Information gathered by a scientist about the toxicity of chemical X and chemical Y showed that they had individual safe limits for fish at particular concentrations. But when they were used together at the safe concentrations, there were extensive fish kills. This is an example of _________

Answer:

1

Explanation:

Trigonal planar geometry is shown by the compounds where hybridization of central atom is [tex]sp^2[/tex].

In [tex]sp^2[/tex] hybridization, three hybrid orbitals are equally spaced at an angle of 120°.

Some of the compounds having [tex]sp^2[/tex] hybridization are [tex]CO_3^{2-}[/tex], [tex]SO_3[/tex], etc

In [tex]sp^2[/tex] hybridizationm, one s-orbital and 2 p-orbitals are involved.

Total no. of orbitals present in p-subshell is 3.

As 2 is involved in [tex]sp^2[/tex] hybridization, therefore no. of unhybridized orbital in [tex]sp^2[/tex] hybridization is 1.

Answer:

molar mass is [tex]6.39\times10^3\ g/mol[/tex]

Explanation:

Osmotic pressure is related with concentration as follows:

[tex]\pi =CRT[/tex]

Where, C is concentration or molarity , R is gas constant and T is temperature.

Osmostic pressure given is 0.0766 atm

R is [tex]0.0821 L \ atm \ K^{1}mol^{-1}[/tex]

T = 25 + 273 = 298 K

Rearrange the above equation to calculate concetration of the solution as follows:

C = P/RT

[tex]C=\frac{0.0766}{0.0821 \times 298} \\=0.00313\ M[/tex]

molarity = moles/volume in L

moles = molarity × volume in L

volume = 5.00 mL = 0.005 L

moles = 0.00313 × 0.005

           =[tex]1.565 \times 10^{-5}[/tex]

[tex]molar\ mass=\frac{mass\ in\ g}{moles} \\=\frac{0.1\ g}{1.565\times 10^{-5}mol} \\=6.389\times 10^3\ g/mol\\[/tex]

Therefore, molecular mass of protein is [tex]6.39\times10^3\ g/mol[/tex]

Answer:

1. at any given location, the physical properties of a substance do not change.

2. Each substance has a unique set of physical properties

Explanation:

The physical property of a substance is a feature of a substance that can be noticed or measured without changing the key identity of the substance. Physical properties include density, hardness, melting point, colour etc...

Answer:

1. at any given location, the physical properties of a substance do not change

3. each substance has a unique set of physical properties

Explanation:

1. The physical properties of a substance do not change provided that the physical conditions in the niche environment remain same.

3. Every substance on Earth has a unique set and arrangement of atoms and molecules that give them their own unique properties.

Answer:

productivity and water depth

Explanation:

The productivity and the depth of water are both equally important as it directly affects the accumulation of biogenic sediments such as the siliceous ooze and calcareous ooze. In the equator and the coastal upwelling areas, and at the site of divergence of oceans, there occurs a high rate and amount of productivity, and these are considered to be the primary productivity.  

The siliceous oozes are a good indicator of extensively high productivity in comparison to the carbonate oozes. The main reason behind this is that the silica can be easily dissolved in the surface water. On the other hand, the carbonates dissolve at a relatively lower ocean water depth, so there requires a high amount of surface productivity in order to allow these siliceous oozes to reach the ocean bottom.

Thus, the water depth and productivity, both are considered as the limiting factor in determining the accumulation of biogenic oozes.

Answer : The amount of heat required is, [tex]1.004\times 10^6J[/tex]

Explanation :

Formula used :

[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]

where,

q = heat required = ?

m = mass of water = 30.0 kg = 3000 g

c = specific heat of water = [tex]4.184J/g^oC[/tex]

[tex]T_{final}[/tex] = final temperature = [tex]95^oC[/tex]

[tex]T_{initial}[/tex] = initial temperature = [tex]15^oC[/tex]

Now put all the given values in the above formula, we get:

[tex]q=3000g\times 4.184J/g^oC\times (95-15)^oC[/tex]

[tex]q=1004160J=1.004\times 10^6J[/tex]

Thus, the amount of heat required is, [tex]1.004\times 10^6J[/tex]

Final answer:

To raise the temperature of 30.0 kg of water from 15°C to 95°C, 10,041,600 joules of heat energy are required. This is calculated using the product of the mass of the water, the specific heat capacity of water, and the change in temperature.

Explanation:

The amount of heat transfer required to change the temperature of a substance can be calculated using the formula Q = mcΔT, where Q is the heat energy in joules, m is the mass of the substance in kilograms, c is the specific heat capacity (for water it's 4,184 J/kg/°C), and ΔT is the change in temperature in degrees Celsius.

For the given problem of raising the temperature of 30.0 kg of water from 15°C to 95°C, the temperature change (ΔT) is 95°C - 15°C = 80°C. The calculation would be as follows: Q = (30.0 kg) × (4,184 J/kg/°C) ×(80°C).

Calculation:

Therefore, the heat required is 10,041,600 joules.

Answer:

T = 3,3-dimethylbut-1-ene

U = 2,3-dimethylbut-2-ene

Explanation:

The chemical reactions for the two reactions and the mechanisms of reaction are shown in the two attachments to this answer.

Hope it helps!

Answer:

The given statement is false.

Explanation:

Endothermic reactions are defined as the reactions in which energy of products is more than the energy of the reactants. In these reactions, energy is absorbed by the system.

The total enthalpy of the reaction [tex](\Delta H)[/tex] comes out to be positive.

Exothermic reactions are defined as the reactions in which energy of reactants is more than the energy of the products. In these reactions, energy is released by the system.

The total enthalpy of the reaction [tex](\Delta H)[/tex] comes out to be negative.

Answer: solution

Explanation:

The osmotic pressure of a solution formed by dissolving 44.3 mg of aspirin in 0.358 L of water at 25 ∘C is 0.01633 atm. This is calculated using the formula for osmotic pressure 'Pi = n/V RT', inserting the required values including the mole of aspirin, the volume, the gas constant, and temperature.

To calculate the osmotic pressure, we need to use the formula Pi = n/V RT, where 'n' is the number of moles of the solute, 'V' is the volume in liters, 'R' is the gas constant (0.08206 L atm/mol K), and 'T' is the temperature in Kelvin. Firstly, we need to find the molar mass of aspirin (C9H8O4) which is approximately 180.16 g/mol. Consequently, we can determine the mole of aspirin used as 44.3 mg / 180.16 g/mol = 0.000246 moles. Now knowing all values, plug them into the formula: Pi = 0.000246 moles /0.358L * 0.08206 atm mol^-1K^-1 * 298.15K which equals 0.01633 atm, the osmotic pressure of the solution.

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For the osmotic pressure of an aspirin solution, first convert the mass of aspirin to moles, find the molarity, convert the temperature to Kelvin, and then apply the osmotic pressure formula. The osmotic pressure is found to be approximately 0.0168 atm.

The question asks to determine the osmotic pressure of a solution containing aspirin at a certain temperature. The osmotic pressure can be calculated using the formula:

\(\Pi = MRT\)

Where:

\(\Pi\) is the osmotic pressure,

\(M\) is the molarity of the solution (moles of solute per liter of solution),

\(R\) is the gas constant (0.0821 L atm K^{-1} mol^{-1}), and

\(T\) is the temperature in Kelvin.

First, convert the mass of aspirin to moles using the molecular weight of aspirin (C9H8O4), which is 180.16 g/mol:

\(44.3 mg = 0.0443 g\)

\(0.0443 g \times \dfrac{1 mol}{180.16 g} \approx 2.46 \times 10^{-4} mol\)

Next, find the molarity \(M\) by dividing the number of moles by the volume in liters:

\(M = \dfrac{2.46 \times 10^{-4} mol}{0.358 L} \approx 6.87 \times 10^{-4} M\)

Now convert the temperature to Kelvin:

\(T = 25 \degree C + 273.15 = 298.15 K\)

Finally, calculate the osmotic pressure:

\(\Pi = (6.87 \times 10^{-4} M)(0.0821 L atm K^{-1} mol^{-1})(298.15 K)\)

\(\Pi \approx 0.0168 atm\)

Therefore, the osmotic pressure of the aspirin solution is approximately 0.0168 atm.

Volume of the gas at 300K will be 525l itres.

According to Charles' law, the volume of a gas is inversely proportional to its absolute temperature when the pressure of the gas is constant.

Here the V/T is given to be constant and equal to 1.75.

The temperature of the gas = T = 300K.

Let the volume of the gas be V liters.

So according to the question,

V  / T = 1.75.

So,V / 300 = 1.75.

So, V = 300 × 1.75.

Or, V = 525 litres.

So, the volume of the gas at 300K will be 525 litres.

The Volume of the gas at 300K will be 525l liters.

As per Charles' law, the volume of gas should be inversely proportional to the absolute temperature at the time when the pressure of the gas is constant.

Here the V/T should be constant and equivalent to 1.75.

The temperature of the gas = T = 300K.

Also, we assume that the volume of the gas is V liters.

So,

V  / T = 1.75.

V / 300 = 1.75.

V = 300 × 1.75.

V = 525 litres.

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To find the mass of the copper sample, we use the formula for heat transfer Q = mcΔT. By substituting the given values of specific heat capacity of copper, the amount of heat applied and the change in temperature into the formula and solving for mass, we find the mass of the copper sample to be 29.54g.

The subject of this question is physics, specifically dealing with the concept of heat transfer. To determine the mass of the copper sample, we'll use the formula for heat transfer: Q = mcΔT, where 'Q' is the amount of heat transferred, 'm' is the mass of the substance, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature.

In this case, we know that the specific heat capacity of copper is 0.39 J/g °C, the amount of heat applied (Q) is 265 J and the change in temperature ΔT is 23.0 °C. By substituting these values into the formula, we can solve for 'm' (mass). Rearranging the formula gives us m = Q / (cΔT). Substituting the given values, we find m = 265 J / (0.39 J/g°C * 23.0°C) gives us the mass of the copper sample to be 29.54 g.

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Final answer:

To prepare a 0.0150 M solution of strontium nitrate in 2.000 L, you would need to use 4.49 grams of strontium nitrate.

Explanation:

To calculate the mass of strontium nitrate required to prepare the solution, we can use the formula:

moles = concentration x volume

First, convert the concentration to moles per liter:

moles/liter = concentration = 0.015 M

Next, multiply the moles per liter by the volume in liters to find the moles of strontium nitrate:

moles = (0.015 M) x (2.000 L) = 0.030 mol

Finally, calculate the mass of strontium nitrate using the molar mass:

mass = moles x molar mass = 0.030 mol x 149.6 g/mol = 4.49 g

Answer:

PART A: The LDF occurs between all molecules. Dispersion forces result from shifting electron clouds, which cause weak, temporary dipole.

PART B: Dipole dipole operates only between polar molecules. This is when two polar molecules get near each other and the positively charged portion of the molecule is attracted to the negatively charged portion of another molecule.

PART C: Dipole dipole and in some cases hydrogen bonding operate between the hydrogen atom of a polar bond and a nearby small electronegative atom. Only if the atom bonded to it were F, O or N it would be hydrogen bonding. Otherwise it is dipole dipole.

All molecules - Dispersion force.

Polar molecules - Dipole–Dipole interaction.

The hydrogen atom of a polar bond and a nearby small electronegative atom  - Dipole–Dipole and Hydrogen Bonding.

Intermolecular force is a force that connect molecules to each other.

There are four types of forces present

Thus, the bonds present are: All molecules - Dispersion force.

Polar molecules - Dipole–Dipole interaction.

The hydrogen atom of a polar bond and a nearby small electronegative atom  - Dipole–Dipole and Hydrogen Bonding.

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Answer:

D. remain unchanged

Explanation:

A Zero-order reaction is defined as a chemical reaction wherein the rate doesn't vary with the increase or decrease in the concentration of the reactants.

Thus, If you double the concentration of that reactant, the reaction rate will:

D. remain unchanged

I hope it helps!

Answer:

Explanation:

Sodium chloride consits of sodium cations (positive ions), Na⁺, and chloride anions (negative ions), CL⁻.

Pure sodium chloride is packed in crystals: sodium ions and chloride ions are packed together and the ions are in fixed positions. There are not free electrons that can move. Thus, sodium chloride doesn't conduct electricity, because there are no electrons or ions which are free to move.

In aqueous solution, sodium chloride units dissociates into their ions:

         [tex]NaCl\rightarrow Na^++Cl^-[/tex]

Those ions are freely to move in the solution, and such they are charge carriers, which conduct the electricity.

As explained above, in solid sodium chloride, the ions cannot move and there is not flow of current.

That is why solid pure salt of NaCl does not conduct electricity and the solutions of NaCl do conduct electricity.

The image attached show both diagrams. In the diagram A, the ions are packed together, showing that they cannot move. In the diagram B, the ions are dissolved in water, showing that they can move and carry the charge, allowing the flow of current.

Answer:

The ion present in the sample solution is Mn2+

Explanation:

(a) No precipitate formed when an aqueous solution of NaCl was added to the sample solution.

Most chlorides are quite soluble in water except for he  H g 2 C l 2, that  is an insoluble salt. Since no precipitation formed upon addition of NaCl,  H g 2 +  was not present in the sample.

(b) No precipitate formed when an aqueous solution of Na2SO4 was added to the sample solution.

The  B a S O 4  is an insoluble salt too. Since no precipitation formed upon addition of N a 2 S O 4 ,  the ion B a 2 +  was not present in the sample.

(c) A precipitate formed when the sample solution was made basic with NaOH.

As the M n ( O H ) 2 is insoluble in this last solution, the precipitate formed corresponds to B a S O 4 and H g 2 C l 2 compounds , therefore it can be concluded that the present ion in the sample solution is Mn+2.

The density of the rock sample is 10.75 g/cm³.

The density of an object can be calculated using the formula:

Density = mass/volume

In this case, we have the mass of the rock sample (8.6 g) and the change in volume of the water in the graduated cylinder (13.6 mL - 12.8 mL = 0.8 mL). To calculate the density, we need to convert the volume from mL to cm³, since the mass is in grams.

1 mL = 1 cm³, so the volume of the rock sample in cm³ is 0.8 cm³.

Now, we can calculate the density:

Density = mass/volume

Density = 8.6 g / 0.8 cm³

Density = 10.75 g/cm³

Therefore, the density of the rock sample is 10.75 g/cm³.

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The density of the rock sample is found to be 10.75 g/mL by dividing its mass, 8.6 g, by the volume of water it displaced, which is 0.8 mL.

To determine the density of the rock sample, we use the formula density = mass/volume. The mass of the rock is given as 8.6 g. The volume displaced by the rock is the change in the water level in the graduated cylinder: 13.6 mL - 12.8 mL = 0.8 mL.

Now, we can calculate the rock's density:

Density = Mass / Volume

Density = 8.6 g / 0.8 mL

Density = 10.75 g/mL

Therefore, the density of the rock sample is 10.75 g/mL.

Answer:

The correct answer is Dust

Explanation

Dust is a dry dirt in powder form usually found on surfaces of items in a building, it comprises of very small particles of soil, sand and sometimes includes toxic substances, skin cells,bacteria, soil particles, particles of clothing material, tiny pieces of dead insects and pollen

Answer: Dust

Explanation:

Dust is a solid matter with fine particles, it mostly hangs in the atmosphere and are produced from various sources such as pollutions, loose soil which are lifted through aeolian process. Continuous exposure to dust in house holds or work place can make someone sick and affect one's health. Apart from that, it also reduces comfort and can even affect the efficient functioning of home or office appliances and gadgets.

Answer:

[tex]\%\ Composition\ of\ sulfur=40.1\ \%[/tex]

Explanation:

Percent composition is percentage by the mass of element present in the compound.

Given , Mass of sulfur= 32.1 amu

Mass of oxygen = 16.0 amu

Mass of sulfur trioxide [tex]SO_3[/tex] = 32.1 amu + 3*16.0 amu = 80.1 amu

[tex]\%\ Composition\ of\ sulfur=\frac{Mass_{sulfur}}{Mass_{SO_3}}\times 100[/tex]

[tex]\%\ Composition\ of\ sulfur=\frac{32.1\ amu}{80.1\ amu}\times 100=40.1\ \%[/tex]

Final answer:

To determine the percent by mass of sulfur in sulfur trioxide (SO3), calculate the molar mass of the compound and divide the atomic mass of sulfur by the total molar mass. The percent by mass of sulfur in SO3 is found to be 40.0%.

Explanation:

To find the percent by mass of sulfur in sulfur trioxide (SO3), we first calculate the molar mass of the compound. We have one sulfur atom and three oxygen atoms in SO3.

Next, we calculate the molar mass of SO3:

Molar mass of SO3 = Molar mass of S + 3 × Molar mass of O
= 32.07 amu + 3 × 16.00 amu
= 32.07 amu + 48.00 amu
= 80.07 amu

To find the percent composition of sulfur, we divide the molar mass of sulfur by the total molar mass of the compound and multiply by 100:

Percent by mass of S in SO3 = (Molar mass of S / Molar mass of SO3) × 100%
= (32.07 amu / 80.07 amu) × 100%
= 40.0% (to the nearest tenth of a percent)

Answer:

C. Infinitely Many Solutions

Explanation:

No solution case : This is the case when all given variables are not equal to any constant, for example: there is one row of zeros in matrix e.g 0=3. matrix B don't have any zero row. So, Not True.

One Solution case: This is the case when all variables are independent variables like if they are equal to some constant. e.g x=1,y=2,z=4 , Matrix B have more than one variable in first row due to which it made equation look like x+y=-5. so matrix B can't have only one solution. So, Not True.

Infinitely Many Solutions case: when there is one or more variables which is not equal to any constant and acting as linearly dependent variable, then that matrix have infinite solutions. Matrix B have that variable which is linearly dependent as show in the attachment solution. So, True.  

Answer:  A  Explanation:

Answer:

The correct answer is B. exogenous

Explanation:

Let us try to describe exogenous and endogenous variables an exogenous variable value is influenced only by factors outside a model or system and is forced onto the model, while a change in an exogenous variable is known as an exogenous change. Also an endogenous variable is one whose value is influenced only by the system or model under study.

Answer:

He responded by saying their business definition was far too narrow.

Explanation:

Williams Bryan relates the battle for free silver with the revolution in the United States.

His iconic "Cross of Gold" address ignited the Democratic National Convention in 1896 where he begged the american people not to be "crucified on the cross of gold." He was said this because of the Republicans plan to introduce a firm gold standard and abolish silver coinage.

Answer:

Endpoint is point in which indicator change the color, and equivalence point is point in which there is equal amount of two chemicals. Endpoint is after Equivalence point.

Explanation:

To know the end point we use phenolphthalein as indicator. Therefore, endpoint is point in which indicator change the color, and equivalence point is point in which there is equal amount of two chemicals.

Titration is a technique by which we know the concentration of unknown solution using titration of this solution with solution whose concentration is known. To know the end point we use phenolphthalein as indicator. End point is a point where completion of reaction happen.

Endpoint is point in which indicator change the color, and equivalence point is point in which there is equal amount of two chemicals. Endpoint is after Equivalence point.

Therefore, endpoint is point in which indicator change the color, and equivalence point is point in which there is equal amount of two chemicals.

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Final answer:

The concentration of Na+ in a 1.15 M Na2SO3 solution is 2.30 M, and the concentration of SO32- is 1.15 M, as Na2SO3 dissociates entirely into these ions in solution.

Explanation:

To calculate the concentration of all species in a 1.15 M Na2SO3 (sodium sulfite) solution, we need to consider the ionization of sulfurous acid (H2SO3), which isn't explicitly present but relates to the anion derived from Na2SO3. The ionization constants given are Ka1 = 1.4 × 10−2 for the first dissociation (H2SO3 to HSO3−) and Ka2 = 6.3 × 10−8 for the second dissociation (HSO3− to SO32−). Although, given the context, we aren't going through the complete step-by-step equilibrium calculation for Ka1 and Ka2, we understand that Na2SO3 dissociates completely in solution to Na+ and SO32−. Therefore, initially, the concentration of Na+ is 2.30 M (since each Na2SO3 unit yields two Na+ ions), and the concentration of SO32− is 1.15 M.

The concentrations in the 1.15 M Na2SO3 solution are approximately:

- [tex][H2SO3] = [HSO3^-] = 1.15 M,[/tex]

- [tex][SO3^2-] = 2.10 \times 10^-5 M,[/tex]

- [tex][Na^+] =1.15 M, and[/tex]

- [tex][OH^-] = 8.70 \times 10^-14 M.[/tex]

Given:

- Initial concentration of [tex]\(Na_2SO_3\) = 1.15 M[/tex]

- [tex]\(K_a1\)[/tex] for sulfurous acid [tex]= \(1.4 \times 10^{-2}\)[/tex]

- [tex]\(K_a2\)[/tex]for sulfurous acid [tex]= \(6.3 \times 10^{-8}\)[/tex]

We can assume that the ionization of [tex]\(Na_2SO_3\)[/tex] into [tex]\(HSO_3^-\)[/tex] and  [tex]\(Na^+\)[/tex] is small due to the weak acid nature of sulfurous acid. So, the initial concentration of [tex]\(H_2SO_3\)[/tex] is approximately equal to 1.15 M.

Now, we approximate the concentration of [tex]\(H^+\)[/tex] ions to be approximately equal to the concentration of [tex]\(H_2SO_3\)[/tex], which is 1.15 M.

Using [tex]\(K_a2\)[/tex], we calculate the concentration of [tex]\(SO_3^{2-}\)[/tex] ions:

[tex]\[ [SO_3^{2-}] = \frac{[H^+]^2}{K_a2} \approx \frac{(1.15)^2}{6.3 \times 10^{-8}} \approx 2.10 \times 10^{-5} \, \text{M} \][/tex]

Since [tex]\(HSO_3^-\)[/tex] ions concentration is approximately equal to [tex]\(H^+\)[/tex] ions concentration and [tex]\(Na^+\)[/tex] concentration is the same as [tex]\(Na_2SO_3\)[/tex], we have:

- [tex]\([H_2SO_3] = [HSO_3^-] \approx 1.15 \, \text{M}\)[/tex]

- [tex]\([SO_3^{2-}] \approx 2.10 \times 10^{-5} \, \text{M}\)[/tex]

- [tex]\([Na^+] \approx 1.15 \, \text{M}\)[/tex]

The concentration of [tex]\(OH^-\)[/tex] ions can be calculated using the [tex]\(K_w\)[/tex]expression:

[tex]\[ [OH^-] = \frac{K_w}{[H^+]} \approx \frac{1.0 \times 10^{-14}}{1.15} \approx 8.70 \times 10^{-14} \, \text{M} \][/tex]

Answer:

The answer to your question is below

Explanation:

Reaction

           N₂ + 3H₂   ⇔   2 NH₃ + energy

a) The concentration of NH₃, if the concentration of NH₃ increases, the reaction will move to the left.

b) Diminishing the temperature, no more energy will be released and the reaction will move to the reactants.

Answer:

The answer to your question is letter B.

Explanation:

Water boils at sea level at 100°C at higher heights the boiling point diminishes. That means that at higher heights the time needed to boil water will be lower than at sea level.

In this problem at mountain B is only 3 minutes needed to boil water that means that the height is higher here.

Answer:

The correct answer is B. Mountain X has a higher elevation than Mountain Y

Explanation:

The boiling point of a liquid reduces as the altitude where the boling takes place increases. With every 500 ft increase in altitude the boiling point of water devreses by 0.5 °C as such the boiling point of water at 8000 ft is just 92 °C. To compenasate cooking at altidude require the use of a pressure cooker as it is hard to boil items such as potatoes at very high altitudes of thousands of feet

Answer:

India had a greater contact and net work with other areas or territory.

Explanation:

This lead to the introduction of new ideas and they also learnt new techniques.

Answer:

1) Gas and Liquid

Explanation:

hope it helps

Answer:

The answer to this can be arrived at by clculating the mole fraction of atoms higher than the activation energy of 10.0 kJ by pluging in the values given into the Arrhenius equation. The answer to this is 20.22 moles of Argon have energy equal to or greater than 10.0 kJ

Explanation:

From Arrhenius equation showing the temperature dependence of reaction rates.

[tex]K = Ae^{\frac{Ea}{RT} }[/tex] where

k = rate constant

A = Frequency or pre-exponential factor

Ea   =       energy of activation

R = The universal gas constant

T = Kelvin absolute temperature

we have

[tex]f = e^{\frac{Ea}{RT} }[/tex]

Where

f = fraction of collision with energy higher than the activation energy

Ea = activation energy = 10.0kJ = 10000J

R = universal gas constant = 8.31 J/mol.K

T = Absolute temperature in Kelvin = 400K

In the Arrhenius equation k = Ae^(-Ea/RT), the factor A is the frequency factor and the component e^(-Ea/RT) is the portion of possible collisions with high enough energy for a reaction to occur at the a specified temperature  

Plugging in the values into the equation relating f to activation energy we get

[tex]f = e^{\frac{10000J}{(8.31J/((mol)(K)))(400K)} }[/tex] or f = [tex]e^{3.01}[/tex] = 20.22 moles of argon have an energy of 10.0 kJ or greater

The question asks for the fraction of argon atoms at 400 K with an energy of 10.0 kJ or higher, which would require integrating the Maxwell-Boltzmann distribution, but key information is missing for a definitive answer.

To calculate the fraction of argon atoms in a gas sample at 400 K with an energy of 10.0 kJ or greater, the principles of statistical mechanics and the Maxwell-Boltzmann distribution are applied. The Maxwell-Boltzmann distribution gives the fraction of particles at a certain energy level within a system at thermal equilibrium. However, we are missing some key information like the Maxwell-Boltzmann distribution function for the particular conditions set by the question, which is essential for performing such a calculation.

Typically, the calculation would require integrating the partition function over all energy states equal to or greater than 10.0 kJ. Without the exact form of the energy distribution, it is not possible to provide an accurate answer. The question pertains to a high level of physical chemistry or statistical mechanics typically studied at the college or university level.