A saturated solution of barium chloride at 30 degrees Celsius contains 150 g water. How much additional barium chloride can be dissolved by heating this solution to 60 degrees Celsius?

Answer : 1) CuO is a basic salt and therefore reacts with sulfuric acid to give copper (II) sulfate and water.

2) CuO is very stable and therefore does not dissociate in water and therefore cannot react with the sulfate ion in potassium sulfate.

In first case, [tex]CuO_{(s)} + H_{2}SO_{4}_{(aq)} ----> CuSO_{4}_{(s)} + H_{2}O_{(g)}[/tex]

Wherein when CuO was made to react with potassium sulphate it didn't had any product because CuO was not ionised by water in presence of potassium sulphate.

The lattice energy of CuO (s) is very high, so it does not dissolve in water to give its ions. But it is a Bronsted–Lowry base thus it can react with the Bronsted–Lowry acid such as sulphuric acid but not with Bronsted–Lowry base [tex]{{\mathbf{K}}_{\mathbf{2}}}{\mathbf{S}}{{\mathbf{O}}_{\mathbf{4}}}[/tex].

Further Explanation:

The definition of acids and bases can be expressed in many ways based on different theories, which are as follows:

• According to Arrhenius theory, acid is defined as the one which produces hydrogen ions in a solution, while the base is defined as the one which produces hydroxide ions in a solution.

• According to Bronsted–Lowry theory, the acid in the reaction donates a proton while a base is one that accepts a proton.

• According to Lewis theory, the acid in the reaction accepts a pair of electrons while a base donates a pair of electrons.

Lattice energy is termed as the amount of energy released when the ions that exist in gaseous state combine to form compound. The lattice energy of a compound is inversely related to the size of the ions present in it.

The size of [tex]{\text{C}}{{\text{u}}^{2+}}[/tex]and [tex]{{\text{O}}^{2-}}[/tex]is small and therefore lattice energy of CuO(s) is very high. Thus it does not dissolve in water to give its ions. But since it is a Bronsted–Lowry base thus, it can accept the hydrogen ions from sulphuric acid and form [tex]{\text{CuS}}{{\text{O}}_{\text{4}}}[/tex]and [tex]{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex].

Therefore, the complete reaction is,

[tex]{\text{CuO}}\left(s\right)+{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left({aq}\right)\to{\text{CuS}}{{\text{O}}_{\text{4}}}\left({aq}\right)+{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)[/tex]

But [tex]{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left({aq}\right)[/tex] is not a Bronsted–Lowry acid. Therefore CuO (s) can not react with [tex]{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left({aq}\right)[/tex]in a way as it does with the sulphuric acid.

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Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Acids and bases

Keywords: Acids, bases, lattice energy, CuO, k2so4, Bronsted–Lowry theory, proton acceptor, proton donor, h2so4, k2so4.

The number of moles of aluminum, sulfur, and oxygen in 9.00 moles of aluminum sulfate are 18.00 moles of Al, 27.00 moles of S, and 108.00 moles of O, respectively.

To calculate the number of moles of aluminum (Al), sulfur (S), and oxygen (O) atoms in 9.00 moles of aluminum sulfate,
Al₂(SO₄)₃, we need to consider the stoichiometry of the compound. Aluminum sulfate contains 2 atoms of aluminum, 3 atoms of sulfur, and 12 atoms of oxygen per formula unit.

For aluminum (Al):
2 moles of Al are in 1 mole of Al₂(SO₄)₃, so in 9.00 moles of Al₂(SO₄)₃, there are 2 x 9.00 moles = 18.00 moles of Al.

For sulfur (S):
3 moles of S are in 1 mole of Al₂(SO₄)₃, so in 9.00 moles of Al₂(SO₄)₃, there are 3 x 9.00 moles = 27.00 moles of S.

For oxygen (O):
12 moles of O are in 1 mole of Al₂(SO₄)₃, so in 9.00 moles of Al₂(SO₄)₃, there are 12 x 9.00 moles = 108.00 moles of O.

Therefore, the number of moles of Al, S, and O atoms in 9.00 moles of Al₂(SO₄)₃ are 18.00, 27.00, 108.00 respectively, separated by commas.

Final answer:

An oxygen atom gains two electrons during a chemical reaction to form an oxide ion, achieving a stable electron configuration with ten electrons overall.

Explanation:

During a chemical reaction, an oxygen atom typically gains two electrons to achieve a stable electron configuration with a total of ten electrons. This is because a neutral oxygen atom has six valence electrons and needs two more to fill its outer shell, forming an oxide ion (O2-) with an electron configuration of 1s² 2s² 2p¶. In the context of oxidation-reduction (redox) reactions, the principle that the number of electrons lost must equal the number of electrons gained is fundamental, ensuring that charge is conserved.

The Planck's equation  is used to determine  the energy released as an electron moves to a lower orbital.

According to the Bohr's model of the atom, energy is released when an electron moves from a higher to a lower energy level.

The magnitude of this energy released is obtained by the Plank's equation; E = h × v.

This energy often appears as a photon of light.

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6.023*1023 He atoms=1 mole He

1.23×1024 helium atoms=1.23×1024/6.023*1023 =2.042 mole He

one mole He=4.0 gm

2.042 mole He=2.042*4.0=8.168 gm

answer in grams of Helium is 8.168 gm

the right answer is "Two atoms of silver are needed to complete the reaction." because two atoms of silver are needed to complete the reaction. The silver atoms are used to produce silver sulfide. There must be two atoms of silver on each side of the arrow to balance the equation. Atoms are never lost or gained in a chemical reaction.

the 2 in front of 2Ag in the chemical equation indicates that two atoms of silver are produced by this reaction. Therefore, B) Two atoms of silver are produced by this reaction.

The number in front of a chemical formula in a balanced chemical equation represents the coefficient, which indicates the ratio of moles of each substance involved in the reaction.

In this case, the 2 in front of 2Ag tells us that two moles of silver (Ag) are involved in the reaction.

Given the options provided:

B) Two atoms of silver are produced by this reaction.

Therefore, the 2 in front of 2Ag in the chemical equation indicates that two atoms of silver are produced by this reaction.

Forgetting to include the 12 H2O in formula mass calculations will result in a falsely high percent yield. Proper percent yields range from 0% to 100%, and accurate calculations are essential for true values.

If a student forgets to include the 12 H2O when calculating the formula mass of a compound, their calculation of the formula mass will be significantly lower than it should be. Consequently, if they use this incorrect lower mass to calculate the percent yield, they would end up with a percent yield that is mistakenly higher than the actual value. At extremes, it might even exceed 100%, which is typically a clear sign of an error in calculation or the presence of impurities which might be solvents like water. The correct calculation of formula mass including the hydrate water is essential to accurately determine the percent composition of the compound and subsequently the percent yield from a given reaction.

It is important to recognize that proper percent yields are between 0% and 100%; yields greater than 100% usually point towards experimental errors or contaminants within the product. A high percent yield, for instance, 80%-90%, is generally deemed good to excellent, while a 50% yield is considered fair. Including all components such as water in hydrates is crucial for correct percent composition calculations.

Final answer:

Resonance structures differ in electron arrangement around a fixed atom layout, while structural isomers have different atom arrangements and properties, like butane and isobutane. Electron-pair geometry can differ from molecular structure based on lone electron pairs' presence.

Explanation:

The structures mentioned, where all atoms are in the same relative positions but the distribution of electrons is different, describe resonance structures. These are Lewis electron structures that showcase different arrangements of electrons around a set of atoms that do not move. In contrast, structural isomers share the same chemical formula but have different physical placement of atoms and/or chemical bonds, leading to different molecular structures and properties. An example of structural isomers are butane and isobutane (C4H10), each having unique uses due to their differing arrangements. Additionally, the difference between electron-pair geometry and molecular structure depends on whether there are lone electron pairs around the central atom in a molecule.

Correct option is c) Resonance Structures describe molecules with the same atom arrangement but different electron distributions

In chemistry, structures where all atoms are positioned the same relative to one another, but the distribution of electrons around them differs, are known as resonance strcutures.

Resonance structures have the same arrangement of atoms but show different arrangements of electrons.

These forms are different ways of representing a single molecule and are connected by a double-headed arrow, indicating that while the electron distribution may vary, the core structure remains the same.

Thus the correct option is c) resonance structures

Complete question is - Structures with all atoms in the same relative position to one another, but the distribution of electrons around them is different is referred to as a)Condensed formulas b)Skeletal formulas c) Resonance structures d) Lewis structures

The answer is A. Na (apex answer)

We have to take some data like the energy needed for the formation of CO2, H2O, CO

We know that Go = H - TS

1kJ/mol = 238.846 cal/mol

C + 1/2O2 ------> CO                    Go1= -26700 - 20.95 T cal/mol

H2 + 1/2 O2 -------> H2O              Go2=-58900 + 13.1 T cal/mol

C + O2 ----------> CO2                 Go3= -94200 - 0.2 T cal/mol

Now the reaction gives

                     H2   + CO2 ------> H2O + CO

Now Go4 = -8600 - 7.65 T cal/mol

At T( K )= 900oC = 900 + 273 = 1173 K , Go4= -8693 +7.65 X 1173 = 373.45 cal

Go4 = -RT ln K

ln K = (-373.45/ -(1.986 X 1173))

K = e0.160 = 1.173

                 H2      +       CO2 ------>        H2O +    CO

intial mole   0.25       0.25 ------->        X   +     0.5

After reaction (0.25 -X)     (0.25-X)             X          (0.5+X)

Now calculate for X, we know that K = product / reactant

K = (0.5+ X)* X / (0.25-X ) * (0.25-X)                     now K= 1.17

So,    1.173(0.0625- 0.5X-X2) = 0.5X- X2

0.0733- 1.0865X+ 0.173X2= 0

Calculate the value of X using quadratic equation

value of X = 6.81 % =0.068

So P(H2O)= 0.068

Total pressure = P(CO) + P(CO2) + P(H2) + P(H2O)=1

Now putting the value of X in the following

P(H2) = P(CO2)= 0.25- 0.068= 0.182

P(CO) = 0.5- 0.068= 0.568

The composition of the equilibrium gas mixture in the furnace can be determined by using thermochemistry and equilibrium principles, setting up equations based on the initial gas composition and the balanced reactions, and solving for the equilibrium constants.

This is a thermochemistry and equilibrium problem in chemistry. The balanced equations for the reactions in the furnace are:
CO + H2 = CO2 + H2O
CO2 + H2 = CO + H2O
The first equation represents oxidation of carbon monoxide and the second equation represents the reduction of carbon dioxide. For a given reaction, the quantity of a product at equilibrium is determined by the Gibbs free energy, which in turn is dependent on the temperature, pressure, and initial concentrations/moles of the reactants.

In this scenario, the initial volume percentages translate into mole fractions since gases in a mixture occupy volume proportionally to their mole fractions. The equilibrium concentrations (or in this case mole fractions) of each gas can be determined by setting up an expression for each reaction's equilibrium constant (K) in terms of partial pressures and then solving the system of equations represented by the equilibrium constants and the initial quantities of gases, which are conserved.

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The heat released when 7.15 g CaO(s) is added to 152 g H2O(l) is -8.28 kJ.

According to the given equation and the enthalpy change for the reaction (ΔHrxn), we can calculate the amount of heat released when 7.15 g of CaO(s) is added to 152 g of H2O(l). First, we need to convert the masses of CaO and H2O to moles. The molar mass of CaO is 56.08 g/mol and the molar mass of H2O is 18.02 g/mol. So, the number of moles of CaO is 7.15 g / 56.08 g/mol = 0.1275 mol and the number of moles of H2O is 152 g / 18.02 g/mol = 8.4417 mol.

Next, we can use the stoichiometric coefficients of the balanced equation to determine the moles of products formed. From the equation, we see that 1 mole of CaO reacts to form 1 mole of Ca(OH)2. So, the moles of Ca(OH)2 formed is also 0.1275 mol.

Finally, we can use the enthalpy change value (ΔHrxn) to calculate the heat released. Since the reaction releases -64.8 kJ/mol, we can multiply this value by the number of moles of Ca(OH)2 formed (0.1275 mol) to get the heat released: -64.8 kJ/mol * 0.1275 mol = -8.28 kJ.

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Silver (Ag) is the substance most likely to be malleable among the listed materials because it is a metal with characteristics of ductility and malleability, unlike the brittle ionic compound sodium chloride or the gaseous nitrogen and propane. Hence, correct option B.

Among sodium chloride (NaCl), silver (Ag), nitrogen, and propane, the substance that is likely malleable is silver (Ag). Malleability is a property characteristic of metals that allows them to be hammered or rolled into thin sheets without breaking. Sodium chloride, being an ionic compound, is hard, brittle, and not malleable. Nitrogen is a diatomic gas at room temperature, and propane is a hydrocarbon gas, neither of which has malleable properties. Silver, however, is a metal known for its excellent malleability and ductility. It is soft enough to be worked into various shapes and maintains its structural integrity when manipulated.

The first step is to determine each compound's total molar mass by adding up all of its constituent parts.

Compounds are defined as anything made up of similar molecules with atoms from two or more different chemical elements. Chemical connections that are challenging to break are created when the elements interact with one another.

Chlorobenzene, the chemical on the left, has a mass of 112 and is composed of 5 carbons (12 g/mol), 1 chlorine (35 g/mol), and 5 hydrogens (1 g/mol). The right-hand chemical (2-chloropentane), which consists of 3 carbons, 1 chlorine, and 7 hydrogens, has a mass of 78. The mass loss for chlorobenzene is 112 - 77 = 35. The mass loss for 2-chloropropane is 78 - 77 = 1. The loss of a hydrogen ion results in a loss of one unit. This is typical and is referred to as the "M-1 peak." Therefore, the dehydrogenated molecular ion is represented by the 77 m/z fragment.

Thus, the first step is to determine each compound's total molar mass by adding up all of its constituent parts. '

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The stability of an excited-state ion, in this case, the H⁻₂ ion (diatomic hydrogen anion), depends on several factors, including the energy of the excited state, the electronic configuration, and the propensity for the ion to release energy and return to a lower energy state.

In general, when an electron is excited to a higher-energy molecular orbital, the resulting ion is not stable in the long term. Excited states are typically higher in energy and have higher potential energy compared to the ground state. As a result, the excited-state ion is often in an energetically unfavorable condition.

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The excited-state H₂²- ion is unlikely to be stable because its ground state already has a bond order of zero, indicating instability, and excitation to an antibonding orbital further destabilizes it.

The stability of the H₂²- ion in an excited state can be assessed using a molecular orbital energy-level diagram. When an electron in this ion is excited to a higher energy level, such as from a bonding to an antibonding orbital due to the absorption of light, the bond order decreases, affecting stability. In the case of H₂²-, the bond order in the ground state is already zero, indicating that the ion is not stable. Exciting an electron to an antibonding orbital would not change the bond order but would further destabilize any remnant bonding interactions. Thus, it would be unlikely for the excited-state H₂²- ion to be stable.

In a methyl tert-butyl ether sample that contains 69.0 moles of hydrogen, there are 5.75 moles of oxygen.

Methyl tert-butyl ether is an organic compound that can be represented through the semi condensed formula (CH₃)₃COCH₃ or through the condensed formula C₅H₁₂O. As we can see, in methyl tert-butyl ether the molar ratio of H to O is 12:1. The number of moles of oxygen in a sample that contains 69.0 moles of H are:

[tex]69.0 mol H \times \frac{1mol O}{12 mol H} = 5.75 mol O[/tex]

In a methyl tert-butyl ether sample that contains 69.0 moles of hydrogen, there are 5.75 moles of oxygen.

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Answer: The correct option is 3 ohms.

Solution:

When resistors are connected in parallel the total resistance is given by :

[tex]\frac{1}{R_{total}}=\sum_{i=1}^n\frac{1}{R_1}+\frac{1}{R_2}+...\frac{1}{R_n}[/tex]

[tex]R_1=12 ohm[/tex]

[tex]R_2=4 ohm[/tex]

[tex]\frac{1}{R_{total}}=\frac{1}{12}+\frac{1}{4}=3 ohm[/tex]

The total resistance of the two devices is 3 ohm .

A battery stores chemical potential energy, which is converted into electric energy when it is connected to a circuit. The energy is then used to power the circuit and is eventually transformed into other forms of energy.

A battery stores chemical potential energy. When it is connected in a circuit, a chemical reaction takes place inside the battery which converts chemical potential energy to electric energy. This electric energy powers the electrons to move through the circuit, and the energy is converted into other forms like heat and light by the circuit elements. The battery goes flat when all its chemical potential energy has been converted into other forms of energy.

Using the principles of Hooke's Law and potential energy, we calculate the spring constant when a 61 kg cheerleader stands on the spring, causing it to compress. When a second cheerleader stands on the first, causing further compression, we can calculate their combined weight.

The subject of this question is Physics, specifically the principles of Hooke's Law and the potential energy associated with a spring. In this case, we will use Hooke's law (F = kx) to figure out the spring constant first. The Spring constant (k) can be determined by dividing the force (which is equal to the weight of the cheerleader) by the compression in the spring. Hence, when a cheerleader of 61 kg stands on the spring, her weight would exert a force equal to mass x gravity, i.e, 61 kg x 9.8 m/s² = 598.8 N. As this force causes a compression of 5.8 cm or 0.058 m in the spring, the spring constant k can be calculated by 598.8 N / 0.058 m which is approximately 10324 N/m.

When the second cheerleader stands on the shoulders of the first, the spring compresses an additional 4.2 cm (or 0.042 m). The combined force exerted on the spring is the weight of both cheerleaders. If we denote x as the unknown weight of the second cheerleader, then the new force is (61kg + x kg) * 9.8m/s². This force leads to a compression of 0.058m + 0.042m = 0.1m. By again using Hooke's law, we can say (61kg + x kg) * 9.8m/s² = 10324 N/m * 0.1 m. Solving this equation will yield the weight of the second cheerleader.

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The atomic number (Z) uniquely identifies a chemical element. In an uncharged atom, the atomic number is also equal to the number of electrons.

The atomic number, Z, should not be confused with the mass number, A, which is the number of nucleons, the total number of protons and neutrons in the nucleus of an atom.

In this video Kristine Born explains this two concepts in more detail.

Given that,

The concentration of TRIS = 0.30 M

The concentration of TRIS+ = 0.60 M

Kb = 1.2 x 10^-6

pKb = -log Kb = - log (1.2 x 10^-6) = 5.920

Now, by using the Hendersonn equation,

pH = pKa + log TRIS+/TRIS = 5.920 + log (0.60/0.30) = 6.221

pOH=14-pH=14-6.221 = 7.779

Given:

Buffer system : Tris/TrisH+

[Tris] = 0.30 M

[TrisH+] = 0.60 M

pKb = 5.92

To determine:

pH of the buffer

Explanation:

The pH of a buffer can be obtain using the Henderson-Hasselbalch equation:

pH = pKa + log[Base]/[Acid]

In this case the conjugate base = [Tris]

Acid = [TrisH+]

Now, pKa = 14-pKa = 14-5.92 = 8.08

pH = 8.08 + log[0.30]/[0.60] = 7.778

Ans: pH of the buffer = 7.78

In CO, the oxidation number of C is +2 and that of O is -2. In H2, the oxidation number of H is 0. In CH3OH, the oxidation number of C is -2, that of O is -2, and that of H is +1.

To assign an oxidation number to each element in the reaction of CO(g) + 2H2(g) → CH3OH(g), we follow the standard rules for oxidation states.

In CO, carbon is more electropositive compared to oxygen, so while oxygen typically has an oxidation number of -2, carbon must balance this with a +2.

In H2, as an elemental form, hydrogen has an oxidation number of 0.

In CH3OH (methanol), the carbon atom is bonded to one oxygen atom and three hydrogen atoms. Following the rules, oxygen has an oxidation number of -2, and each hydrogen atom has an oxidation number of +1. Since carbon is bonded to four hydrogen atoms and one oxygen atom, the total for hydrogen is +3 (3*+1) and for oxygen is -2. To balance the charges, carbon must have an oxidation number of -2 in CH3OH.

The oxidation numbers in CO are +2 for carbon and -2 for oxygen. Hydrogen in H2 has an oxidation number of 0. In CH3OH, carbon's oxidation number is -2, oxygen's is -2, and hydrogen's is +1 for each atom.

The oxidation numbers for the elements in the given compounds and molecule can be assigned following certain rules. Let's go through each substance in the reaction CO(g) + 2H2(g) → CH3OH(g).

Therefore, in CO, the oxidation number of C is +2, and that of O is -2. In H2, the oxidation number of H is 0. In CH3OH, the oxidation number of C is -2, that of O is -2, and that of H is +1.

The purpose of washing the product with NaOH is simply to neutralize any acid which remained or leaked after the 1st initial separation. The NaOH base reacts with the acid to form neutralization reaction products which are soluble in water.

The distilled product in an SN2 experiment is washed with 5% NaO for the purpose of extracting and neutralizing acidic impurities to produce a pure product. This relates to the use of NaOH during titration experiments where it is used to neutralize an acid and find its acidity.

In the SN2 experiment, the distilled product was washed with 5% NaOH for the purpose of neutralization and extraction of acidic impurities. During the reaction performed, there might be some acidic byproducts which can affect the final result or the purity of the product. The washing step with sodium hydroxide, which is a base, neutralizes them and helps to obtain a pure product. This is similar to the provided titration examples where an acid is neutralized by a base. For example, in titration, HCl is neutralized by NaOH.A closely related real-world application of this concept is the use of NaOH (sodium hydroxide) during titration experiments. Here, NaOH is used to determine the acidity of an unknown solution. As indicated in the information, different solutions and volumes can be used to reach the so-called 'equivalence point' or the point at which the acid is fully neutralized by the base.

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