how many moles of NH, will be produced if 3.5 moles of N2, are reacted completely

The elements in the human body tend to be located in different regions of the periodic table. This is because the human body is made up of a wide range of elements, each with its own properties, uses, and functions. Overall, the elements in the human body tend to be located in different regions of the periodic table depending on their properties and functions.

Most of these elements are found in the first four rows of the periodic table, which are also known as the main group elements. These elements include hydrogen, carbon, nitrogen, oxygen, phosphorus, and sulfur, which are all essential for life. They are located in different regions of the periodic table, with hydrogen in the first row, carbon and nitrogen in the second row, oxygen in the third row, and phosphorus and sulfur in the fourth row. Other important elements in the human body include sodium, potassium, calcium, magnesium, iron, and zinc, which are located in the lower regions of the periodic table. Overall, the elements in the human body tend to be located in different regions of the periodic table depending on their properties and functions.

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Answer:

B

Explanation:

Pressure is directly proportional to temperature

The graph correctly showing the pressure - temperature relationship is figure B. Temperature and pressure is in direct proportionality.

According to Gay-Lussacs law, at constant volume of gases, the pressure of a gas is directly proportional to the temperature. Thus we can write it as:

P/T = a constant.

Gay-Lussacs law is supportive of kinetic molecular theory. As per kinetic theory of gases, as the temperature increases, kinetic energy of particles increases results in forceful collisions of energetic particles to the wall of container and in between them.

Therefore, the temperature - pressure graph will show a linear relationship.  The figure B represents a linear curve and hence it shows the correct relation.

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Answer:

After 36 minutes, there will be a population of 800,000 bacteria.

Explanation:

After the first 18 minutes, it will double to 400,000. Then, at 36 minutes, it will have doubled again, giving you 800,000.

The solubility of methyl bromide in water at 25°C and a partial pressure of 300 mm Hg can be calculated using Henry's Law. The Henry's Law constant for methyl bromide is given as 0.159 mol/(L atm) at 25°C. By applying the equation for Henry's Law, the solubility of methyl bromide in water can be determined.

Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The equation for Henry's Law is written as:

S = k * P

Where S is the solubility of the gas in the liquid, k is the Henry's Law constant, and P is the partial pressure of the gas. In this case, we are given the Henry's Law constant for methyl bromide as 0.159 mol/(L atm) at 25°C. The partial pressure of methyl bromide is given as 300 mm Hg.

Substituting the values into the equation, we have:

S = 0.159 mol/(L atm) * (300 mm Hg)

To convert mm Hg to atm, we divide by the conversion factor of 760 mm Hg/atm:

S = 0.159 mol/(L atm) * (300 mm Hg / 760 mm Hg/atm)

Simplifying the equation, we find:

S ≈ 0.0628 mol/L

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The heat of vaporization decreases as the strength of the attractive intermolecular forces increases.

The heat of vaporization is the amount of energy required to convert a substance from its liquid phase to its gaseous phase at a constant temperature. As the strength of the attractive intermolecular forces increases, it becomes more difficult for the molecules to overcome these forces and transition into the gas phase.

This means that a greater amount of energy (heat) is required to break the intermolecular forces and vaporize the substance. Therefore, the heat of vaporization increases as the strength of the attractive intermolecular forces increases.

On the other hand, the boiling point of a liquid (B), the vapor pressure of a liquid (C), and the viscosity of a liquid (D) all tend to increase as the strength of the attractive intermolecular forces increases. Higher intermolecular forces result in higher boiling points, lower vapor pressures, and higher viscosity due to the stronger interactions between molecules.

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Answer:

1. (NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)

2. Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)

Explanation:

The dissociation of ammonium sulphide, (NH₄)₂S when dissolved in water is given in the equation below:

(NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)

However very little S²- ions are present in solution due to the very basic nature of the S²- ion (Kb = 1 x 105).

The ammonium ion being a better proton donor than water, donates a proton to sulphide ion to form hydrosulphide ion which exists in equilibrium with aqueous ammonia.

S²- (aq) + NH₄+ (aq) ⇌ SH- (aq) + NH₃ (aq)

Aqueous solutions of ammonium sulfide are smelly due to the release of hydrogen sulfide and ammonia, hence, their use in making stink bombs.

2. The reaction between aluminium nitrate and sodium phosphatein aqueous solution is a double decomposition reaction whish results in the precipitation of insoluble aluminium phosphate. The equation of the reaction is given below :

Al(NO₃)₃ (aq) + Na₃PO₄ (aq) ----> AlPO₄ (s) + 3 NaNO₃ (aq)

The net ionic equation is given below:

Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)

Answer:

no they dont

Explanation:

Titration is the process of determining the amount of a substance in a solution by measuring the volume of a solution with a known concentration that is required to react with it. The answer to the given question is option d) 5.25.

In the titration of a weak base with a strong acid, the pH at the half-equivalence point can be calculated as follows: At the half-equivalence point, we have equal moles of the weak base and the strong acid. As a result, we get a solution that contains the weak base, its conjugate acid, and water. In the solution, there is an equilibrium between the weak base and its conjugate acid. This equilibrium has an acid dissociation constant, Ka. It's given by:

Ka = [H+][A–]/[HA]

The pKa is calculated by taking the negative logarithm of Ka:

pKa = -log(Ka)

At the half-equivalence point, [HA] = [A–] and the expression for pKa becomes:

pKa = -log([H+])

Therefore, the pH at the half-equivalence point is:

pH = 1/2 (pKb + pKa)

Given that pKb = 7.95 for the weak base, we can calculate the pKa:

pKw = 14 (at 25°C)

pKw = pKa + pKb

14 = pKa + 7.95

pKa = 6.05

Therefore, the pH at the half-equivalence point is:

pH = 1/2 (7.95 + 6.05)

pH = 1/2 (14)

pH = 7

At the half-equivalence point, the pH of the solution is equal to 7. Therefore, option d) 5.25 is incorrect.

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In short, the Rydberg equation and the Bohr equation are two different forms of the same fundamental relationship, describing the behavior of energy levels and wavelengths in hydrogen or hydrogen-like systems.

What is Rydberg Equation ?

The Rydberg equation is an empirical relationship equation expressed by Balmer and Rydberg.

The Rydberg equation and the Bohr equation are closely related and can be considered different forms of the same mathematical expression. Let's explore them and compare their similarities.

Rydberg equation:

The Rydberg equation is an empirical formula that describes the wavelengths of spectral lines emitted or absorbed by hydrogen or hydrogen species. Is given:

1/λ = R* (1/n₁2 - 1/n₂2)

where λ is the wavelength of emitted or absorbed light, R is the Rydberg constant, and n1 and n₂ are integers representing the principal quantum numbers of the respective energy levels.

Bohr equation:

Bohr's equation, derived from Bohr's model of the hydrogen atom, relates the energy levels of an electron in a hydrogen-like atom to its principal quantum number. Is given:

E = -13.6 eV/n2

where E is the energy of the electron, -13.6 eV is the ionization energy of hydrogen, and n is the principal quantum number.

Now let's compare the two equations:

1/λ = R * (1/n₁2 - 1/n₂2) (Rydberg equation)

E = -13.6 eV/n² (Bohr equation)

We can observe that the Bohr equation gives the energy of the electron in terms of the principal quantum number, while the Rydberg equation gives the wavelength of light emitted or absorbed in terms of the principal quantum numbers.

We can use the relationship between energy and wavelength to make the connection between these two equations:

E = h*c/A

where h is Planck's constant and c is the speed of light.

Combining this relation with the Bohr equation, we have:

-13.6 eV / n² = h * c / λ

Rearranging the equation:

1/λ = (-13.6 eV / n²) / (h * c)

By comparing this equation with the Rydberg equation, we can see that the term (-13.6 eV / n²) / (h * c) is equivalent to the Rydberg constant R. Therefore, we can rewrite the equation as:

1/λ = R* (1/n₁2 - 1/n₂2)

This shows that the Rydberg equation and the Bohr equation are practically the same, just expressed in slightly different forms. The Rydberg constant R in the Rydberg equation includes the constant terms such as the ionization energy, Planck's constant, and the speed of light present in the Bohr equation.

In short, the Rydberg equation and the Bohr equation are two different forms of the same fundamental relationship, describing the behavior of energy levels and wavelengths in hydrogen or hydrogen-like systems.

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To compute the flux of [tex]F(x, y, z) = x^3 i + y^3 j + z^3 k[/tex] over the surface Σ of the sphere [tex]x^2 + y^2 + z^2 = 9[/tex] using the Divergence Theorem, set up and evaluate the triple integral in spherical coordinates:Flux = ∫₀²π ∫₀ᴨ ∫₀³ ([tex]3p^4[/tex] sin(φ)) dρ dθ dφ

To compute the flux of the vector field [tex]F(x, y, z) = x^3 i + y^3 j + z^3 k[/tex] over the surface Σ, which is the surface of the sphere [tex]x^2 + y^2 + z^2 = 9[/tex], we can apply the Divergence Theorem.

The Divergence Theorem states that the flux of a vector field across a closed surface is equal to the triple integral of the divergence of the vector field over the volume enclosed by the surface.

First, let's calculate the divergence of F:

div(F) = (∂/∂x)([tex]x^3[/tex]) + (∂/∂y)(y^3) + (∂/∂z)([tex]z^3[/tex])

= [tex]3x^2 + 3y^2 + 3z^2[/tex]

Now, we need to set up the triple integral using spherical coordinates.

In spherical coordinates, the volume element is given by [tex]dV = P^2[/tex] sin(φ) dρ dθ dφ, where ρ is the radial distance, θ is the azimuthal angle, and φ is the polar angle.

The surface Σ represents the boundary of the volume enclosed by the sphere. In spherical coordinates, the equation of the sphere [tex]x^2 + y^2 + z^2 = 9[/tex]becomes [tex]p^2 = 9[/tex].

The unit outward normal vector on the surface of the sphere can be expressed as n = (ρ/3)p, where p is the unit vector in the radial direction.

Using the Divergence Theorem, the flux (F · n) over the surface Σ is equal to the triple integral of the divergence of F over the volume enclosed by Σ:

Flux = ∭V (div(F)) dV

= ∭V ([tex]3p^2[/tex]) dV

= ∫₀²π ∫₀ᴨ ∫₀³ (3[tex]p^2[/tex]) [tex]p^2[/tex] sin(φ) dρ dθ dφ

Here, the limits of integration are as follows:

ρ: 0 to 3

θ: 0 to 2π

φ: 0 to π

Now, we can calculate the flux by evaluating the triple integral:

Flux = ∫₀²π ∫₀ᴨ ∫₀³ ([tex]3p^4[/tex] sin(φ)) dρ dθ dφ

Evaluating this triple integral will give us the flux of the vector field F over the surface Σ.

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We want to take inventory of the right and left side

Right: S=1 O = 2 Li = 2 Se= 1

Left: S= 1 Se=2 Li= 2 O=1

Lets balance out each side because we see we are off by 1 oxygen on the left

Add a coefficient of 2 on the Li2O

Add a coefficient of 2 on the right Li2Se

Now we have So2+ 2Li2Se ---> SSe2+ 2Li2O

or

The equation is already balanced, assuming that there is supposed to be a yields symbol between 2Li2Se and SSe2.

To find out whether or not this equation is balanced, make a little T-chart with the left side of the equation on one side and the left on the other. Next to each element, write down the amount they start off with and make changes as you add coefficients.

Hope this helps!

Answer:

See explanation

Explanation:

The postulates of the kinetic theory of matter are;

Answer: 46.66 Grams

Explanation:

Credits to jasminealexismacias, Enjoy your CK12

N2 +H2 →2 NH3

"Reactants Products Nitrogen 2 2 Hydrogen 2 6 Since NH3 is multiplied by a coefficient of 2 there are now 2 nitrogens and 6 hydrogens. The 6 hydrogens come from the 2 multiplied by the subscript of 3."

CO 2 (s) → CO 2 (g)  produces an increase in the entropy of the system

Define entropy.

Entropy is the measurement of the amount of thermal energy per unit of temperature in a system that cannot be used for productive work. Entropy is a measure of a system's molecular disorder or unpredictability since work is produced by organised molecular motion.

Because a higher temperature increases the kinetic energy of molecules and, as a result, unpredictability, the entropy of the system rises with temperature. When a reaction generates more molecules than it started with, entropy typically rises. When a reaction creates fewer molecules than it began with, entropy typically decreases.

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If 5.00 mol of hydrogen gas and 1.20 mol of oxygen gas react, the limiting agent is O₂.

The number of moles of water produced according to the equation is 1.20 mol.

To determine the limiting reactant, we need to compare the moles of hydrogen gas (H₂) and oxygen gas (O₂) and determine which reactant is present in a lower stoichiometric ratio.

From the information, we have:

Moles of H₂ = 5.00 mol

Moles of O₂ = 1.20 mol

The balanced equation for the reaction between hydrogen gas and oxygen gas to form water (H₂O) is:

2H₂(g) + O₂(g) -> 2H₂O(g)

According to the stoichiometry of the balanced equation, the ratio of H₂ to O₂ is 2:1. This means that for every 2 moles of H₂, we need 1 mole of O₂ to completely react.

Calculating the stoichiometric ratio for the given amounts:

Moles of H₂ / Coefficient of H₂ = 5.00 mol / 2 = 2.50 mol

Moles of O₂ / Coefficient of O₂ = 1.20 mol / 1 = 1.20 mol

Comparing the calculated stoichiometric ratios, we see that the mole ratio of H₂ (2.50 mol) is greater than the mole ratio of O₂ (1.20 mol). This means that the H₂ is in excess, and O₂ is the limiting reactant.

Therefore, the limiting reactant is O₂.

To determine the number of moles of water (H₂O) produced according to the balanced equation, we can use the stoichiometry:

For every 2 moles of H₂O, we need 1 mole of O₂. Since O₂ is the limiting reactant, the number of moles of H₂O produced is equal to the moles of O₂:

nH₂O = 1.20 mol

Therefore, the number of moles of water produced according to the equation is 1.20 mol.

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It would take the same number of moles of Ar(g) approximately 6.0 min to effuse through the same membrane.

The Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass (i.e., the larger the molar mass of a gas, the slower it will effuse). Therefore, we can use this law to find the answer to the given problem. Here are the steps to solve the problem:

Step 1: Calculate the molar mass of Br2(g) and Ar(g)

The molar mass of Br2(g) is:1 × 2 + 79.904 × 2 = 159.808 g/mol

The molar mass of Ar(g) is:39.95 g/mol

Step 2: Calculate the ratio of the square roots of the molar masses

Ratio of the square roots of molar masses = sqrt(molar mass of Ar(g)) / sqrt(molar mass of Br2(g))= sqrt(39.95) / sqrt(159.808)= 0.25

Step 3: Calculate the time required for Ar(g) to effuse through the membrane

We can use the ratio of the square roots of molar masses to find the time required for Ar(g) to effuse through the same membrane.

Time for Ar(g) to effuse = (ratio of the square roots of molar masses) × (time for Br2(g) to effuse) = 0.25 × 24.0 min = 6.0 min

Therefore, it would take the same number of moles of Ar(g) approximately 6.0 min to effuse through the same membrane.

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The mass of H2 needed is approximately 1.09 g.among the given options, the closest value is:C. 1.10 g H2

To determine the mass of H2 needed to react with 8.75 g of O2, we need to use the balanced equation and stoichiometry. The balanced equation is:

[tex]O_2(g) + 2H_2(g)[/tex] → [tex]2H_2O(g)[/tex]

From the equation, we can see that 1 mole of O2 reacts with 2 moles of H2. To calculate the mass of H2, we need to convert the mass of O2 to moles using its molar mass and then use the mole ratio to find the corresponding mass of H2.

1. Calculate the number of moles of O2:

  Moles of O2 = Mass of O2 / Molar mass of O2

The molar mass of O2 is 32 g/mol.

Moles of O2 = 8.75 g / 32 g/mol

2. Use the mole ratio to find the moles of H2:

  Moles of H2 = Moles of O2 × (2 moles H2 / 1 mole O2)

3. Calculate the mass of H2:

  Mass of H2 = Moles of H2 × Molar mass of H2

The molar mass of H2 is 2 g/mol.

Now, let's perform the calculations:

Moles of O2 = 8.75 g / 32 g/mol ≈ 0.2734 mol

Moles of H2 = 0.2734 mol × (2 moles H2 / 1 mole O2) ≈ 0.5468 mol

Mass of H2 = 0.5468 mol × 2 g/mol ≈ 1.0936 g

Rounded to three significant figures, the mass of H2 needed is approximately 1.09 g.Among the given options, the closest value is:

C. 1.10 g H2.

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If Ca(ox) precipitate was present in the solution after centrifugation, the result would be artificially low for the percentage mass of Fe.

Centrifugation is a technique used to separate solid particles from a liquid solution. In this case, the Fe(III)-oxalate solution was centrifuged to remove any solid precipitates, ensuring that only the supernatant (liquid portion) was analyzed.

If Ca(ox) precipitate was present in the solution, it would also be pelleted along with the Fe(III) precipitate during centrifugation. To determine the effect on the percentage mass of Fe, we need to consider the calculation used to determine the mass of Fe in the sample.

Assuming the experiment aims to determine the percentage mass of Fe in the Fe(III)-oxalate solution, the typical calculation involves measuring the mass of the Fe precipitate after it is dried and then dividing it by the initial mass of the sample.

Let's say the initial mass of the sample is M and the mass of the Fe precipitate obtained after drying is m(Fe). The percentage mass of Fe would be calculated as:

% mass of Fe = (m(Fe) / M) * 100

However, if Ca(ox) precipitate is present in the solution, it would contribute to the mass of the obtained precipitate. This would result in an artificially low measurement of the mass of Fe precipitate and, consequently, a lower percentage mass of Fe in the calculation.

If Ca(ox) precipitate is present in the Fe(III)-oxalate solution after centrifugation, it would lead to an artificially low percentage mass of Fe. The presence of Ca(ox) would contribute to the mass of the obtained precipitate, reducing the measured mass of Fe and affecting the overall calculation.

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Answer:

are u brocken

Explanation:

To calculate the colony-forming units per milliliter (CFU/mL), you need to know the volume plated and the number of colonies counted.

In this case, you plated 0.1 mL of urine and observed 279 colonies after incubation.

CFU/mL can be calculated using the following formula:

CFU/mL = (Number of Colonies / Volume Plated) × Dilution Factor

Since you plated 0.1 mL of urine, the volume plated is 0.1 mL. The dilution factor is assumed to be 1 since no dilution was mentioned.

CFU/mL = (279 colonies / 0.1 mL) × 1

= 2790 CFU/mL

So, there are 2790 CFU/mL of urine.

To calculate the CFU per 100 mL, you can use the following formula:

CFU per 100 mL = CFU/mL × Volume

Since you want to calculate the CFU per 100 mL, the volume is 100 mL.

CFU per 100 mL = 2790 CFU/mL × 100 mL

= 279,000 CFU

Therefore, there are 279,000 CFU per 100 mL of urine.

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Answer:

Calorie

Explanation:

This can mesure thermal energy

The relationship between temperature and flux in a carrier ionophore is generally described by the Arrhenius equation, which relates the rate of a chemical reaction to temperature.

In the context of ionophores, which are molecules that facilitate the transport of ions across cell membranes, the flux refers to the rate or magnitude of ion transport.

According to the Arrhenius equation, the rate of a reaction or flux is exponentially dependent on temperature. The equation is typically represented as:

k = A * exp(-Ea / (RT))

In this equation, k represents the rate constant or flux, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

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The partial pressure inside a gas mixture shall consist of the notional pressure of that constituent gas if the whole quantity of its starting material alone was occupied at the same temperature. The partial gas pressure is a measure of thermodynamic action in the particles of a gas, and the calculation can be defined as follows:

Given:

[tex]\to \bold{ P_T=3.50 \ atm}\\\\\to \bold{P_B=1.25 \ atm}\\\\[/tex]

To find:

partial pressure=?

Solution:

Using formula: [tex]\bold{P_T=P_A+P_B}\\\\[/tex]

                   [tex]\to \bold{3.50=P_A+1.25}\\\\\to \bold{P_A=3.50-1.25}\\\\\to \bold{P_A=2.25}\\\\[/tex]

As we know that pressure is not negative and as [tex]\bold{P_T}[/tex] is total pressure so, it has a large value, and[tex]\bold{ P_A , P_B}[/tex] is partial.

Therefore, the final answer is "Option C".

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Explanation:

Elements heavier than Helium are synthesized in a number of environments. For elements that are lighter than Iron, those elements are synthesized during various phases in the evolution of massive stars. For elements heavier than Iron, one needs quite a bit of energy input to form these heavy elements.

Answer: 110

Explanation:

At 25 ∘C, the calculated value of ΔG∘ for the reaction 3H2(g) + Fe2O3(s) -> 2Fe(s) + 3H2O(g) is 138.4 kJ/mol. Since ΔG∘ is positive, the reaction is non-spontaneous under standard conditions at this temperature. A positive ΔG∘ indicates that the reaction requires energy input to occur.

To calculate ΔG∘ at 25 ∘C for the reaction 3H2(g) + Fe2O3(s) -> 2Fe(s) + 3H2O(g) using standard free energies of formation, we need to subtract the sum of the standard free energies of formation of the reactants from the sum of the standard free energies of formation of the products.

The standard free energies of formation for the given compounds at 25 ∘C can be looked up in reference tables. The values are as follows:

ΔG∘f(H2(g)) = 0 kJ/mol

ΔG∘f(Fe2O3(s)) = -824.2 kJ/mol

ΔG∘f(Fe(s)) = 0 kJ/mol

ΔG∘f(H2O(g)) = -228.6 kJ/mol

Using these values, we can calculate ΔG∘ for the reaction:

ΔG∘ = (2 * ΔG∘f(Fe(s)) + 3 * ΔG∘f(H2O(g))) - (3 * ΔG∘f(H2(g)) + ΔG∘f(Fe2O3(s)))

ΔG∘ = (2 * 0 kJ/mol + 3 * (-228.6 kJ/mol)) - (3 * 0 kJ/mol + (-824.2 kJ/mol))

ΔG∘ = -685.8 kJ/mol + 824.2 kJ/mol

ΔG∘ = 138.4 kJ/mol

At 25 ∘C, the calculated value of ΔG∘ for the reaction 3H2(g) + Fe2O3(s) -> 2Fe(s) + 3H2O(g) is 138.4 kJ/mol. Since ΔG∘ is positive, the reaction is non-spontaneous under standard conditions at this temperature. A positive ΔG∘ indicates that the reaction requires energy input to occur.

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Answer:

when population increases the amount of carbon dioxide also increases as population use oxygen and release carbon dioxide

Carbon dioxide is a major greenhouse gas. The increase in the population increases the carbon dioxide amount.

The main product released from the respiratory process of organisms, especially animals is carbon dioxide. The increase in their population will increase this product.

The increased population will increase the demand for the burning of fossil fuel, pollution, and respiration, and hence the product of these activities, carbon dioxide will increase in the atmosphere.

Therefore, the carbon dioxide will increase with an increase in the population.

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The pH of a buffered solution will begin to change significantly when the concentration of added strong acid or base is greater than the capacity of the buffer.

This capacity is determined by the buffer's concentration and the dissociation constant of its acid-base pair.

When a buffered solution is subjected to small amounts of strong acid or base, it should retain its pH value because the buffer will react with the added ions to produce an excess of weak acid or base ions, keeping the pH constant.

As the concentration of strong acid or base added to the solution increases, however, the capacity of the buffer is eventually exceeded, and the pH of the solution will change significantly.

The capacity of a buffer depends on its concentration and on the acid dissociation constant (Ka) of the weak acid component and the base dissociation constant (Kb) of the weak base component.

This can be calculated using the Henderson-Hasselbalch equation:pH = pKa + log ([A-]/[HA])where pH is the pH of the buffer solution, pKa is the dissociation constant of the weak acid component, [A-] is the concentration of the weak base component, and [HA] is the concentration of the weak acid component.

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