31.40cm³ of 0.15moldm^-3 hydrochloric acid, HCl, is neutralised by 20.0cm³ of potassium hydroxide solution, KOH. Calculate molarity of potassium hydroxide.

The mass fraction of HCl in the final solution is 0.294%.

We need to determine the mass of the solution after mixing the two components.

First, we need to calculate the volume of the 10 g of pure water added to the 25 mL of 9.87 M HCl solution:

Density of the HCl solution = 1.179 g/mL

Volume of the HCl solution = 25 mL = 0.025 L

Mass of the HCl solution = Density x Volume = 1.179 g/mL x 0.025 L = 0.0295 g

Volume of water added = 10 g / (1 g/mL) = 10 mL = 0.01 L

Total volume of the final solution = 25 mL + 10 mL = 35 mL = 0.035 L

To calculate the concentration of the final solution, we can use the following equation:

M1V1 = M2V2

Where

Rearranging the equation to solve for M2, we get :

M2 = (M1V1) / V2

Substituting the values we have, we get:

M2 = (9.87 mol/L x 0.025 L) / 0.035 L = 7.06 mol/L

Now, we can calculate the mass of the final solution:

Mass of the HCl solution = 0.0295 g

Mass of water added = 10 g

Total mass of the final solution = 0.0295 g + 10 g = 10.0295 g

Finally, we can calculate the mass fraction of the HCl in the final solution:

Mass fraction of HCl = (mass of HCl) / (total mass of solution) x 100%

Mass fraction of HCl = (0.0295 g / 10.0295 g) x 100% = 0.294%

Therefore, the mass fraction of HCl in the final solution is 0.294%.

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Answer:

aluminium reacts with sulfuric acid to produce aluminium sulfate and hydrogen gas according to the following equation: H2SO4 (aq) + Al (s) = Al2 (SO4)3 (aq) + H2 (g).

Explanation:

The titration of acetic acid with KOH is a weak acid-strong base titration. At the beginning of the titration (part a), we have only acetic acid in the solution, and its concentration is 0.200 M. As we add KOH, it reacts with acetic acid to form acetate and water:

CH3COOH + KOH → CH3COOK + H2O

The acetate ion is the conjugate base of acetic acid and can be considered a weak base. We can use the following equation to calculate the pH of the resulting solution at each point of the titration:

pH = pKa + log([A^-]/[HA])

where pKa is the acid dissociation constant of acetic acid (1.8 × 10^-5), [A^-] is the concentration of acetate ion, and [HA] is the concentration of undissociated acetic acid.

a) At the beginning of the titration (0.0 mL of KOH added), the solution contains only acetic acid. Therefore, [HA] = 0.200 M and [A^-] = 0 M.

pH = pKa + log([A^-]/[HA])

pH = -log(1.8 × 10^-5) + log(0/0.200)

pH = 2.40

The pH of the solution is 2.40.

b) When 50.0 mL of 0.100 M KOH is added, we have added 5.00 mmol of KOH. This amount of KOH reacts with 5.00 mmol of acetic acid, and the remaining 0.050 mol - 0.005 mol = 0.045 mol of acetic acid remains in the solution. At the same time, 0.005 mol of acetate ion is formed.

[HA] = 0.045 mol / 0.100 L = 0.450 M

[A^-] = 0.005 mol / 0.100 L = 0.050 M

pH = pKa + log([A^-]/[HA])

pH = -log(1.8 × 10^-5) + log(0.050/0.450)

pH = 4.41

The pH of the solution is 4.41.

c) When 100.0 mL of 0.100 M KOH is added, we have added 10.00 mmol of KOH. This amount of KOH reacts with 10.00 mmol of acetic acid, and there is no acetic acid remaining in the solution. At the same time, 0.010 mol of acetate ion is formed.

[HA] = 0 mol / 0.100 L = 0 M

[A^-] = 0.010 mol / 0.100 L = 0.100 M

pH = pKa + log([A^-]/[HA])

pH = -log(1.8 × 10^-5) + log(0.100/0)

pH = 4.74

The pH of the solution is 4.74.

d) When 110.0 mL of 0.100 M KOH is added, we have added 11.00 mmol of KOH. This amount of KOH reacts with 10.00 mmol of acetic acid, and there is an excess of 1.00 mmol of KOH in the solution. This excess KOH completely dissociates to give 1.00 mmol of OH^- ion. At

Explanation:

That the mass can neither be created nor destroyed in a chemical reaction.

Yes, carbon dioxide (CO₂) is involved in environmental problems such as the Greenhouse Effect. However, CO₂ is not directly linked to the Ozone Hole, Acid Rain, Air Pollution, or Water Pollution.

Carbon dioxide is a greenhouse gas that naturally occurs in the Earth's atmosphere, but human activities such as burning fossil fuels (coal, oil, and gas), deforestation, and land use changes have significantly increased its concentration, trapping more heat in the atmosphere and leading to global warming.

To curb the amount of excess carbon dioxide, individuals and governments can take several actions, such as: Reducing energy consumption by using energy-efficient appliances, turning off lights and electronics when not in use, and using public transportation, walking or cycling instead of driving, Switching to clean, renewable energy sources such as solar, wind, and hydropower to reduce the use of fossil fuels, Promoting afforestation and reforestation to absorb carbon dioxide from the atmosphere and restore natural ecosystems, Encouraging sustainable agriculture practices that reduce carbon emissions from livestock and soil.

By taking such actions, we can curb the amount of excess carbon dioxide and mitigate the impact of global warming and climate change.

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At 95.7 °C, the flask's partial pressure of air is 2.741 atm.

One of two methods can be used to compute partial pressures: 1) Use PV = nRT to calculate the individual pressure of each gas in a mixture. 2) Determine the proportion of pressure from the total pressure that may be assigned to each individual gas by using the mole fraction of each gas.

PV = nRT

n = PV/RT

where P = 0.9150 atm, V = 250 mL = 0.250 L, T = 21.1 °C + 273.15 = 294.25 K, and R = 0.08206 L atm/mol K.

n = (0.9150 atm)(0.250 L)/(0.08206 L atm/mol K)(294.25 K) = 0.0111 mol

n = PV/RT

where P = 0.9150 atm, V = 5 mL = 0.005 L, T = 21.1 °C + 273.15 = 294.25 K, and R = 0.08206 L atm/mol K.

n = (0.9150 atm)(0.005 L)/(0.08206 L atm/mol K)(294.25 K) = 0.000173 mol

[tex]n_total = n_air + n_ethanol = 0.0111 mol + 0.000173 mol = 0.01127 mol[/tex]

[tex]P_air = X_air * P_total[/tex]

[tex]X_air = n_air / n_total = 0.0111 mol / 0.01127 mol = 0.983[/tex]

[tex]P_air = X_air * P_total = 0.983 * 2.791 atm = 2.741 atm[/tex]

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Both the initial heat transfer when the piston initially contacts the stops and the overall heat transfer throughout the procedure are -11,172 kJ.

h1 = 3279.1 kJ/kg

h2 = 751.6 kJ/kg

Q = ΔU = m(u2 - u1) (u2 - u1)

where m is the steam's mass.

We may use the ideal gas law to determine the mass of the steam:

PV = mRT

PV/(RT) = (3.5 MPa) (0.33 m3)/(0.287 kJ/kg-K) (380.4 K) = 4.47 kg for the formula m.

We can now determine the rate of heat transfer:

Q is equal to m(u2 - u1) = (4.47 kg)(751.6 kJ/kg - 3279.1 kJ/kg) = -11,172 kJ

Heat is leaving the system, as indicated by the negative sign.

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A sample of ammonia gas has a final temperature of 313.25K.

The Combined Gas Law is a combination of the Ideal Gas Law and Charles’s Law. It states that the ratio of the pressure of a gas, its volume, and its temperature is constant, provided the amount of gas and the molar mass of the gas remain the same. The combined gas law states that the ratio of the product of pressure and volume of a gas is equal to the product of the pressure and volume of the same gas at a different temperature.

Rearranging the equation to solve for T2, we have:

[tex]T2 = \frac{(P1V1T1)}{P2V2}[/tex]

Substituting the given values, P1 has value 3.84 atm,V1 has value 250mL,P2 has value 5.84 atm,V2 has value 215mL and T1 has value (35+273) degrees, we have:

[tex]T2 = \frac{(3.84 * 250 * 35 + 273) }{ (5.84 * 215)}\\T2 = 313.25 K[/tex]

Therefore,The final temperature of a sample of ammonia gas is 313.25K

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Answer:More purple

Explanation: Its more purple

Copper ions  and acetate ions make up the chemical known as solid copper (II) acetate . It will dissolve and split into its component ions when placed in water. Moreover, copper (II) acetate is a strong electrolyte, which means that when dissolved in water, it almost entirely separates into ions.

Pb(OAc) 23H₂O, a colorless or white efflorescent monoclinic crystalline material, is the trihydrate that is created when it reacts with water.

Cu₂(OAc)4(H₂O) is a more bluish-green crystalline solid than anhydrous copper(II) acetate. Copper acetates have been utilized as fungicides and green pigments in some way since antiquity. Copper acetates are currently utilized as reagents for  the synthesis of various inorganic and organic compounds.

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Selection is the process by which we filter out irrelevant information and focus on the information that is most relevant to us.

For example, when we walk down a busy street, we may not pay attention to every detail of every building or person we pass by. Instead, we might focus on things that stand out, such as a bright billboard or an interesting conversation nearby.

Regarding stereotypes, one common example is the belief that all individuals from a certain race, gender, or religion share certain characteristics or behaviors. This stereotype is often based on limited or inaccurate information and can lead to biased judgments and discrimination. It's essential to recognize and challenge our own stereotypes to avoid unfairly treating others based on their group membership.

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The number of moles of acetic acid which has a molar mass of 60.05 g/mol present in 16.84 grams of acetic acid is 3.56.

Given the molar mass of Acetic acid (M) = 60.05 g/mol.

The mass of acetic acid (m) = 16.84g

Let the number of moles = n

A mole is a unit of measurement used in chemistry to measure amount of substance. It is calculated as molar mass by mass of the given substance. The mole is defined as the amount of a substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of carbon-12. One mole of a substance contains this number of particles, and can be used to calculate the mass of a given sample of the substance.

n = 60.05/16.84 = 3.56 moles

Hence the required number of moles are 3.56.

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Value of "n" in the hydrate formula Na₂CO₃.nH₂O is 0.874, or approximately 7/8. The formula for the hydrated compound is Na₂CO₃.7/8H₂O.

A substance that contains water molecule(s) within its structure is known as a hydrate.

As , mass of water lost = Mass of hydrated sample - Mass of anhydrous sample.

So, Mass of water lost = 25.00 g - 9.257 g

Mass of water lost = 15.743 g

As, moles of water lost = (Mass of water lost) / (Molar mass of water)

So moles of water lost = 15.743 g / 18.015 g/mol

moles of water lost = 0.874 mol

0.874 mol H₂O / 1 mol Na₂CO₃ = n / 1

n = 0.874 mol H₂O/ 1 mol Na₂CO₃

n = 0.874

Therefore, value of "n" in the hydrate formula Na₂CO₃.nH₂O is 0.874, or approximately 7/8. The formula for the hydrated compound is  Na₂CO₃.7/8H₂O

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Answer:

All of these choices are correct

The blood alcohol level in mass percent is 0.0956% if 8.33 mL of 0.04988 M [tex]K2Cr2O7[/tex] is required for titration of a 9.9950 g sample of blood.

To calculate the blood alcohol level in mass percent, we need to first determine the amount of ethanol present in the blood sample. This can be done by using a redox titration with potassium dichromate [tex](K2Cr2O7)[/tex], which oxidizes the ethanol to acetic acid.

The balanced chemical equation for the reaction is:

[tex]C2H5OH + 2Cr2O7^2- + 16H+ → 2CO2 + 4Cr^3+ + 11H2O[/tex]

From the balanced equation, we can see that the stoichiometric ratio between [tex]K2Cr2O7[/tex] and ethanol is 2:1. Therefore, the moles of ethanol in the blood sample can be calculated as:

moles of ethanol = 0.5 × moles of [tex]K2Cr2O7[/tex]

The moles of [tex]K2Cr2O7[/tex]can be determined from its concentration and volume used in the titration:

[tex]moles of K2Cr2O7 = 0.04988 mol/L × 8.33 × 10^-3 L = 4.15 × 10^-4 mol[/tex]

Substituting this value into the equation above, we get:

moles of ethanol = 0.5 × 4.15 × 10^-4 mol = 2.075 × 10^-4 mol

The mass of ethanol in the blood sample can be calculated from its molar mass:

[tex]mass of ethanol = 2.075 × 10^-4 mol × 46.07 g/mol = 9.551 × 10^-3 g[/tex]

Finally, the blood alcohol level in mass percent can be titration calculated as:

mass percent = mass of ethanol / mass of blood sample × 100%

mass percent = [tex]9.551 × 10^-3 g / 9.9950 g × 100%[/tex]

mass percent = 0.0956%

Therefore, the blood alcohol level in mass percent is 0.0956%.

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6.86 g of hydrochloric acid would be neutralized by 5.50 g of magnesium hydroxide.

The balanced chemical equation for the reaction between magnesium hydroxide and hydrochloric acid is:

Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l)

From the equation, we can see that 1 mole of Mg(OH)₂ reacts with 2 moles of HCl. We can use the molar mass of Mg(OH)₂ to convert its mass to moles:

moles of Mg(OH)₂ = mass / molar mass = 5.50 g / 58.32 g/mol = 0.0942 mol

Since 1 mole of Mg(OH)₂ reacts with 2 moles of HCl, we know that twice as many moles of HCl will be neutralized:

moles of HCl neutralized = 2 × moles of Mg(OH)₂ = 2 × 0.0942 mol = 0.1884 mol

Finally, we can use the molar mass of HCl to convert the moles of HCl to grams:

mass of HCl neutralized = moles of HCl × molar mass = 0.1884 mol × 36.46 g/mol = 6.86 g

Therefore, 6.86 g of hydrochloric acid would be neutralized by 5.50 g of magnesium hydroxide.

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Solution and Justification: 2.5103mol 25 mL at 0.100 M NaOH N an O H solution contain 2.5 10 3 mol of NaOH N an O H.

The amount of NaOH that needs to be dissolved to make 25 ml or 0.5M NaOH solution is 0.02*25, or 0.5g. In order to create a 0.5M NaOH solution, 0.5g or NaOH should be mixed in 25 ml of water.

For instance, if someone asked me to estimate how many moles are present in 20 ml of a 1M NaOH solution, I would normally reply that the answer is 20 mmol since 20 ml/1000 ml/L x 1M (= 0.02 moles = 20 mol) is the formula. Uncomplicated! You receive an injection of one mole of NaOH. The measurement is in mol / l.

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The range of paracetamol dosages expressed in mg per kg body weight for children who weigh between 20 and 30 pounds and are recommended a dosage of 1.5 teaspoons of Carfemol is between 165.5 mg/kg and 248.1 mg/kg.

What is  Paracetamol?

Paracetamol, also known as acetaminophen, is a common over-the-counter pain reliever and fever reducer. It is used to treat mild to moderate pain, such as headaches, menstrual cramps, toothaches, and backaches. Paracetamol is also used to reduce fever, such as in the case of flu or common cold.

To calculate the range of paracetamol dosages expressed in mg per kg body weight for children who weigh between 20 and 30 pounds and are recommended a dosage of 1.5 teaspoons of Carfemol, we need to convert the weight of the child to kilograms and then use the dosage and concentration of paracetamol in Carfemol to calculate the dosage in mg/kg.

Converting 20 pounds to kilograms:

20 pounds = 20 / 2.205 = 9.07 kg

Converting 30 pounds to kilograms:

30 pounds = 30 / 2.205 = 13.61 kg

Using the dosage recommended for a child of 1.5 teaspoons of Carfemol, we can calculate the dosage of paracetamol for a child weighing between 20 and 30 pounds:

Dosage of paracetamol for a child weighing 9.07 kg = 1.5 teaspoons * 75 mg/0.50 teaspoons = 2,250 mg

Dosage of paracetamol for a child weighing 13.61 kg = 1.5 teaspoons * 75 mg/0.50 teaspoons = 2,250 mg

Now, we can calculate the range of paracetamol dosages expressed in mg/kg body weight:

For a child weighing 9.07 kg: 2,250 mg / 9.07 kg = 248.1 mg/kg

For a child weighing 13.61 kg: 2,250 mg / 13.61 kg = 165.5 mg/kg

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the answer is 0.652 mol.

How to solve this problem?

To solve this problem, we need to use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the absolute temperature.

First, we need to convert the temperature to Kelvin by adding 273.15 to the Celsius temperature:

T = 105 + 273.15 = 378.15 K

Next, we can rearrange the ideal gas law to solve for the number of moles:

n = PV/RT

Substituting the given values, we get:

n = (170.0 kPa)(3.35 L)/(8.31 J/mol K)(378.15 K)

n = 0.652 mol

Therefore, the answer is 0.652 mol.

In physics and chemistry, volume refers to the amount of three-dimensional space that a substance or object occupies. It is a physical quantity that is usually measured in units of cubic meters (m³) or its derived units such as liters (L) or milliliters (mL). The volume of a substance or object can be calculated by measuring its dimensions (length, width, and height) and applying the appropriate formula, such as V = l × w × h for a rectangular solid. In the case of a gas, the volume can be determined by measuring the container that holds the gas or by using other techniques such as the displacement method, where the volume of a gas is determined by measuring the volume of liquid that it displaces. The volume of a substance is an important parameter that affects its properties and behavior, such as its density, pressure, and temperature.

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Answer:

To calculate the amount of solute (NaNO3) that will be left after the saturated solution is evaporated to dryness, we need to use the solubility of NaNO3 in water at 30 degrees Celsius. The solubility of NaNO3 in water at 30 degrees Celsius is 88.0 g/100 mL. First, we need to convert the 50 grams of water to milliliters. The density of water at 30 degrees Celsius is 0.996 g/mL. Therefore: 50 g ÷ 0.996 g/mL = 50.2 mL So, we have 50.2 mL of water in which the solubility of NaNO3 is 88.0 g/100 mL. To calculate the amount of NaNO3 that will be left after the solution is evaporated to dryness, we can

The amount of sodium acetate in grams required to prepare a buffer solution having pH 5.1 with one liter of 0.2N acetic acid solution is 91.8 g. The moles of acetic acid is 0.2 and the moles of acetate is 1.12. The Ka of acetic acid is 1.8 x 10^4.

Moles of acetic acid = (0.2 x 1000)/1000= 0.2 mol

pH of buffer = 5.1

[Acetate]/[Acetic acid] = 10^(pH - pKa)

[Acetate]/[Acetic acid] = 10^(5.1 - 4.75)

[Acetate]/[Acetic acid] = 5.62

Moles of acetate = 0.2 x 5.62 = 1.12 mol

Mass of sodium acetate = 1.12 x 82.03 = 91.8 g

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The solution's volume is 100 divided by 0.971 to get 102.98 mL. The number of moles of the solute (acetone) per litre of the solution is its molarity. Divide the mass by the molar mass to get the number of moles of acetone: 1.41 g / 58.08 g/mol = 0.0243 mol.

Molarity (M), which is determined by dividing the solute's mass in moles by the volume of the solution in litres, is the most widely used unit to express solution concentration: litres of solution/moles of solute equals M. One litre of a solution with a 1.00 molar concentration (1.00 M) contains 1.00 moles of solute.

The proportion of the solute that is dissolved in a solution is indicated by the solution's concentration. This formula can be used to determine a solution's concentration: Concentration is calculated as Volume of Solute multiplied by 100 and Volume of Solution (ml).

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The water in this experiment absorbs 20,920 J of energy.

What is Energy?  

Energy is the capacity of a physical system to perform work. It can exist in many forms, such as kinetic energy (the energy of motion), potential energy (energy that is stored), thermal energy (energy in the form of heat), electrical energy, and more. Energy can be transformed from one form to another, but it cannot be created or destroyed, only transferred or converted.

To calculate the energy absorbed by the water, we can use the formula:

Q = m * c * ΔT

Where Q is the energy absorbed, m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature.

Given:

Mass of water (m) = 200.0 g

Specific heat of water (c) = 4.18 J/g°C

ΔT = T₂ - T₁ = 45.1°C - 20.1°C = 25.0°C

Using the formula:

Q = 200.0 g * 4.18 J/g°C * 25.0°C = 20,920 J

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Answer:

88.55%

Explanation:
4.9995 g of crude material contain 4.4275 g of NaOH, therefore  

4.4275/4.9995 = 88.55%

⋅

According to Henry's law, the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid.

Therefore, we can use the solubilities given to calculate the amount of nitrogen and oxygen gas dissolved in 1.2 L of water at 25 °C and 1 atm pressure.

The solubility of oxygen gas at 25 °C and 1 atm is 36.0 mg/L, and the solubility of nitrogen gas at 25 °C and 1 atm is 14.7 mg/L. Using these values and the given partial pressures of oxygen and nitrogen in the air above the water, we can calculate the amount of nitrogen and oxygen gas dissolved in the water at 25 °C.

Next, we can use the ideal gas law to calculate the volume of gas that will bubble out of the water upon warming to 50 °C. Assuming the gas bubbles out at constant pressure, we can use the ideal gas law to calculate the volume of gas released.

Using these calculations, the total volume of nitrogen and oxygen gas that should bubble out of 1.2 L of water upon warming from 25 °C to 50 °C is approximately 13.3 mL.

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As may be shown, the molar entropy from the question is 85.1 kJ/K.mol.

By dividing the substance's overall entropy by the quantity of moles present, one can determine a substance's molar entropy. By measuring a substance's heat capacity or by examining its thermodynamic behavior under various circumstances, one can empirically ascertain the entropy of that substance.

Entropy = ΔH/T

= 1.6 * 10^4 * 10^3J/mol/187.95 K= 85.1kJ/K.mol

b) If we have the Br- ion;

Rate = 5/1 * 2.7 * 10^-3 mol/s

= 1.35 * 10^-2 mol/s

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(a) Partial pressure of Ne is 470 torr. (b) Partial pressure of of CF₄ is 110 torr. (c) Total pressure is 580 torr. (d) Mole fraction of Ne is 0.621.

(a) The initial pressure of Ne is 470 torr, so the partial pressure of Ne after mixing is also 470 torr.

(b) The initial pressure of CF₄ is 110 torr, so the partial pressure of CF₄ after mixing is also 110 torr.

(c) The total pressure is the sum of the partial pressures of the two gases:

Total pressure = partial pressure of Ne + partial pressure of CF₄

Total pressure = 470 torr + 110 torr

Total pressure = 580 torr

(d) To find the mole fraction of Ne, we need to know the number of moles of Ne and CF₄. We can use the ideal gas law to find the number of moles of each gas:

PV = nRT

n = PV/RT

For Ne:

n = (470 torr x 1.50 L)/(0.0821 L·atm/mol·K x 298 K)

n = 19.25 mol

For CF₄:

n = (110 torr x 2.50 L)/(0.0821 L·atm/mol·K x 298 K)

n = 11.72 mol

The total number of moles is:

nTotal = nNe + nCF₄

nTotal = 19.25 mol + 11.72 mol

nTotal = 30.97 mol

The mole fraction of Ne is:

XNe = nNe/nTotal

XNe = 19.25 mol/30.97 mol

XNe = 0.621

Therefore, the mole fraction of Ne is 0.621.

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The correct answer is a) KBr.

KBr is an ionic compound composed of a metal (K) and a non-metal (Br). When this compound is added to water, the polar water molecules surround the ions in the solid and separate them, which leads to the compound dissolving in water.

What is Ionic Compound?

An ionic compound is a chemical compound composed of ions held together by electrostatic forces called ionic bonds. Ions are atoms or molecules that have gained or lost one or more electrons, giving them a positive or negative charge. In an ionic compound, a positively charged ion (cation) and a negatively charged ion (anion) are attracted to each other to form a stable compound.

CO2, CH4, and O2 are nonpolar molecules, and therefore, do not dissolve well in water. CO2 and O2 are gases at room temperature and pressure, while CH4 is a gas at room temperature but can be liquefied under pressure.

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The experiment on determining the effect of soil type on plant growth can be improved by controlling variables such as the amount of sunlight and the pH of water.

Option 1, "Change the amount of water given to each plant," is not a way to improve the experiment because it introduces a new variable that can affect plant growth. If the student wants to test the effect of soil type, then the plants should receive the same amount of water, which is enough to keep the soil moist, but not so much that it causes waterlogging.

Option 2, "Make sure the plants receive the same amount of sunlight," is a way to improve the experiment because sunlight is a factor that can affect plant growth, and therefore, it is important to control the amount of sunlight that each plant receives.

Option 3, "Make sure the plants receive water with the same pH," is also a way to improve the experiment because the pH of water can affect plant growth, and therefore, it is important to keep the pH of water the same for all plants.

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