calculate de broglie wavelength of an electron of a lamborghini at its top speed of 350 km/h

The molecular formula of the compound, given that the compound made up of N and O contains 30.4% N is N₂O₄

First, we shall obtain the molar mass of the compound. Details below:

The mole of the gas is obtained as follow:

PV = nRT

1 × 0.974 = n × 0.0821 × 273

Divide both sides by 24.0553

n = 0.974 / (0.0821 × 273)

n = 0.043 mole

Thus, the molar mass is obtained as:

Molar mass = mass / mole

Molar mass = 4 / 0.043

Molar mass = 93 g/mol

Next, we shall obtain the empirical formula of the compound. details below:

Divide by their molar mass

N = 30.4 / 14 = 2.171

O = 69.6 / 16 = 4.35

Divide by the smallest

N = 2.171 / 2.171 = 1

O = 4.35 / 2.171 = 2

Thus, the empirical formula is NO₂

Finally, we shall obtain the molecular formula of the compound. This is shown below:

Molecular formula = empirical × n = mass number

[NO₂]n = 150

[14 + (2 × 16)]n = 150

46n = 93

Divide both sides by 46

n = 93 / 46

n = 2

Molecular formula = [NO₂]n

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

Thus, the molecular formula of the compound is N₂O₄

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Carboxylic acids are organic acids that contain the carboxyl functional group (–COOH) as their structural feature.  the structures of the carboxylic acids with the formula C6H12O2 that contain an ethyl group branching off the main chain are pentanoic acid and 3-methylbutanoic acid.

They can be found in various organic materials such as fruits, fats, and oils. The structure(s) of carboxylic acids with the formula C6H12O2 that contain an ethyl group branching off the main chain can be represented as follows:Two isomers can be possible for the given formula C6H12O2. They are pentanoic acid and 3-methylbutanoic acid.Pentanoic acid has a straight-chain of five carbon atoms (pentane) with a carboxyl group at one end and an ethyl group branching off from the fourth carbon atom. The structure of pentanoic acid is as follows:3-Methylbutanoic acid is a branched-chain carboxylic acid in which the carboxyl group is attached to the third carbon atom of a four-carbon chain, with an ethyl group attached to the second carbon atom. The structure of 3-methylbutanoic acid is as follows:Therefore, the structures of the carboxylic acids with the formula C6H12O2 that contain an ethyl group branching off the main chain are pentanoic acid and 3-methylbutanoic acid.

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Non-acetone nail polish remover, which typically contains ethyl acetate, does undergo evaporation. Ethyl acetate is a volatile organic compound with a relatively low boiling point.

It is known for its ability to evaporate quickly, making it an effective solvent in nail polish removers. The evaporation process occurs due to the intermolecular forces present in the ethyl acetate molecules. In ethyl acetate, there are two main intermolecular forces at play: dipole-dipole interactions and London dispersion forces. The presence of an oxygen atom and a carbonyl group in the molecule leads to a partial positive charge on the carbon and partial negative charge on the oxygen. This polarity allows for dipole-dipole interactions between neighboring ethyl acetate molecules.

Additionally, ethyl acetate molecules experience London dispersion forces, which arise from temporary fluctuations in electron distribution that induce temporary dipoles. These temporary dipoles can induce similar dipoles in neighboring molecules, leading to attractive forces. As the temperature increases, the kinetic energy of the molecules also increases. This results in an increased likelihood of molecules escaping from the liquid phase and transitioning into the gas phase through evaporation.

Therefore, due to the intermolecular forces present in ethyl acetate and its relatively low boiling point, non-acetone nail polish remover containing ethyl acetate does undergo evaporation.

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Answer:

50000 dollars or 5 e5

Explanation:

Mr. Garibay has [tex]5.0 * 10^4[/tex] dollars in his bank.

Scientific notation represents the data in the format of expanded number or with e character.

For instance, 5.0 * 10^4 dollars = 50000 dollars or 5 e5

The maximum electrical work that the cell can accomplish if 80.0 g of Sn is consumed is 87215 Joules (J).

To determine the maximum electrical work that the voltaic cell can accomplish, we need to calculate the maximum cell potential (E°cell) and use it to find the maximum electrical work (Wmax).

First, let's write the half-reactions for the oxidation and reduction processes

Oxidation half-reaction: Sn(s) → Sn₂+(aq) + 2e⁻

Reduction half-reaction: I₂(s) + 2e⁻ → 2I⁻(aq)

Now, we can determine the standard reduction potentials (E°red) for each half-reaction from the ALEKS Data tab

E°red(Sn₂+/Sn) = -0.14 V

E°red(I₂/I-) = 0.54 V

The standard cell potential (E°cell) can be calculated using the formula:

E°cell = E°red(reduction) - E°red(oxidation)

E°cell = 0.54 V - (-0.14 V) = 0.68 V

The maximum electrical work (Wmax) can be calculated using the equation

Wmax = nF E°cell

Where

n = moles of electrons transferred (determined from the balanced equation)

F = Faraday's constant (96485 C/mol)

From the balanced equation, we can see that 2 moles of electrons are transferred per mole of Sn consumed.

Molar mass of Sn = 118.71 g/mol

Number of moles of Sn consumed = 80.0 g / 118.71 g/mol = 0.674 mol

n = 2 moles of electrons/mol of Sn consumed = 2 × 0.674 mol = 1.348 mol

Now we can calculate Wmax

Wmax = (1.348 mol)(96485 C/mol)(0.68 V) = 87215 J

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When, unknown acid, HA, having a percent dissociation of 2.16%. If the initial concentration of the acid is 0.084 M. Then, the pH of the solution is approximately 2.74. Option C is correct.

To determine the pH of the solution, we first need to calculate the concentration of H⁺ ions in the solution. The percent dissociation is given as 2.16%, which means that 2.16% of the initial concentration of the acid dissociates into H⁺ ions.

Percent dissociation = (concentration of dissociated acid / initial concentration of acid) × 100

2.16% = (concentration of H⁺ / 0.084 M) × 100

Let's solve for the concentration of H⁺;

2.16/100 = concentration of H⁺ / 0.084

0.0216 = concentration of H⁺ / 0.084

Concentration of H⁺ = 0.0216 × 0.084

Concentration of H⁺ = 0.0018144 M

Now that we have the concentration of H⁺ ions, we can calculate the pH using the formula;

pH = -log10(concentration of H⁺)

pH = -log10(0.0018144)

pH ≈ 2.74

Therefore, the pH of the solution is approximately 2.74.

Hence, C. is the correct option.

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Entropy is an indicator of the randomness of the system. Entropy is a measure of the amount of energy in a system that is unavailable for doing useful work. The greater the entropy, the more randomized the system is, indicating that it is less likely to be able to do useful work.

In this case, we need to rank the following substances from the least to greatest in terms of standard entropy (1 as the least and 4 as the greatest) CO(l) = 2. CO(s) = 1. CO2(g) = 4. CO(g) = 3.Carbon monoxide has a liquid and solid phase at standard pressure, but its gas phase has a higher standard entropy because it is more randomized. CO2 is a more disordered and randomized system than CO because it is a gas. CO has a liquid and solid phase, but they are less disordered than the gas phase because the molecules are more structured. Therefore, the correct answer to the question is: CO(l) = 2. CO(s) = 1. CO2(g) = 4. CO(g) = 3.

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The mole fraction of oxygen gas in air from table 5.3 of the textbook is 0.2095.

In the textbook table 5.3, the mole fraction of oxygen gas in air is 0.2095. The mole fraction refers to the number of moles of a substance in a given solution divided by the total number of moles of all components present in the solution. For air, the other components include nitrogen, carbon dioxide, and other trace gases.

In conclusion, the mole fraction of oxygen gas in air from table 5.3 of the textbook is 0.2095.

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Based on the Lewis diagrams of NH3 and NF3, the correct prediction is that NH3 is a stronger base than NF3.

NH3 has a lone pair of electrons on the nitrogen atom that is available for donation to a proton, making it a Lewis base. In contrast, NF3 has a similar lone pair on nitrogen, but the presence of electronegative fluorine atoms reduces the electron density on nitrogen, making the lone pair less available for donation. The stronger electron availability in NH3 allows it to more readily accept a proton, making it a stronger base compared to NF3.

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An electrochemical cell consists of two half-cells separated by a salt bridge or porous membrane that allows ions to flow freely between the two halves.

One half-cell contains an oxidizing agent, which is responsible for accepting electrons, while the other half-cell contains a reducing agent, which is responsible for donating electrons. In an electrochemical cell, the overall reaction must be balanced so that no charge accumulates in either half-cell. The equation that represents the net cell reaction for the given electrochemical cell is as follows.Co(s) + 2Ag⁺(aq) → Co⁺²(aq) + 2Ag(s)The given electrochemical cell consists of the following half-reactions:Anode: Co(s) → Co²⁺(aq) + 2e⁻Cathode: 2Ag⁺(aq) + 2e⁻ → 2Ag(s)The Co metal oxidizes to Co²⁺ ions at the anode, producing two electrons. The Ag⁺ ions are reduced to Ag metal at the cathode, receiving two electrons. These two half-reactions combine to yield the net cell equation.

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Answer:

D [polymers]

Explanation:

The joining of monomers (small molecules) is polymerization.

The hydroxide ion concentration in the given aqueous solution is 2.86 × 10-12 M.

The given aqueous solution has a hydronium ion concentration of 3.50 × 10-3 M. To calculate the hydroxide ion concentration, the following steps need to be followed:

Step 1: Write the balanced chemical equation for the dissociation of water:

H2O(l) ⇌ H+(aq) + OH-(aq)

Step 2: Write the expression for the equilibrium constant for this reaction:

Kw = [H+(aq)][OH-(aq)]

Step 3: Substitute the value of

Kw (1.0 × 10-14 M2 at 25°C) and the given hydronium ion concentration (3.50 × 10-3 M) in the expression to solve for hydroxide ion concentration:

[OH-(aq)] = Kw/[H+(aq)] = (1.0 × 10-14 M2) ÷ (3.50 × 10-3 M) = 2.86 × 10-12 M

Therefore, the hydroxide ion concentration in the given aqueous solution is 2.86 × 10-12 M.

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Answer:

1. 7 (a neutral solution)

Answer: 10-7= 0.0000001 moles per liter

2. 5.6 (unpolluted rainwater)

Answer: 10-5.6 = 0.0000025 moles per liter

3. 3.7 (first acid rain sample in North America)

Answer: 10-3.7 = 0.00020 moles per liter

The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.

Explanation:

The more soluble compounds in water are: (a) Nickel(II) hydroxide over magnesium hydroxide, b) Copper(II) sulfide over lead(II) sulfide, (c) Magnesium fluoride over silver sulfate.

Solubility is the ability of a substance to dissolve in a solvent, such as water. It depends on several factors, including the nature of the solute and the solvent, as well as their respective chemical properties.

(a) Magnesium hydroxide (Mg(OH)2) and nickel(II) hydroxide (Ni(OH)2) are both metal hydroxides. However, nickel(II) hydroxide is more soluble in water than magnesium hydroxide. This is because nickel(II) hydroxide forms a more stable complex with water molecules, resulting in better solvation and higher solubility.

(b) Lead(II) sulfide (PbS) and copper(II) sulfide (CuS) are both metal sulfides. Copper(II) sulfide is more soluble in water than lead(II) sulfide. Copper(II) sulfide has a smaller lattice energy and forms a more stable complex with water, leading to higher solubility compared to lead(II) sulfide.

(c) Silver sulfate (Ag2SO4) and magnesium fluoride (MgF2) are both ionic compounds. However, magnesium fluoride is more soluble in water than silver sulfate. This is due to the higher lattice energy of silver sulfate and the stronger ion-dipole interactions between magnesium fluoride and water molecules, resulting in greater solubility for magnesium fluoride.

The more soluble compounds in water are:

(a) Nickel(II) hydroxide over magnesium hydroxide

(b) Copper(II) sulfide over lead(II) sulfide

(c) Magnesium fluoride over silver sulfate.

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This means that if you know how many moles of solute you have in the target solution, you also know how many moles of solute were present in the stock solution sample.

Use the molarity and volume of the target solution to determine how many moles of hydrochloric acid,

HCl

, you need in that solution

c

=

n

V

⇒

n

=

c

⋅

V

n

HCl

=

0.10 M

⋅

500

⋅

10

−

3

L

=

0.050 moles HCl

Now the question is - what volume of stock solution would contain this many moles of hydrochloric acid?

c

=

n

V

⇒

V

=

n

c

V

stock

=

0.050

moles

12

moles

L

=

0.0041667 L

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the volume of the target solution

V

stock

=

4.2 mL.follow me

Answer:

HCl

Explanation:

is a strong acI'd,,By contrast a weak acid like acetic acid (CH3COOH) does not dissociate well in water

Answer:

Gravitational potential energy depends on an object's weight and its height above the ground (GPE = weight x height).Explanation:

The complete balanced overall ionic equation for sodium iodide (NaI) dissolving in water can be written as follows: NaI(s) → Na+(aq) + I-(aq)

In this equation, NaI (s) represents solid sodium iodide, Na+(aq) represents sodium ions in the aqueous solution, and I-(aq) represents iodide ions in the aqueous solution.

When solid sodium iodide is added to water, it dissociates into its constituent ions, sodium ions (Na+) and iodide ions (I-). This dissociation occurs due to the attraction between the positive and negative charges in the water molecules and the ions of the solid.

The resulting solution contains sodium ions and iodide ions, both of which are now freely mobile in the aqueous medium. This dissociation process is reversible, meaning that if the solution is evaporated, the ions can recombine to form solid sodium iodide again.

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The electron configurations for each of the following ions are as follows:

A. Cl⁻: 1s² 2s² 2p⁶ 3s² 3p⁶

B. K⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

C. P3⁻: 1s² 2s² 2p⁶ 3s² 3p⁶

D. Mo³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

E. V³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²

Electronic configuration refers to the arrangement of electrons in the energy levels, orbitals, and sub-orbitals of an atom or ion. It describes the distribution of electrons among the various energy levels and subshells within an atom. The electronic configuration provides information about the organization and stability of an atom, as well as its chemical properties. It is typically represented using a notation that indicates the number of electrons in each energy level and subshell, such as the superscript notation (e.g., 1s² 2s² 2p⁶) used to represent the electronic configuration of an atom.

Therefore, the electron configurations for each of the following ions are as follows:

A. Cl⁻: 1s² 2s² 2p⁶ 3s² 3p⁶

B. K⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

C. P3⁻: 1s² 2s² 2p⁶ 3s² 3p⁶

D. Mo³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

E. V³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²

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Answer:

The answer is maybe B hope it helps

The substances which are made up of only one kind of particle and have a fixed or constant structure is defined as the pure substance. Here One block has a greater volume than the other block so that the blocks are composed of different substances. The correct option is B.

The amount of space occupied by any three dimensional solid is known as the volume. These solids can be a cube, a cuboid, a cone, a cylinder or a sphere. It is simply how much space an object or substance takes up.

It is possible to have the volume without mass, such as an enclosed vacuum. The capacity of a container is not necessarily the same as its volume. It is the interior volume of a vessel. If you measure the exterior dimensions of the container, its volume is greater than its capacity.

Here the blocks are composed of different substances is indicated by the difference in the volume of the blocks.

Thus the correct option is B.

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Calcium carbonate crystals can be distinguished from bacteria based on several key factors. Firstly, their physical characteristics differ significantly.

Calcium carbonate crystals have a distinct geometric shape, such as rhomboids, hexagons, or prisms, which can be observed under a microscope. In contrast, bacteria are living microorganisms that possess cellular structures, such as membranes, cytoplasm, and genetic material.

Secondly, the size of calcium carbonate crystals tends to be larger and more uniform compared to the varied sizes of bacteria. Additionally, calcium carbonate crystals are inert structures, lacking the metabolic activities and biological functions exhibited by bacteria.

By considering these factors and employing microscopic examination, it is possible to differentiate calcium carbonate crystals from bacteria with a reasonable degree of accuracy.

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Given,Initial volume, Vi = 10-3 m3Pressure, P = 10^4 PaTemperature, T1 = 25°CAvogadro's Number, NA = 6.022×1023 atoms/molAtomic weight of Argon gas, m = 0.040 kg/mole

Explanation: 1) What is the number density of atoms in this chamber?Number density is given by:N/V = PNAT1V1 = 10^4×6.022×1023/8.314×298×10-3N/V = 2.4299 × 1024 atoms/m3Therefore, the number density of atoms in the chamber is N/V = 2.4299 × 1024 atoms/m3

2) What is the new temperature, T?Volume of the container is changed from V1 to V2Pressure remains constantTemperature of the gas changes from T1 to T2Since the expansion is free expansion, the internal energy of the gas remains constantFor an ideal gas,U = (3/2)Nk(T2 - T1)Where k is the Boltzmann constant or the gas constant divided by the Avogadro number k = R/NA = 8.314/6.022×1023 = 1.381×10-23 JK-1Therefore, U = (3/2)PV(T2 - T1)/kV1 = (3/2)(P/NA)(T2 - T1)V1/kV2 = V1 × 3 = 3×10-3m3T2 = T1 × V1/V2T2 = 25 × 10-3/3 = 8.33°CThus, the new temperature T is T = 8.33°C

3) What is the new pressure, P?According to Boyle's Law, P1V1 = P2V2P2 = P1V1/V2P2 = 10^4×10-3/(3×10-3)P2 = 3333 PaTherefore, the new pressure is P2 = 3333 Pa

4) How much work is needed to do this?In the compression process, work is done on the system.W = -∫PdVWhere, P = P(V) is the pressure as a function of the volume V.The compression is done slowly and isothermal, which means that the temperature remains constant at T1 = 25°CSo the ideal gas law,PV = NkTTemperature remains constant during the compression,So, P = NkT/V = nRT/VWhere n is the number of moles of gas and R is the molar gas constantWe have seen before thatN/V = P/kTRearranging this expression gives us N = (PV/kT)Therefore,W = -∫PdV = -∫(nRT/V)dV = -nRT ln(Vf/Vi)The amount of gas remains constant, so n is constant.The final volume is Vf = Vi = 10-3 m3W = -nRT ln(Vf/Vi)W = -PV ln(Vf/Vi)Since Vf/Vi = 1/3,W = -PV ln(1/3)W = PV ln(3)W = 10^4 × 10-3 × 0.040 × 8.31 × ln(3)W = -106.6 JThus, the amount of work needed to compress the gas back to its initial volume is W = -106.6 J.

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To calculate the required flow rate of the entering steam and the rate of heat transfer from water to methanol, we can use the principles of energy balance and heat transfer.

Given data:

Temperature of saturated steam entering the heat exchanger (T1): 273.3°C

Temperature of methanol vapor entering the heat exchanger (T2): 70.0°C

Temperature of methanol vapor leaving the heat exchanger (T3): 252.9°C

Temperature of water leaving the heat exchanger (T4): 90.0°C

Flow rate of methanol vapor (Q2): 6530 standard liters per minute

Step 1: Calculate the required flow rate of entering steam (Q1) in m³/min.

We need to determine the mass flow rate of methanol vapor (m2) and the mass flow rate of water (m4) to perform the energy balance.

To convert the flow rate of methanol vapor from standard liters per minute to m³/min, we use the ideal gas law:

PV = nRT

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

Using the ideal gas law, we can calculate the volume of methanol vapor (V2) in m³/min:

V2 = (Q2 * R * T2) / (P * 1000)

Assuming the pressure is constant, we can write the equation for the methanol vapor mass flow rate (m2) as:

m2 = V2 * ρ2

where ρ2 is the density of methanol vapor at temperature T2.

Step 2: Calculate the mass flow rate of water (m4) in kg/min.

We can determine the mass flow rate of water based on its specific heat capacity (Cp) and the temperature change (ΔT = T3 - T4).

Using the equation:

Q = m4 * Cp * ΔT

where Q is the heat transferred from water to methanol, we can solve for m4:

m4 = Q / (Cp * ΔT)

Step 3: Calculate the required flow rate of entering steam (Q1) in m³/min.

To perform the energy balance, we assume that the heat transferred from the steam to the methanol is equal to the heat transferred from the water to the methanol:

m1 * H1 = m2 * H2 + m4 * H4

where m1 is the mass flow rate of entering steam, H1 is the enthalpy of saturated steam at temperature T1, H2 is the enthalpy of methanol vapor at temperature T2, and H4 is the enthalpy of water at temperature T4.

Rearranging the equation to solve for m1:

m1 = (m2 * H2 + m4 * H4) / H1

Step 4: Calculate the rate of heat transfer from water to methanol (Q) in kW.

We can now substitute the values of m2, m4, and H2, H4 into the energy balance equation to calculate Q:

Q = m2 * (H2 - H3)

where H3 is the enthalpy of methanol vapor at temperature T3.

Finally, we can convert the flow rates from kg/min to m³/min by dividing by the density of water and methanol vapor, respectively.

Conclusion:

The required flow rate of entering steam (Q1) can be calculated using the mass flow rates and enthalpies, and the rate of heat transfer from water to methanol (Q) can be determined using the enthalpies.

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Answer:

Option 2 and 4 are correct

Explanation:

The reactants in the attached image have more enthalpy and hence less stability as they are more reactive. Thus, Product is more stable than the reactants.

This is an addition reaction in which two reactants add up to form the product.

Very less activation energy is required as the reactants themselves are unstable, possess high energy and hence are very reactive.

Reactants have more energy than the products.  

In the presence of 0.10 M NH₄NO₃ or 0.10 M NaCH₃COO, the solubility of barium phosphate is expected to be similar to its solubility in pure water. In the presence of 0.10 M Na₃PO₄ or 0.10 M Ba(CH₃COO)₂, the solubility of barium phosphate is likely to be reduced compared to its solubility in pure water.

To determine the effect of each aqueous solution on the solubility of barium phosphate, we need to consider the common ion effect and the solubility product constant (KSP) of barium phosphate (Ba₃(PO₄)₂).

0.10 M NH₄NO₃

NH₄NO₃ does not contain any common ions with barium phosphate. Therefore, the solubility of barium phosphate is likely to be similar to its solubility in pure water. The correct answer is 2) similar solubility as in pure water.

0.10 M Na₃PO₄

Na₃PO₄ introduces PO₄⁻ ions, which are common ions with barium phosphate. According to the common ion effect, the presence of a common ion reduces the solubility of a compound. Therefore, the solubility of barium phosphate is expected to be reduced in the presence of Na₃PO₄. The correct answer is 3) less soluble than in pure water.

0.10 M Ba(CH₃COO)₂

Ba(CH₃COO)₂ introduces Ba₂⁺ ions, which are common ions with barium phosphate. Again, according to the common ion effect, the solubility of barium phosphate is likely to be reduced in the presence of Ba(CH₃COO)₂. The correct answer is 3) less soluble than in pure water.

0.10 M NaCH₃COO

NaCH₃COO does not contain any common ions with barium phosphate. Therefore, the solubility of barium phosphate is likely to be similar to its solubility in pure water. The correct answer is 2) similar solubility as in pure water.

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the alkyl halide (CH3)2CHCH2CH2C! is likely the least reactive in an E2 reaction among the provided options.

The reactivity of alkyl halides in an E2 (elimination bimolecular) reaction is influenced by the stability of the carbocation intermediate formed during the reaction. In general, more substituted alkyl halides form more stable carbocations, resulting in increased reactivity in E2 reactions.

Let's analyze the given options:

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5.0 mol Al produces mass up to 255 g  of Al2O3.

The balanced equation for the reaction is:4Al + 3O2 → 2Al2O3. We have to find the mass of Al2O3 produced when 5.0 mol Al produces up to 2.5 mol Al2O3. To find the mass of Al2O3 produced, we have to follow the steps given below: Step 1: Calculate the number of moles of Al2O3 produced when 5.0 mol Al produces up to 2.5 mol Al2O3.We can use the stoichiometric coefficients of Al and Al2O3 to calculate the number of moles of Al2O3 produced. According to the balanced equation,1 mol Al produces 0.5 mol Al2O3(2.5 mol Al2O3 / 5 mol Al) = 0.5 mol Al2O3Therefore, 5.0 mol Al produces up to 0.5 * 5.0 = 2.5 mol Al2O3Step 2: Calculate the mass of Al2O3 produced when 5.0 mol Al produces up to 2.5 mol Al2O3.To calculate the mass of Al2O3 produced, we can use the molar mass of Al2O3 and the number of moles of Al2O3 produced. The molar mass of Al2O3 is 102 g/mol. Mass of Al2O3 produced = Number of moles of Al2O3 produced * Molar mass of Al2O3= 2.5 mol * 102 g/mol= 255 g. Therefore, 5.0 mol Al produces up to 255 g of Al2O3.

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The compounds HCOOH and HCOONa, HF and NaF form a buffer in aqueous solution.

A buffer solution is composed of a weak acid and its conjugate base (or a weak base and its conjugate acid) that can resist changes in pH when small amounts of acid or base are added. Based on this definition, the set of compounds that would form a buffer in aqueous solution is:

HCOOH and HCOONa

HCOOH (formic acid) is a weak acid, and HCOONa (sodium formate) is its conjugate base. Together, they can act as a buffer system in aqueous solution.

The other compounds listed do not form a buffer system:

NaF and KF - These are salts, not weak acid-conjugate base pairs.

HBr and NaBr - These are both strong acids, not weak acid-conjugate base pairs.

HF and NaF - This is a weak acid-conjugate base pair and can form a buffer system.

HF and KCN - These are not weak acid-conjugate base pairs.

KF and KOH - These are both strong bases, not weak acid-conjugate base pairs.

NaBr and KBr - These are salts, not weak acid-conjugate base pairs.

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15. Lead (II) chloride

16. potassium chloride

17. Lithium oxide

18. Arsenious trioxide

19.Phosphorus tribromide

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The gas will fill the entire 25 mL of the container. Hope this helps

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