Give the IUPAC name for: image attached ​

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 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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Answer:

All of these choices are correct

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

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

Answer:

88.55%

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

4.4275/4.9995 = 88.55%

⋅

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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The contagious disease described with symptoms of fever, pain, redness, and swelling of the throat and tonsils is most likely to be strep throat, which is caused by a bacterial infection called Streptococcus pyogenes.

Strep throat is a common bacterial infection that affects people of all ages but is most commonly found in children between the ages of 5 and 15. The symptoms of strep throat can range from mild to severe, and it can be treated with antibiotics. If left untreated, it can lead to more serious conditions such as rheumatic fever or kidney damage. The common cold and the flu are viral infections, while asthma is a chronic respiratory condition and not a contagious disease caused by bacterial infection. The contagious disease described with symptoms of fever, pain, redness, and swelling of the throat and tonsils is most likely to be strep throat, which is caused by a bacterial infection called Streptococcus pyogenes.

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The balanced chemical equation for the given reaction is:

8OH^- + 2MnO₂ + 16H⁺ + 3Cl₂ --> 2MnO₄^- + 6Cl^- + 4H₂O

To balance a chemical equation, you need to adjust the coefficients of the reactants and products so that the number of atoms of each element is the same on both sides of the equation. Start by identifying the elements present in the reactants and products, and then adjust the coefficients until the equation is balanced.

To balance the charges in the given chemical equation:

i. Let's assign oxidation numbers to each element in the equation:

Hydrogen (H) has an oxidation number of +1.

Chlorine (Cl) has an oxidation number of -1.

Manganese (Mn) has an oxidation number of +4 in MnO₂ and +7 in MnO₄^-.

Oxygen (O) has an oxidation number of -2.

ii. Now we can balance the charges by adding electrons (e^-) to the appropriate side of the equation:

8OH^- + 2MnO₂ + 6Cl^- --> 2MnO₄^- + 6Cl^- + 4H₂O + 8e^-

iii. Next, we can simplify the equation by canceling out the common ions on both sides:

8OH^- + 2MnO₂+ 6Cl^- --> 2MnO₄^- + 6Cl^- + 4H₂O + 8e^-

8OH^- + 2MnO₂ --> 2MnO₄^- + 4H₂O + 8e^-

iv. Finally, we can write the balanced chemical equation by adding H+ ions to the appropriate side to balance the equation:

8OH^- + 2MnO₂ + 16H+ --> 2MnO₄^- + 6Cl^- + 4H₂O

Therefore, the balanced chemical equation for the given reaction is:

8OH^- + 2MnO₂ + 16H+ + 3Cl₂ --> 2MnO₄^- + 6Cl^- + 4H₂O

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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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The energy (in joules) of 1 mole of photons with this frequency is 1.99 x 10⁻¹⁴ J per photon.

Should a substance happen to have a lot of electrons in a higher level, and a lower level is mostly empty  then a photon can cause an electron to transfer from a higher state to a lower one. This change releases energy and creates a new photon, in addition to the one which caused the transfer. This photon can in turn induce more electrons to fall to a lower state.

use formula

The energy of 1 mole of photons with the given frequency can be calculated using the following equation:

Energy (J) = Avogadro's number x Plank's Constant x Frequency

Therefore,

Energy (J) = 6.02 x 10²³ x 6.626 x 10⁻³⁴ x 5.40 x 10⁴

Energy (J) = 1.99 x 10⁻¹⁴ J

Therefore, the energy of 1 mole of photons with a frequency of 5.40 x 10⁴ Hz is 1.99 x 10⁻¹⁴ J, expressed in scientific notation.

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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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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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Explanation:

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

Answer:

mass of salt = 70 grams
mass of water = 280 grams

Explanation:

It's about Mass percentage

Mass percentage = [tex]\frac{Solute}{solution } \times 100 \%[/tex]

Given that:

By substituting with given data in mass percentage formula:

[tex]20\%= \frac{solute}{350}[/tex]

then, solute mass = [tex]\frac{350 \times 20}{100} = 70 g[/tex]

So, mass of water = 350 - 70 = 280 grams


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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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Answer:

0.61 V

Explanation:

a. The two metals in this voltaic cell are cadmium (Cd) and copper (Cu). Cadmium is oxidized (loses electrons) and copper is reduced (gains electrons).

b. Half-reactions:

Cathode: Cu2+(aq) + 2e- → Cu(s)

Anode: Cd(s) → Cd2+(aq) + 2e-

Overall reaction: Cd(s) + Cu2+(aq) → Cd2+(aq) + Cu(s)

c. The anode is where oxidation occurs, so the anode is the cadmium electrode. The cathode is where reduction occurs, so the cathode is the copper electrode.

d. Standard cell potential (Eᵒcell) can be calculated using the standard reduction potentials (Eᵒred) for each half-reaction.

Eᵒcell = Eᵒred(cathode) - Eᵒred(anode)

Eᵒred(Cu2+(aq) + 2e- → Cu(s)) = +0.34 V (from tables)

Eᵒred(Cd(s) → Cd2+(aq) + 2e-) = -0.40 V (from tables)

Eᵒcell = +0.34 V - (-0.40 V) = +0.74 V

e. The standard free energy change (ΔG°) can be calculated from the standard cell potential (Eᵒcell) using the following equation:

ΔG° = -nFEᵒcell

where n is the number of electrons transferred in the balanced equation (2), and F is the Faraday constant (96,485 C/mol).

ΔG° = -2 x 96,485 C/mol x 0.74 V = -141,800 J/mol = -141.8 kJ/mol

f. The cell diagram for this voltaic cell is:

Cd(s) | Cd(NO3)2 (0.10 M) || Cu(NO3)2 (0.20 M) | Cu(s)

g. The Nernst equation can be used to calculate the cell potential (E) under non-standard conditions, where the concentrations of the species involved are not at their standard state. The Nernst equation is:

E = Eᵒcell - (RT/nF) x ln(Q)

where R is the gas constant (8.314 J/mol-K), T is the temperature in Kelvin, n is the number of electrons transferred (2), F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient, which is the ratio of the concentrations of the products over the concentrations of the reactants raised to their stoichiometric coefficients.

At non-standard conditions, the concentrations of the species involved are:

[Cd2+] = 0.10 M

[Cu2+] = 0.20 M

The reaction quotient (Q) is:

Q = ([Cd2+]/[Cu2+])^2 = (0.10 M / 0.20 M)^2 = 0.25

Plugging in the values:

E = 0.74 V - (8.314 J/mol-K x 298 K / (2 x 96,485 C/mol)) x ln(0.25) = 0.61 V

Therefore, the cell potential at non-standard conditions is 0.61 V.

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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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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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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[tex]57.27[/tex] grams of potassium chlorate were used in the reaction to produce [tex]15.7 L[/tex] of  [tex]O_2[/tex] gas at STP.

The balanced chemical equation for the decomposition of potassium chlorate is:

[tex]2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)[/tex]

From the equation, we can see that 2 moles of [tex]KClO_3[/tex] produce [tex]3[/tex] moles of  [tex]O_2[/tex] . Therefore, the number of moles of  [tex]KClO_3[/tex] can be calculated as:

moles of  [tex]KClO_3 =[/tex] (moles of [tex]O_2[/tex] produced) [tex]\times (2/3)[/tex]

At STP (standard temperature and pressure), [tex]1[/tex] mole of any gas occupies [tex]22.4 L[/tex] of volume. Therefore,  [tex]15.7 L[/tex] of O2 at STP corresponds to:

moles of [tex]O_2[/tex] produced [tex]= (15.7 L) / (22.4 L/mol) = 0.7018 mol[/tex]

Substituting this value into the above equation, we get:

moles of [tex]KClO3 = (0.7018 mol) \times (2/3) = 0.4679 mol[/tex]

The molar mass of [tex]KClO_3[/tex] is [tex]122.55 g/mol[/tex]  . Therefore, the mass of [tex]KClO_3[/tex] used in the reaction can be calculated as:

mass of  [tex]KClO_3[/tex] = (moles of [tex]KClO_3[/tex]) x (molar mass of [tex]KClO_3[/tex])

[tex]= (0.4679 mol) \times (122.55 g/mol)[/tex]

[tex]= 57.27 g[/tex]

Therefore, [tex]57.27[/tex] grams of potassium chlorate were used in the reaction to produce [tex]15.7 L[/tex] of  [tex]O_2[/tex] gas at STP.

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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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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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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 percent yield of the reaction is approximately 43.6%.

We need to compare the actual yield of the reaction with the theoretical yield of the reaction.

First, we need to determine the theoretical yield of the reaction, which is the amount of lead(II) oxide that would be formed if all of the lead reacted with the oxygen. We can use stoichiometry and the molar mass of each substance to calculate the theoretical yield.

The balanced chemical equation for the reaction is:

2 Pb(s) + O₂(g) → 2 PbO(s)

The molar mass of Pb is 207.2 g/mol and the molar mass of PbO is 223.2 g/mol.

From the given information, we know that 451.4 g of Pb reacted with an excess of O₂ to form 338.4 g of PbO. We can use this information to calculate the amount of PbO that would be formed if all of the Pb reacted:

451.4 g Pb × (1 mol Pb / 207.2 g Pb) × (2 mol PbO / 2 mol Pb) × (223.2 g PbO / 1 mol PbO) = 776.8 g PbO

So, the theoretical yield of PbO is 776.8 g.

Now, we can calculate the percent yield of the reaction:

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

From the given information, the actual yield is 338.4 g of PbO.

Percent yield = (338.4 g / 776.8 g) × 100% ≈ 43.6%

Therefore, the percent yield of the reaction is approximately 43.6%.

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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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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).

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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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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