Which of the following exhibits the weakest intermolecular forces? A) H2​O
B) NH3​
C) He D) HCl

The balanced chemical equation is: 2 CO + O2(g) = 2 CO2. According to this equation, 2 moles of carbon monoxide (CO) react with 1 mole of oxygen gas (O2) to produce 2 moles of carbon dioxide (CO2). Therefore, the correct answer is:b. 2 mol

According to the balanced chemical equation, 2 moles of carbon monoxide (2 CO) react with 1 mole of oxygen gas (O2) to form 2 moles of carbon dioxide (2 CO2). Therefore, the answer is option b, which is 2 mol. This means that for every 1 mole of oxygen gas, we need 2 moles of carbon monoxide to react completely. It is important to note that in any chemical reaction, the balanced equation tells us the stoichiometry or the ratio of the number of moles of reactants and products involved. This information is useful in determining the amount of reactants needed or the amount of products formed in a reaction.
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Acetic acid (CH3COOH) is an ingredient in vinegar. To find out how much acetic acid is present in the vinegar, titration of the acetic acid with a well-known sodium hydroxide solution will be done.

The NaOH is added to the sample of vinegar until all acetic acid is exactly absorbed (reacted off). At this stage, the reaction is complete and no additional NaOH is needed. This is known as the equivalent point of titration. According to the balanced chemical equation, one mole of acetic acid reacts with exactly 1 mole of NaOH.

When barium chloride and sulfate ions react, a precipitate of insoluble barium chloride is formed. This precipitate is then precipitated in the presence of sulfate ions, resulting in the formation of barium sulfate which is highly exothermic and can be further titrated thermometrically. Thermometrically titrated barium chloride allows for a fast and precise analysis that is fully automated.

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74.5 grams of lead (II) nitrate were reacted to produce 62.6 grams of lead (II) chloride.

To determine the mass of lead (II) nitrate that was reacted when 62.6 grams of lead (II) chloride is produced, we need to use the stoichiometry of the balanced chemical equation and calculate the molar masses of the compounds involved.

The balanced chemical equation for the reaction is:

2Pb(NO3)2 + 2NaCl → 2PbCl2 + 2NaNO3

From the equation, we can see that 2 moles of Pb(NO3)2 react to produce 2 moles of PbCl2. Therefore, the molar ratio of Pb(NO3)2 to PbCl2 is 1:1.

First, let's calculate the molar mass of PbCl2 and Pb(NO3)2:

Molar mass of PbCl2 = Atomic mass of Pb + 2 × Atomic mass of Cl

= 207.2 g/mol + 2 × 35.45 g/mol

= 278.1 g/mol

Molar mass of Pb(NO3)2 = Atomic mass of Pb + 2 × (Atomic mass of N + 3 × Atomic mass of O)

= 207.2 g/mol + 2 × (14.01 g/mol + 3 × 16.00 g/mol)

= 331.2 g/mol

Next, we can calculate the moles of PbCl2 produced:

Moles of PbCl2 = Mass of PbCl2 / Molar mass of PbCl2

= 62.6 g / 278.1 g/mol

≈ 0.225 mol

Since the molar ratio of Pb(NO3)2 to PbCl2 is 1:1, the moles of Pb(NO3)2 reacted will also be 0.225 mol.

Finally, to find the mass of Pb(NO3)2 that was reacted, we can use the moles and molar mass:

Mass of Pb(NO3)2 = Moles of Pb(NO3)2 × Molar mass of Pb(NO3)2

= 0.225 mol × 331.2 g/mol

≈ 74.5 g

Therefore, approximately 74.5 grams of lead (II) nitrate were reacted to produce 62.6 grams of lead (II) chloride.

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The volume occupied by 12.6 g of argon gas at a pressure of 1.19 atm and a temperature of 304 K can be calculated using the ideal gas law: PV = nRT.

First, we need to convert the mass of argon to moles. The molar mass of argon is 39.95 g/mol, so 12.6 g of argon is equal to 0.315 mol.
Next, we can plug in the values:
(1.19 atm) V = (0.315 mol) (0.0821 L•atm/mol•K) (304 K)
Solving for V, we get V = 8.74 L. Therefore, 12.6 g of argon gas at a pressure of 1.19 atm and a temperature of 304 K occupies a volume of 8.74 L.
To find the volume occupied by 12.6 g of argon gas at 1.19 atm and 304 K, we can use the ideal gas law formula: PV = nRT. First, we need to convert the mass of argon (Ar) to moles (n) by dividing by its molar mass (39.95 g/mol). So, n = 12.6 g / 39.95 g/mol ≈ 0.315 mol.
Now, we can plug the values into the formula:
(1.19 atm) x V = (0.315 mol) x (0.0821 L·atm/mol·K) x (304 K)
Next, solve for V:
V ≈ (0.315 x 0.0821 x 304) / 1.19 ≈ 6.45 L
Thus, the volume occupied by 12.6 g of argon gas under the given conditions is approximately 6.45 L.

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The molality of sodium carbonate in the given solution is 2.19 mol/kg.

To find the molality of sodium carbonate in the given solution, we need to use the formula:
molality = moles of solute / mass of solvent (in kg)
First, let's calculate the moles of sodium carbonate present in 510g of Na2CO3:
molar mass of Na2CO3 = 2(23) + 12 + 3(16) = 106 g/mol
moles of Na2CO3 = 510g / 106 g/mol = 4.81 mol
Next, we need to convert the mass of ethylene glycol to kg:
mass of ethylene glycol = 2.2×10^3 g = 2.2 kg
Now, we can calculate the molality of sodium carbonate:
molality = 4.81 mol / 2.2 kg = 2.19 mol/kg
It is important to note that molality is a useful unit for expressing concentrations in solutions as it does not depend on the temperature or the volume of the solution, but rather on the mass of the solvent.

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Approximately 31.1 L of [tex]CO_2[/tex] gas will be generated when 58.0 L of oxygen gas reacts according to the given balanced equation.

To determine the volume of gas generated when 58.0 L of oxygen gas reacts according to the given balanced equation, we need to consider the stoichiometry of the reaction.

From the balanced equation:[tex]CH_3CH_2OH (l) + 3O_2 (g) -- > 2CO_2 (g) + 3H_2O (l)[/tex]

We can see that for every 3 moles of [tex]O_2[/tex] consumed, 2 moles of [tex]CO_2[/tex] are produced. Therefore, we need to determine the number of moles of [tex]O_2[/tex] present in the initial 58.0 L volume.

Using the ideal gas law, PV = nRT, we can rearrange the equation to solve for moles:

n = PV / RT

At STP (Standard Temperature and Pressure), the values are:

P = 1 atm

V = 58.0 L

R = 0.0821 L·atm/(mol·K)

T = 273.15 K

n = (1 atm)(58.0 L) / (0.0821 L·atm/(mol·K) × 273.15 K)

≈ 2.25 mol

Since the stoichiometric ratio between [tex]O_2[/tex] and [tex]CO_2[/tex] is 3:2, we can determine the number of moles of [tex]CO_2[/tex] produced:

moles of [tex]CO_2[/tex] = (2/3) × moles  = (2/3) × 2.25 mol ≈ 1.50 mol

V = nRT / P

n = 1.50 mol

R = 0.0821 L·atm/(mol·K)

T = 273.15 K

P = 1 atm

V = (1.50 mol)(0.0821 L·atm/(mol·K))(273.15 K) / (1 atm) ≈ 31.1 L

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The molar solubility of AgBr in a solution containing 0.120 M NaBr is approximately 7.31 * 10^{-7} M(option A).

To determine the molar solubility of AgBr, we need to consider the common ion effect. The presence of NaBr in the solution provides the common ion (Br-) that affects the solubility of AgBr.

The solubility product constant (Ksp) expression for AgBr is given as:

Ksp = [Ag+][Br-]

Since the molar solubility of AgBr is denoted as "s," we can write the expression:

Ksp = (s)(0.120 + s)

Using the given Ksp value of 5.35 * 10^{-13} and the concentration of NaBr as 0.120 M, we can set up an equation: 5.35 * 10^{-13} = (s)(0.120 + s)

Solving this equation will give the value of "s," which represents the molar solubility of AgBr in the presence of 0.120 M NaBr. The calculated value is approximately 7.31 * 10^{-7} M.

Therefore, the molar solubility of AgBr in the solution is approximately 7.31 * 10^{-7} M (option A).

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There are two statements that correctly describe the process by which an ionic compound dissolves in water.

Firstly, the positive and negative ions dissociate from each other. Secondly, the negative ions are attracted to the partially negative O atom of the H2O while the positive ions are attracted to the partially positive H atom of the H2O. The attraction between the H2O molecules and the ions is stronger than the attraction of the ions for each other, which results in the compound dissolving completely into individual ions that remain in solution. The compound does not form pairs of oppositely charged ions that remain tightly attached.
In the process of an ionic compound dissolving in water, the positive and negative ions dissociate from each other. The negative ions are attracted to the partially positive H atoms of the H2O, while the positive ions are attracted to the partially negative O atom of the H2O. The attraction between the H2O molecules and the ions is stronger than the attraction of the ions for each other, allowing the compound to dissolve. The statement about forming pairs of tightly attached oppositely charged ions is incorrect, as ions disperse in water instead.

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The octet rule states that atoms tend to gain, lose, or share electrons in order to have a full outer shell of eight electrons.

In the compound NO, the nitrogen atom does not follow the octet rule because it only has seven valence electrons. In XeF4, the xenon atom does not follow the octet rule because it has twelve valence electrons. In OPBr3, the phosphorus atom does not follow the octet rule because it has ten valence electrons. In BF3, the boron atom does not follow the octet rule because it only has six valence electrons. In ICl2, the iodine atom does not follow the octet rule because it only has seven valence electrons. It's important to note that some elements, such as hydrogen and helium, only need two valence electrons to have a full outer shell.

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The correct option is (2) may occupy half of the tetrahedral holes in the lattice. This arrangement allows for a 1:1 ratio between the cations and anions, maintaining the chemical formula of MX.

In a simple cubic lattice, the anions (Xn-) occupy the lattice points, forming a cubic arrangement. The cations (Mn+) can occupy the vacant spaces in the lattice, which are referred to as holes.

In this case, the MX compound has the cations (Mn+) and anions (Xn-) in a 1:1 ratio. The cations can occupy two types of holes: cubic holes and tetrahedral holes.

Cubic Holes: Each cubic hole is surrounded by eight anions, forming a cube. In a simple cubic lattice, there is one cubic hole at the center of each edge and one cubic hole at the center of each face. The number of cubic holes is equal to the number of lattice points. If the cations occupy all of the cubic holes, the ratio of cations to anions becomes 1:1, which is not consistent with the formula MX. Therefore, the cations cannot occupy all of the cubic holes.

Tetrahedral Holes: Each tetrahedral hole is surrounded by four anions, forming a tetrahedron. In a simple cubic lattice, there is one tetrahedral hole at the center of each face diagonal. The number of tetrahedral holes is twice the number of lattice points. If the cations occupy half of the tetrahedral holes, the ratio of cations to anions becomes 1:1, consistent with the formula MX. Therefore, the cations may occupy half of the tetrahedral holes.

Based on the arrangement of anions and the cations in a simple cubic lattice, the cations in the MX compound can occupy half of the tetrahedral holes. This arrangement allows for a 1:1 ratio between the cations and anions, maintaining the chemical formula of MX.

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Your answer: The product of the nuclear reaction in which 28Si is subjected to neutron capture followed by alpha emission is D) 25Mg.

The product of the nuclear reaction in which 28Si is subjected to neutron capture followed by alpha emission is 25Mg. In this reaction, 28Si captures a neutron to become 29Si, which then undergoes alpha emission to produce 25Mg. This is a type of nuclear transmutation, where one element is transformed into another through nuclear reactions. The entire process can be described as follows: 28Si undergoes neutron capture to become 29Si, which then undergoes alpha emission to produce 25Mg, a lighter and more stable isotope. This reaction is important in understanding nucleosynthesis, the process by which elements are formed in the universe.
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The vibration frequency for a C-Cl bond would be lower compared to the frequency of vibration for a C-O bond.

The vibrational frequencies of bonds are determined by the masses of the atoms involved and the strength of the bond. In general, heavier atoms and stronger bonds result in lower vibrational frequencies. The atomic mass of chlorine (Cl) is greater than that of oxygen (O), and the C-Cl bond is generally stronger than the C-O bond. Therefore, based on this information, we can conclude that the vibration frequency for a C-Cl bond would be lower than the vibration frequency for a C-O bond.

To further support this conclusion, we can consider the typical range of vibrational frequencies for different types of bonds. Carbon-oxygen (C-O) bonds typically have vibrational frequencies in the range of around 1000-1400 cm-1. On the other hand, carbon-chlorine (C-Cl) bonds tend to have lower vibrational frequencies, typically falling within the range of 600-800 cm-1. This suggests that the vibration frequency for a C-Cl bond would indeed be lower than the vibration frequency for a C-O bond. Therefore, the correct answer is B: lower.

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The formula for iron pyrite, also known as fool's gold, is FeS2.

This means that it consists of one iron atom and two sulfur atoms. It is called fool's gold because it has a metallic luster and is often mistaken for real gold by amateur gold miners. Iron pyrite is an important mineral as it is a source of sulfur and also contains iron, which is a valuable metal used in many industries. However, it is not considered a reliable source of iron as it often contains impurities and is difficult to extract. In addition, it can also cause environmental problems if not properly managed as it can release sulfuric acid when exposed to air and water.

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Among the given complexes, [Co(NH3)5NO2]Cl2 will not show geometrical isomerism. This is because it has an octahedral geometry with five ammine (NH3) ligands and one nitro (NO2) ligand, resulting in no possibility of cis-trans isomerism. The other complexes can exhibit geometrical isomerism due to the presence of different ligands.

The complex compounds that show geometrical isomerism have a different spatial arrangement of ligands around the central metal atom due to the presence of a chiral center. In the given options, only [Pt(NH3)2Cl2] will not show geometrical isomerism as it has only two types of ligands, and the arrangement of these ligands around the central metal atom is symmetrical. On the other hand, [Cr(NH3)4Cl2]Cl, [Co(en)2Cl2]Cl, and [Co(NH3)5NO2]Cl2 all have chiral centers and can exhibit geometrical isomerism.
Your answer: c. [Co(NH3)5NO2]Cl2
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French fries are not a monosaccharide, disaccharide, or polysaccharide. Monosaccharides are single sugar molecules such as glucose and fructose, while disaccharides are composed of two sugar molecules linked together such as sucrose and lactose. Therefore, French fries are a complex carbohydrate and not a monosaccharide or disaccharide.

Polysaccharides are complex carbohydrates made up of many sugar molecules linked together such as starch and cellulose.
French fries are made from potatoes, which are a complex carbohydrate that contains starch. Starch is a polysaccharide made up of many glucose molecules linked together. When potatoes are fried, the high temperature causes some of the starch to break down into simpler sugars, such as glucose and fructose. However, the overall composition of French fries is still primarily complex carbohydrates, rather than simple sugars.
In summary, French fries are a complex carbohydrate and not a monosaccharide or disaccharide.

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a. The balanced equation for the reaction between magnesium (Mg) and chlorine gas (Cl₂) is:
Mg + Cl₂ → MgCl₂
b. The limiting reactant, chlorine gas (Cl₂) is the limiting reactant in this reaction.

For the reaction between magnesium and chlorine gas, the balanced equation is:
Mg + Cl2 -> MgCl2
To determine which substance limits the reaction, we need to calculate the number of moles of each substance.
The molar mass of magnesium is 24.31 g/mol, so 2.0 g of magnesium is equal to 0.0822 moles.
The molar mass of chlorine is 35.45 g/mol, so 5.0 g of chlorine gas is equal to 0.1409 moles.
To find the limiting reactant, we compare the number of moles of each substance. In this case, magnesium is the limiting reactant because there are fewer moles of magnesium (0.0822) than chlorine (0.1409).
In 100 words, we can say that the balanced equation for the reaction between magnesium and chlorine gas is Mg + Cl2 -> MgCl2. To determine the limiting reactant, we need to calculate the number of moles of each substance. 2.0 g of magnesium is equal to 0.0822 moles and 5.0 g of chlorine gas is equal to 0.1409 moles. Since there are fewer moles of magnesium, it is the limiting reactant. This means that the reaction will stop when all of the magnesium is used up and there will be some excess chlorine gas left over. It is important to know the limiting reactant in order to calculate the maximum amount of product that can be formed.
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The compound that is insoluble in water based on solubility rules is BaSO4 (barium sulfate).

The compound that is insoluble in water based on solubility rules is BaSO4 (barium sulfate). According to the general solubility rule, sulfates (SO4^2-) are typically soluble except for a few exceptions, and barium sulfate is one of those exceptions. Barium sulfate is considered insoluble in water and forms a precipitate when mixed with water or aqueous solutions.

On the other hand, the rest of the compounds listed have different solubilities in water:

Ca(NO3)2 (calcium nitrate) and AgNO3 (silver nitrate) are both soluble in water.

MgSO4 (magnesium sulfate) is soluble in water.

FeSO4 (ferrous sulfate) is also soluble in water.

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The value of ΔG°rxn for the given reaction is (b) +121.8 kJ.

The value of ΔG°rxn for the given reaction can be determined using the equation ΔG°rxn = ΔH° - TΔS°, where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change.

Given that ΔH° = +179.2 kJ and ΔS° = +160.2 J/K, we need to ensure that the units are consistent. Converting ΔS° to kJ/K, we have ΔS° = +0.1602 kJ/K.

Substituting these values into the equation, we have:

ΔG°rxn = +179.2 kJ - (358 K * 0.1602 kJ/K)

ΔG°rxn = +179.2 kJ - 57.3396 kJ

ΔG°rxn = +121.8604 kJ

Therefore, the value of ΔG°rxn for the given reaction at 358 K is approximately +121.9 kJ.

Among the provided answer choices, the closest value to +121.9 kJ is (b) +121.8 kJ.

Hence, the correct answer is (b) +121.8 kJ.

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The sum of the coefficients of phosphoric acid and ammonium hydroxide is 6.

The chemical equation for the reaction between phosphoric acid (H₃PO₄) and ammonium hydroxide (NH₄OH) is as follows:

H₃PO₄ + NH₄OH → (NH₄)₃PO₄ + H₂O

To find the sum of the coefficients, we add up the coefficients of all the compounds involved in the balanced equation:

1 + 1 + 3 + 1 = 6

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12.5%. οf a sample wοuld be left after 24.06 day

Iοdine-131 was discοvered by Glenn Seabοrg and Jοhn Livingοοd in 1938 at the University οf Califοrnia, Berkeley.

Its radiοactive decay half-life is rοughly eight days. It has a bearing οn nuclear energy, medical diagnοsis, and natural gas prοductiοn.

Tο determine the percentage οf a sample remaining after a certain time periοd, we can use the fοrmula fοr expοnential decay:

N(t) = N₀ * [tex](1/2)^{(t / T1/2)[/tex]

Where:

N(t) is the amοunt remaining after time t

N₀ is the initial amοunt

T₁/₂ is the half-life

In this case, we want tο find the percentage remaining, which can be calculated by dividing the remaining amοunt by the initial amοunt and multiplying by 100:

Percentage remaining = (N(t) / N₀) * 100

Given that the half-life οf iοdine-131 is 8.02 days, we can calculate the percentage remaining after 24.06 days:

Percentage remaining = (N(24.06) / N₀) * 100

Nοw, let's plug in the values:

Percentage remaining =[tex](0.5^{(24.06 / 8.02)})[/tex] * 100

Percentage remaining ≈ 12.5%

Therefοre, the cοrrect answer is B. 12.5%.

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The balanced nuclear equation for the nuclear fission of 235U into 92Kr and 141Ba can be written as follows:

235U + 1n → 92Kr + 141Ba + 2(1n)

In this equation, a neutron (1n) collides with a uranium-235 (235U) nucleus, resulting in the fission of the uranium nucleus. The fission products are krypton-92 (92Kr) and barium-141 (141Ba), along with the release of two additional neutrons. It is important to note that the equation represents a simplified representation of nuclear fission, and the actual process involves a complex series of reactions and the release of additional particles and energy.

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The pressure of the vapor (Pv) and liquid (PL) phases are zero at equilibrium.

To show that the conditions for vapor-liquid equilibrium at constant N (number of moles), T (temperature), and V (volume) are given by Gv = GL and Pv = PL, we can use the Gibbs free energy (G) as the thermodynamic potential.

At equilibrium, the chemical potential (μ) of the vapor (v) and liquid (L) phases are equal. The chemical potential is related to the Gibbs free energy by the equation:

μ = G / N

Since the total number of moles (N) is constant, we can write:

Gv = Nμv

GL = NμL

Now, let's consider the pressure (P) and volume (V) of the vapor and liquid phases. The pressure is related to the chemical potential by:

Pv = - (∂Gv / ∂V)T,N

PL = - (∂GL / ∂V)T,N

Since the volume (V) is constant, the partial derivatives (∂Gv / ∂V)T,N and (∂GL / ∂V)T,N are both zero. Therefore, we have:

Pv = 0

PL = 0

Combining the equations Gv = Nμv and GL = NμL, and Pv = PL = 0, we can conclude that at vapor-liquid equilibrium, Gv = GL and Pv = PL.

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The underlying assumptions of the Kinetic Molecular Theory (KMT) of gases are partially reflected in the simulation.

Due to its depiction of a simulation of individual particles moving freely inside the container, it reflects the earlier idea that the particles of a gas are small and widely separated. The particles exhibit unpredictable velocities and change position with time, representing the idea that the particles of a gas are always in a state of random motion.

The simulation also demonstrates the idea of ​​elastic collisions, as the particles collide with the walls of the container and with each other without any permanent damage. However, neither the ratio of the average kinetic energy to the temperature nor the absence of attractive or repulsive forces between the particles are clearly demonstrated by the simulations.

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The lightest pseudoscalar isovector mesons are the pions. There are three types of pions: π+, π0, and π-.

Pions primarily decay through the weak interaction, specifically the decay of a quark-antiquark pair within the meson. The decay modes of pions are as follows:

π+ decays into a muon (μ+) and a muon neutrino (νμ).

π+ -> μ+ + νμ

π- decays into an antimuon (μ-) and an antimuon neutrino (νμ-bar).

π- -> μ- + νμ-bar

π0 decays into two photons (γ).

π0 -> γ + γ

These decay modes conserve charge, lepton flavor, and baryon number. The weak interaction is responsible for these decays, which involve the transformation of one type of quark into another and the emission of appropriate leptons or photons. Pions are crucial in mediating the strong nuclear force and are involved in various interactions within atomic nuclei.

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The molality of the solution containing 30.0 g of naphthalene dissolved in 500.0 g of toluene is 0.468 mol/kg.

The molality of the solution can be calculated by dividing the moles of solute (naphthalene) by the mass of the solvent (toluene) in kilograms. In this case, 30.0 g of naphthalene is dissolved in 500.0 g of toluene.

To find the molality (m) of the solution, we need to calculate the moles of naphthalene and convert the mass of toluene to kilograms.

The molar mass of naphthalene (C10H8) is 128.18 g/mol. To find the moles of naphthalene, we divide the mass by the molar mass:

moles of naphthalene = \frac{30.0 g }{128.18 g/mol }= 0.234 mol.

Next, we convert the mass of toluene to kilograms:

mass of toluene = 500.0 g = \frac{500.0 g }{ 1000} = 0.500 kg.

Finally, we calculate the molality:

molality (m) = \frac{moles of solute }{ mass of solvent in kg}

molality =\frac{ 0.234 mol }{ 0.500 kg} = 0.468 mol/kg.

Therefore, the molality of the solution containing 30.0 g of naphthalene dissolved in 500.0 g of toluene is 0.468 mol/kg.

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The two Group B cations that form insoluble hydroxides when NH3 is added but dissolve when excess NaOH is added are Al3+ and Cr3+.

When NH3 is added to a solution containing Al3+ and Cr3+ ions, it forms insoluble hydroxides, Al(OH)3 and Cr(OH)3, respectively. These hydroxides are not very soluble and precipitate out of the solution. However, when excess NaOH is added, it reacts with the insoluble hydroxides, forming soluble complex ions. The resulting compounds, Na[Al(OH)4] and Na[Cr(OH)4], are soluble in water.

This behavior is due to the amphoteric nature of aluminum (Al) and chromium (Cr) ions. They can act as both acids and bases, forming different soluble complexes depending on the pH conditions. In the presence of NH3, they act as acids and form insoluble hydroxides. With excess NaOH, they act as bases and form soluble complex ions.

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The final temperature of the mixture is approximately -9.88°C of a 100.0 g sample of copper at 100.0 Celsius is added to 50.0g water at 26.5 degrees Celsius.

To determine the final temperature of the mixture, we can use the principle of conservation of energy, assuming no heat is lost to the surroundings.

The heat gained by the water can be calculated using the formula:

Q = m * c * ΔT, where Q is the heat gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the water:

Q_water = (50.0 g) * (4.18 J/g·°C) * (T_f - 26.5°C)

For the copper:

Q_copper = (100.0 g) * (0.385 J/g·°C) * (T_f - 100.0°C)

Since the total heat gained by the water is equal to the total heat lost by the copper (Q_water = -Q_copper), we can set up the equation:

(50.0 g) * (4.18 J/g·°C) * (T_f - 26.5°C) = -(100.0 g) * (0.385 J/g·°C) * (T_f - 100.0°C)

Now, we can solve for T_f, the final temperature of the mixture. By simplifying and rearranging the equation:

(50.0 g * 4.18 J/g·°C - 100.0 g * 0.385 J/g·°C) * T_f = -50.0 g * 4.18 J/g·°C * 26.5°C + 100.0 g * 0.385 J/g·°C * 100.0°C

T_f = (-50.0 g * 4.18 J/g·°C * 26.5°C + 100.0 g * 0.385 J/g·°C * 100.0°C) / (50.0 g * 4.18 J/g·°C - 100.0 g * 0.385 J/g·°C)

Calculating the values inside the parentheses:

T_f = (-5535 J + 3850 J) / (209 J - 38.5 J)

T_f = (-1685 J) / (170.5 J)

T_f ≈ -9.88°C

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The new gas volume is approximately 24,488 cm³.

To solve this problem, we can use the combined gas law, which relates the initial and final states of a gas sample when pressure, volume, and temperature change while the amount of gas remains constant.

The combined gas law equation is:

(P1 * V1) / T1 = (P2 * V2) / T2

Where:

P1 = Initial pressure

V1 = Initial volume

T1 = Initial temperature (which remains constant)

P2 = Final pressure

V2 = Final volume (what we need to find)

T2 = Final temperature (which remains constant)

We are given:

P1 = 1.26 × 10^3 atm

V1 = 54.3 cm³

P2 = 2.77 atm

Since the temperature remains constant, T1 = T2, we can simplify the equation to:

(P1 * V1) = (P2 * V2)

Now we can plug in the values:

(1.26 × 10^3 atm) * (54.3 cm³) = (2.77 atm) * V2

Solving for V2, we get:

V2 = (1.26 × 10^3 atm * 54.3 cm³) / (2.77 atm)

V2 ≈ 24,488 cm³

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Among the compounds listed, [tex]Ca(ClO_4)_2[/tex] will be more soluble in acidic solution than in pure water.

The solubility of a compound depends on its interaction with the solvent molecules. In the case of acidic solutions, the presence of excess hydrogen ions (H+) affects the solubility of certain compounds.

a) [tex]PbCl_2[/tex]: Lead(II) chloride ( [tex]PbCl_2[/tex]) is a sparingly soluble salt in pure water. In acidic solutions, the solubility of  [tex]PbCl_2[/tex]is not significantly affected because there are no specific interactions between lead ions and hydrogen ions.

b) FeS: Iron(II) sulfide (FeS) is insoluble in both pure water and acidic solutions. Its solubility is not influenced by the presence of acid.

c)  [tex]Ca(ClO_4)_2[/tex] : Calcium perchlorate  [tex]Ca(ClO_4)_2[/tex]  is more soluble in acidic solutions than in pure water. The perchlorate anions (ClO4-) in the compound can undergo acid-base reactions with the excess hydrogen ions in the acidic solution, increasing its solubility.

d) CuI: Copper(I) iodide (CuI) is insoluble in both pure water and acidic solutions. It does not exhibit significant solubility changes in the presence of acid.

Therefore, among the given options,  [tex]Ca(ClO_4)_2[/tex]  is the compound that will be more soluble in an acidic solution compared to pure water due to acid-base interactions between the perchlorate anions and hydrogen ions in the solution.

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The =3 level has three sublevels: s, p, and d.

There are nine orbitals in the =3 level.

The maximum number of electrons in the =3 level is 18.

The =3 level has three sublevels. There are three sublevels in total, labeled as s, p, and d. Each sublevel can hold a certain number of orbitals and electrons.

In the =3 level, the s sublevel has 1 orbital, the p sublevel has 3 orbitals, and the d sublevel has 5 orbitals. The total number of orbitals in the =3 level is 1 + 3 + 5 = 9 orbitals.

The maximum number of electrons in each orbital is 2, according to the Pauli exclusion principle. Therefore, in the =3 level, with a total of 9 orbitals, the maximum number of electrons is 9 x 2 = 18 electrons.

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