A 0.682-gram sample of an unknown weak monoprotic organic acid, HA, was dissolved in sufficient water to make 50.0 mL of solution and was titrated with a 0.135 M NaOH solution. After the addition of 10.6 mL of base, a pH of 5.65 was recorded. The equivalence point was reached after the addition of 27.4 mL of the 0.135 M NaOH.
a. Calculate the number of moles of acid in the original sample.
b. Calculate the molar mass of the organic acid.
c. Calculate the molarity of the unreacted HA remaining in the solution at pH = 5.65.

Answers

Answer 1

a. The number of moles of acid in the original sample is 0.00369. b. The molar mass of the organic acid is 0.135  M. c. The molarity of the unreacted HA remaining in the solution at pH 5.65 is 0.045 M

Calculation:

a. The equivalence point was reached after the addition of 27.4 mL of the 0.135 M NaOH.a.

Moles of NaOH = M × V = 0.135 M × 27.4 mL = 0.00369 moles

Using the balanced equation, we find that the number of moles of HA is equal to the number of moles of NaOH at the equivalence point. HA + NaOH → NaA + HOH0. 00369 moles of NaOH are needed to react with 0.00369 moles of HA.

b. Molar mass of HA = (mass of HA) / (number of moles of HA) = 0.682 g / 0.00369 moles = 184.7 g/molc. Calculate the molarity of the unreacted HA remaining in the solution at pH = 5.65.The pH of the solution was 5.65 after 10.6 mL of NaOH were added.

c. To calculate the molarity of the remaining HA, we first need to find the pKa of the acid.

pH = pKa + log([A-]/[HA])5.65 = pKa + log([A-]/[HA]). We know that at the equivalence point, [A-] = [HA] / 2.

Therefore,[A-] = 0.00369 moles / 2 = 0.00185 moles[Ligand] = (moles of ligand) / (liters of solution). We need to find [HA] in moles/L, so we need to find [A-] in moles/L. We can use the molarity of the NaOH solution to do this. [NaOH] = 0.135 M

moles of NaOH = [NaOH] × (liters of solution)moles of NaOH = 0.135 M × 0.0106 L.

moles of NaOH = 0.00144 moles

moles of HA at pH = 5.65 = moles of HA initially - moles of NaOH added = 0.00369 moles - 0.00144 moles

= 0.00225 moles[HA] = 0.00225 moles / 0.050 L = 0.045 M

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Related Questions

if a sample of a hydrate contains 0.02mol of anhydrous salt and 0.1mol of water, how many water molecules are present in one formula unit of the hydrate (ie. what is z in the formula )?

Answers

Answer : There are 5 water molecules per formula unit of the hydrate.

In order to calculate the number of water molecules in a hydrate, we first need to understand what a hydrate is. A hydrate is a compound that contains water molecules bound within its crystal structure. The water molecules are referred to as “water of hydration” and are typically present in a fixed ratio to the other molecules in the compound.

The formula for a hydrate can be written as: AxBy * zH2O, where x and y represent the number of ions in the anhydrous salt and z represents the number of water molecules per formula unit. In order to calculate z, we need to use the information provided in the question. The question tells us that we have 0.02 mol of anhydrous salt and 0.1 mol of water in the sample. we need to divide the number of moles of water by the number of moles of anhydrous salt.

0.1 mol of water / 0.02 mol of anhydrous salt = 5. This means that for every mole of anhydrous salt, there are 5 moles of water. Therefore, the formula for the hydrate can be written as: AxBy * 5H2O. This means that there are 5 water molecules per formula unit of the hydrate. Therefore, z is equal to 5.

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which of the combinations below will produce an insoluble salt? a) ba(oh)2 hcl b) mnso4 pb(no2)2 c) h2so4 albr3

Answers

The combination that will produce an insoluble salt is b) MnSO4 Pb(NO2)2.

A salt is a chemical compound made up of cations (positively charged ions) and anions (negatively charged ions) (negatively charged ions). The ions must be combined in such a way that the sum of the charges is zero. NaCl is the most well-known saltand it is made up of sodium cations (Na+) and chloride anions (Cl-).MnSO4 Pb(NO2)2 is the answer since both of these elements are soluble. MnSO4 is a soluble substance that is sometimes used in the production of ceramics.

MnSO4 is often used as a nutritional supplement for animals since it is a good source of manganese. Pb(NO2)2 is a powder that is bright yellow, it has a molar mass of 325.2 g/mol. It is made up of two NO2 anions (negatively charged ions) and one Pb2+ cation (positively charged ion).The formation of insoluble salts can occur when the cations and anions in a reaction solution bind to create a new solid. Since the newly formed solid is insoluble, it settles to the bottom of the solution and can be separated from the liquid through filtration. The insoluble salt that is formed is a white or colorless substance that appears as a powder.

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organic molecules are those that contain at least multiple choice carbon. carbon and oxygen. carbon and hydrogen. carbon, oxygen, and hydrogen.

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Organic molecules are those that contain carbon and often hydrogen atoms bonded together, and they are the building blocks of life.

Carbon is an element that is essential to life on Earth and is the central atom in organic compounds. It can form covalent bonds with other elements such as hydrogen, oxygen, nitrogen, and sulfur.

Carbon has the unique ability to form long chains of molecules, branched structures, and rings that are essential to the structure and function of organic molecules.

Organic molecules include carbohydrates, lipids, proteins, and nucleic acids. Carbohydrates are sugars and starches that provide energy to living organisms.

Lipids are fats and oils that are important for insulation and energy storage. Proteins are complex molecules that carry out many functions in the body, such as catalyzing chemical reactions and providing structure to cells.

Nucleic acids are DNA and RNA, which carry genetic information and are essential for the synthesis of proteins.

Oxygen is another element that is essential to life on Earth. It is often found in organic molecules, especially in carbohydrates and lipids.

Oxygen is important for respiration, the process by which living organisms use energy stored in organic molecules to carry out cellular processes.

In respiration, oxygen reacts with organic molecules such as glucose to produce carbon dioxide, water, and energy in the form of ATP.

Organic molecules contain carbon and often hydrogen atoms bonded together, and they are the building blocks of life.

Carbon has the unique ability to form long chains of molecules, branched structures, and rings that are essential to the structure and function of organic molecules.

Oxygen is another element that is often found in organic molecules and is important for respiration.

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the radioactive decay of c14 which is used in estimating the age of archaeological samples follows first order kinetics with a half-life of 5725 years at 300k. if a sample of c114 initially contains 0.0035 mol of c14, how many moles remain after 2500 years.

Answers

the radioactive decay of c14 which is used in estimating the age of archaeological  after 2500 years, 0.0027 mol of c14 remain in the sample.

The amount of c14 remaining after 2500 years can be calculated using the first-order rate equation:

N(t) = N0 * e^(-kt)

where N0 is the initial amount of c14, N(t) is the amount remaining after time t, k is the decay constant, and e is the base of the natural logarithm. The half-life of c14 is given as 5725 years, which means that k can be calculated as:

k = ln(2)/t1/2 = ln(2)/5725

Substituting the values given in the problem, we get:

k = ln(2)/5725 = 1.21 * 10^-4 /year

Now, we can use the rate equation to find the amount of c14 remaining after 2500 years:

N(2500) = 0.0035 * e^(-1.21*10^-4 * 2500) = 0.0027 mol

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which phase change will have a more dramatic increase in entropy? select the statement that best explains why.

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Answer: Phase change from solid to gas will have a more dramatic increase in entropy.

This is because gas has the highest entropy of all phases. Gas has the highest entropy because its molecules are moving randomly, and it has the greatest amount of disorder. In addition, the transition from solid to gas involves both increasing temperature and changing the arrangement of particles from an ordered solid to a disordered gas. This results in a significant increase in entropy.

Phase transition refers to the process of changing from one phase of matter to another. When a substance changes from one phase to another, its entropy changes. Entropy refers to the degree of disorder or randomness in a system, and it is related to the number of ways that a system can be arranged. When the degree of disorder increases, the entropy also increases.

In summary, phase change from solid to gas has a more dramatic increase in entropy. This is because gas has the highest entropy of all phases, and the transition from solid to gas involves both increasing temperature and changing the arrangement of particles from an ordered solid to a disordered gas, resulting in a significant increase in entropy.



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A pie can be cut into eight slices. What is the minimum number of pies you would need if you were to serve a slice of pie with each cup of hot chocolate in item 6? How many slices of pie would be left over?

Answers

(a) We would need 7 pies to serve a slice of pie with each cup of hot chocolate.

(b) There would be 6 slices of pie left over.

What is number of pies that will be left over?

From item 6, we know that there are 50 cups of hot chocolate to be served.

Since each pie can be cut into 8 slices, we would need to serve 50/8 = 6.25 pies.

Since we cannot serve a fractional pie, we would need to round up to the next whole number of pies, which is 7.

To find out how many slices of pie would be left over, we need to calculate the total number of slices of pie and subtract the number of slices used to serve the hot chocolate.

Total number of slices of pie = 7 pies x 8 slices per pie = 56 slices

Number of slices used to serve the hot chocolate = 50 slices

Therefore, the number of slices of pie left over would be:

56 slices - 50 slices = 6 slices

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Indicate if the following pairs of compounds could be separated via liquid-liquid extraction.First, draw the structures of the compounds, next determine whether they are acids or bases or neutral. Finally, look up their pKa (or pKb) values and indicate which aqueous solution would separate them or if they are inseparable. Assume that you can utilize aqueous HCl, NaOH, or NaHCO3 for your extractions. For each case that will not work, give the reason.You don't have to draw structure, just explain if they are able to be separated or not and with what and why.1. picric acid and phenol2. salicyclic acid and phenol3. triethylamine and diethylamine4. 3-nitrobenzoic acid and 2-nitrobenzoic acid5. benzylamine and aniline

Answers

Picric acid and phenol, Salicylic acid and phenol & Benzylamine and aniline can be separated using liquid-liquid extraction but Triethylamine and diethylamine & 3-nitrobenzoic acid and 2-nitrobenzoic acid cannot be separated using liquid-liquid extraction.

1. Picric acid and phenol can be separated using liquid-liquid extraction. Picric acid is a stronger acid (pKa ~0.4) than phenol (pKa ~10). Adding aqueous NaOH will deprotonate picric acid and make it soluble in the aqueous layer, while phenol remains in the organic layer. Then, the two compounds can be separated.
2. Salicylic acid and phenol can also be separated using liquid-liquid extraction. Salicylic acid (pKa ~3) is more acidic than phenol (pKa ~10). Adding aqueous NaHCO3 will deprotonate salicylic acid, making it soluble in the aqueous layer, while phenol remains in the organic layer. The compounds can then be separated.
3. Triethylamine and diethylamine cannot be easily separated via liquid-liquid extraction, as both are bases (pKb values are similar). Aqueous HCl, NaOH, or NaHCO3 will not be effective in separating these compounds. Alternative separation methods, like distillation, may be needed.
4. 3-nitrobenzoic acid and 2-nitrobenzoic acid cannot be separated using liquid-liquid extraction, as they have similar acidity (pKa values are close) and will react similarly with HCl, NaOH, or NaHCO3. Alternative separation methods, like chromatography, should be considered.
5. Benzylamine and aniline can be separated using liquid-liquid extraction. Benzylamine is a weaker base (pKb ~4.2) than aniline (pKb ~9.4). Adding aqueous HCl will protonate aniline, making it soluble in the aqueous layer, while benzylamine remains in the organic layer. The two compounds can then be separated.

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Dark waters movie
What is the significance of the call from the Kigers?

Answers

Answer: In the movie Dark Waters, the call from the Kigers is significant because it leads to the discovery of a link between unexplained cattle deaths and pollution caused by the chemical company DuPont.

Explanation: In the movie Dark Waters, the call from the Kigers is the key moment that sets off the plot. The Kigers, who are farmers in West Virginia, call Robert Bilott, a corporate defense attorney, and ask for his help in investigating the strange deaths of their cattle. Bilott is reluctant to take on the case at first, but he eventually agrees to visit the Kigers' farm and see the situation for himself.

During his visit, Bilott discovers that the Kigers are just one of many families in the area who have experienced unexplained deaths and illnesses among their livestock, as well as health problems among their own family members. Bilott begins to suspect that the cause of these health issues is pollution from a nearby chemical plant owned by DuPont, a multinational chemical company.

Bilott takes on the case and begins a long and difficult legal battle against DuPont, uncovering evidence that the company had long known about the dangers of the chemicals it was using - specifically a substance called PFOA, which was used in the production of Teflon - but had covered up the evidence and misled regulators and the public about the risks.

In the end, the call from the Kigers is significant because it leads to the discovery of a link between unexplained cattle deaths and pollution caused by DuPont, and sets off a series of events that ultimately lead to the exposure of corporate wrongdoing and the pursuit of justice for those affected by the pollution. The Kigers' call is a catalyst for change, prompting Bilott to take action and exposing the truth about a powerful and deceitful corporation.

Hope this helps, and have a great day!

potassium nitrate is used for a variety of applications, including fertilizer, rocket fuel, and fireworks. how many formula units of potassium nitrate are in a 25 g sample?

Answers

There are 1.49 × 10²³ formula units of potassium nitrate in a 25 g sample.

One formula unit is defined as the simplest formula of a substance, which indicates the relative amounts of the elements in the molecule. As a result, the number of formula units in a sample can be calculated by dividing the sample's mass by the substance's molar mass.

The molecular formula of potassium nitrate is KNO3. It contains one potassium atom (K), one nitrogen atom (N), and three oxygen atoms (O). The atomic masses of the elements can be used to calculate the molar mass of the compound.

One potassium atom has a molar mass of 39.1 g/mol, one nitrogen atom has a molar mass of 14.0 g/mol, and three oxygen atoms have a combined molar mass of 48.0 g/mol.

The molar mass of KNO3 = (1 × 39.1 g/mol) + (1 × 14.0 g/mol) + (3 × 16.0 g/mol) = 101.1 g/mol.

Now, on dividing the sample's mass (25 g) by the molar mass of potassium nitrate (101.1 g/mol), a value of 0.247 mol is obtained. The Avogadro constant can be used to convert moles into formula units. The Avogadro constant, 6.022 × 10²³ formula units per mole, represents the number of formula units in one mole of a substance.

The number of formula units = (0.247 mol) × (6.022 × 10²³ formula units/mol) = 1.49 × 10²³ formula units.

Therefore, there are 1.49 × 10²³ formula units of potassium nitrate in a 25 g sample.

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if the rate constant for a reaction triples when the temperature rises from 25 oc to 65 oc, what is the activation energy of the reaction? give answer in kj/mole.

Answers

The activation energy of the reaction, given that the rate constant has tripled when the temperature rose from 25 °C to 65 °C, is 42.6 kJ/mole.


Activation energy is the minimum energy required for a reaction to take place. It is calculated using the Arrhenius equation, which states that the rate constant, k, is proportional to the exponential of negative activation energy (Ea) divided by the gas constant (R) multiplied by the absolute temperature (T).

As the rate constant has tripled when the temperature increased, the activation energy can be calculated as Ea = -R * (1/T2 - 1/T1).

Plugging in the given temperature values of 25 °C and 65 °C and the gas constant, R, the activation energy is 42.6 kJ/mole.

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calculate the molarity of a solution prepared by mixing 100.0 ml of the solution made in number 3 with 900.0 ml of 0.0250 m nacl.

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The molarity of the solution prepared by mixing 100.0 ml of the solution made in number 3 with 900.0 ml of 0.0250 m NaCl is 0.1225 M.

We first calculate the moles of NaCl present in 900.0 ml of 0.0250 m NaCl solution.The formula to calculate the moles of solute is given as:

Moles of solute = molarity x volume (in liters)

So, the moles of NaCl in 900.0 ml of 0.0250 m NaCl solution would be:

Moles of NaCl = 0.0250 x (900.0/1000) = 0.0225 mol

Calculate the total volume of the mixed solution.The total volume of the mixed solution would be the sum of the volumes of the two solutions used in the mixing process.Total volume of mixed solution = 100.0 ml + 900.0 ml = 1000.0 ml or 1.0 L

Calculate the total number of moles of NaCl in the mixed solution.Total moles of NaCl in the mixed solution = moles of NaCl in 900.0 ml of 0.0250 m NaCl solution + moles of NaCl in 100.0 ml of the solution made in number 3

Total moles of NaCl in the mixed solution = 0.0225 mol + 0.100 mol = 0.1225 mol

Calculate the molarity of the mixed solution.The molarity of the mixed solution would be the number of moles of solute present in the solution per liter of solution.

Molarity of the mixed solution = Total moles of NaCl in the mixed solution / Total volume of the mixed solution

Molarity of the mixed solution = 0.1225 mol / 1.0 L = 0.1225 M

Therefore, the molarity of the solution prepared by mixing 100.0 ml of the solution made in number 3 with 900.0 ml of 0.0250 m NaCl is 0.1225 M.

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A mixture of 90. 0grams of ch4 and 10. Ograms of argon has a pressure of 250 torr under the condition of constant temperature and pressure the partial pressure of ch4 is

Answers

The partial pressure of CH₄ in the mixture is 239 torr.

We can use the mole fraction of methane (CH4) to calculate its partial pressure in the mixture. First, we need to convert the masses of each component into moles:

moles of CH₄ = 90.0 g / 16.04 g/mol = 5.61 mol

moles of Ar = 10.0 g / 39.95 g/mol = 0.250 mol

Next, we can calculate the total moles of gas in the mixture,

total moles = moles of CH₄ + moles of Ar = 5.61 mol + 0.250 mol = 5.86 mol

Now we can calculate the mole fraction of CH₄,

mole fraction of CH₄ = moles of CH₄ / total moles = 5.61 mol / 5.86 mol = 0.957

Finally, we can use the mole fraction and total pressure to calculate the partial pressure of CH₄,

partial pressure of CH₄ = mole fraction of CH₄ x total pressure = 0.957 x 250 torr = 239 torr

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a solution of cacl2 cacl 2 in water forms a mixture that is 31.5% 31.5 % calcium chloride by mass. if the total mass of the mixture is 195.4 g, 195.4 g, what masses of cacl2 cacl 2 and water were used?

Answers

The masses of calcium chloride (CaCl2) and water used to form the mixture are 61.18 g and 134.22 g, respectively.

The mass of calcium chloride (CaCl2):
The percentage of calcium chloride (CaCl2) in the mixture is 31.5%.


Multiply the total mass of the mixture (195.4 g) by 31.5% to find the mass of calcium chloride (CaCl2) in the mixture:
Mass of calcium chloride (CaCl2) = (195.4 g) x (31.5%) = 61.18 g

The mass of water:
Subtract the mass of calcium chloride (CaCl2) from the total mass of the mixture (195.4 g) to find the mass of water in the mixture:


Mass of water = (195.4 g) - (61.18 g) = 134.22 g

Therefore, masses of calcium chloride (CaCl2) and water used to form the mixture are 61.18 g and 134.22 g, respectively.

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a sample of xe takes 75 seconds to effuse out of a container. an unknown gas takes 37 seconds to effuse out of the identical container under identical conditions. what is the most likely identity of the unknown gas?

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The most likely identity of the unknown gas that effuses taking 37s is Oxygen(O₂).


Since the unknown gas effuses out faster, it must be lighter than Xe.

The most likely identity of the unknown gas can be determined using Graham's Law of Diffusion. According to this, the time taken for effusion/diffusion of two different gases under identical conditions is directly proportional to the square roots of their densities or molecular masses. It is given as:

t₂/t₁ = √(M₂/M₁)

where t₂,t₁ are the times taken and M₂, M₁ are the molecular masses.

This ratio is determined by the ratio of the molecular weights of the unknown gas and the sample of Xe. The heavier the molecular weight, the slower the rate of effusion.



Rearranging and plugging in the values as t₂= 75s, t₁= 37s,  M₁= 131g (for Xe), we get M₂ as follows:

M₂= (37/75)² x 131 = 31.8 ≈ 32g

32g corresponds to the molecular weight of O₂ and it is lighter than Xe.

Therefore, the unknown gas that effuses out of the container faster than the sample of Xe, resulting in the unknown gas taking 37 seconds, and the sample of Xe taking 75 seconds is oxygen(O₂).

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why is it important not to dilute the initial sample befoe it has been loaded onto the chromatography column

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It is important not to dilute the initial sample before loading it onto the chromatography column because this can negatively impact the separation and resolution of the components in the sample.

Dilution can lead to a decrease in the concentration of the components in the sample, which can result in poor separation and overlap of the peaks. Additionally, dilution can cause loss of the target compound or impurities in the sample due to adsorption onto the walls of the container used for dilution.

By keeping the sample concentrated and loading it directly onto the chromatography column, the chances of obtaining a clear separation and good resolution of the components in the sample are increased

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A gas sample at constant pressure and temperature filled with Helium gas had a volume of 221 mL and 4.00 moles. If the volume is increased to 500 ml what is the number of moles of Helium gas that could occupy the container? 0.11 K 9.05 kPa 0.11 kPa 9.05 mol​

Answers

The number of moles of Helium gas that could occupy the container when the volume is increased to 500 mL is 9.05 mol.

What is the number of moles of the gas?

We can use the combined gas law to solve this problem:

(P1 x V1) / (n1 x T1) = (P2 x V2) / (n2 xT2)

where;

P is pressure, V is volume, n is number of moles, and T is temperature.

We know that the pressure and temperature are constant, so we can simplify the equation to:

V1/n1 = V2/n2

Solving for n2, we get:

n2 = (V2n1) / V1

Plugging in the values, we get:

n2 = (500 mL * 4.00 mol) / 221 mL

n2 = 9.05 mol

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write an equation for each acid or base showing its ionization in water, and write the equilibrium constant expression for the weak acid or base

Answers

The equation for the ionization of a weak acid in water is HA + H₂O ⇌ H₃O⁺ + A⁻, and the equilibrium constant expression for this reaction is K = [H₃O⁺ ][A⁻]/[HA].

The ionization of a weak base in water is B + H₂O ⇌ OH⁻ + BH+, and the equilibrium constant expression for this reaction is K = [OH⁻][BH⁺]/[B].

Weak acids and bases partially dissociate into their ions in aqueous solutions. For a weak acid, HA, the equilibrium expression for its ionization is HA + H₂O ⇌  H₃O⁺  + A⁻, and the corresponding equilibrium constant expression is K = [ H₃O⁺ ][A-]/[HA].

The same process happens with a weak base, B, where the equilibrium expression is B + H₂O ⇌ OH⁻ + BH⁺, and the corresponding equilibrium constant expression is K = [OH⁻][BH⁺]/[B]. Thus, the equations for the ionization of both weak acids and bases and the corresponding equilibrium constant expressions can be

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onsider a process in which an ideal gas is compressed to one-fourth of its original volume at a constant temperature. calculate the entropy change per mole of gas.

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The entropy change per mole of gas is -1.387R.

The entropy change per mole of gas in a process in which an ideal gas is compressed to one-fourth of its original volume at a constant temperature can be calculated as follows:

Let us denote the original volume as V₁, the final volume as V₂, and the number of moles of the gas as n. The entropy change can be calculated using the formula:

ΔS = nR ln (V₂/V₁)

Therefore, the entropy change per mole of gas is given by:

ΔSper mole = R ln (V₂/V₁)


In this case, V₁ = 4V₂ and so,

ΔSper mole = R ln (1/4) = - R ln 4 = -2.303 R log 4 = -1.387R

Thus, the entropy change per mole of gas when an ideal gas is compressed to one-fourth of its original volume at a constant temperature is -1.387R.

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question every atom in the universe emits energy in the form of a nucleus. responses true true false

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The given statement "every atom in the universe emits energy in the form of a nucleus" is False.

In the universe, every atom does not emit energy in the form of a nucleus. It is not true in the case of every atom in the universe. But it is true that every atom in the universe emits energy.

According to the Bohr model of the atom, an electron orbiting an atomic nucleus emits radiation when it changes its energy level. The radiation emitted by the electron is in the form of a photon of electromagnetic energy. This is a spontaneous process and it is called spontaneous emission. It can be said that every atom in the universe emits energy.

Therefore, it is false that every atom in the universe emits energy in the form of a nucleus.

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the identity of an unknown monoprotic organic acid is determined by titration. a 0.173 g sample of the acid is titrated with 0.157 m naoh. what is the molar mass of the compound if 6.12 ml of the naoh solution is required to neutralize the sample?

Answers

The molar mass of the unknown monoprotic organic acid is 180.0 g/mol. by titration. If 6.12 ml of the naoH solution is required to neutralize the sample.

In order to determine the molar mass of the unknown monoprotic organic acid, follow the steps given below:

Step 1:

Calculate the number of moles of NaOH used in the titration by using the formula given below:

n(NaOH) = M(NaOH) × V(NaOH)

= 0.157 mol/L × 0.00612 L

= 9.62 × 10^-4 mol

Step 2:

Calculate the number of moles of the acid used in the titration by using the formula given below:

n(acid) = n(NaOH)

= 9.62 × 10^-4 mol

Step 3:

Calculate the mass of the acid used in the titration by using the formula given below:

mass(acid) = n(acid) × M(acid) = 0.173 gM(acid) = mass(acid) / n(acid)

= 0.173 g / 9.62 × 10^-4 mol

= 180.0 g/mol

Therefore, the molar mass of the unknown monoprotic organic acid is 180.0 g/mol.

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raising solvent temperature causes solvent-solute collisions to become group of answer choices more frequent and more energetic. less frequent and less energetic. less frequent and more energetic. more frequent and less energetic.

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When raising solvent temperature, solvent-solute collisions become more frequent and more energetic.

In chemistry, a solvent is a substance capable of dissolving another substance, usually a solid, liquid, or gas, to produce a homogeneous solution (mixture). The most common solvent is water, although there are other solvents that are widely used in many different industries. In a solvent, a solute is a substance that dissolves. It is usually a solid, but it can also be a liquid or a gas.

When a solute dissolves in a solvent, it forms a homogeneous solution.The solute will dissolve in the solvent when they collide. If the solute is in the solid-state, a solvent-solute collision may only occur if the solute dissolves in the solvent. The rate and frequency of solvent-solute collisions are impacted by a variety of factors, including solvent temperature. When solvent temperature is increased, the kinetic energy of solvent molecules is also increased, resulting in more frequent and energetic collisions.

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Why do you think only two drops of phenolphthalein are used in these titrations? (Hint: Phenolphthalein is a weak acid.)

Answers

Phenolphthalein is a commonly used indicator in acid-base titrations because it changes color at a pH around 8.2-10.0.

Phenolphthalein itself is a weak acid and has a specific equilibrium between its acidic and basic forms. When added to an acidic solution, it is predominantly in the acidic form and colorless. As the titration progresses and the solution becomes more basic, the equilibrium shifts towards the basic form which is pink.

The amount of indicator used in the titration should be kept to a minimum to avoid affecting the accuracy of the results. Using too much indicator can affect the stoichiometry of the reaction, leading to inaccurate results.

Therefore, only a small amount of phenolphthalein, typically two drops, is used to minimize its impact on the titration while still providing a clear visual indication of the endpoint.

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in which scenario would we be unable to treat the first and second ionization of a diprotic acid as completely separate reactions?

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The first and second ionization of a diprotic acid cannot be treated as completely separate reactions when the reaction is taking place in an environment with a fixed pH.

The second ionization of the acid is dependent on the concentration of the ions produced from the first ionization.

If the pH is fixed, then the concentration of the first ionization is also fixed, so the second ionization will not occur completely independently.

For example, a diprotic acid such as oxalic acid can be completely ionized in two steps. In the first ionization, the hydrogen ions of the oxalic acid are replaced with hydroxide ions, forming the oxalate ion:

H2C2O4 + 2H2O → H3O+ + HC2O4–

In the second ionization, the oxalate ion is further dissociated, forming two separate anions and hydronium ions:

HC2O4– + H2O → H3O+ + C2O4–2

However, in an environment with a fixed pH, the second ionization will not take place as the concentration of oxalate ions from the first ionization is fixed.

Therefore, the two ionizations must be treated together in order to accurately predict the final concentrations of the products.

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Answer:

The first ionization constant is greater than the second ionization constant by only a factor of 10.

Explanation:

The two ionization constants must differ by a factor of at least 20 in order to treat the first and second ionizations as chemically (and mathematically) distinct.

If I have 6.00 moles of gas held at a temperature of 93.5 C and in a container with a volume of 41.7 liters, what is the pressure of the gas (ka)?

Answers

The pressure of the gas is approximately 4.57 atm or 438.629 kPa

What is the pressure of the gas (ka)?

The Ideal gas law or general gas equation states that "the pressure multiplied by volume is equal to moles multiply by the universal gas constant multiply by temperature.

It is expressed as;

PV = nRT

Where P is pressure, V is volume, n is the amount of substance, T is temperature and R is the ideal gas constant ( 0.08206 Latm/molK )

Given that;

P = pressure of the gas (in atm) = ?V = volume of the gas (in L) = 41.7 Ln = number of moles of gas = 6R = the ideal gas constant (0.08206 L.atm/mol.K)T = temperature of the gas (in Kelvin) 93.5°C

First, we need to convert the temperature to Kelvin:

T (K) = T (Celsius) + 273.15

T (K) = 93.5 + 273.15

T (K) = 366.65 K

Now we can substitute the given values into the formula:

PV = nRT

P = nRT / V

P = ( 6 × 0.08206 × 366.65 ) / 41.7

P = 4.33 atm

Convert to kPa by multiplying the pressure value by 101.3

P = ( 4.33 × 101.3 ) kPa

P = ( 4.33 × 101.3 ) kPa

P = 438.629 kPa

The pressure is approximately 4.57 atm or 438.629 kPa.

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cobalt(ii) chloride is dissolved in ethanol, and then water is added. what is the co(ii) complex equilibrium reaction? equilibrium reaction:

Answers

The equilibrium reaction for the formation of cobalt(II) complex when cobalt(II) chloride is dissolved in ethanol and then water is added is given by the following equation:

CoC₂l + 4 ethanol → Co(C₂H₅OH)₄Cl₂


When the cobalt(II) chloride is dissolved in ethanol, a cobalt(II) complex is formed. The complex is a tetrahedral molecule with four ethanol molecules attached to the cobalt ion. When water is added, it causes the equilibrium reaction to shifting to the right, with more of the cobalt(II) complex being formed. This is because the water molecules can displace the ethanol molecules from the complex, allowing the complex to form. The reaction can be expressed as:

CoC₂H₅OH)₄Cl₂ + 4 H₂O ↔ Co(H₂O)₄Cl₂ + 4 C₂H₅OH



In conclusion, the equilibrium reaction for the formation of cobalt(II) complex when cobalt(II) chloride is dissolved in ethanol and then water is added can be given as:

CoCl₂ + 4 ethanol → Co(C₂H₅OH)₄Cl₂ + 4 H₂O ↔ Co(H₂O)₄Cl₂ + 4 C₂H₅OH.

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A fluorinated organic gas in a cylinder is com- pressed from an initial volume of 910 mL at 156 Pa to 490 mL at the same temperature. What is the final pressure?
Answer in units of Pa.

Answers

The problem can be solved using Boyle's Law. The final pressure of the gas in the cylinder is 289.31 Pa.

What is Boyle's Law?

Boyle's law is a gas law that describes the relationship between the pressure and volume of a gas at a constant temperature. Boyle's Law states that the pressure and volume of a gas are inversely proportional when temperature is held constant. Mathematically, it 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.

We can plug in the given values to solve for the final pressure:

P₁ = 156 Pa

V₁ = 910 mL = 0.91 L

V₂ = 490 mL = 0.49 L

P₁V₁ = P₂V₂

156 Pa × 0.91 L = P₂ × 0.49 L

P₂ = (156 Pa × 0.91 L) / 0.49 L

P₂ = 289.31 Pa

Therefore, the final pressure is 289.31 Pa.

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Calculate the pH of a 0.050 M solution of hydroxylamine, NH2OH. Kb = 6.6 x 10^-9

Answers

The pH of hydroxylamine will be 8.76.

The first step is to write the balanced equation for the reaction of hydroxylamine with water:

NH₂OH + H₂O ⇌ NH₃OH⁺ + OH⁻

The Kb expression for this reaction is:

Kb = [NH₃OH⁺][OH⁻] / [NH₂OH]

We are given the Kb value as 6.6 x 10⁻⁹, so we can use this to find the concentration of hydroxylamine that has been deprotonated:

Kb = [NH₃OH⁺][OH⁻] / [NH₂OH]

6.6 x 10⁻⁹ = x² / (0.050 - x)

Assuming that x is very small compared to 0.050, we can simplify the expression as follows:

6.6 x 10⁻⁹ = x² / 0.050

x² = 3.3 x 10⁻¹⁰

x = 5.7 x 10⁻⁶ M

Now that we have the concentration of hydroxide ions, we can use this to find the pH of the solution:

pOH = -log[OH-] = -log(5.7 x 10⁻⁶) = 5.24

pH = 14.00 - pOH = 8.76

Therefore, the pH of a 0.050 M solution of hydroxylamine is 8.76.

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How many moles are in 1.2 x 10^24 formula units of Li₂SO4? (round your answer to the nearest tenths place)

Answers

In 1.2 x [tex]10^{24}[/tex] formula units of [tex]Li_{2} (SO)_{4}[/tex], there are roughly 1.993 moles of

[tex]Li_{2} (SO)_{4}[/tex].

How many moles of  [tex]Li_{2} (SO)_{4}[/tex] are contained in 1.2 x [tex]10^{24}[/tex] formula units?

Using Avogadro's number, or 6.022 x [tex]10^{23}[/tex] molecules/mol, we can calculate the number of moles of Li2SO4 in 1.2 x [tex]10^{24}[/tex]formula units.

First, we need to figure out how many moles of [tex]Li_{2} (SO)_{4}[/tex]  are needed to equal 1.2 x [tex]10^{24}[/tex]  formula units:

Formula units equal 6.022 x [tex]10^{23}[/tex] per mole of [tex]Li_{2}(SO)_{4}[/tex].

As a result, there are: 1.2 x [tex]10^{24}[/tex] moles of [tex]Li_{2}(SO)_{4}[/tex] in the formula units.

1.993 moles are equal to 1.2 x [tex]10^{24}[/tex] formula units / 6.022 x [tex]10^{23}[/tex] formula units/mol.

Hence, 1.2 × [tex]10^{24}[/tex] formula units of [tex]Li_{2} (SO)_{4}[/tex] contain about 1.993 moles.

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How many moles of glucose C6H12O6 can react with 15.7 moles of oxygen? C6H12O6 + 6O2 -----------> 6CO2 + 6H2O

Answers

2.62 moles of glucose can react with 15.7 moles of oxygen. The balanced chemical equation for the combustion of glucose is:

C6H12O6 + 6O2 → 6CO2 + 6H2O

From the equation, we can see that for every mole of glucose that reacts, 6 moles of oxygen are required. Therefore, the number of moles of glucose that can react with 15.7 moles of oxygen can be calculated as follows:

Number of moles of glucose = (Number of moles of oxygen) / 6

Number of moles of glucose = 15.7 / 6

Number of moles of glucose = 2.62

Therefore, 2.62 moles of glucose can react with 15.7 moles of oxygen.

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4. a laboratory experiment calls for 0.150 m hno3. what volume of 0.150 m hno3 can be prepared form 0.350 l of 1.98 m hno3?

Answers

The volume of 0.150 M HNO3 that can be prepared from 0.350 L of 1.98 M HNO3 is 0.07112 L, or approximately 71.12 mL (since 1 L = 1000 mL).

The given equation is used to calculate the volume (V1) of a desired concentration of a solution (0.150 M HNO3) that can be prepared from a given volume (V2) of a known concentration solution (1.98 M HNO3), using the ratios of their concentrations (C1 and C2).

Let's break down the calculation step by step using the given values:

V2 (given volume) = 0.350 L

C1 (desired concentration) = 0.150 M

C2 (known concentration) = 1.98 M

Plugging these values into the equation, we get:

V1 (0.150 M HNO3) = V2 (1.98 M HNO3) x (C1 (0.150 M) / C2 (1.98 M))

V1 = 0.350 L x (0.150 M / 1.98 M)

V1 = 0.350 L x 0.0758

V1 = 0.07112 L

Therefore, the volume of 0.150 M HNO3 that can be prepared from 0.350 L of 1.98 M HNO3 is 0.07112 L, or approximately 71.12 mL (since 1 L = 1000 mL).

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