what is the ph after 0.195 mol of naoh is added to the buffer from part a? assume no volume change on the addition of the base. express the ph numerically to three decimal places.

Multiplying the slope (-2905 k) by the gas constant (0.008314 kJ/mol K) gives the activation energy: 24.1 kJ/mol.

The slant of the best-fit line for the diagram of ln(k) versus 1/T is equivalent to - Ea/R, where Ea is the actuation energy for the response, R is the gas consistent, and T is the temperature in Kelvin. To decide the actuation energy, we really want to improve this condition to address for Ea.

Ea = - slant x R

We realize that the slant of the best-fit line is - 2905 K, and R is 8.314 J/(mol·K). In any case, the slant should be changed over completely to units of J/(mol·K) by duplicating by 1000, since we need the actuation energy in units of kJ/mol. Accordingly:

Ea = - (- 2905 K x 8.314 J/(mol·K)) x (1/1000 kJ/J)

Ea = 24.1 kJ/mol

The initiation energy for the response is 24.1 kJ/mol. This worth addresses the base energy expected for the reactants to defeat the energy hindrance and structure items. The higher the initiation energy, the more slow the response rate, as well as the other way around.

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In the carbon reaction (light-independent reaction), carbon dioxide molecules are reduced by the addition of electrons carried from the light reaction by NADPH (Nicotinamide Adenine Dinucleotide Phosphate).

NADPH is a coenzyme that carries high-energy electrons and hydrogen ions, which are produced during the light-dependent reactions of photosynthesis, to the stroma of chloroplasts where carbon dioxide fixation occurs.

In the stroma, the electrons and hydrogen ions from NADPH are used to reduce carbon dioxide to form carbohydrates, such as glucose, through a series of enzyme-catalyzed reactions known as the Calvin Cycle.

The Calvin Cycle is the process by which plants and other photosynthetic organisms convert carbon dioxide and other inorganic molecules into organic compounds that can be used for energy and growth.

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Answer: exothermic reaction

Explanation: To create more heat, shift to the product side with the exothermic reaction. An exothermic reaction produces heat. The equation is:

N2(g) + 3H2(g) <-> 2NH3(g) + heat. It's exothermic because heat is a product.

The number of moles, atoms, and mass in the problems are calculated as below.

(8.30 x 10^23 atoms) / (6.022 x 10^23 atoms/mol) = 1.38 mol

(5.5 x 10^23 atoms) / (6.022 x 10^23 atoms/mol) = 0.914 mol

(3.4 mol) x (6.022 x 10^23 atoms/mol) = 2.05 x 10^24 atoms

(2.0 mol) x (6.022 x 10^23 atoms/mol) = 1.21 x 10^24 atoms

(2.5 mol) x (197.34 g/mol) = 493.35 g

(8.3 x 10^28 atoms) / (6.022 x 10^23 atoms/mol) = 1.38 x 10^6 mol

(1.38 x 10^6 mol) x (23.95 g/mol) = 3.30 x 10^7 g

(5 x 10^25 atoms) / (6.022 x 10^23 atoms/mol) = 8.30 mol

(5.67 mol) x (6.022 x 10^23 atoms/mol) = 3.42 x 10^24 atoms

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To find the percent ionization of the acid, we need to first calculate the initial concentration of the acid (HA) before it dissociates.

Since the solution is originally 0.532 M in acid (HA), we can assume that the initial concentration of HA is also 0.532 M.

Next, we need to calculate the concentration of the conjugate base (A-) at equilibrium. We can use the equation for the dissociation of an acid:
HA + H2O ⇌ H3O+ + A-

We know that the hydronium concentration at equilibrium is 0.112 M, so the concentration of the conjugate base is also 0.112 M.

To calculate the percent ionization of the acid, we use the equation:
% ionization = (concentration of dissociated acid / initial concentration of acid) x 100

We can find the concentration of dissociated acid (H3O+) by subtracting the concentration of the conjugate base (A-) from the hydronium concentration:

[H3O+] = 0.112 M - 0 M = 0.112 M

Plugging in the values, we get:
% ionization = (0.112 M / 0.532 M) x 100 = 21.05%

Therefore, the percent ionization of the acid is 21.05%.

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The percent of ionization of an acid in solution of 0.532 M in acid HA i and have a hydronium concentration of 0.112 M is equals to the 21.1%.

The ionization of acids results hydrogen ions, thus, that's why compounds act as proton donors.

Molarity of solution = 0.532 M

At Equilibrium, hydronium concentration = 0.112 M

As we know, concentration is defined as the number of moles of substance in a litre of solution, that most of time concentration is replaced by molarity. So, concentration of acid solution, [ H A] = 0.532 M

Chemical reaction, [tex]HA (aq) + H_2O -> H_3O^{ +}+A^{-}[/tex]

percent of ionization of the acid =

[tex] \frac{ [ H_3O^{+}] }{ [ HA]} × 100 [/tex]

= (0.112/0.532) × 100

= 21.1%

Hence, required value is 21.1%.

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The actuation method triggered by a product-of-combustion detector in a CO2 system is typically referred to as "automatic actuation."

In this method, the product-of-combustion detector senses the presence of a fire or heat source and sends a signal to the CO2 system, causing it to automatically release CO2 to extinguish the fire.

This actuation method is commonly used in areas where fires may start unexpectedly, such as in server rooms, electrical substations, and other critical infrastructure facilities. It provides a fast and effective response to fires, helping to minimize damage and protect personnel.

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The amount of air required to burn 1 gram of paper is 17.22 grams. This is because paper is made up of pure cellulose which is a compound of 6 carbon atoms, 10 hydrogen atoms, and 5 oxygen atoms (C6H10O5).

To burn this compound, the oxygen from the air must combine with the carbon and hydrogen atoms from the paper. For every 1 mole of C6H10O5, 12 moles of oxygen are required.

Since 1 mole of oxygen has a mass of 32 grams, 12 moles of oxygen would have a mass of 384 grams.

Since 1 gram of paper has 1 mole of C6H10O5, 384 grams of oxygen is required to burn 1 gram of paper.

Since air is composed of approximately 21% oxygen, the amount of air required to burn 1 gram of paper is 17.22 grams (384/21 = 17.22).

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The new volume of the gas should be 41.41 L when its pressure drops to 0.1 atm, not 7 L as stated in the original statement, This statement is incorrect.

What is new volume?

According to Boyle's Law, the pressure and volume of a gas are inversely proportional, meaning that as one increases, the other decreases, as long as the temperature and amount of gas remain constant. Therefore, if the pressure of a gas decreases, its volume should increase, and vice versa.

Using Boyle's Law, we can calculate the initial volume of the gas when its pressure drops to 0.1 atm:

P1V1 = P2V2

(0.73 atm)(5.64 L) = (0.1 atm)(V2)

V2 = (0.73 atm)(5.64 L) / (0.1 atm) = 41.41 L

Therefore, the new volume of the gas should be 41.41 L when its pressure drops to 0.1 atm, not 7 L as stated in the original statement.

What is Boyle's Law?

Boyle's Law is a gas law named after the Irish chemist Robert Boyle. It states that the pressure of a gas is inversely proportional to its volume, provided that the temperature and amount of gas remain constant. Mathematically, Boyle's Law can be expressed as:

P1V1 = P2V2

where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively. This relationship means that if the volume of a gas is reduced (at constant temperature and amount), the pressure will increase proportionally, and vice versa. Boyle's Law is often applied in situations where the pressure and volume of a gas need to be controlled, such as in the design of engines and pneumatic systems.

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if the pressure is held constant in a system, the combined gas law reduces to Gay-Lussac's law or the pressure-temperature law:

V₁/T₁ = V₂/T₂

where V₁ and T₁ are the initial volume and temperature of the gas, and V₂ and T₂ are the final volume and temperature, respectively.

According to this law, the volume and temperature of a gas are directly proportional to one another for a set amount of gas at a constant pressure. In other words, a gas's volume will grow if its temperature is raised while its pressure remains constant, and vice versa if its temperature is lowered.

A formula known as the combined gas law connects a gas's pressure, volume, and temperature. Three gas laws—Boyle's law, Charles law, and Gay-Lussac law—are combined to create it.

As long as the quantity of gas and the number of gas particles are held constant, this formula can be used to determine the change in one variable (pressure, volume, or temperature) while the other two variables are known.

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10 kg.cm/s² is equivalent to 0.1 N when converted into newton.

The unit of force in the International System of Units (SI) is the newton (N). One Newton is defined as the amount of force required to accelerate a mass of one kilogram at a rate of one meter per second squared (1 N = 1 kg⋅m/s² ).

10 kg⋅cm/s²  can be converted to newtons using the following formula:

1 N = 1 kg⋅m/s²

First, we need to convert cm to meters, as the unit of force is in newtons, which is based on meters.

1 cm = 0.01 m

Therefore, 10 kg⋅cm/s² can be converted to:

10 kg × 0.01 m/s² = 0.1 kg⋅m/s²

Now, using the formula:

1 N = 1 kg⋅m/s²

We can convert 0.1 kg⋅m/s² to newtons:

0.1 kg⋅m/s² = 0.1 N

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when writing the rate law for the reaction mechanism with equilibrium preceding the rate-determining step, the rate law will have constructed by using data about the : rate constant for the rate  step and the rate constant for the reverse reaction for equilibrium concentration.

The steps in the rate determining  law are as :

1. The sum of all the elementary step in the reaction mechanism should yield the overall reaction equation.

2. The rate law for the determining steps will agree that with the experimentally determine the rate law.

Therefore, the rate constant for the rate determining step, the rate constant for the reverse reaction and that the concentration of the product for the equilibrium are all consider as the major steps.

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The partial pressure of [tex]O_2[/tex] is -146 mmHg

The given problem involves a gas mixture consisting of [tex]N_2, CO_2[/tex], and [tex]O_2[/tex] at a total pressure of 730 mmHg. The partial pressure of [tex]CO_2[/tex] is given as 182 mmHg, and it is also given that there are twice as many moles of [tex]N_2[/tex] as there are of [tex]CO_2[/tex].

To solve the problem, we need to use Dalton's law of partial pressures, which states that the total pressure of a mixture of gases is equal to the sum of the individual gases. We can also use the mole fraction concept, which is the ratio of the number of moles of a gas to the total number of moles of all gases in the mixture.

Let x be the number of moles of [tex]CO_2[/tex] in the mixture.

Then, the number of moles of [tex]N_2[/tex] is 2x. Therefore, the number of moles of [tex]O_2[/tex] is (total number of moles) - (number of moles of [tex]CO_2[/tex]) - (number of moles of [tex]N_2[/tex]), which is x/2.

We can now use the mole fraction concept to calculate the mole fractions of each gas. The mole fraction of

[tex]CO_2[/tex] is

x/(2x + x + x/2) = 2x/5x = 0.4.

Similarly, the mole fraction of

[tex]N_2[/tex] is 2x/(2x + x + x/2) = 4x/5x = 0.8.

The mole fraction of

[tex]O_2[/tex] is (x/2)/(2x + x + x/2) = x/5x = 0.2.

Finally, we can use Dalton's law of partial pressures to calculate the partial pressure of oxygen:

Total pressure = P([tex]N_2[/tex]) + P([tex]CO_2[/tex]) + P([tex]O_2[/tex])

730 mmHg = P([tex]N_2[/tex]) + 182 mmHg + P([tex]O_2[/tex])

Substituting the mole fraction and pressure values, we get:

730 mmHg = (0.8)(730 mmHg) + (0.4)(730 mmHg) + P([tex]O_2[/tex])

730 mmHg = 584 mmHg + 292 mmHg + P([tex]O_2[/tex])

P([tex]O_2[/tex]) = 730 mmHg - 876 mmHg

P([tex]O_2[/tex]) = -146 mmHg

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The classification of the given reaction is "Redox" which is the correct answer based on the chemical equation of the reaction given in the question.

The classification for the given reaction is a double displacement reaction.
The given reaction is:

2I¯(aq) + Cl2(aq) → I2(aq) + 2Cl¯(aq)

This reaction involves the exchange of ions between the reactants, and it can be classified as a "Redox" reaction (which stands for reduction-oxidation). In this reaction, the iodide ions (I¯) are oxidized, as they lose electrons to form I2, while the chlorine (Cl2) is reduced, as it gains electrons to form chloride ions (Cl¯).
To summarize, the classification of the given reaction is "Redox."

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The classification for the reaction 2I¯(aq) + [tex]Cl_{2}[/tex](aq) → [tex]I_{2}[/tex](aq) + 2Cl¯(aq) is a redox reaction.

Redox reactions involve the transfer of electrons from one species to another, resulting in changes in oxidation states (the charge or electron density on an atom).  The reaction can be classified as a redox (reduction-oxidation) reaction. In this reaction, iodine (I¯) is being oxidized, and chlorine ( [tex]Cl_{2}[/tex]) is being reduced. To identify such reactions, we can

1. Identify the initial oxidation states: I¯ has an oxidation state of -1, and [tex]Cl_{2}[/tex] has an oxidation state of 0.
2. Identify the final oxidation states:  [tex]I_{2}[/tex] has an oxidation state of 0, and Cl¯ has an oxidation state of -1.
3. Observe the change in oxidation states: I¯ is oxidized from -1 to 0, while [tex]Cl_{2}[/tex] is reduced from 0 to -1.

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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)

Answer:

To find the number of molecules of lithium chloride in 78.40 g, we need to first convert the mass to moles using the molar mass of lithium chloride:

moles = mass / molar mass

moles = 78.40 g / 42.39 g/mol

moles = 1.849 mol

Now we can use Avogadro's number to convert from moles to molecules:

molecules = moles x Avogadro's number

molecules = 1.849 mol x 6.022 x 10^23 molecules/mol

molecules = 1.111 x 10^24 molecules

Therefore, there are 1.111 x 10^24 molecules of lithium chloride in 78.40 g.

Explanation:

The both local and the global effects of burning petroleum in the car engines are smog and the global warming.

The Global effects defines to the various effects at which the actions of the individuals, the businesses, and the governments will be on the environment and the society at the large. The Global effects will leads to the changes to the climate, the water cycle, the biodiversity, and the food production, and the other natural systems.

The Smog is the form of the air pollution and will be created by the reaction of the sunlight and with the emissions from the car exhausts.

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Chromatography can be used to separate chlorophyll into a variety of pigment colours.

The leaf contains the chlorophyll molecule, which can be isolated using paper chromatography. Based on the distance that pigment molecules on the paper in a nonpolar solvent travelled, the paper chromatography separates the pigments in the leaf.

During paper chromatography, pigments separate because of their various polarity. Being polar, cellulose interacts more with polar pigments than non-polar pigments. Capillarity and absorbance are essential for chromatography.

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Question:

Using chromatography you can separate chlorophyll into several different pigment colors. Which statement is true about chlorophyll in this context, and what is its role in photosynthesis?

The pressure within a gas container that is sealed would rise as the temperature climbed. The right answer is B.

This is easily understood by picturing how the gas particles in the container move more energetically as the temperature rises. This indicates that there are more contacts between them and the container's sidewalls, increasing the pressure.

The pressure inside a closed container of gas is brought on by collisions between the gas's particles and the container's walls. The pressure of the gas increases with the frequency of collisions.

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As a voltaic cell operates, the reaction progresses and electrons flow from the anode to the cathode. This flow of electrons drives the reduction and oxidation reactions that occur at the two electrodes.

In a voltaic cell, the reactants are located at the anode and the products are located at the cathode. As the reaction proceeds, the concentration of reactants decreases at the anode, while the concentration of products increases at the cathode. This means that the ratio of products to reactants will increase as the cell operates.

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Verify that the number of atoms of each element is equal on both sides of the equation and, if the equation contains ions, that the charges are balanced equation.

The number and type of each atom in balanced chemical equations are the same on both sides of the equation. The simplest whole number ratio must be used as the coefficients in a balanced equation. In chemical processes, mass is always preserved.

Each element must have the same number of atoms on the left as it has on the right. You must add integers to the left of one or more equations to balance an imbalanced equation.

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, the molar concentration of the solution is 2.202 M. molar concentration of a solution that contains 45.0 g of nacl dissolved in 350.0 ml of water

To calculate the molar concentration of a solution, we need to first determine the number of moles of the solute present in the solution, and then divide that by the volume of the solution in liters.

The molar mass of NaCl is 58.44 g/mol. Therefore, the number of moles of NaCl in 45.0 g can be calculated as:

mole= mass / molar mass = 45.0 g / 58.44 g/mol = 0.7709 mol

Next, we need to convert the volume of the solution from milliliters to liters:

volume = 350.0 ml = 0.3500

Finally, we can calculate the molar concentration (M) of the solution as:

M = moles / volume = 0.7709 mol / 0.3500 L = 2.202 M

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AgCl will be more soluble in pure water than in 0.500 M NaCl solution. This is because the high concentration of Cl- ions in the NaCl solution will decrease the solubility of AgCl due to the common ion effect.

The solubility of solid silver chloride (AgCl) will be affected by the presence of other ions in the solution. When AgCl is added to pure water, it will dissociate into its constituent ions, Ag+ and Cl-.

However, in the presence of 0.500 M NaCl, the concentration of Cl- ions in the solution will increase. This increase in Cl- concentration will shift the equilibrium of AgCl dissociation towards the formation of more AgCl, making it less soluble.

The presence of other ions in the solution can affect the solubility of a solute, and this phenomenon is an important consideration in many chemical reactions and processes.

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Molecules with polar functional groups behave differently than those with nonpolar functional groups due to their distinct properties. Polar functional groups contain an unequal distribution of electron density, leading to the presence of partial positive and negative charges. This results in stronger intermolecular forces, such as hydrogen bonding, dipole-dipole interactions, and increased solubility in polar solvents like water.

On the other hand, nonpolar functional groups have an equal distribution of electron density, which means there are no partial charges. These molecules experience weaker intermolecular forces, like van der Waals or London dispersion forces. Consequently, they tend to be less soluble in polar solvents but more soluble in nonpolar solvents, like hydrocarbons.

In summary, the presence of polar or nonpolar functional groups impacts a molecule's properties, including intermolecular forces and solubility in different solvents.

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It is  observed that the temperature varies greatly at the three different locations - North Pole, South Pole, and Equator.

The  possible claim for this variation in temperature could be the difference in the amount of solar radiation received by these regions

The equator is described as  a circle of latitude that divides a spheroid, such as Earth, into the northern and southern hemispheres.

The North and South Poles experience much lower temperatures compared to the Equator while the equator experiences much higher temperatures.

This assertion is supported by the graph's data, which shows that while the temperature in the North and South Poles continues to be low, it is constantly high at the equator.

A greater sensitivity to variations in solar radiation is also indicated by the fact that the change in temperature over time is far greater in the poles than it is at the Equator.

In conclusion, it can be inferred that the variance in solar energy received at these places is mostly to blame for the fluctuation in temperature.

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Okay, let's solve this step-by-step without using quadratics:

1) The equilibrium constant Kc = 0.36 means the equilibrium lies to the left. So there will be more N2O4 than NO2 at equilibrium.

2) The initial concentration of N2O4 is 0.1 mol dm^-3. Let's call this [N2O4]initial.

3) At equilibrium, the concentrations of N2O4 and NO2 will be [N2O4]equil and [NO2]equil respectively.

4) We know the equilibrium constant expression for this reaction is:

Kc = ([NO2]equil)^2 / [N2O4]equil

5) Setting this equal to 0.36 and plugging in 0.1 for [N2O4]initial, we get:

0.36 = ([NO2]equil)^2 / (0.1 - [NO2]equil)

6) Simplifying, we get:

0.036 = [NO2]equil^2

7) Taking the square root of both sides, we get:

[NO2]equil = 0.06 mol dm^-3

So the equilibrium concentration of NO2 is 0.06 mol dm^-3.

Let me know if you have any other questions! I can also provide a more step-by-step explanation if needed.

The standard potential of the cathode half-cell reaction for CO₂ electroreduction is -0.610 V. This means that energy input is required to drive the reaction in the forward direction.

The free energy required for the reaction can be calculated using the equation ∆G = -nFE, where n is the number of electrons transferred, F is the Faraday constant, and E is the standard potential of the reaction.

For the CO₂ electroreduction reaction, n = 2, and at room temperature, F is approximately 96.485 kJ/mol·V.

Thus, the free energy required for the reaction is

∆G = -(2)(96.485 kJ/mol·V) × (-0.610 V) = 117.4 kJ/mol.

In addition, the energy input/output can also be affected by the efficiency of the electrocatalytic system. Thus, while the calculated free energy can provide a rough estimate, the actual energy input/output for the CO₂ electroreduction reaction will depend on several factors and needs to be experimentally determined.

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The complete question is:

Electrocatalytic CO₂ reduction is a promising strategy to close the anthropogenic carbon cycle by converting CO₂ and H₂O to fuels or other valuable small molecules. A half cell reaction and standard potential (vs. SHE in aqueous electrolyte at pH 7) is:

Cathode half cell reactions is CO₂⁺ + 2H⁺ + 2e⁺ ⇄ HCOOH is -0.610 V

How does the energy that you calculated qualitatively compare with the free energy that is actually required for this reaction (more/less energy input/output) and why?

The results for potassium were elevated because the lack of anticoagulant caused hemolysis, which led to the release of potassium from red blood cells.

When a blood specimen is drawn without an anticoagulant, it can cause hemolysis, which is the rupture of red blood cells and the release of their contents into the serum or plasma. Hemolysis can occur due to trauma during the venipuncture or by using a needle that is too small or too large.

One of the main constituents released from red blood cells during hemolysis is potassium, which can lead to falsely elevated potassium levels in the blood. This occurs because potassium is normally present in high concentrations inside the red blood cells, and hemolysis releases this intracellular potassium into the serum or plasma.

Therefore, in this case, the lack of anticoagulant during the blood draw caused hemolysis, which led to the release of potassium from the red blood cells, resulting in falsely elevated potassium levels of 6.0 mmol/L. To obtain accurate potassium levels, blood specimens should be drawn using appropriate techniques and anticoagulants to prevent hemolysis.

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In the second step of this reaction, proton transfer occurs after the initial nucleophilic attack. The pattern of curved arrow pushing represents the movement of electron pairs during the process.

The nucleophile donates its lone pair of electrons to form a bond with the electrophile, while the proton is transferred to a nearby base, resulting in the breaking of the bond between the electrophile and the leaving group. This arrow pushing pattern helps to visualize the electron flow and the formation of new bonds during the reaction.

Based on the terms provided, the pattern of curved arrow pushing for the second step of this reaction (proton transfer nucleophilic attack) would involve the transfer of a proton from the acidic site to the nucleophile, followed by the nucleophilic attack on the electrophilic site. This is commonly represented as a curved arrow pointing from the proton donor (acidic site) to the proton acceptor (nucleophile) and then to the electrophilic site. The exact details of the curved arrow pushing would depend on the specific reaction and the structures involved.

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The second step of a reaction involving proton transfer and nucleophilic attack, the pattern of curved arrow pushing would be as follows:

1. Nucleophilic attack: In this step, the electron-rich nucleophile donates a pair of electrons to the electron-poor electrophile, forming a new bond between them. The curved arrow starts at the nucleophile's electron pair and points towards the electrophile.

2. Proton transfer: Following the nucleophilic attack, a proton (H+) is transferred from one atom to another, typically facilitated by a base. The curved arrow starts at the base's electron pair and points towards the hydrogen atom being transferred. After that, another curved arrow starts at the bond between the hydrogen atom and the atom it is attached to, pointing towards the atom that will be accepting the proton.

In summary, the pattern of curved arrow pushing in the second step of this reaction involves nucleophilic attack followed by proton transfer.

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The buffered solution that would be more resistant to pH change when the strong acid or the strong base is added is the buffer b - acidic buffer.

The acidic buffer will be the most resistant to the pH change when the both of the concentrations of the acid and the base are equal. The Acid buffer contains the acidic pH and which is prepared by the mixing of the weak acid and its salt with the strong base.

The basic buffer has the basic pH and which is prepared by the mixing of the weak base and its salt with the strong acid. Therefore, the buffer will be more resistant to the change in the pH.

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This question is incomplete, the complete question is :

which buffered solution, buffer a or buffer b, would be more resistant to ph change when a strong acid or a strong base is added? justify your answer.

buffer a - basic buffer

buffer b - acidic buffer