Initial rate of reaction = [tex]k[O_{3}]_{0}[NO]_{0}[/tex]   , where k is rate constant, [tex][O_{3}]_{0}[/tex] is the initial concentration of [tex]O_{3}[/tex] and [tex][NO]_{0}[/tex] is the initial concentration of NO.

Here, k = [tex]4.09\times 10^{6}M^{-1}.s^{-1}[/tex], [tex][O_{3}]_{0}=5.84\times 10^{-6}M[/tex] and [tex][NO]_{0}=8.65\times 10^{-5}M[/tex]

So, initial rate of reaction = [tex](4.09\times 10^{6}M^{-1}.s^{-1})\times (5.84\times 10^{-6}M)\times (8.65\times 10^{-5}M)[/tex]

= [tex]2.07\times 10^{-3}M.s^{-1}[/tex]

So, initial rate of reaction is [tex]2.07\times 10^{-3}M.s^{-1}[/tex]

Answer:

1) increases

2) smaller

Explanation:

Generally, as electron- electron repulsion increases and more electrons are added to the atom while Z is held constant, the electron cloud size is increased. The size of the anion formed is usually measured as the size of this extended electron cloud. Hence the larger electron cloud means a larger anion size compared to the size of the neutral atom.

For a cation, the converse is true and the cation is found to be smaller than the neutral atom.

This question is incomplete, I got the complete one from google as below:

I−>I>I+

1. As Z stays constant and the number of electrons increases, the electron-electron repulsions ______ (increases or decreases), and the anion becomes larger.

2. The reverse is true for the cation, which becomes ____ (smaller or larger) than the neutral atom.

Answer:

1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger.

2.The reverse is true for the cation, which becomes smaller than the neutral atom.

Explanation:

1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger. This is because in anions of the same atoms, the net force of attraction on electrons decreases.

2. The reverse is true for the cation, which becomes smaller than the neutral atom. This is because in cations of the same atoms, the net force of attraction on electrons increases.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below for the step by step explanation to the question above.

Answer:

Naming ionic compounds with transition metals isn't too hard either. They are named like the binary compounds, with the cation first, then the anion with -ide added to it, but you have to take into account the variations of the metal ions.

Explanation:

Answer:

Kw= = [H3O+][OH-] = [H+][OH-].

Explanation:

In chemistry, the term 'equilibrium'' is a period or time in which the rate at which the product is been produced or generated is equal to the quantity of the reactants reacting. In order way, we can say the equilibrium is when forward Reaction is equal to the backward Reactions.

The general equilibrium expression for the self-ionization of water is given below;

Kw = [H3O+][OH-] = [H+][OH-].

The kw is now equal to 1.001x10-14 at 25°C. Kw which is the equilibrium constant is also refer to as the dissociation constant of water.

Answer:

Concentration of water at equilibrium is 0.1177 M.    

Explanation:

Balanced equation: [tex]CH_{4}(g)+H_{2}O(g)\rightleftharpoons CO(g)+3H_{2}(g)[/tex]

Equilibrium concentration of [tex]CH_{4}[/tex], [[tex]CH_{4}[/tex]] = [tex]\frac{0.231}{0.669}[/tex] M = 0.345 M

Equilibrium concentration of CO, [CO] = [tex]\frac{0.276}{0.669}[/tex] M = 0.413 M

Equilibrium concentration of [tex]H_{2}[/tex], [[tex]H_{2}[/tex]] = [tex]\frac{0.207}{0.669}[/tex] M = 0.309 M

Equilibrium constant for the given reaction in terms of concentration, [tex]K_{c}[/tex] is expressed as:           [tex]K_{c}=\frac{[CO][H_{2}]^{3}}{[CH_{4}][H_{2}O]}[/tex]

                          [tex]\Rightarrow [H_{2}O]=\frac{[CO][H_{2}]^{3}}{[CH_{4}].K_{c}}=\frac{(0.413)\times (0.309)^{3}}{(0.345)\times (0.30)}= 0.1177[/tex]

Hence, concentration of water at equilibrium is 0.1177 M                          

Answer:

4.83% of acetic acid in the vinegar

Explanation:

Answer:

It connects the two half‑reactions.

Explanation:

A salt bridge is a laboratory device used to connect the oxidation and reduction of half-cells of a galvanic cell. It prevents the cell from rapidly running its reaction to equilibrium.

Salt bridge or a porous plate connects the solutions of the half-cells that allow ions to pass from one solution to the other. This process balances the charges of the solutions and allows the reaction to continue.

The function of the salt bridge with respect to the electrochemical cell should be that it connected the two -half reactions.

It is the device that should be used in the laboratory for connecting the oxidation and the reduction of half-cells. It also prevents the cell from suddenly run from its reaction to the equilibrium.  Also, it connect the half cell solution where it probide the permission for ions from one solution to another solution. Moreover, there is balancing of the solutions charges due to this it permits the reaction for continuing it.

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Answer:

This reaction is favorable, and is likely regulated.

Explanation:

The equation to calculate delta G (dG) of a reaction is dG = dGo' + RTln [initial P]/[initial R]. You could use just dG = RTln [initial P]/[initial R] if (and that's a big IF) dGo' is zero, meaning that the reaction is at equilibrium when we have equal amounts of [P] and [R] (which is rarely the case). What really matters is the ratio of Q ([initial P]/[initial R]) to Keq ([P at equilibrium]/[R at equilibrium]), meaning how far off are we from equilibrium.

If Q=Keq, we are already at equilibrium (EQ).

If Q<Keq, we are not yet at EQ, having relatively more [R], or less [P] than under EQ conditions. This means the reaction will move forward to produce more P until EQ is achieved (dG is therefore NEGATIVE).

If Q>Keq, we are also off EQ, but we have relatively more [P], or less [R] than under EQ conditions. This means the reaction will move backwards to produce more R until EQ conditions are achieved (dG is therefore POSITIVE).

Try to understand these equations below (they say what I tried to describe in words)

dGo' = -RTlnKeq (under "standard conditions", i.e. we try to figure out how a reaction "behaves" if we start out with the same molar concentrations of R and P)

dG = dGo' + RTlnQ Q=[initial P]/[initial R]   or

dG = -RTlnKeq + RTlnQ or

dG = RTlnQ - RTlnKeq   or

dG = RTln Q/Keq

Answer:

0.15 M

Explanation:

Step 1: Write the neutralization reaction

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the moles of HCl that reacted

10 mL of 0.15 M HCl was used. The moles of HCl that reacted are:

[tex]0.010L \times \frac{0.15mol}{L} = 1.5 \times 10^{-3} mol[/tex]

Step 3: Calculate the moles of NaOH that reacted

The molar ratio of NaOH to HCl is 1:1. Then, the moles of NaOH that reacted are 1.5 × 10⁻³ moles.

Step 4: Calculate the concentration of NaOH

1.5 × 10⁻³ moles of NaOH are in 10.3 mL of solution. The molarity of NaOH is:

[tex]\frac{1.5 \times 10^{-3} mol}{10.3\times 10^{-3}L} =0.15 M[/tex]

I think the answer is B...

I am not sure

Answer:

B

Explanation:

Hurricanes are very powerful, circular storms that begin over warm water. These storms have heavy rains, very high winds, and low atmospheric pressures. They are fueled by a strong cycle of warmer, moister air flowing upward from near the ocean's surface and being replaced by cooler, drier air from the surrounding area. This cycle gets faster and stronger as it is fed by the ocean's heat and water evaporating from the ocean's surface.

So, the first step in hurricane formation occurs when warm ocean water evaporates and then condenses, forming a low pressure area below.

Answer:

See explaination

Explanation:

To calculate the second law of efficiency, temperatures here should be in Kelvin → K = ºC + 273.15 or Rankin = 460 + ºF. 2. Second Law Efficiency Second Law efficiency is a measure of how much of the theoretical maximum (Carnot) you achieve, or in other words, a comparison of the system's thermal efficiency to the maximum possible efficiency.

See attachment for the step by step solution of the given problem.

To solve this problem, we'll follow these steps:

a. Determine the entropy change per unit mass, (s2 - s0): The entropy change per unit mass can be calculated using the following equation:

[tex]\[ \Delta s = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \][/tex]

where:

- [tex]\( c_p \)[/tex] is the specific heat at constant pressure,

- [tex]\( R \)[/tex]is the gas constant,

- [tex]\( T_1 \) and \( T_2 \)[/tex] are the initial and final temperatures in Kelvin,

-[tex]\( P_1 \) and \( P_2 \)[/tex]are the initial and final pressures.

Given values:

- [tex]\( c_p = 1.009 \) kJ/kg·K[/tex]

- [tex]\( R = 0.287 \) kPa·m³/kg·K[/tex]

-[tex]\( T_1 = 28 + 273.15 \) K[/tex]

- [tex]\( T_2 = 231 + 273.15 \) K[/tex]

- [tex]\( P_1 = 137 \) kPa[/tex]

-[tex]\( P_2 = 830 \) kPa[/tex]

Substitute the values into the entropy change equation to calculate[tex]\( \Delta s \) per unit mass.[/tex]

[tex]\[ \Delta s = 1.009 \ln\left(\frac{231 + 273.15}{28 + 273.15}\right) - 0.287 \ln\left(\frac{830}{137}\right) \][/tex]

b. Determine the volume change per unit mass, (v2 – v0): The volume change per unit mass can be calculated using the equation:

[tex]\[ \Delta v = R \ln\left(\frac{V_2}{V_1}\right) \][/tex]

where:

-[tex]\( V_1 \) and \( V_2 \)[/tex] are the initial and final volumes.

Given values:

- [tex]\( V_1 = 3.6 \) L[/tex]

- [tex]\( V_2 \)[/tex] can be calculated using the ideal gas law:

[tex]\( V_2 = \frac{nRT_2}{P_2} \),[/tex] where \( n \) is the number of moles, [tex]\( R \)[/tex] is the gas constant,[tex]\( T_2 \)[/tex] is the final temperature in Kelvin, and [tex]\( P_2 \)[/tex] is the final pressure.

c. Determine the exergy of the air at the initial and the final states: Exergy [tex](\( X \))[/tex] is given by the equation:

[tex]\[ X = h - h_0 - T_0(s - s_0) \][/tex]

where:

- [tex]\( h \)[/tex] is the enthalpy per unit mass,

-[tex]\( h_0 \)[/tex] is the enthalpy at a reference state,

- [tex]\( T_0 \)[/tex] is the temperature at the reference state,

- [tex]\( s \)[/tex]  is the entropy per unit mass,

- [tex]\( s_0 \)[/tex] is the entropy at the reference state.

Given values:

-[tex]\( T_0 = 15 + 273.15 \)[/tex]K (temperature of the surroundings)

- [tex]\( P_0 = 112 \) kPa[/tex] (pressure of the surroundings)

We'll need to calculate [tex]\( h \), \( h_0 \), \( s \), and \( s_0 \)[/tex] for both initial and final states.

d. Determine the minimum work that must be supplied to accomplish this compression process: The minimum work required [tex](\( W_{\text{min}} \))[/tex]for an isentropic process can be calculated using the following equation:

[tex]\[ W_{\text{min}} = h_1 - h_2 \][/tex]

where:

-[tex]\( h_1 \)[/tex] is the enthalpy at the initial state,

- \( h_2 \) is the enthalpy at the final state.

e. Determine the second law efficiency of this process: The second law efficiency[tex](\( \eta \))[/tex] is given by the ratio of the actual work done[tex](\( W_{\text{actual}} \)) to the minimum work (\( W_{\text{min}} \)[/tex]):

[tex]\[ \eta = \frac{W_{\text{actual}}}{W_{\text{min}}} \][/tex]

Given values:

- [tex]\( W_{\text{actual}} = 1.9 \) kJ[/tex]

Now, let's go step by step and calculate each part of the problem.

a. Entropy Change per Unit Mass[tex](\( \Delta s = s_2 - s_0 \))[/tex]:

Using the given specific heats [tex]\( c_p \), \( R \)[/tex], temperatures [tex]\( T_1 \), \( T_2 \)[/tex], pressures [tex]\( P_1 \), and \( P_2 \)[/tex]:

[tex]\[ \Delta s = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \][/tex]

Substitute the values:

[tex]\[ \Delta s = 1.009 \ln\left(\frac{231 + 273.15}{28 + 273.15}\right) - 0.287 \ln\left(\frac{830}{137}\right) \][/tex]

[tex]\[ \Delta s \approx 1.009 \ln(1.842) - 0.287 \ln(6.0657) \][/tex]

[tex]\[ \Delta s \approx 0.6087 \, \text{kJ/kg·K} \][/tex]

b. Volume Change per Unit Mass [tex](\( \Delta v = v_2 - v_0 \))[/tex]:

First, calculate [tex]\( V_2 \)[/tex] using the ideal gas law:

[tex]\[ V_2 = \frac{nRT_2}{P_2} \][/tex]

where [tex]\( n \)[/tex] is the number of moles of air, which can be calculated using the initial conditions:

[tex]\[ n = \frac{P_1V_1}{RT_1} \][/tex]

Substitute the values and calculate[tex]\( V_2 \)[/tex]:

[tex]\[ n = \frac{137 \times 3.6}{0.287 \times (28 + 273.15)} \][/tex]

[tex]\[ n \approx 1.932 \, \text{kg} \][/tex]

[tex]\[ V_2 = \frac{1.932 \times 0.287 \times (231 + 273.15)}{830} \][/tex]

[tex]\[ V_2 \approx 1.476 \, \text{L/kg} \][/tex]

Now, calculate [tex]\( \Delta v \)[/tex] :

[tex]\[ \Delta v = R \ln\left(\frac{V_2}{V_1}\right) \][/tex]

[tex]\[ \Delta v \approx 0.287 \ln\left(\frac{1.476}{3.6}\right) \][/tex]

[tex]\[ \Delta v \approx -0.5305 \, \text{m}^3/\text{kg} \][/tex]

The negative sign indicates compression.

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Answer:

Chloroethane is denser and has a higher boiling point

Explanation:

The density of a gas depends directly on the molar mass of the gas. This means that as the molar mass increases, density increases and vice versa.

Having said that, we can easily see that the molar mass of chloroethane (64.51 g/mol) is greater than the molar mass of ethane (30.07 g/mol). Hence we expect that chloroethane is denser than ethane as established above.

In the absence of other strong intermolecular forces, the higher the molecular mass of a substance the greater its boiling point. Thus the boiling point of chloroethane is higher than that of Ethane since they both have weak Van der Waals forces holding their molecules together in the gaseous state.

The enthalpy change when 1.6 kg of methane gas is heated from a temperature of 280 K to 320 K is 140.6 kJ.

The subject this question pertains to is heat capacity and enthalpy change in the field of Chemistry. Simply put, when the temperature of an amount of substance changes, the change in enthalpy (∆H), can be calculated using the formula, ∆H = nCp∆T, where n is the number of moles, Cp is the heat capacity at constant pressure, and ∆T is the change in temperature.

To calculate the answer to your question, we need to convert the mass of methane gas to moles because the molar heat capacity is given. The molar mass of methane (CH4) is approximately 16.04 g/mol. Thus, we have: n (number of moles) = mass (in kg) / molar mass (in kg/mol) = 1.6 kg / 0.01604 kg/mol = 99.75 mol.

The next step is to plug our numbers into the formula: ∆H = nCp∆T = 99.75 mol * 35.31 JK-1mol-1 * (320 K - 280 K) = 140.6 kJ.

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Answer:

The correct answer is (c) 1 × 10⁻⁹ M

Explanation:

Hydroxide ion = OH⁻

Hydrogen ion = H⁺

The autoionization equilibrium of water at 25ºC has a water constant Kw which is expressed as follows:

Kw = [H⁺] x [OH⁻]= 1 x 10⁻¹⁴

If we know the concentration of hydroxide ion ([OH⁻]) we can calculate the hydrogen ion concentration ([H⁺]) as follows:

[H⁺] = Kw/[OH⁻]= (1 x 10⁻¹⁴)/1 x 10⁻⁵ M= 1 x 10⁻⁹ M

The hydrogen ion concentration of a solution with a hydroxide ion concentration of 1 × 10-5 M is 1 × 10-9 M.

In chemistry, the product of the concentration of hydrogen ions [H+] and the concentration of hydroxide ions [OH-] in a solution always equals 1 × 10–14 M² at a temperature of 25°C (this is known as the ion product of water).

Since the concentration of hydroxide ions [OH-] in your solution is given as 1 × 10-5 M, you can calculate the concentration of hydrogen ions [H+] by dividing the ion product of water, 1 × 10-14 M², by the concentration of hydroxide ions [OH-]. Therefore, the hydrogen ion concentration [H+] in your solution would be 1 × 10-14 M² / 1 × 10-5 M = 1 × 10-9 M.

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