Which of the following pictures is a chemical change, and which is a
physical change? Explain in the boxes next to the picture.

Plz explain

The volume of 72.3 g of carbon dioxide gas at 22.00 °C and 875 mmHg is 48.32 L.

The given values for the carbon dioxide gas are:

Mass of CO₂ (m) = 72.3 g

Temperature (T) = 22.00 °C or 295 K

Pressure (P) = 875 mmHg or 115.99 kPa

To calculate the volume (V) in liters, we use the ideal gas law equation which is given by:PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

To solve for V, we rearrange the equation as follows:V = nRT/PWe need to determine n, the number of moles of carbon dioxide. To do this, we use the formula:n = m/M

where m is the mass of  CO₂ and M is the molar mass of  CO₂, which is 44.01 g/mol.n = 72.3 g/44.01 g/moln = 1.64 mol

Now that we know n, we can substitute the given values and solve for V.V = nRT/PV = (1.64 mol)(0.08206 L•atm/mol•K)(295 K)/(115.99 kPa)

Note that we convert the pressure from mmHg to kPa by dividing by 7.50062.V = 48.32 L

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The redox reactions that occur spontaneously in the forward direction are 2Ag+(aq) + Ni(s) → 2Ag(s) + Ni2+(aq) and Ca2+(aq) + Zn(s) → Ca(s) + Zn2+(aq).

In a redox reaction, the spontaneity of the forward direction is determined by the reduction potentials of the species involved. The reactions 2Ag+(aq) + Ni(s) → 2Ag(s) + Ni2+(aq) and Ca2+(aq) + Zn(s) → Ca(s) + Zn2+(aq) occur spontaneously because the reduction potentials of Ag+ and Ca2+ are higher than those of Ni2+ and Zn2+ respectively. This means that Ag+ and Ca2+ have a greater tendency to be reduced and gain electrons, while Ni and Zn have a greater tendency to be oxidized and lose electrons. On the other hand, the reactions 2Cr(s) + 3Pb2+(aq) → 2Cr3+(aq) + 3Pb(s) and Sn(s) + Mn2+(aq) → Sn2+(aq) + Mn(s) do not occur spontaneously because the reduction potentials of Cr3+ and Sn2+ are lower than those of Pb2+ and Mn2+ respectively.

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Carbon 14 dating is a widely used radiometric dating method for determining the age of archaeological and paleontological specimens up to 50,000 years old.

Here are the assumptions that are made when carbon 14 dating is used:

1. The rate of carbon-14 production in the upper atmosphere is constant over time.

2. The ratio of carbon-14 to carbon-12 in the atmosphere has been constant over time.

3. Carbon-14 is readily absorbed by living organisms and is incorporated into their tissues in proportion to its concentration in the atmosphere.

4. Once an organism dies, the carbon-14 in its tissues decays at a constant rate.

5. The rate of carbon-14 decay has been constant over time.

6. The amount of carbon-14 remaining in a sample can be accurately measured.

7. The sample has not been contaminated with carbon from a different source.

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The observed color changes in the cobalt solutions can be explained in terms of Le Chatelier's Principle, which states that when a system at equilibrium is subjected to a stress, it will adjust to minimize the effect of that stress.

Here, experiment, we start with a solution of CoCl2, which appears as a clear pink color. The pink color is due to the presence of hydrated cobalt(II) ions [Co(H2O)6]2+ in the solution. This complex absorbs certain wavelengths of light, resulting in the observed color.

When HCl is added to the solution, it introduces additional chloride ions (Cl-) into the system. According to Le Chatelier's Principle, the increased concentration of chloride ions will shift the equilibrium towards the formation of the complex [CoCl4]2-, which is dark blue in color.

The shift occurs because the system tries to counteract the stress caused by the increase in chloride ions by favoring the reaction that consumes the excess chloride ions.

Finally, when water (H2O) is added, it dilutes the solution. This decrease in concentration again perturbs the equilibrium, and Le Chatelier's Principle predicts a shift back towards the formation of the hydrated cobalt(II) ions [Co(H2O)6]2+, leading to the restoration of the clear pink color.

In summary, the changes in color observed in the cobalt solutions can be explained by Le Chatelier's Principle, as the system adjusts to counteract the stress caused by changes in concentration.

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In the example of distilling miscible liquids, two compounds commonly used are ethanol and water.

One of the common example for distilling miscible liquids are ethanol and water. These two substances can mix in any ratio since they are miscible. Because ethanol and water have different molecular structures, when heated, the boiling temperatures of the two liquids vary.

Compared to water, ethanol has a lower boiling point. The ethanol vaporizes first when the combination is heated, leaving the water behind. After collecting and condensing the vapor, pure ethanol is produced. Distillation is a common practice in many different industries, including the creation of alcoholic beverages and ethanol for fuel.

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The grams of H₂ produced, if the chemical reaction produces 129 grams of  AlCl₃ are 2.92 grams. (Option b)

To determine the grams of H₂ produced, we need to use the balanced equation and the molar ratios between AlCl₃ and H₂.

From the balanced equation:

2 moles of AlCl₃ react with 3 moles of H₂

To find the moles of AlCl₃ produced:

129 grams AlCl₃ x (1 mole AlCl₃ / molar mass AlCl₃) = moles of AlCl₃

Now, using the molar ratios, we can determine the moles of H₂ produced:

moles of AlCl₃ x (3 moles H₂ / 2 moles AlCl₃) = moles of H₂

Finally, we can convert the moles of H₂ back to grams:

moles of H₂ x (molar mass H₂ / 1 mole H₂) = grams of H₂

Let's calculate it:

Given:

Mass of AlCl₃ produced = 129 grams

Molar mass of AlCl₃:

Al: 26.98 g/mol

Cl: 35.45 g/mol x 3 = 106.35 g/mol

Total molar mass = 26.98 g/mol + 106.35 g/mol = 133.33 g/mol

Calculations:

moles of AlCl₃ = 129 g AlCl₃ / 133.33 g/mol = 0.9676 moles AlCl₃

moles of H₂ = 0.9676 moles AlCl₃ x (3 moles H₂ / 2 moles AlCl₃) = 1.4514 moles H₂

grams of H₂ = 1.4514 moles H₂ x (2.02 g/mol / 1 mole H₂) = 2.93 grams of H₂

Therefore, the correct answer is b. 2.92 grams.

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To find the molarity of ethanol in beer that is 5% ethanol by volume, we first need to calculate the volume of ethanol in the beer. Since the density of ethanol is 0.789 g/cm³ and the beer is 5% ethanol by volume, we can assume that 100 mL of beer contains 5 mL of ethanol.

Volume of ethanol in 100 mL of beer = 5 mL

Next, we convert the volume of ethanol to its mass using its density:

Mass of ethanol in 100 mL of beer = volume × density = 5 mL × 0.789 g/cm³ = 3.945 g

Now we can calculate the number of moles of ethanol using its molecular weight:

Moles of ethanol = mass / molecular weight = 3.945 g / 46 g/mol ≈ 0.0857 mol

Finally, we convert the moles of ethanol to molarity by dividing by the volume of beer in liters:

Molarity of ethanol in beer = moles / volume (in liters) = 0.0857 mol / 0.1 L = 0.857 M

B. To find the molarity of ethanol in a person at the legal blood alcohol limit of 0.08 (80 mg/100 mL), we can follow a similar approach. The person's blood contains 80 mg of ethanol in 100 mL.

Moles of ethanol = mass / molecular weight = 80 mg / 46 g/mol = 1.74 mmol

Convert the moles to molarity by dividing by the blood volume in liters (100 mL = 0.1 L):

Molarity of ethanol in blood = moles / volume (in liters) = 1.74 mmol / 0.1 L = 17.4 M

C. To determine the number of 12-oz (355 mL) bottles of beer a 70 kg person can drink and remain under the legal limit, we need to calculate the total amount of ethanol in the person's blood at the legal limit and then find how many bottles it would take to reach that amount.

Ethanol content in blood = 80 mg/100 mL = 0.08 g/100 mL

Total ethanol content in blood = 0.08 g/100 mL × 40 L = 32 g

Now, we need to calculate the number of moles of ethanol:

Moles of ethanol = mass / molecular weight = 32 g / 46 g/mol ≈ 0.696 mol

Assuming each bottle of beer contains the same amount of ethanol as calculated in part A (0.0857 mol), we can calculate the number of bottles a person can drink:

Number of bottles = Moles of ethanol in blood / Moles of ethanol per bottle = 0.696 mol / 0.0857 mol ≈ 8.12 bottles

Therefore, a 70 kg person could drink approximately 8 bottles of beer and remain under the legal limit.

D. To find the time required for a 70 kg person's blood alcohol level to fall below the legal limit (160 mg/100 mL to 80 mg/100 mL), we can use the constant rate of ethanol metabolism given.

The difference in ethanol concentration is 160 mg/100 mL - 80 mg/100 mL = 80 mg/100 mL

Moles of ethanol to be metabolized = mass / molecular weight = 80 mg / 46 g/mol = 1.74 mmol

Ethanol metabolism rate for a 70 kg person = 120 mg/hour/kg = 120 mg/hour × 70 kg = 8400 mg/hour

Time required to metabolize the ethanol = moles of ethanol / (ethanol metabolism rate / 1000) = 1.74 mmol / (8400 mg/hour / 1000) = 0.207 hours ≈ 12.4 minutes

Therefore, it would take approximately 12.4 minutes for the blood alcohol level to fall below the legal limit for a 70 kg person at twice the legal limit.

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The mass percent of NaOH in the given solution is 0.0596%.Thus, the required mass percent of solute in the given solutions are:(a) 7.47%(b) 8.96%(c) 10.92%(d) 0.0596%.

(a) Given, Mass of solute NaCl = 5.63 gMass of solvent H2O = 69.8 gThe mass percent of solute in NaCl solution is calculated using the following formula:Mass percent of solute = (Mass of solute / Mass of solution) × 100Now, the mass of solution is the sum of the mass of solute and solvent.Mass of solution = Mass of solute + Mass of solvent= 5.63 g + 69.8 g= 75.43 gMass percent of solute = (5.63 / 75.43) × 100= 7.47 %Hence, the mass percent of NaCl in the given solution is 7.47%.(b) Given, Mass of solute LiBr = 3.28 gMass of solvent H2O = 33.3 gThe mass percent of solute in LiBr solution is calculated using the following formula:Mass percent of solute = (Mass of solute / Mass of solution) × 100Now, the mass of solution is the sum of the mass of solute and solvent.Mass of solution = Mass of solute + Mass of solvent= 3.28 g + 33.3 g= 36.58 gMass percent of solute = (3.28 / 36.58) × 100= 8.96 %

Hence, the mass percent of LiBr in the given solution is 8.96%.(c) Given, Mass of solute KNO3 = 36.7 gMass of solvent H2O = 299 gThe mass percent of solute in KNO3 solution is calculated using the following formula:Mass percent of solute = (Mass of solute / Mass of solution) × 100Now, the mass of solution is the sum of the mass of solute and solvent.Mass of solution = Mass of solute + Mass of solvent= 36.7 g + 299 g= 335.7 gMass percent of solute = (36.7 / 335.7) × 100= 10.92 %Hence, the mass percent of KNO3 in the given solution is 10.92%.(d) Given, Mass of solute NaOH = 2.8 × 10⁻³ gMass of solvent H2O = 4.7 g

The mass percent of solute in NaOH solution is calculated using the following formula:Mass percent of solute = (Mass of solute / Mass of solution) × 100Now, the mass of solution is the sum of the mass of solute and solvent.Mass of solution = Mass of solute + Mass of solvent= 2.8 × 10⁻³ g + 4.7 g= 4.70028 gMass percent of solute = (2.8 × 10⁻³ / 4.70028) × 100= 0.0596 %Hence, the mass percent of NaOH in the given solution is 0.0596%.Thus, the required mass percent of solute in the given solutions are:(a) 7.47%(b) 8.96%(c) 10.92%(d) 0.0596%.

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The percent ionization in a 0.300 m solution of formic acid is 0.00403%.

Given data: Formic acid (HCOOH), Ka = 1.78 × 10⁻⁴

Molarity of the solution (M) = 0.300 m

Moles of HCOOH, initial = M × volume = 0.300 × 1000 = 300 mol/L

Let x be the moles of HCOOH ionized . Let's write down the ionization reaction:

HCOOH + H₂O ↔ H₃O⁺ + HCOO⁻

Initial (mol/L): 300 0 0 0

Change (mol/L):

-x +x +x +x

Equilibrium (mol/L): 300 - x x x x

We know that

Ka = [H₃O⁺][HCOO⁻]/[HCOOH]

Let's write down the equation for Ka using the initial and equilibrium concentrations of the species.

Ka = x²/(300-x)

Ka = 1.78 × 10⁻⁴

Solve for x.

x = √[Ka × (300 - x)]

x = √[(1.78 × 10⁻⁴) × (300 - x)]

x = 0.0121 (approx)

The percent ionization can be calculated as follows:

Percent ionization = (moles ionized/initial moles) × 100= x/300 × 100= 0.0121/300 × 100= 0.00403% (approx)

Hence, the percent ionization in a 0.300 m solution of formic acid is 0.00403%.

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The required partial pressure of neon in mmHg) in the tank is 272 mmHg.

The partial pressure of neon in the tank is 152 mmHg.A mixture of neon, argon, and helium gases are contained in a tank. The total pressure in the tank is 600. mmHg, while the partial pressures of neon and argon are 120 torr and 0.20 atm, respectively.To calculate the partial pressure of neon in mmHg, we need to convert the pressure of argon to mmHg:0.20 atm x 760 mmHg/atm = 152 mmHgNext, add the partial pressures of neon and argon:120 torr + 152 mmHg = 272 mmHg

Therefore, the answer is 272 mmHg, option (b) is correct.

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A reactant-favored reaction is a reaction where the equilibrium position lies more towards the side of the reactants, meaning there is a higher concentration of reactants compared to products at equilibrium.

a. A + B → C + D

This reaction is a forward reaction where reactants (A and B) are being converted into products (C and D). It is not a reactant-favoured reaction because the reactants are being consumed to form products.

b. C + D → A + B

This reaction is the reverse of the previous reaction. It is also not a reactant-favored reaction because the reactants (C and D) are being consumed to form the products (A and B).

c. C + D ↔ A + B

This reaction is a reversible reaction indicated by the double arrow. In a reversible reaction, the equilibrium position can be towards the side of the reactants or towards the side of the products depending on the relative concentrations of the reactants and products. Without further information about the concentrations, we cannot determine if this reaction is reactant-favoured or product-favoured.

d. A + B ↔ C + D

This reaction is also a reversible reaction. Similar to the previous case, without information about the concentrations, we cannot determine if this reaction is reactant-favoured or product-favoured.

In summary, among the given reactions, it cannot be definitively determined which one is a reactant-favored reaction without additional information.

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The maximum number of electrons in this sublevel is 3 ×2 = 6, where n = 2, l = 1. The maximum number of electrons in this sublevel is 5 × 2 = 10 for 3d. The maximum number of electrons in this sublevel is 1 × 2 = 2 for 4s.

A.

n = 2, l = 1 , n = 2 (second energy level) and l = 1 (p sublevel).

In the p sublevel, there are three orbitals: px, py, and pz.

Each orbital can hold a maximum of 2 electrons (one with spin-up and one with spin-down).

Therefore, the maximum number of electrons in this sublevel is 3 × 2 = 6.

B.

For this quantum number 3d, 

n = 3 (third energy level) and l = 2 (d sublevel).

In the d sublevel, there are five orbitals: dxy, dxz, dyz, dx2-y2, and dz2.

Each orbital can hold a maximum of 2 electrons.

Therefore, the maximum number of electrons in this sublevel is 5 × 2 = 10.

C.

For this quantum number 4s, 

n = 4 (fourth energy level) and l = 0 (s sublevel).

In the s sublevel, there is one orbital: 4s.

The 4s orbital can hold a maximum of 2 electrons.

Therefore, the maximum number of electrons in this sublevel is 1 ×2 = 2.

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The order of the substituents according to CIP sequence rules is as follows: 1. -Br 2. -CH2CH2Br 3. -C equivalence N 4. -CH2Br

The Cahn-Ingold-Prelog (CIP) sequence rules describe how to assign the absolute configuration of a chiral center to an enantiomer. The ranking of substituents can be done with the help of CIP sequence rules. The sequence rules are as follows:At first, the priority of substituents is determined by atomic number. The higher the atomic number, the higher the priority. If a molecule has isotopes, the one with a higher atomic mass takes priority.Next, if the atoms in two substituents have the same atomic number, the atoms in each substituent are compared, going atom by atom down the chains of atoms until a difference is found. When a difference is found, the substituent with the atom of higher atomic number is given the higher priority.The steps for ranking the given set of substituents are as follows:As we can see in the given set of substituents, the most common atoms are carbon and bromine. The carbon has an atomic number of 6, and bromine has an atomic number of 35.5. Hence, Bromine has a higher atomic number than Carbon. Therefore, bromine gets the highest priority among the given substituents.Now, we have to compare the other substituents to the highest priority substituent (Bromine).If we compare -CH2Br with -CH2CH2Br, both substituents have the same atoms up to the second carbon. After that, -CH2Br has a single carbon atom, whereas -CH2CH2Br has two carbon atoms. The substituent with more carbon atoms is given higher priority. Therefore, -CH2CH2Br is ranked higher than -CH2Br.In -C equivalence N, nitrogen has an atomic number of 7, which is higher than the atomic number of carbon in -CH2Br and -CH2CH2Br. Therefore, -C equivalence N is ranked third.Lastly, -Br is ranked the lowest among the substituents.The order of the substituents according to CIP sequence rules is as follows: 1. -Br 2. -CH2CH2Br 3. -C equivalence N 4. -CH2Br.

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Photosynthesis caused the amount of carbon dioxide to decrease and oxygen to increase. The carbon dioxide was absorbed by the oceans and converted into sedimentary rocks.

Hydrogen predominated in the early atmosphere of Earth, which was composed of gases that originated in the solar nebula. Over the course of time, there was a dramatic shift in the atmosphere, which was caused by a variety of processes including life, weathering, and volcanic activity. As a result of photosynthesis, the amount of carbon dioxide decreased, while the amount of oxygen increased. Carbon dioxide was taken up by the oceans, where it was transformed into sedimentary rocks. When the Earth was younger, its atmosphere had a higher concentration of carbon dioxide and water vapor but a lower oxygen content than it has now.

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for the left column:

for the right column:

The opposite electrons must have an opposite spin direction to occupy the same orbital. This is important in assigning Identical electron configurations, as it allows you to determine how and where the electrons should be assigned.

The electron configuration is described as the distribution of electrons of an atom or molecule in atomic or molecular orbitals.

In this, we consider the energy ordering of orbitals, the spin direction of electrons, and the concept of identical electron configurations.

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Pyridine is a weak base, and when dissolved in water, it accepts protons from water molecules to form hydroxide ions (OH⁻) and the pyridinium ion (C₅H₅NH⁺).

To calculate the pH of the solution, we need to determine the concentration of the pyridinium ion, which is equal to the concentration of hydroxide ions.

C₅H₅N + H₂O ⇌ C₅H₅NH⁺ + OH⁻

The equilibrium constant expression for this reaction is:

Kw = [C₅H₅NH⁺][OH⁻] / [C₅H₅N]

Since the concentration of hydroxide ions is given as 2.15 × 10⁻⁴ M, we can assume that the concentration of pyridinium ion is also 2.15 × 10⁻⁴ M.

(10⁻¹⁴) = (2.15 × 10⁻⁴)(2.15 × 10⁻⁴) / [C₅H₅N]

Solving for [C₅H₅N], we find [C₅H₅N] = 6.9 × 10⁻⁸ M.

Now, we can use the concentration of pyridine to calculate the pOH of the solution:

pOH = -log10([OH⁻]) = -log10(2.15 × 10⁻⁴) ≈ 3.67

Finally, we can calculate the pH using the relation pH + pOH = 14:

pH = 14 - pOH ≈ 10.33

Therefore, the pH of the aqueous solution of pyridine is approximately 10.33.

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The correct label for the salts is:

a) NaCN = acidic, NH4F = basic, KCN = neutral

- NaCN: Sodium cyanide (NaCN) is a salt of a weak acid (HCN) and a strong base (NaOH). The weak acid (HCN) partially dissociates in water, producing cyanide ions (CN-) and a small amount of H+ ions, making the solution slightly acidic.

- NH4F: Ammonium fluoride (NH4F) is a salt of a weak base (NH3) and a strong acid (HF). The weak base (NH3) partially reacts with water to produce ammonium ions (NH4+) and hydroxide ions (OH-), making the solution slightly basic.

- KCN: Potassium cyanide (KCN) is a salt of a weak acid (HCN) and a strong base (KOH). Similar to NaCN, KCN produces cyanide ions (CN-) and a small amount of H+ ions in water, resulting in a neutral solution.

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Water is a better solvent for CH3OH and C6H6 due to their polar nature, while carbon tetrachloride is a better solvent for CaCl2 and Br2 due to their nonpolar nature, matching the nonpolar nature of carbon tetrachloride.

The solubility of a solute in a particular solvent depends on the intermolecular interactions between the solute and solvent molecules. The choice of a better solvent between water and carbon tetrachloride depends on the solute in question. For CH3OH (methanol) and C6H6 (benzene), water is a better solvent due to its polar nature. Methanol contains a hydroxyl group (-OH) that can form hydrogen bonds with water molecules, while benzene is slightly polar and can dissolve to some extent in water. However, for CaCl2 (calcium chloride) and Br2 (bromine), carbon tetrachloride is a better solvent. These solutes are nonpolar, and carbon tetrachloride, being nonpolar as well, can effectively dissolve them.

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To gravimetric factor for converting BaSO₄ to sulfite (SO₃) is 1.

To gravimetric factor for converting BaSO₄ to sulfite (SO₃) is determined based on the stoichiometry of the reaction.

The balanced chemical equation for the reaction is:

BaSO₄ + 4 H₂ → BaS + 4 H₂O + SO₃

From the equation, the stoichiometry shows that for every 1 mole of BaSO₄, we obtain 1 mole of SO₃.

Therefore, the gravimetric factor is 1.

This means that if we have a known mass of BaSO₄, we can directly convert it to an equivalent mass of SO₃ using the gravimetric factor of 1.

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The major product formed is nitrobenzene.

When benzene reacts with tert-butyl bromide and AICI3 it produces tert-butylbenzene as a major product. The reaction occurs via an electrophilic substitution reaction. When bromine reacts with a nail in the presence of benzene, the aromatic compound will undergo electrophilic substitution. The major product formed is bromobenzene.  When iodine reacts with HNO3 in the presence of benzene, the electrophilic substitution occurs and the major product formed is nitrobenzene. The major product formed when benzene reacts with carbon monoxide, HCl, and AICI3/CUCI is benzaldehyde, produced via the Gattermann-Koch reaction. Nitric acid and sulfuric acid are nitrating agents that cause benzene to undergo electrophilic substitution. The major product formed is nitrobenzene.

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Approximately 8.786 grams of potassium alum should be produced. we need to consider the stoichiometry of the reaction and the molar masses of the compounds involved.

The balanced equation for the reaction is:

2Al + 2KOH + H₂SO₄ → KAl(SO₄)₂·12H₂O + 3H₂

From the equation, we can see that 2 moles of aluminum react to produce 1 mole of potassium alum. Therefore, we need to calculate the number of moles of aluminum.

Molar mass of Al = 26.98 g/mol

Number of moles of Al = Mass of Al / Molar mass of Al = 1.000 g / 26.98 g/mol = 0.03706 mol

According to the stoichiometry, 2 moles of aluminum will produce 1 mole of potassium alum. Therefore, the number of moles of potassium alum produced is half the number of moles of aluminum:

Number of moles of KAl(SO₄)₂·12H₂O = 0.03706 mol / 2 = 0.01853 mol

Finally, we can calculate the mass of potassium alum using the molar mass of KAl(SO₄)₂·12H₂O:

Molar mass of KAl(SO₄)₂·12H₂O = 474.38 g/mol

Mass of KAl(SO₄)₂·12H₂O = Number of moles of KAl(SO₄)₂·12H₂O × Molar mass of KAl(SO₄)₂·12H₂O = 0.01853 mol × 474.38 g/mol = 8.786 g

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The correct order of increasing density at STP among the given gases is; NH₃ < N₂ < Kr < N₂O₄. Option D is correct.

To determine the order of increasing density at STP among the given gases, we need to consider their molar masses and the behavior of gases under standard conditions.

The molar masses of gases are as follows;

N₂O₄; 92.01 g/mol

N₂; 28.01 g/mol

Kr; 83.80 g/mol

NH₃; 17.03 g/mol

At STP (Standard Temperature and Pressure), gases behave ideally, meaning they have similar volumes and occupy the same amount of space. The density of a gas will be directly proportional to its molar mass. Therefore, the gas with the lowest molar mass will have the lowest density.

Comparing the molar masses of the given gases, we can determine the order of increasing density;

NH₃ < N₂ < Kr < N₂O₄

Hence, D. is the correct option.

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The final pressure of the gas, when the volume is increased from 12.8 L to 25.6 L, can be calculated using Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

In the first scenario, the initial pressure is 0.987 atm and the initial volume is 12.8 L. The product of pressure and volume is constant (P₁V₁ = P₂V₂), so we can calculate the final pressure (P₂) using the equation:

P₂ = (P₁ * V₁) / V₂

Plugging in the values, we have:

P₂ = (0.987 atm * 12.8 L) / 25.6 L = 0.494 atm

Therefore, the final pressure of the gas, when the volume is increased to 25.6 L, is 0.494 atm. Hence, the correct answer is option c) 0.494 atm.

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The ideal gas law formula is used to determine the missing values in questions. When dealing with problems that require solving for missing values using the ideal gas law formula, always ensure that all values are expressed in the correct units and temperature is converted to kelvin.

The ideal gas law formula is represented as PV = nRT, where P represents pressure, V represents volume, n represents the number of moles of gas, T represents the temperature in kelvin, and R represents the universal gas constant. Solve for missing values using the ideal gas law formula:1. 10°C, 5. 5 L, 2 mol, __ atm.The temperature must be converted to kelvin first: T(K) = T(°C) + 273.15K = 10°C + 273.15 = 283.15KPV = nRT

Rearrange the equation to isolate P: P = nRT / V

Substitute the given values:

P = (2 mol)(0.0821 L•atm/mol•K)(283.15K) / 5.5 L

: P = 8.28 atm

2. __ °C, 8. 3 L, 5 mol, 1. 8 atm.The equation PV = nRT can be rearranged to T = PV / nRThe temperature must be converted to kelvin first: T(K) = T(°C) + 273.15T = PV / nR

Substitute the given values: T = (1.8 atm)(8.3 L) / (5 mol)(0.0821 L•atm/mol•K)T(K) = T +

: T = 332 K or 59°C

The temperature must be converted to kelvin first:

T(K) = T(°C) + 273.15K

= 12°C + 273.15

= 285.15

KPV = nRT

Solve for n by rearranging the equation: n = PV / RT

Substitute the given values: n = (1.2 atm)(3.4 L) / (0.0821 L•atm/mol•K)(285.15K): n = 0.141 mol

The ideal gas law formula is used to determine the missing values in questions. When dealing with problems that require solving for missing values using the ideal gas law formula, always ensure that all values are expressed in the correct units and temperature is converted to kelvin.

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a) Malonic acid will exist primarily as an anion in aqueous solution at pH 5.4.

b) This option is not valid.

c) This option is not valid.

d) Alloxanic acid would not primarily exist as an anion at pH 5.4.

e) Propanoic acid will exist primarily as an anion in aqueous solution at pH 5.4.

f) Glyoxylic acid will exist primarily as an anion in aqueous solution at pH 5.4.

To determine which weak acid would not exist primarily as an anion in aqueous solution at pH 5.4, we need to compare the pKa values of the acids with the pH value.

The pKa value represents the negative logarithm of the acid dissociation constant (Ka), and it indicates the strength of the acid. A lower pKa value indicates a stronger acid.

Let's analyze each option:

a. Malonic acid, Ka = 1.5 x 10^-3, pKa = 2.8: The pKa of malonic acid is lower than pH 5.4, indicating that it will exist primarily as an anion in aqueous solution at pH 5.4.

b. None would be anionic: This option suggests that none of the acids would exist primarily as an anion at pH 5.4. However, this statement contradicts the given information, as weak acids do exist as anions to some extent in their dissociated form in aqueous solution. Therefore, this option is not valid.

c. All would be anionic: This option suggests that all the weak acids would exist primarily as anions at pH 5.4. While weak acids do exist as anions to some extent, it is not necessarily the case that all weak acids will be predominantly in their anionic form at a specific pH. Therefore, this option is not valid.

d. Alloxanic acid, Ka = 2.3 x 10^-7, pKa = 6.6: The pKa of alloxanic acid is higher than pH 5.4, indicating that it will exist primarily in its neutral (non-anionic) form in aqueous solution at pH 5.4. Therefore, alloxanic acid would not primarily exist as an anion at pH 5.4.

e. Propanoic acid, Ka = 1.4 x 10^-5, pKa = 4.9: The pKa of propanoic acid is lower than pH 5.4, indicating that it will exist primarily as an anion in aqueous solution at pH 5.4.

f. Glyoxylic acid, Ka = 6.6 x 10^-4, pKa = 3.2: The pKa of glyoxylic acid is lower than pH 5.4, indicating that it will exist primarily as an anion in aqueous solution at pH 5.4.

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The replacement of a hydrogen atom in CH[tex]_{4}[/tex] by a chlorine atom causes an increase in boiling point because chlorine is more electronegative than hydrogen, resulting in a stronger dipole-dipole attraction between molecules.

When a hydrogen atom in CH[tex]_{4}[/tex] is replaced by a chlorine atom, the resulting molecule becomes CH[tex]_{3}[/tex]Cl. Chlorine is more electronegative than hydrogen, meaning it has a higher affinity for electrons. This causes the chlorine atom to pull the shared electrons closer to itself, creating a partial negative charge. The hydrogen atom in CH[tex]_{4}[/tex], on the other hand, is less electronegative, resulting in a partial positive charge.

The difference in electronegativity between chlorine and hydrogen leads to a stronger dipole-dipole attraction between CH[tex]_{3}[/tex]Cl molecules compared to CH[tex]_{4}[/tex] molecules. This increased intermolecular force requires more energy to break the attractive forces and convert the substance from a liquid to a gas, resulting in a higher boiling point.

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Answer:Endothermic, because heat is absorbed.

Explanation:

The given reaction, hydrogen gas + iodine gas → hydrogen iodide gas, is classified as exothermic because it releases heat energy during the formation of the product. Option 4.

The given reaction, hydrogen gas (H2) + iodine gas (I2) → hydrogen iodide gas (HI), is an exothermic reaction because it releases heat. Exothermic reactions are characterized by the release of energy in the form of heat, which means that the products of the reaction have lower energy compared to the reactants.

In this particular reaction, hydrogen gas and iodine gas combine to form hydrogen iodide gas. This process involves the breaking of covalent bonds in the reactants and the formation of new covalent bonds in the product.

The energy released during bond formation is greater than the energy required to break the existing bonds, resulting in a net release of energy in the form of heat.

To determine the classification of the reaction, it is necessary to consider the change in enthalpy (∆H). If ∆H is negative, it indicates an exothermic reaction, while a positive ∆H value would indicate an endothermic reaction.

Given that the reaction is exothermic, it means that the formation of hydrogen iodide gas is accompanied by the release of heat energy. This can be observed experimentally as a temperature increase in the surroundings.

The reaction releases energy in the form of heat due to the stabilization of the product, hydrogen iodide, which is more stable than the reactants, hydrogen gas and iodine gas. Option 4 is correct.

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The concentration of Ni²⁺ if the cell is allowed to run to equilibrium at 25°C is 19.7 M.

To calculate the concentration of Ni²⁺ at equilibrium, we need to compare the standard reduction potentials of the two half-reactions and use the Nernst equation.

Given the reduction potentials:

Zn²⁺ + 2e⁻ → Zn    E⁰ = -0.76 V

Ni²⁺ + 2e⁻ → Ni    E⁰ = -0.23 V

The overall cell reaction is:

Zn²⁺ + Ni → Zn + Ni²⁺

E⁰(cell) = E⁰(cathode) - E⁰(anode)

E⁰(cell) = (-0.23 V) - (-0.76 V)

E⁰(cell) = 0.53 V

The Nernst equation relates the cell potential (Ecell) to the standard reduction potentials and the concentrations of the species involved:

Ecell = E⁰cell - (0.059 V/n) log(Q)

where E⁰cell is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient.

In this case, the reaction quotient (Q) can be expressed as:

Q = [Ni²⁺] / [Zn²⁺]

At equilibrium, the cell potential (Ecell) is zero.

[tex]0 = 0.53 V - ( \frac{0.0592 V}{2}) log(Q)[/tex]

Since Ecell = 0, we can rearrange the equation to solve for Q:

log(Q) = 1

This implies that Q = 1.

Substituting Q = 1 into the reaction quotient equation:

17.96 = [Ni²⁺] / [Zn²⁺]

Given that the concentration of Zn²⁺ is 1.0 M, we can solve for [Ni²⁺]:

17.96 = [Ni²⁺] / 1.10 M

[Ni²⁺] = 19.7 M

Therefore, the concentration of Ni²⁺ at equilibrium is 19.7 M.

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A total of 22.41 grams of aluminum will react fully with 1.25 moles of Cl₂.

Given that 1.25 moles of Cl₂ will react fully with aluminum. We need to find the grams of aluminum required to react fully with it.Molar mass of Cl₂ = 35.5 × 2 = 71 g/mol

Molar mass of Al = 27 g/molNow, using the balanced chemical equation:2Al + 3Cl₂ → 2AlCl₃Moles of Al required to react fully with 1.25 moles of Cl₂ can be calculated as follows:

Moles of Cl₂ = given mass / molar mass

=> 1.25 moles = given mass of Cl₂ / 71 g/mol

=> Given mass of Cl₂ = 1.25 × 71 = 88.75 g

Moles of Al required = 2/3 × moles of Cl₂ = 2/3 × 1.25 = 0.83 moles

Now, we can find the grams of aluminum required as follows:Grams of aluminum required = moles of Al × molar mass of Al = 0.83 × 27 = 22.41 g

Therefore, 22.41 grams of aluminum will react fully with 1.25 moles of Cl₂.The balanced chemical equation for the reaction of aluminum with chlorine is:2Al + 3Cl₂ → 2AlCl₃

To calculate the amount of aluminum required to react fully with 1.25 moles of Cl₂, we first need to calculate the number of moles of Cl₂. Given the molar mass of Cl₂ as 71 g/mol, 1.25 moles of Cl₂ correspond to a mass of:1.25 moles x 71 g/mol = 88.75 g

Now, we can use the stoichiometric coefficients of the balanced chemical equation to determine the number of moles of Al required:2 mol Al / 3 mol Cl₂ x 1.25 mol Cl₂ = 0.83 mol Al

Finally, we can use the molar mass of Al to calculate the mass of Al required:0.83 mol x 27 g/mol = 22.41 g

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The given table contains a list of reagents and possible reactions for the synthesis of given compounds from butanenitrile. The compounds include:

1. Butan-1-ol: Butanenitrile can be reduced by using lithium aluminum hydride (LiAlH4) in dry ether, and then hydrolysis of intermediate to get butan-1-ol.

2. Butanoic acid: Butanenitrile can undergo hydrolysis by sodium hydroxide (NaOH) to get butanoic acid.

3. Butanal: Butanenitrile can be reduced by using lithium aluminum hydride (LiAlH4) in dry ether and then hydrolysis of intermediate to get butanal.

4. But-2-enenitrile: Butanenitrile can be treated with sodium amide (NaNH2) in liquid ammonia (NH3) to get but-2-enenitrile.

Therefore, to synthesize the given compounds from butanenitrile, the appropriate reagents from the table can be used according to the desired reaction.

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