what will be the ph change when 20.0 ml of 0.100 m naoh is added to 80.0 ml of a buffer solution consisting of 0.166 m nh3 and 0.186 m nh4cl? (assume that there is no change in total volume when the two solutions mix.)

Answer:

OH-

Explanation:

Answer:

C. H2O is the base and H3O+ is the conjugate acid

Explanation:

According to Bronsted-Lowry acid-base theory, an acid is a substance that loses an hydrogen ion or proton (H+) while a base is a substance that gains an hydrogen ion (H+) or proton. Furthermore, this theory states that, the molecule formed when an acid donates its proton is called the CONJUGATE BASE, while the molecule formed when the base accepts proton is called CONJUGATE ACID.

In this question, the following equation is given:

NH4+(aq) + H2O(aq) ⇌NH3(aq) + H3O+ (aq)

Water (H2O) is the base in this equation because according to Bronsted-Lowry acid-base theory, it accepts an hydrogen ion (H+) while hydroxonium ion (H3O+) is the conjugate acid.

The third law of thermodynamics describes the entropy of a: solid.

The third law of thermodynamics states that the entropy of a pure crystalline substance approaches zero as the temperature approaches absolute zero (0 Kelvin or -273.15 degrees Celsius). This law implies that at absolute zero, a perfectly ordered and pure crystalline solid will have zero entropy.

The third law of thermodynamics is not specific to liquids or gases but applies to solids. In a solid, the molecules are highly ordered and have fixed positions in a regular lattice structure. As the temperature decreases towards absolute zero, the thermal motion of the molecules reduces, and the system becomes more ordered, resulting in a decrease in entropy.

In contrast, liquids and gases have higher entropy compared to solids at absolute zero because their molecules have more freedom of movement and are not as tightly arranged. Therefore, the third law of thermodynamics specifically addresses the entropy of solids and does not apply to liquids or gases.

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Answer:

the answer is absolutely A

Explanation:

Answer:

Explanation:

One can solve this in two ways. The first is to calculate the moles of NaCl. This would be (25g NaCl/(58.5 g/mole NaCl) = 0.437 moles of NaCl. The balanced equation tells us that the number of moles of Cl2 required is 2X that of the moles NaCl produced. That means that 2(0.437 moles) = 0.8of NaCl ar produced.

The half life of uranium is 10 minutes. Then, the 48 g of uranium sample will decay to 0.21 g after 78 minutes.

Heavy unstable radioactive isotopes undergo nuclear decay by the emission of charged particles. The nuclear decay is a first order reaction.

thus, decay constant k = 1/t ln W0/Wt.

The half life time of the sample = 10 minute.

decay constant = 0.693/ t1/2

k = 0.693/10 = 0.069 min⁻¹

then,W0 = 48 g

we have to find the amount  Wt.

time of decay t = 78 min

ln 48 g/Wt =  0.069 min⁻¹ × 78 min

Wt = 0.12 g.

Therefore, the mass of uranium sample left after the decay will be 0.12 g.

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CIF will tend to increase as the reaction proceeds toward equilibrium.

Given that Kp is equal to 20 at 2500 K, we can calculate the initial concentrations of CIF using the ideal gas law. Let's assume the initial volume is 1 liter for simplicity.

For Cl2:

P(Cl2) = 0.18 atm

n(Cl2) = P(Cl2) * V / (RT) = 0.18 mol

For F2:

P(F2) = 0.31 atm

n(F2) = P(F2) * V / (RT) = 0.31 mol

For CIF:

P(CIF) = 0.92 atm

n(CIF) = P(CIF) * V / (RT) = 0.92 mol

Based on the balanced equation, for every 1 mole of CIF, 1 mole of Cl2 and 1 mole of F2 are consumed. Therefore, the initial moles of CIF are equal to the initial moles of Cl2 and F2.

Since the initial concentrations of CIF, Cl2, and F2 are the same, and the reaction is not at equilibrium, we can conclude that CIF will tend to increase as the reaction proceeds toward equilibrium. This is because the reaction favors the formation of CIF, as indicated by the value of Kp. As CIF forms, the concentrations of Cl2 and F2 decrease, driving the reaction in the forward direction to restore equilibrium.

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Answer:

6

Explanation:

If there are 6 electrons and it has a balanced charge, there also must be six protons.

The procedural steps in order to get iodinate salicylamide are as follows:

The Electrophilic aromatic substitution is known to be a kind of an organic reaction where an atom is said to be added or attached to a kind of aromatic structure (that is made up of hydrogen) is said to be replaced by an electrophile.

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#Use Avogadro

Number of atoms = mol × 6×10²³

7.28 × 10²⁵ = mol × 6×10²³

mol = (7.28×10²⁵) : (6×10²³)

mol = 121.3mol

Answer: By the very laws of the universe, matter cannot be created or destroyed, the Big Bang cannot have happened by its own power. There was a creator involved.

The specific gravity measured by the urine chemical reagent strip method is a measure of the density of the urine compared to water.

What is specific gravity?
The ratio of a substance's density (mass per unit volume) to the density of a specific reference material is known as its relative density or specific gravity. When determining a substance's specific gravity, liquids are typically compared to the densest form of water (at 4 °C or 39.2 °F); for hydrocarbons, the reference point is air at room temp (20 °C or 68 °F). Scientific jargon favours the term "relative density," which is frequently abbreviated as "RD," while deprecating the term "specific gravity." A substance is less dense than the reference if its relative density is less than 1, and it is denser than the reference if it is greater.

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Answer:

MgCl₂(aq) + Al₂(CO₃)₂(aq) --> MgCO₃(s) + AlCl₃(aq)

Explanation:

Magnesium Chloride - MgCl₂

Aluminium Carbonate - Al₂(CO₃)₃

Magnesium carbonate - MgCO₃

Aluminium Chloride - AlCl₃

The equation of this reaction is given as;

MgCl₂(aq) + Al₂(CO₃)₃(aq) --> MgCO₃(s) + AlCl₃(aq)

Answer:

The earlier bog bodies that were discovered in the 1600s might have not been preserved properly due to a lack of knowledge on how to preserve them or a lack of awareness of their significance. It is also possible that they might have decayed and decomposed over time and not survived till the present day. However, the bog bodies found after the late 1800s were preserved and studied extensively due to the increasing awareness and understanding of their historical and archaeological significance.

Explanation:

Hope this helped!! Have a great day/night!!

Answer:

Velocity Equation in these calculations:

Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v.

Explanation:

To determine the mass of CaSO₄ produced in a chemical reaction, you need to know the balanced equation and the molar masses of the reactants and products is 6.123 g.

The given equation provides the necessary information to determine the stoichiometry of the reaction. Let's assume the balanced equation is:

CaCl₂ + Na₂SO₄ → CaSO₄ + 2 NaCl

To calculate the mass of CaSO₄ produced, you need to convert the given mass of each reactant (5.00 g) to moles using their molar masses. Then, based on the stoichiometric coefficients in the balanced equation, you can determine the moles of CaSO₄ produced. Finally, you convert the moles of CaSO₄ to mass using its molar mass.

For example, let's calculate the mass of CaSO₄ produced using CaCl₂ as the limiting reactant. The molar mass of CaCl₂ is 110.98 g/mol, and the molar mass of CaSO₄ is 136.14 g/mol.

First, calculate the moles of CaCl₂: 5.00 g / 110.98 g/mol = 0.045 mol.

Using the stoichiometry, you can see that the mole ratio between CaCl₂ and CaSO₄ is 1:1. Therefore, the moles of CaSO₄ produced is also 0.045 mol.

Finally, calculate the mass of CaSO₄: 0.045 mol × 136.14 g/mol = 6.123 g.

So, when 5.00 g of each reactant is combined, approximately 6.123 g of CaSO₄ is produced based on the given equation.

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The volume of the nitrous oxide gas is 1729.3 Liters

The number of moles of gas present in 3.40 kg of nitrous oxide is determined from the formula below:

Numbers of moles = mass/molar mass

the mass of nitrous oxide = 3.40 kg or 3400 g

the molar mass of nitrous oxide = 44.013 g/mol

Moles of gas = 3400 / 44.013

Moles of gas = 77.25 moles

Using the ideal gas equation to determine the volume of the gas:

PV= nRT

V = nRT/P

where;

V = 77.25 * 0.082 * 273 / 1

The volume of the gas = 1729.3 Liters

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Explanation:

Wyzant

CHEMISTRY

Zoey C. asked • 06/05/19

Calcium hydroxide reacts with sodium phosphate to produce calcium phosphate and sodium hydroxide. If 100 grams of calcium hydroxide reacts with 100 grams of sodium phosphate,

determine the amount of each chemical present when the reaction is complete

Grams calcium hydroxide =?

Grams of sodium phosphate=?

Grams of calcium phosphate=?

Grams of sodium hydroxide?

Mass reactants=?

Mass after reaction?

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J.R. S. answered • 06/06/19

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3Ca(OH)2 + 2Na3PO4 ==> Ca3(PO4)2 + 6NaOH ... balanced equation

First, find the limiting reactant:

moles Ca(OH)2 present = 100 g x 1 mole/74.09 g = 1.35 moles (divided by 3-->0.45)

moles Na3PO4 present = 100 g x 1 mole/163.9 g = 0.610 moles (divided by 2-->0.31)

LIMITING reactant is Na3PO4

There will be zero grams Na3PO4 left after the reaction

For grams Ca(OH)2 left: 0.610 mol NaP x 3 mol Ca(OH)2/2 mol NaP = 0.915 mol Ca(OH)3 used up

Grams left over: 1.35 mol - 0.915 mol = 0.435 mol x 74.09 g/mol = 32 g left over (30 g if use 1 sig. fig.)

For grams of calcium phosphate left: 0.610 mol NaP x 1 mol CaP/2 mol NaP = 0.305 mol CaP formed

Grams at end of rx: 0.305 moles CaP x 310 g/mol = 94 g (90 g if use 1 sig. fig.)

For grams NaOH left: 0.610 mol NaP x 6 mol NaOH/2 mol NaP = 1.83 moles NaOH formed

Grams at end of rx: 1.83 moles NaOH x 40 g/mol = 73 g (70 g if use 1 sig. fig.)

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According to stoichiomtry, 19.99 grams of calcium hydroxide will be needed to completely react with 29.5 grams of sodium phosphate.

Stoichiometry is a process of the determination of proportions of elements or compounds that are present in a chemical reaction. The related relations are based on law of conservation of mass and law of combining weights and volumes.

Stoichiometry is often  used in quantitative analysis for measuring concentrations of substances which are  present in the sample.

As per the stoichiometry of the the given balanced chemical equation,\(222.27\ g\) calcium hydroxide gives \(327.88 \ g\) sodium phosphate, thus,\(29.5 \ g\) sodium phosphate requires,

\(\dfrac{29.5}{327.88}\times{222.27}\\=19.99 \ g\)

Thus, 19.99 grams of calcium hydroxide will be needed to completely react with 29.5 grams of sodium phosphate.

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Answer:

True

Explanation:

Phislosophy does not accept something even though it is a khown fact unless it can proves it's true .

Descartes sais that in philosophy we must doubt the facts we khow , we must start from the beginning to reach the truth

Trial c. 1 gram of powdered calcium reacting with 20 ml of 1.5 M acid is most likely to be the most rapid.

The reaction rate between a reactant and an acid is influenced by several factors, including the concentration of the acid and the surface area of the reactant. In this case, we are comparing the effects of concentration and physical form (powdered vs. chunk) on the reaction rate.

The higher the concentration of the acid, the greater the number of acid particles available for collisions with the calcium particles. In trial c, the acid concentration is 1.5 M, which is higher than the concentrations in trials a and d (5.0 M > 1.5 M). A higher acid concentration means more acid particles are present to interact with the calcium, increasing the likelihood of successful collisions and faster reaction rates.

Additionally, in trial c, the calcium is in powdered form. Powdered calcium has a larger surface area compared to chunk calcium, as it is broken down into smaller particles. A larger surface area exposes more calcium particles to the acid, allowing for a greater number of acid-calcium collisions and enhancing the reaction rate. In contrast, trials a and d involve chunk calcium, which has a smaller surface area.

Therefore, the combination of a higher acid concentration and the larger surface area provided by powdered calcium in trial c makes it the most likely trial to exhibit the most rapid reaction.

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To calculate the standard Gibbs free energy change (ΔG°rxn) for the given reaction, we can use the equation:ΔG°rxn = ΔH°rxn - TΔS°rxn, Given: ΔH°f (kJ/mol) values:HNO3(aq): -207.0 kJ/mol, NO(g): 91.3 kJ/mol, NO2(g): 33.2 kJ/mol and H2O(l): -285.8 kJ/mol.

S° (J/mol∙K) values:

HNO3(aq): 146.0 J/mol∙K

NO(g): 210.8 J/mol∙K

NO2(g): 240.1 J/mol∙K

H2O(l): 70.0 J/mol∙K

Let's calculate the ΔH°rxn:

ΔH°rxn = [3 × ΔH°f(NO2(g))] + [ΔH°f(H2O(l))] - [2 × ΔH°f(HNO3(aq))] - [ΔH°f(NO(g))]

ΔH°rxn = [3 × 33.2 kJ/mol] + [-285.8 kJ/mol] - [2 × (-207.0 kJ/mol)] - [91.3 kJ/mol]

ΔH°rxn = 99.6 kJ/mol - 285.8 kJ/mol + 414.0 kJ/mol - 91.3 kJ/mol

ΔH°rxn = 136.5 kJ/mol

Calculate the ΔS°rxn:

ΔS°rxn = [3 × S°(NO2(g))] + [S°(H2O(l))] - [2 × S°(HNO3(aq))] - [S°(NO(g))]

ΔS°rxn = [3 × 240.1 J/mol∙K] + [70.0 J/mol∙K] - [2 × 146.0 J/mol∙K] - [210.8 J/mol∙K]

ΔS°rxn = 720.3 J/mol∙K + 70.0 J/mol∙K - 292.0 J/mol∙K - 210.8 J/mol∙K

ΔS°rxn = 287.5 J/mol∙K

Now, we can calculate ΔG°rxn using the equation:

ΔG°rxn = ΔH°rxn - TΔS°rxn

If we assume a standard temperature of 298 K, we can substitute the values: ΔG°rxn = 136.5 kJ/mol - (298 K * 0.2875 kJ/mol∙K)

ΔG°rxn = 136.5 kJ/mol - 85.57 kJ/mol

ΔG°rxn ≈ 50.93 kJ/mol

The calculated ΔG°rxn is positive (+50.93 kJ/mol). Therefore, based on the given options, the closest answer is: +50.8 kJ

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Answer:

Percent yield = 88.23%

Explanation:

Given data:

Percent yield of ammonia = ?

Actual yield = 15 g

Mass of nitrogen = 14 g

Solution:

Chemical equation:

N₂ + 3H₂      →     2NH₃

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 14 g/ 28 g/mol

Number of moles = 0.5 mol

Now we will compare the moles of ammonia and nitrogen.

         N₂          :         NH₃

          1            :         2

         0.5        :        2/1×0.5 = 1 mol

Mass of ammonia/Theoretical yield

Mass = number of moles × molar mass

Mass = 1 mol × 17 g/mol

Mass = 17 g

Percent yield:

Percent yield = ( actual yield / theoretical yield )× 100

Percent yield =  (15 g/ 17 g)× 100

Percent yield = 88.23%

The Haber process is a complex industrial process, and several factors must be taken into consideration when determining the optimum temperature.

What is Harber process?

Ammonia (NH3) is created chemically by the Haber process from nitrogen gas (N2) and hydrogen gas (H2). The technique was created in the early 20th century by a German chemist by the name of Fritz Haber, hence the name.

With the aid of a catalyst, usually iron, the Haber process involves the reaction of nitrogen gas and hydrogen gas at high pressure and high temperature. The reaction's chemical equation is as follows:

                   N2 + 3H2 ⇌ 2NH3

In the Haber process, the creation of ammonia (NH3) is an exothermic reaction. Le Chatelier's principle states that an exothermic reaction is more advantageous at lower temperatures. In order to maximise the ammonia yield, one might therefore infer that the Haber process should be carried out at low temperatures.

But because the Haber process is a sophisticated industrial procedure, choosing the ideal temperature requires careful thought of a number of variables. While a low temperature may promote the forward reaction, it may also cause the reaction to proceed more slowly, resulting in less ammonia being produced. The pressure and concentration of the reactants also have an impact on reaction rate.

The Haber process has been successfully used at temperatures between 400 and 450°C, based on the catalyst employed. Due to the low temperature favouring the forward reaction, the reaction rate is high enough at this temperature to achieve an acceptable ammonia production rate while also allowing for a good yield of ammonia. Moreover, the inclusion of a catalyst (often iron) can speed up the reaction even more and boost ammonia output.

In conclusion, even if the forward reaction in the Haber process may be favoured by a low temperature, this does not mean that this temperature is the best one because other variables that affect the reaction rate and production rate may also be at play. By testing, it has been discovered that a catalyst and a temperature of roughly 400–450°C offer the greatest ammonia yield and reaction rate balance.

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To determine the specific heat of a liquid, measure its initial and final temperatures after heating it using a known heat source and calculate the specific heat using the formula Q = mcΔT. Q is the heat gained or lost, m is the mass, c is the specific heat, and ΔT temperature change.

In ordr to determine the specific heat of a liquid using the equipment in the lab, one can heat a known mass of the liquid to a known temperature and then transfer it to a container of known mass and specific heat capacity. The temperature change of the container and liquid can be measured and used to calculate the specific heat of the liquid.

Alternatively, the liquid can be heated using a known power source, and the temperature change of the liquid can be measured to determine its specific heat.

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On the analysis of the first and second laws of thermodynamics, we can only conclude that the process satisfies the first law (energy conservation) but it cannot be determined whether it satisfies the second law (entropy).

To determine whether the given process is possible, we need to analyze whether it satisfies the first and second laws of thermodynamics.

First Law of Thermodynamics (Energy Conservation):

The first law states that energy is conserved in a closed system. For an adiabatic process (no heat transfer), the first law can be expressed as:

ΔQ = ΔU + ΔW

Where:

ΔQ is the heat transferred to the system (in this case, zero),

ΔU is the change in internal energy, and

ΔW is the work done on the system.

Since ΔQ = 0 for an adiabatic process, the first law simplifies to:

ΔU = -ΔW

To analyze whether the process satisfies the first law, we need to compare the change in internal energy (ΔU) with the work done (ΔW).

For an ideal gas, the change in internal energy can be expressed as:

ΔU = nCpΔT

Where:

n is the number of moles of gas,

Cp is the molar heat capacity at constant pressure, and

ΔT is the change in temperature.

For the given process, we have two equal flows, so n is the same for both flows.

For Flow 1:

ΔU₁ = nCp(70°C - 20°C)

For Flow 2:

ΔU₂ = nCp(-30°C - 20°C)

Since the gas is claimed to be separated into two equal flows, the change in internal energy for each flow should be equal in magnitude but opposite in sign.

Therefore, ΔU₁ = -ΔU₂

Now, let's analyze the work done on the system.

For an adiabatic process, the work done can be expressed as:

ΔW = C(ΔP/γ)

Where:

C is a constant,

ΔP is the change in pressure, and

γ is the adiabatic index (specific heat ratio) for the gas.

Since the process involves expansion, ΔP = P₃ - P₁ = 0.1 MPa - 0.8 MPa = -0.7 MPa (note the negative sign due to the expansion).

Now, let's compare ΔU and ΔW:

ΔU₁ = -ΔU₂

nCp(70°C - 20°C) = -nCp(-30°C - 20°C)

50°C = 50°C

The change in internal energy is equal and opposite for both flows, satisfying the first law of thermodynamics.

Second Law of Thermodynamics (Entropy):

The second law of thermodynamics states that in an isolated system, the total entropy of the system cannot decrease.

For an adiabatic process, the entropy change can be expressed as:

ΔS = nCp ln(T₂/T₁) + nCp ln(T₃/T₂)

Where:

ΔS is the change in entropy.

For the given process, let's calculate the entropy change:

ΔS = nCp ln(T₂/T₁) + nCp ln(T₃/T₂)

= nCp ln(70°C/20°C) + nCp ln((-30°C)/(70°C))

= nCp ln(3.5) + nCp ln(-0.42857)

The natural logarithm of a negative value is undefined, which means that ln(-0.42857) is not a valid calculation. Therefore, the entropy change cannot be determined for this process.

Since the entropy change cannot be determined or verified, it is not possible to conclude whether the given process satisfies the second law of thermodynamics.

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The electron configuration of a carbon anion having with -2 charge is 1s² 2s² 2p⁶.

The electron configuration for a neutral carbon atom is 1s² 2s² 2p².

When a carbon atom gains two extra electrons to form a -2 charge (carbon anion), these electrons will occupy the available orbitals in the order of increasing energy levels, following the Aufbau principle and Hund's rule.

To accommodate the additional two electrons, the electron configuration of the carbon anion (-2 charge) would be;

1s² 2s² 2p⁶

In this configuration, the 1s orbital is fully filled with two electrons, the 2s orbital is fully filled with two electrons, and all three 2p orbitals are fully filled with six electrons.

Hence, the electron configuration for a carbon anion with a -2 charge is 1s² 2s² 2p⁶.

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Answer:

Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements. All elements are pure substances.

Explanation:

PLEASE MARK ME AS BRAINLIEST

Answer:

Because it was based on mystical belief instead of the scientific method (which had not been codified for most of alchemy's existence). It is completely wrong, even if it stumbled on techniques which are still useful.

Explanation:

3 moles⋅22.4 L/1 mole=67.2 L Likewise, 0.5 moles of gas will occupy half the volume 1 mole occupies 0.5 moles⋅22.4 L/1 mole = 11.2 L

Standard Temperature and Pressure conditions imply a temperature of 273.15 K and a pressure of 1 atm. When these conditions are met, 1 mole of any ideal gas occupies exactly 22.4 L. Keep in mind that ideal gas particles are assumed to have no volume of their own, that's why, for example, at STP 1 mole of helium gas will occupy the same volume as, say, 1 mole of chlorine gas.

However, as you can see in the picture, the two gases will have different masses because of the difference in their molar mass. The balloon filled with helium will weigh less than the one filled with chlorine, despite the fact that they occupy the same volume.

So, if 1 mole occupies 22.4 L, the immediate conclusion is that a bigger number of moles will occupy more than 22.4 L, and a smaller number of moles will occupy less than 22.4 L.

In your case, 3 moles of gas will occupy 3 times more volume than 1 mole of gas.

3 moles⋅ 22.4 L/1 mole=67.2 L

Likewise, 0.5 moles will occupy half the volume 1 mole occupies

0.5 moles⋅ 22.4 L/1 mole=11.2 L

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The amount of dry NH₄HCl needed to prepare a buffer solution that has a pH of 8.56 is 188.29 grams.

To prepare the buffer solution with a pH of 8.56, first, we need to calculate the concentration of NH₄⁺ ions using the Henderson-Hasselbalch equation:

pH = pKa + log([NH₄⁺]/[NH₃])

Since we have pH and [NH₃], we need to find pKa using Kb for ammonia:

Kb = 1.8 × 10⁻⁵
Kw = 1 × 10⁻¹⁴ (ion product of water)

Ka = Kw / Kb = 1 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰
pKa = -log(Ka) = 9.26

Now, we can use the Henderson-Hasselbalch equation:

8.56 = 9.26 + log([NH₄⁺]/(0.800))

Rearranging to find [NH₄⁺]:

[NH₄⁺] = 0.800 * 10^(9.26 - 8.56) = 0.800 * 10^(0.70) = 1.6 M

Now, we can calculate the grams of NH₄Cl needed to achieve this concentration:

1.6 mol/L * 2.20 L = 3.52 mol of NH₄Cl

Finally, convert moles to grams using the molar mass of NH₄Cl (53.49 g/mol):

3.52 mol * 53.49 g/mol = 188.29 g

So, you need to add 188.29 grams of dry NH₄Cl to the 2.20 L of 0.800 M NH₃ solution to prepare the buffer with a pH of 8.56.

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Answer:

10-9 Millimeters/liters

Explanation:

Because 10-9 M Is more than 10-8 M

I hope this is correct