Which statement is true about a polyatomic ion? it forms metallic bonds with other ions. It forms covalent bonds with other ions. It is made of atoms that are covalently bonded together. It has a charge that is distributed over only part of the ion.

D. Peridotite is probably closest in chemical composition to the upper mantle.

Peridotite, a coarse-grained, darkish-coloured, heavy, intrusive igneous rock that carries as a minimum 10 percentage olivine, different iron- and magnesia-wealthy minerals (normally pyroxenes), and now no longer greater than 10 percentage feldspar. Uses - as a supply of precious ores and minerals, inclusive of chromite, platinum, nickel and valuable garnet; diamonds are acquired from mica-wealthy peridotite (kimberlite) in South Africa. Peridotite is the overall call for the ultrabasic or ultramafic intrusive rocks, darkish inexperienced to black in color, dense and coarse-grained texture, frequently as layered igneous complex.

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Complete question-

_____________ is probably closest in chemical composition to the upper mantle.

A. Granite

B. Shale

C. Andesite

D. Peridotite

The mass, in grams, of zinc electroplated out of a solution containing zinc ion if the solution is electrolyzed for 1 hour and 9 minutes at a current of 0.407 amps would be 0.571 grams.

According to Faraday's laws of electrolysis, the amount of substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity that passes through the cell.

Q = It

1 hour 9 minutes = 60 minutes + 9 minutes = 69 minutes

69 minutes = 69 x 60 seconds = 4140 seconds

Thus:

Q = It = 0.407 A x 4140 s = 1685.88 C

The molar ratio between zinc and the number of electrons transferred during the electrolysis to determine the moles of zinc deposited:

2 mol e- → 1 mol Zn

The moles of zinc deposited can be calculated as:

moles Zn = Q / (2 x 96,485 C/mol e-) = 8.744 x 10^-3 mol

mass Zn = moles x molar mass

= 8.744 x 10^-3 x 65.38

= 0.571 g

In other words, approximately 0.571 grams of zinc would be electroplated out of the solution.

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We can plug them into the Henderson-Hasselbalch equation to calculate the pH 2.14

Ascorbic acid, also known as Vitamin C, is a naturally occurring water-soluble vitamin found in citrus fruits, green vegetables, and other fruits and vegetables. It is an essential nutrient for humans, playing an important role in the growth and repair of tissues, and helping to protect the body from free radicals and other environmental toxins.

Ka1 for H₂C₆H₆O₆ = 7.5 x 10⁻⁵
The initial concentration of H₂C₆H₆O₆ is 0.0942M.
We can calculate the pH of the solution by using the Henderson-Hasselbalch equation:
pH = pKa1 + log ([C₆H₆O₆²⁻] / [H₂C₆H₆O₆)
Since we don't know the concentrations of H₂C₆H₆O₆ and C₆H₆O₆²⁻, we can use the law of mass action to solve for them.
At equilibrium,
[H₂C₆H₆O₆] * [C₆H₆O₆²⁻] = Ka1 * [H₂C₆H₆O₆]²
[C₆H₆O₆²⁻] = Ka1 * [H₂C₆H₆O₆]
Substituting the initial concentration of H₂C₆H₆O₆:
[C₆H₆O₆²⁻] = (7.5 x 10⁻⁵) * (0.0942) = 7.08 x 10⁻⁶ M
Now that we have the concentrations of both species, we can plug them into the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa1 + log ([C₆H₆O₆²⁻] / [H₂C₆H₆O₆])
pH = -log(7.5 x 10⁻⁵) + log (7.08 x 10⁻⁶ / 0.0942) = 2.14

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Using the Henderson-Hasselbalch equation the pH of the system is 2.14

The concentration of an acidic or basic species, the conjugate base or acid, and the pH of a solution are all connected by the Henderson-Hasselbalch equation.

We know that;

pH = pKa1 + log ([C₆H₆O₆²⁻] / [H₂C₆H₆O₆)

Then;

[H₂C₆H₆O₆] * [C₆H₆O₆²⁻] = Ka1 * [H₂C₆H₆O₆]²

[C₆H₆O₆²⁻] = Ka1 * [H₂C₆H₆O₆]

Therefore;

[C₆H₆O₆²⁻] = (7.5 x 10⁻⁵) * (0.0942) = 7.08 x 10⁻⁶ M

And we have that;

pH = -log(7.5 x 10⁻⁵) + log (7.08 x 10⁻⁶ / 0.0942)

pH = 2.14

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The amount of CO2 released by each barrel of crude oil is 150 pounds of CO2 per million BTUs multiplied by 5.8 million BTUs, which equals 870,000 pounds of CO2.

Barrels are cylindrical containers used for storing and transporting materials such as oil, wine, beer, and other liquids. Barrels are usually made of metal, usually steel, or wood, which is often used for storing alcohol. Barrels come in many different sizes and shapes, with the most common being the 55 gallon drum. Barrels have been used for centuries for storing and transporting goods, and are still used today in many industries. Barrels are often used in wineries and breweries to store and age wine and beer. Barrels are also used to transport oil and other hazardous materials, and even as a form of storage for food and other items.

A barrel of crude oil contains approximately 5.8 million BTUs of energy. Therefore, the amount of CO2 released by each barrel of crude oil is 150 pounds of CO2 per million BTUs multiplied by 5.8 million BTUs, which equals 870,000 pounds of CO2.

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The molar solubility of CuCl in a solution containing 0.060 M KCl is calculated as 2 × 10⁻⁵ M

Option C is correct.

The number of ions dissolved per liter of solution is referred to as molar solubility. Here, dissolvability addresses the quantity of particles broke down in a given measure of dissolvable.

K(CuCl)=1.0×10

Concentration of KC1 = 0.050 M

The dissociation of CuCl :  CuCl ------ > Cu + Cl

Total concentration of Cl Ion is (s + 0.05) × M

Presently, consider the normal particle impact, or at least, on the off chance that two solids are disintegrated in an answer having a typical particle, the convergence of the normal particle increments. Because KCl is a strong electrolyte, the chloride ion in KCl has a concentration of 0.050 M.

Hence, the molar solubility of CuCl is mentioned below:

K sp = = (s+0.050) 1 × 10 ^ - 6 = s(s + 0.05)

Since, s < 0.050 M. Therefore, 1 × 10 ⁻⁶= s × 0.05

                        s = 2 × 10⁻⁵

Thus, the molar solubility of CuCl is calculated as 2 × 10⁻⁵ M

Temperature, pressure, and the solid's polymorphic form all affect solubility. Thermodynamic solvency is the convergence of the solute in immersed arrangement in balance with the most steady gem type of the strong compound.

Incomplete question:

Determine the molar solubility of CuCl in a solution containing 0.060 M KCl. K sp (CuCl) = 1.0 × 10 -6.

A. 1.0 × 10-3 M

B.1.7 × 10-5 M

C. 2 × 10⁻⁵ M

D. 1.0 × 10-12 M

E. 0.050 M

F. 6.0 × 10-8 M

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Cr (chromium) and Cu (copper) do not have the expected electron configurations because they achieve greater stability by having a half-filled or fully filled d-subshell.

According to the Aufbau principle, electrons are filled in orbitals following a specific order. However, chromium and copper are exceptions to this rule. Chromium's expected electron configuration is [Ar] 4s2 3d4, but it actually has [Ar] 4s1 3d5 configuration.

Copper's expected electron configuration is [Ar] 4s2 3d9, but its actual configuration is [Ar] 4s1 3d10.

These exceptions occur because having a half-filled (in chromium) or fully filled (in copper) d-subshell provides extra stability due to a lower energy state and better electron repulsion minimization.

Summary: Cr and Cu have unexpected electron configurations because they achieve greater stability by having half-filled (Cr) or fully filled (Cu) d-subshells, which lowers their energy state and minimizes electron repulsion.

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The precipitate after the addition of 6 M HCl would be AgCl, NiS, and HgS. Ba₃(PO₄)₂ and NH₄Cl would remain in solution, while CdS and ZnS would react with the HCl to form NiS and HgS, respectively.

Solution is a term used to describe a process or methodology for addressing a problem or issue. It is a way of resolving an issue or a challenge by finding a workable and effective course of action. Solutions can be applied to a variety of different types of issues, such as technical problems, business challenges, and social issues. Solutions can involve a variety of different approaches, such as technological solutions, policy changes, and organizational changes. Solutions should be tailored to the specific problem and should take into account the goals, resources, and stakeholders involved.

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Secondary structure refers to the arrangement of the covalently bonded atoms in a protein or nucleic acid.

Moleculer is a microservice toolkit for Node.js. It is an open-source framework that enables developers to create, deploy and manage distributed systems in a fast and efficient way.

It is determined by the hydrogen bonding between the atoms, which results in the formation of specific three-dimensional shapes, such as the alpha helix, beta sheet, and loop structures. Hydrogen bonds form between the amide hydrogen atoms of the peptide backbone and carbonyl oxygen atoms of the peptide backbone, as well as between side chain atoms, such as the oxygen and nitrogen atoms of the amino acids. This hydrogen bonding helps to stabilize the secondary structure of the protein or nucleic acid, as the bonds create a lattice-like structure that holds the molecule together.

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Answer:the resistance of a material is determined by its length, cross-sectional area, and resistivity

Explanation:Resistance is a measure of how difficult it is for electric current to flow through a material. There are several quantities that determine the resistance of a piece of material, including its length, cross-sectional area, and the material's resistivity.

Let's start with length. The longer a material is, the more resistance it will have. This is because a longer path means that there is more material for the current to travel through, which increases the likelihood of collisions between electrons and atoms. For example, a long copper wire will have more resistance than a shorter copper wire of the same cross-sectional area.

The cross-sectional area of a material also plays a role in determining resistance. The larger the cross-sectional area of a material, the lower its resistance will be. This is because a larger cross-sectional area means that there is more space for electrons to flow through, reducing the likelihood of collisions. For example, a thick copper wire will have less resistance than a thin copper wire of the same length.

Finally, the material's resistivity is a key factor in determining its resistance. Resistivity is a measure of how strongly a material opposes the flow of electric current. Materials with high resistivity, like rubber or glass, will have a high resistance, while materials with low resistivity, like copper or silver, will have a low resistance.

To calculate the resistance of a material, we can use Ohm's Law, which states that resistance is equal to voltage divided by current. The unit of resistance is the ohm (Ω).

In summary, the resistance of a material is determined by its length, cross-sectional area, and resistivity. By understanding these factors, we can choose materials and design circuits that minimize resistance and improve electrical efficiency.

The molarity of Bi3+ in the resulting solution is 0.050 M.

Assuming that all of the bismuth in the tablet dissolves and forms BiCl3, we can use stoichiometry to determine the molarity of Bi3+ in the resulting solution.

First, we need to convert the mass of Bi2S3 in the tablet to moles. The molar mass of Bi2S3 is 514.16 g/mol, so:

425 mg Bi2S3 x (1 g / 1000 mg) x (1 mol Bi2S3 / 514.16 g) = 8.26 x [tex]10^{4}[/tex] mol Bi2S3

Since each mole of Bi2S3 produces 3 moles of Bi3+ ions upon reaction with HCl, we can determine the number of moles of Bi3+ ions in the solution:

8.26 x [tex]10^{4}[/tex] mol Bi2S3 x (3 mol Bi3+ / 1 mol Bi2S3) = 2.48 x [tex]10^{3}[/tex] mol Bi3+

We diluted the solution to a final volume of 50 mL, or 0.050 L. Thus, the molarity of Bi3+ in the resulting solution is:

Molarity = moles of solute / volume of solution

Molarity = 2.48 x [tex]10^{3}[/tex] mol / 0.050 L = 0.050 M

Therefore, the molarity of Bi3+ in the resulting solution is 0.050 M.

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The balanced chemical equation for the formation of benzene is: 2R-MgBr + R'-Br → 2R-R' + MgBr₂ → R-R' + HBr → R=R' + H₂ → C₆H₆ + HBr.

A chemical equation is a symbolic representation of a chemical reaction which shows the reactants, products, and direction of the reaction. It is written in the form of two or more chemical substances (e.g. reactants) separated by an arrow, with each substance on either side of the arrow represented by a chemical formula. The arrow indicates the direction of the reaction and can represent either the reactants becoming products or the products breaking down into reactants. The equation also shows the number of molecules of each substance that are involved in the reaction.

The formation of benzene during the synthesis of phenylmagnesium bromide is the result of a reverse coupling reaction. Reverse coupling reactions occur when the Grignard reagent is treated with an alkyl halide, such as bromide, in the presence of a strong base. In this case, the Grignard reagent (R-MgBr) reacts with the alkyl halide to produce an alkyl radical, which then couples with the Grignard reagent to form an alkane. This alkane then undergoes dehydrohalogenation to form an alkene, which is then aromatized to produce benzene.

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The liquid that is poured from the vessel without disturbing the solid is called the supernatant. It is the clear liquid layer that sits on top of the solid or sediment after it has settled due to gravity or centrifugation.

The supernatant is separated from the solid or sediment because it has a lower density than the solid or sediment. The process of separating the liquid from the solid or sediment is called decantation. Decantation is a simple separation technique used in various fields, such as chemistry, biology, and environmental science. It is commonly used to separate mixtures of immiscible liquids, such as oil and water, or a solid and liquid, such as sand and water. To perform decantation, the mixture is left undisturbed for a period of time, allowing the solid or sediment to settle to the bottom of the container. The liquid is then carefully poured off, leaving the solid or sediment behind.

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The pKb for hypochlorite ions at 25°C is 6.48.

The Ka expression for hypochlorous acid is:

Ka = [H+][OCl-]/[HOCl]

where [H+] is the hydrogen ion concentration, [OCl-] is the hypochlorite ion concentration, and [HOCl] is the hypochlorous acid concentration.

Since HOCl is an acid and OCl- is its conjugate base, we can write the acid dissociation constant expression for the conjugate base, OCl-:

Kb = [OH-][HOCl]/[OCl-]

where [OH-] is the hydroxide ion concentration.

The product of Ka and Kb for a conjugate acid-base pair is equal to the ion product constant of water, Kw:

Ka x Kb = Kw = 1.0 x 10^-14 (at 25°C)

Substituting the given value of Ka and solving for Kb, we get:

Kb = Kw/Ka = (1.0 × [tex]10^-14[/tex])/(3.0 x[tex]10^-8[/tex]) = 3.33 x 10^-7

The pKb for hypochlorite ions is:

pKb = -log(Kb) = -log(3.33 × [tex]10^-7[/tex]) = 6.48

Therefore, the pKb for hypochlorite ions at 25°C is 6.48.

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To find the [OH⁻] in a 0.20 M oxalic acid (COOH)₂ solution, we need to first determine the dissociation constant (Ka) of oxalic acid and calculate the [H⁺] concentration using the given molarity. Then, we can use the ion product constant of water (Kw) to find the [OH⁻] concentration.This value is closest to option (c), so the answer is c. 1.2 × 10⁻¹³ M

Oxalic acid is a weak diprotic acid with two dissociation steps. However, the first dissociation step contributes significantly more [H⁺] ions than the second step. Thus, we can focus on the first dissociation: (COOH)₂ → H⁺ + (COOH)⁻
The Ka for the first dissociation of oxalic acid is approximately 6.5 × 10⁻⁴. Using the given 0.20 M concentration, we can set up an equilibrium expression to solve for [H⁺]:
Ka = [H⁺][(COOH)⁻] / [(COOH)₂]
Assuming x is the concentration of [H⁺] and [COOH⁻], we get: 6.5 × 10⁻⁴ = x² / (0.20 - x)
Since x is small, we can approximate 0.20 - x ≈ 0.20, which gives: x ≈ √(6.5 × 10⁻⁴ × 0.20) ≈ 0.0113 M

Now that we have the [H⁺] concentration, we can use the ion product constant of water, Kw = 1.0 × 10⁻¹⁴, to find the [OH⁻] concentration: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ = (0.0113)([OH⁻])
[OH⁻] ≈ 8.85 × 10⁻¹³ M

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To find the pH of the solution, we need to first calculate the concentration of H+ ions in the solution using the dissociation constant (Ka) of hydrocyanic acid. Therefore, the correct option is (d) 4.699.

The dissociation reaction of HCN is: HCN ⇌ H+ + CN-

The expression for the Ka is: Ka = [H+][CN-]/[HCN]

We are given the Ka value as 4.0 x 10^-10, the concentration of HCN as 0.445 M, and we assume the concentration of [CN-] is negligible compared to [HCN].

Therefore, we can write: Ka = [H+][CN-]/[HCN] ≈ [H+]^2/[HCN]

Solving for [H+], we get: [H+] = sqrt(Ka x [HCN]) = sqrt(4.0 x 10^-10 x 0.445) = 1.881 x 10^-6 M

Now, we can calculate the pH using the equation: pH = -log[H+]

Substituting the value of [H+], we get: pH = -log(1.881 x 10^-6) ≈ 4.699

Therefore, the correct answer is (d) 4.699.

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Necessary ratio to make buffer solution will 0.39 : 1  [CH₃CO₂⁻]/[CH₃CO₂H] with pH of 4.34 according to Henderson–Hasselbalch Equation,

Option D is correct .

With respect to Henderson–Hasselbalch Equation,

                                   pH  =  pKa + log [Acetate] / [Acetic Acid]

As,                                       =  pKa + log[CH₃CO₂⁻]/[CH₃CO₂H]

          pKa = -log Ka

          pKa = -log (1.8 × 10⁻⁵)

          pKa =  4.74

So ,  pH  =  4.74 + log[CH₃CO₂⁻]/[CH₃CO₂H]

       4.34  =  4.74 + log [Acetate] / [Acetic Acid]

          4.34 - 4.74  = log[CH₃CO₂⁻]/[CH₃CO₂H]

       -0.40  =  log[CH₃CO₂⁻]/[CH₃CO₂H]

Taking Antilog of both sides,

             [Acetate] / [Acetic Acid]  = 0.39 : 1

 Hence , the required ratio to make buffer solution with pH will be 0.39 : 1  

The Henderson-Hasselbalch condition gives a connection between the pH of acids (in watery arrangements) and their pKa (corrosive separation steady). The pH of a cushion arrangement can be assessed with the assistance of this situation when the centralization of the corrosive and its form base, or the base and the comparing form corrosive, are known.

The Henderson-Hasselbalch condition is useful while deciding the pH of an answer utilizing pKa and known centralizations of form base, salt, and corrosive. If the pH is known, it can also be used to determine concentrations of conjugate base, salt, or acid.

Incomplete question :

What is the [CH₃CO₂⁻]/[CH₃CO₂H] ratio necessary to make a buffer solution with a pH of 4.34? K a = 1.8 × 10⁻⁵ for CH₃CO₂H.

A. 2.5:1

B. 0.91:1

C. 1.09:1

D. 0.39:1

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According to the question the pH of the solution is 10.17.

pH is a measure of the acidity or alkalinity of a solution. It is measured on a scale of 0 to 14, with 7 being neutral. Solutions with a pH below 7 are considered acidic and solutions with a pH above 7 are considered basic or alkaline. pH is important to biological processes and environmental chemistry because it affects the availability of certain nutrients, the activity of enzymes, and the growth and activity of microorganisms.

The pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine, CH 3NH 2, with 20.00 mL of 0.10 M methylammonium chloride, CH 3NH 3Cl, can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log [(Conjugate Acid)/(Base)]
where pKa is the acid dissociation constant of the base and the conjugate acid is the salt of the base.
In this case, Kb = 3.70 x 10⁻⁴ for methylamine and the conjugate acid of CH₃NH₃Cl is CH₃NH₂.
Therefore, the pH of the solution is:
pH = -log[3.70 x 10⁻⁴] + log[(0.10 M CH₃NH₂)/(0.10 M CH₃NH₃Cl)]

= 10.17

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Out of the given compounds, aniline is the most basic.

This is because aniline has an unshared electron pair on the nitrogen atom, which can easily accept a proton to form a positively charged ion. This makes it a strong nucleophile and a good Lewis base. In comparison, p-nitroaniline and p-methoxyaniline have electron-withdrawing groups attached to the ring, which reduces their basicity. p-Toluidine is a weaker base than aniline because the methyl group on the nitrogen atom decreases the availability of the lone pair a of electrons on the nitrogen. Therefore, aniline is the most basic among the given compounds.

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There are approximately[tex]6.141 * 10^{21[/tex]molecules  [tex]N_2[/tex] in a 500.0 mL container at 780 mmHg and 135°C.

To determine the number of molecules  [tex]N_2[/tex] in a 500.0 mL container at 780 mmHg and 135°C, we can use the ideal gas law:

PV = nRT

First, we need to convert the volume from milliliters to liters:

V = 500.0 mL = 0.5000 L

Next, we can rearrange the ideal gas law to solve for n:

n = PV/RT

Substituting the values, we get:

n = (780 mmHg)(0.5000 L)/(0.08206 L·atm/(mol·K))(408.15 K) = 0.0102 mol

Finally, we can use Avogadro's number ([tex]6.022 * 10^{23}[/tex] molecules/mol) to convert from moles to molecules:

Number of N2 molecules = [tex]0.0102 mol * 6.022 * 10^{23} molecules/mol[/tex]= [tex]6.141 * 10^{21} molecules[/tex]

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To calculate the hydrolysis constant for the hypochlorite ion, we need to write the chemical equation for the reaction that occurs when the ion reacts with water. The closest answer choice is d) 2.9 × 10−7, which is the correct answer to this question.

The equation is as follows: OCl− + H2O ⇌ HOCl + OH−
In this equation, HOCl is the conjugate acid of the hypochlorite ion, and OH− is the hydroxide ion. The hydrolysis constant, also known as the base dissociation constant, is given by the expression: Kb = [HOCl][OH−] / [OCl−]
where [ ] denotes the concentration of each species in moles per liter. The value of Kb for hypochlorite ion can be calculated using the known values of the equilibrium concentrations of the reactants and products. However, these concentrations are not given in the question. Instead, we can use the relationship between Kb and Ka (the acid dissociation constant) for the conjugate acid HOCl:
Kb = Kw / Ka
where Kw is the ion product constant for water (1.0 × 10−14 at 25°C).
The value of Ka for HOCl is 3.0 × 10−8 at 25°C. Therefore, the value of Kb for OCl− is:
Kb = Kw / Ka = 1.0 × 10−14 / 3.0 × 10−8 = 3.3 × 10−7

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21.2 milliliters of 0.0850 M NaOH are required to titrate 25.0 mL of 0.0720 M HBr to the equivalence point.

The balanced chemical equation for the reaction between NaOH and HBr is [tex]NaOH + HBr → NaBr + H_2O[/tex]

In this reaction, one mole of NaOH reacts with one mole of HBr to produce one mole of NaBr and one mole of water.

We want to find amount of NaOH needed to titrate 25.0 mL of 0.0720 M HBr to the equivalence point, we can use the following formula [tex]M_1V_1 = M_2V_2[/tex] where

[tex]M_1[/tex] is the molarity of the NaOH solution, [tex]V_1[/tex] is the volume of NaOH solution in milliliters required to reach the equivalence point, [tex]M_2[/tex] is the molarity of the HBr solution, and [tex]V_2[/tex] is the volume of the HBr solution in milliliters.

Rearranging the formula to solve for

[tex]V_1[/tex], we get:[tex]V_1 = (M_2 \times V_2) / M_1[/tex]

Substituting the given values, we get:

[tex]V_1 = (0.0720 \: M \times 25.0 \: mL) / 0.0850 \: M[/tex]

[tex]V_1 = 21.2 mL[/tex]

Therefore, the answer is 21.2 milliliters of 0.0850 M NaOH are required to titrate 25.0 mL of 0.0720 M HBr to the equivalence point. The answer is (B) 21.2.

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Without acidifying the solution with H2SO4, the reaction would not have proceeded to completion, and the products would not have formed.

When acid is not added to the solution, the reaction between KMnO4 and oxalic acid does not occur spontaneously. In order to initiate the reaction, H2SO4 is added to provide the required protons for the oxidation of oxalic acid. Without the addition of H2SO4, the reaction between KMnO4 and oxalic acid would not have taken place as there are no protons available to drive the reaction forward. Thus, the products of the reaction would not have been formed in the original solution if it had remained neutral.

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Concentration of remaining copper ions in a solution containing 2.2 × 10^-3 M Cu2+ and 0.33 M LiCN with Kf value 1.0 × 10^25 for Cu(CN)4^2- is 3.8 × 10^-24 M.

What is the concentration of remaining copper ions in a solution containing 2.2 × 10^-3 M Cu2+ and 0.33 M LiCN with Kf value 1.0 × 10^25 for Cu(CN)4^2-?

The formation constant for the complex ion Cu(CN)4^2- is given as Kf = 1.0 × 10^25.

The balanced equation for the reaction between Cu2+ and CN- to form Cu(CN)42- is:

Cu2+ + 4CN- ⇌ Cu(CN)42-

Let x be the concentration of Cu2+ ions that react with CN- ions to form Cu(CN)42-. At equilibrium, the concentration of Cu(CN)42- ions formed will also be x M.

The initial concentration of Cu2+ ions is given as 2.2 × 10^-3 M.

The initial concentration of CN- ions is given as 0.33 M.

Using the formation constant expression for Cu(CN)42- we can write:

Kf = [Cu(CN)42-]/([Cu2+][CN-]^4)

Substituting the values, we get:

1.0 × 10^25 = x/(2.2 × 10^-3)(0.33)^4

Solving for x, we get:

x = 3.8 × 10^-24 M

Therefore, the concentration of copper ions that remains at equilibrium is 3.8 × 10^-24 M. Hence, the correct answer is 3.8 × 10^-24 M.

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The other product formed from the addition of HCl to 3-methyl-1-pentene is 1-chloro-3-methylpentane. This is because the HCl can add to the double bond in two different orientations,

leading to the formation of two possible products. The long answer would involve discussing the mechanism of the reaction and how the different orientations of HCl addition can lead to different products.when HCl is added to 3-methyl-1-pentene, it gives two products. One of them is 2-chloro-3-methylpentane,

as you mentioned. The other product is 3-chloro-3-methylpentane. This occurs due to the addition of HCl across the double bond in the alkene, leading to the formation of two different alkyl halides depending on the position of the chlorine atom.

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The calorimeter is used to measure heat absorbed or released by a chemical reaction and calculate the enthalpy change while following the law of conservation of energy.

What is the role of a calorimeter in measuring the heat of a chemical reaction?

Heat absorbed by the calorimeter is often used to measure the heat of a chemical reaction. The calorimeter measures the heat absorbed or released by the reaction and can be used to calculate the enthalpy change of the reaction.

During a chemical reaction, heat may be absorbed or released depending on whether the reaction is endothermic or exothermic. In an endothermic reaction, heat is absorbed from the surroundings, while in an exothermic reaction, heat is released to the surroundings.

In any chemical reaction, the heat produced or consumed in the reaction must be equal to the heat absorbed or released by the surroundings. The law of conservation of energy holds that energy cannot be generated or destroyed, but can only be transported or changed from one form to another.

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According to the stoichiometry of the reaction, for every 2 moles of S reacted, 3 moles of O2 are consumed. Thus, the limiting reactant will be S, and it will be completely consumed.

The balanced equation shows that 2 moles of S reacts with 3 moles of O2 to produce 2 moles of SO3. Thus, for every 2 moles of S that react, 2 moles of SO3 are produced.

Since there are 5 moles of S initially, it will react with 7.5 moles of O2 (since the ratio is 2:3 for S to O2), producing 5 moles of SO3.

Therefore, after the reaction, all of the S will be consumed, and there will be 0 moles of S left in the reaction vessel.

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The total volume before reaction will be 29.8 L and the volume  after reaction will be 6.54 L.

 CH₄ + H₂O ----- 3H₂ + CO

O.120   0.120

0            0 ---- 3× 0.120   0.120

Total no of moles before reaction =2 × 0.120

                                             = 0.24 mol

Total no. of moles after reaction = 3 × 0.120 + 0.120

                                            = 0.48 mol

        PV= nRT

          V∝ n

         V₁ / n₁ = V₂ / n₂

      14.9 / 0.24 = V₂ / 0.48

            V₂ = 29.8 L

2.         2CO + 2NO ⇒ 2CO₂ + N₂

          0.11          0.11           0         0

           0               0          0.11      0.055

Total no. of moles before reaction =0.11 + 0.11 = 0.22 moles

Total no. of moles  after reaction = 0.11 + 0.055 = 0.0165 mol

                               V₁/n₁ = V₂ / n₂

                8.72/ 0.22 = V₂/ 0.165

                     V₂ = 6.54 L

The quantity of a substance containing the same number of elementary entities—atoms, molecules, ions, electrons, radicals, etc.—is referred to as a mole. as there are particles in 12 grams of carbon - 12. A substance's relative molecular mass in grams is equal to one mole's mass.

We can count atoms and molecules by weighing macroscopically small amounts of matter using the mole concept because atoms and molecules are so small. It establishes a benchmark for determining reaction stoichiometry. It explains the characteristics of gases.

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The energy evolved during the reaction of 32.5 g [tex]B_{2}H_{6}[/tex] and 72.5 g [tex]Cl_{2}[/tex] is -475.24 kJ.

What is the amount of energy evolved during the reaction of 32.5 g [tex]B_{2}H_{6}[/tex] and 72.5 g [tex]Cl_{2}[/tex]?

To determine the amount of energy evolved during the reaction of 32.5 g [tex]B_{2}H_{6}[/tex] and 72.5 g [tex]Cl_{2}[/tex], we need to first determine which reactant is limiting.

First, we need to convert the mass of [tex]B_{2}H_{6}[/tex] into moles:

32.5 g B2H6 ÷ 27.67 g/mol = 1.175 mol [tex]B_{2}H_{6}[/tex]

Next, we need to convert the mass of [tex]Cl_{2}[/tex] into moles:

72.5 g Cl2 ÷ 70.90 g/mol = 1.023 mol Cl2

The stoichiometric ratio between [tex]B_{2}H_{6}[/tex] and [tex]Cl_{2}[/tex] is 1:6, meaning that for every 1 mole of [tex]B_{2}H_{6}[/tex], 6 moles of [tex]Cl_{2}[/tex] are required for complete reaction. Therefore, [tex]Cl_{2}[/tex] is the limiting reagent because there are only 1.023 mol of [tex]Cl_{2}[/tex] available, while 1.175 mol of [tex]B_{2}H_{6}[/tex] are present.

Now that we know [tex]Cl_{2}[/tex] is the limiting reagent, we can use its amount to calculate the amount of energy evolved during the reaction. From the balanced equation, we see that the molar ratio between [tex]B_{2}H_{6}[/tex] and ΔrH° is 1/2, meaning that for every 2 moles of [tex]BCl_{3}[/tex] produced, 1 mole of energy is released.

The amount of [tex]BCl_{3}[/tex] produced from the reaction of 1.023 mol of [tex]Cl_{2}[/tex] is:

(1.023 mol [tex]Cl_{2}[/tex]) x (2 mol [tex]BCl_{3}[/tex] / 6 mol [tex]Cl_{2}[/tex]) = 0.341 mol [tex]BCl_{3}[/tex]

Therefore, the amount of energy evolved from the reaction is:

ΔrH° x n = (-1396 kJ/mol) x (0.341 mol) = -475.24 kJ

So, the energy evolved during the reaction of 32.5 g [tex]B_{2}H_{6}[/tex] and 72.5 g [tex]Cl_{2}[/tex] is -475.24 kJ.

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Therefore, the Ksp of calcium oxalate is 2.28 x 10^-9.

The solubility product constant (Ksp) for calcium oxalate (CaC2O4) can be expressed as follows:

Ksp = [Ca+2][C2O4-2]

where [Ca+2] and [C2O4-2] are the molar concentrations of the respective ions in a saturated solution of calcium oxalate.

In this case, we are given that a saturated solution of calcium oxalate (CaC2O4) holds 0.0061 gram of calcium oxalate in 1 liter of solution. We can use this information to calculate the molar concentration of Ca+2 and C2O4-2 in the solution as follows:

The molar mass of CaC2O4 is 128 g/mol (40 g/mol for Ca+2 and 88 g/mol for C2O4-2). Therefore, the number of moles of CaC2O4 in the solution is:

moles of CaC2O4 = mass of CaC2O4 / molar mass of CaC2O4

moles of CaC2O4 = 0.0061 g / 128 g/mol

moles of CaC2O4 = 4.77 x 10^-5 mol

Since calcium oxalate dissociates into one Ca+2 ion and one C2O4-2 ion, the molar concentration of each ion in the solution is equal to the number of moles of CaC2O4 in the solution:

[Ca+2] = [C2O4-2] = moles of CaC2O4 / volume of solution

[Ca+2] = [C2O4-2] = 4.77 x 10^-5 mol / 1 L

[Ca+2] = [C2O4-2] = 4.77 x 10^-5 M

Finally, we can substitute the molar concentrations of Ca+2 and C2O4-2 into the Ksp expression to find the value of Ksp for calcium oxalate:

Ksp = [Ca+2][C2O4-2]

Ksp = (4.77 x 10^-5 M)(4.77 x 10^-5 M)

Ksp = 2.28 x 10^-9

Therefore, the Ksp of calcium oxalate is 2.28 x 10^-9. The closest option provided in the question is (A) 2.3 x 10^-9.

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