if a chemist has 210 ml of an 85% solution by volume of isopropyl alcohol how much of the volume of that solution is made up by isopropyl alcohol

The balanced chemical equation for the reaction between [tex]HNo_{3}[/tex] and pyridine ([tex]C_{5} H_{5}N[/tex]) is:

[tex]HNo_{3}[/tex] + [tex]C_{5} H_{5}N[/tex]→ [tex]C_{5} H_{5}N[/tex]+[tex]No_{3}[/tex]-

Step 1: Calculate the moles of pyridine present in 2.65 L of 0.0750 M pyridine:

moles of pyridine = (0.0750 mol/L) x 2.65 L = 0.1988 mol

Step 2: Determine the amount of [tex]HNo_{3}[/tex] required to react with all the pyridine present. Since [tex]HNo_{3}[/tex] is a strong acid, it will react completely with pyridine in a 1:1 ratio:

moles of [tex]HNo_{3}[/tex] required = 0.1988 mol

Step 3: Calculate the volume of 0.407 M [tex]HNo_{3}[/tex] required to provide 0.1988 mol of [tex]HNo_{3}[/tex] :

0.407 mol/L = 0.1988 mol / V

V = 0.488 L = 488 mL

Therefore, the volume of 0.407 M [tex]HNo_{3}[/tex] needed to reach the equivalence point is 488 mL.

Step 4: To calculate the pH at the equivalence point, we need to determine the concentration of the resulting salt, [tex]C_{5} H_{5}N[/tex]+[tex]No_{3}[/tex]-. At the equivalence point, moles of pyridine = moles of [tex]HNo_{3}[/tex]. Therefore, the moles of [tex]C_{5} H_{5}N[/tex]+NO3- formed is also 0.1988 mol. The total volume of the solution is 2.65 L + the volume of [tex]HNo_{3}[/tex] added (0.488 L).

Total volume of the solution = 2.65 L + 0.488 L = 3.138 L

Concentration of [tex]C_{5} H_{5}N[/tex]+[tex]No_{3}[/tex]- = moles / volume = 0.1988 mol / 3.138 L = 0.0633 M

Since [tex]C_{5} H_{5}N[/tex]is a weak base and [tex]HNo_{3}[/tex] is a strong acid, the salt [tex]C_{5} H_{5}N[/tex]+[tex]No_{3}[/tex]- is acidic. To calculate the pH, we need to determine the concentration of H+ ions in the solution. The balanced chemical equation for the dissociation of [tex]C_{5} H_{5}N[/tex]+[tex]No_{3}[/tex]- is:

[tex]C_{5} H_{5}N[/tex]+[tex]No_{3}[/tex]- + H2O → [tex]C_{5} H_{5}N H[/tex]+ [tex]HNo_{3}[/tex]+ H+

The equilibrium constant for this reaction is:

Kw / Kb = (H+)([tex]C_{5} H_{5}N[/tex]) / ([tex]C_{5} H_{5}N H[/tex]+[tex]No_{3}[/tex]-)

where Kw is the ion product constant for water (1.0 × 10^-14 at 25°C), and Kb is the base dissociation constant for pyridine (1.7 × 10^-9).

Solving for [H+], we get:

[H+] = (Kw / Kb) x ([tex]C_{5} H_{5}N H[/tex]+[tex]No_{3}[/tex]-) / ([tex]C_{5} H_{5}N[/tex])

[H+] = (1.0 × 10^-14) / (1.7 × 10^-9) x (0.0633 M) / (0.0750 M)

[H+] = 3.33 × 10^-6 M

pH = -log[H+] = -log(3.33 × 10^-6) = 5.48

Therefore, the pH at the equivalence point is 5.48.

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The term "plastic" is often used as a synonym for synthetic polymer.

A synthetic polymer is a big molecule comprised of repeated monomeric building blocks. Long chains made of these monomers are created by chemically bonding them together.

These chains can then be processed to create a range of useful materials. Plastic bags and bottles, high-tech materials used in aircraft, and medical devices—there are many uses for synthetic polymers.

Modern society has been significantly impacted by the usage of synthetic polymers, which offer affordable and adaptable materials in place of more expensive and conventional ones.

Due to the slow breakdown of plastic materials, however, the disposal of plastic trash has also grown to be a significant environmental concern.

Overall, because to the widespread usage of plastic products in our daily lives, the word "plastic" has come to be synonymous with synthetic polymer.

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The determine the half-life for the decomposition of substrate 1, we first need to plot the initial rate data. From the given data, we can see that the initial rate is constant for different initial concentrations of substrate 1. This means that the reaction follows first-order kinetics.

The rates law for a first order reaction rate = k [substrate 1] where k is the rate constant, we can calculate the rate constant for the reaction. From the data, we know that the initial rate is 0.595 m/s when [substrate 1] = 0.5 M. Substituting these values into the rate law, we get 0.595 m/s = k 0.5 M Solving for k, we get k = 1.19 s^-1. Now that we have the rate constant, we can use the half-life formula for a first-order reaction t1/2 = ln (2) / k where ln (2) is the natural logarithm of 2 approximately 0.693. Substituting the given initial concentration of substrate 1 2.31 M and the rate constant 1.19 s^-1 into the formula, we get t1/2 = ln (2) / 1.19 s^-1 / 2.31 M t1/2 = 0.49 s Therefore, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.31 M is 0.49 seconds.

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Mass of Chlorine is0.4963 g Cl- and percentage of chlorine is 101.44% the chromium compound was dissolved in water and excess silver nitrate was added

To solve this problem, we need to use the stoichiometry of the reaction between the chromium compound and silver nitrate to calculate the mass of chloride ion (Cl⁻) in the sample.
The balanced equation for the reaction is:
CrX + 2 AgNO₃ ⇔ Ag₂CrX₄ + 2 AgCl + 2 NO³⁻
where CrX represents the chromium compound and Ag₂CrX₄ represents a silver-chromium compound that remains in solution.
From the equation, we can see that 2 moles of AgCl are formed for each mole of CrX, so we can calculate the number of moles of Cl- in the sample as follows:
1.0042 g AgCl x (1 mol AgCl / 143.32 g AgCl) x (2 mol Cl- / 1 mol AgCl) = 0.01400 mol Cl-
Next, we can use the mass of the sample and the molar mass of CrX to calculate the number of moles of CrX:
0.4892 g CrX x (1 mol CrX / molar mass of CrX) = n mol CrX
We don't need to know the molar mass of CrX to solve the problem, since it will cancel out in the next step.
Finally, we can calculate the mass of Cl- in the sample and the percent Cl-:
Mass of Cl- = 0.01400 mol Cl- x (35.45 g/mol) = 0.4963 g Cl-
Percent Cl- = (0.4963 g Cl- / 0.4892 g sample) x 100% = 101.44%
The percent Cl- is greater than 100% because of a possible error in the weighing or the reaction, or because the sample may contain other sources of chloride ions. However, the calculation shows that most of the chlorine in the sample is present as Cl-.

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The molecular formula C6H14 represents a saturated hydrocarbon with six carbon atoms and 14 hydrogen atoms. To determine the number of structural isomers, we need to consider the different ways in which these atoms can be arranged in a molecule while maintaining the same molecular formula.

One way to approach this problem is to start with the straight-chain structure of n-hexane, which has all six carbon atoms in a row with each carbon atom bonded to two hydrogen atoms. This isomer is called the n-isomer or normal hexane. The other isomers can be obtained by branching or rearranging the carbon chain.The first branched isomer is 2-methylpentane, which has a five-carbon chain with a methyl (CH3) group attached to the second carbon atom. The second branched isomer is 3-methylpentane, which has a five-carbon chain with a methyl group attached to the third carbon atom. The third branched isomer is 2,2-dimethylbutane, which has a four-carbon chain with two methyl groups attached to the second carbon atom.The fourth isomer is 2,3-dimethylbutane, which has a four-carbon chain with one methyl group attached to the second carbon atom and another methyl group attached to the third carbon atom. The fifth isomer is 2,4-dimethylpentane, which has a five-carbon chain with one methyl group attached to the second carbon atom and another methyl group attached to the fourth carbon atom.Therefore, the answer is B. 5, there are five structural isomers possible with the molecular formula C6H14.

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The cell potential of the voltaic cell is 0.8129 V.

The cell potential of a voltaic cell can be calculated using the Nernst equation:

Ecell = E°cell - (RT/nF) ln(Q)

where E°cell is the standard cell potential, R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, n is the number of electrons transferred in the cell reaction, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.

The balanced cell reaction for the[tex]$\mathrm{Mn/Mn^{2+}}$[/tex] [tex]$\mathrm{Cd^{2+}}$[/tex] half-cells can be written as follows:

[tex]\mathrm{Mn^{2+}}$.[/tex]+ + 2e- → Mn (E° = -1.18 V)

[tex]$\mathrm{Cd^{2+}}$[/tex]+ + 2e- → Cd (E° = -0.40 V)

To calculate E°cell, we need to subtract the reduction potential of the anode from the reduction potential of the cathode:

E°cell = E°cathode - E°anode

E°cell = E°Cd - E°Mn

E°cell = (-0.40 V) - (-1.18 V)

E°cell = 0.78 V

Next, we need to calculate the reaction quotient, Q, using the concentrations of the reactants and products:

Q = [[tex]$\mathrm{Cd^{2+}}$[/tex]]/[[tex]$\mathrm{Mn^{2+}}$.[/tex]]

Q = 0.00423 M / 0.28 M

Q = 0.0151

Finally, we can plug in the values into the Nernst equation to calculate the cell potential:

Ecell = E°cell - (RT/nF) ln(Q)

Ecell = 0.78 V - (8.314 J/mol K)(298 K)/(2 mol e-)(96485 C/mol) ln(0.0151)

Ecell = 0.78 V - (-0.0329 V)

Ecell = 0.8129 V

Therefore, the cell potential of the voltaic cell is 0.8129 V.

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The value of standard free energy change (∆G°) for the oxidation of iron by H (at 25 °c) is found to be 20,925 J/mol.

The standard potential for the reduction of Fe³⁺ is -0.036 V. To calculate the standard free energy change (∆G°) for the oxidation of iron by H⁺, we can use the following equation,

∆G° = -nFE°, number of moles of electrons transferred is n, Faraday constant (96,485 C/mol) is F, standard cell potential is E°.

The balanced equation for the oxidation of iron by H⁺ is,

2Fe(s) + 6H⁺(aq) → 2Fe³⁺(aq) + 3H₂(g)

The oxidation of iron by H⁺ involves the transfer of 6 electrons, so n = 6

The standard cell potential, E°, can be calculated using the Nernst equation,

E° = E°(Fe³⁺/Fe²⁺) - (RT/nF) × ln(Q), the gas constant (8.314 J/(mol·K)) is R, temperature in Kelvin (298 K) is T, number of electrons transferred (6) is n, F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient.

At standard conditions, the reaction quotient Q is equal to 1, since the concentrations of all the species in the reaction are 1 M. Therefore, ln(Q) = ln(1)

= 0.

Plugging in the values, we get,

E° = -0.036 V - (8.314 J/(mol·K) × 298 K/6 × 96,485 C/mol) × 0

E° = -0.036 V

Now we can calculate ∆G°,

∆G° = -nFE°

∆G° = -(6 mol e⁻) × (96,485 C/mol) × (-0.036 V)

∆G° = 20,925 J/mol

Therefore, the standard free energy change for the oxidation of iron by H⁺ is 20,925 J/mol at 25 °C.

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Complete question - Calculate ∆G° for the oxidation of Iron by H (at 25 °C). Reduction of Fe³⁺ has a potential of -0.036V. 2Fe(s) + 6H(aq) → 2Fe(aq) + 3H₂(g)

The pH of the aqueous solution that is the mixture of 0.10 M NaNO₂ and 0.20 M Ca(NO₂)₂ is 2.52.

To calculate the pH of the given aqueous solution, we need to first determine the concentration of HNO₂ in the solution. HNO₂ is a weak acid, and its Ka value is given as 4.5 x 10⁻⁴. We can write the dissociation reaction of HNO₂ as:

HNO₂ + H₂O ⇌ H₃O⁺ + NO₂⁻

The equilibrium constant expression for this reaction can be written as:

Ka = [H₃O⁺][NO₂⁻] / [HNO₂]

Assuming that the initial concentration of HNO₂ is negligible compared to the equilibrium concentration, we can simplify the expression as:

Ka = [H₃O⁺]² / [HNO₂]

Solving for [H₃O⁺], we get:

[H₃O⁺] = √(Ka * [HNO₂]) = √(4.5 *10⁻⁴ * 0.10) = 0.015

Now, we can use the concentration of Ca(NO₂)₂ to calculate the concentration of NO₂⁻ in the solution. Ca(NO₂)₂ dissociates into Ca²⁺ and 2NO₂⁻. Since NO₂⁻ is the conjugate base of HNO₂, it can react with H₃O⁺ to form HNO₂ and H₂O. This reaction can be written as:

NO₂⁻ + H₃O⁺ ⇌ HNO₂ + H₂O

The equilibrium constant expression for this reaction can be written as:

Kb = [HNO₂][H₂O] / [NO₂⁻][H₃O⁺]

Since Kb for NO₂⁻ is related to Ka for HNO₂ as:

Ka x Kb = Kw = 1.0 * 10⁻¹⁴

We can use this relation to calculate Kb for NO₂⁻ as:

Kb = Kw / Ka = 1.0 x 10⁻¹⁴ / 4.5 x 10⁻⁴ = 2.22 x 10⁻¹¹

Assuming that the initial concentration of NO₂⁻ is negligible compared to the equilibrium concentration, we can simplify the expression for Kb as:

Kb = [HNO₂][H₂O] / [NO₂⁻]

Solving for [HNO₂], we get:

[HNO₂] = Kb * [NO₂⁻] / [H₂O] = 2.22 * 10⁻¹¹ * (2 * 0.20) / 55.5 = 1.59 * 10⁻¹²

Now, we can use the concentrations of HNO₂ and NO₂⁻ to calculate the pH of the solution using the equation:

pH = -log[H₃O⁺] = -log(√(Ka x [HNO₂] / [NO₂⁻])) = -log(√(4.5 x 10⁻⁴ x 0.10 / (2 x 0.20))) = 2.52

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1. The concentration of H2C6H6O6 is given as 0.0439 M. Assuming complete dissociation, the initial concentration of HC6H6O6- is also 0.0439 M. Therefore, [HC6H6O6-] = 0.0439 M.

2. The pH of the solution is 2.59.

C6H5COOH ⇌ H+ + C6H5COO-

Ka = [H+][C6H5COO-] / [C6H5COOH]

6.5 × [tex]10^{-5}[/tex]= [H+]² / 0.478

[H+] = 0.00255 M

Using the pH formula, we can then calculate the pH of the solution:

pH = -log[H+]

pH = -log(0.00255)

pH = 2.59

3. HCN + H2O ⇌ H3O+ + CN-

Ka = [H+][CN-] / [HCN]

4.9 × [tex]10^{-10}[/tex] = [H+]² / 0.563

[H+] = 1.57 × [tex]10^{-5}[/tex]M

Kw = [H+][OH-]

1.0 × [tex]10^{-14}[/tex] = (1.57 × [tex]10^{-5}[/tex])[OH-]

[OH-] = 6.37 × [tex]10^{-10}[/tex] M

Concentration refers to the amount of a substance that is present in a given volume or mass of another substance. It is a measure of the relative amount of solute present in a solution or mixture. The most common ways of expressing concentration in chemistry are molarity, molality, percent composition, and parts per million.

Molarity, denoted as M, is the number of moles of solute per liter of solution. Molality, denoted as m, is the number of moles of solute per kilogram of solvent. Percent composition is the mass of solute present in a solution expressed as a percentage of the total mass of the solution. Parts per million (ppm) is a measure of the concentration of a solute in a solution, expressed as the number of parts of the solute per million parts of the solution.

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5.4 × [tex]10^22[/tex] oxygen molecules cross the lens in one hour.

To calculate the number of oxygen molecules that cross the lens in one hour, we can use Fick's first law of diffusion, which relates the diffusion rate to the diffusion coefficient, the surface area, and the concentration gradient:

J = -D * A * ΔC/Δx

where J is the diffusion rate (in molecules/s), D is the diffusion coefficient (in [tex]m^2/s[/tex]), A is the surface area (in [tex]m^2[/tex]), ΔC is the concentration difference (in molecules/m^3), and Δx is the thickness of the lens (in m).

First, we need to convert the diameter and thickness of the lens to meters:

d = 14 mm = 0.014 m

h = 40 μm = 4.0 × [tex]10^-5 m[/tex]

The surface area of the lens is:

A = π * [tex](d/2)^2[/tex] = 1.54 × [tex]10^-3 m^2[/tex]

The concentration difference is:

ΔC = (P1 - P2) / (k * T)

where P1 is the partial pressure at the front of the lens, P2 is the partial pressure at the rear, k is the Boltzmann constant (1.38 ×[tex]10^-23[/tex] J/K), and T is the temperature in kelvin.

P1 = 0.2 * 101.3 kPa = 20.26 kPa

P2 = 7.3 kPa

T = 30 + 273.15 K = 303.15 K

ΔC = (20.26 - 7.3) × 1000 / (1.38 × 10^-23 * 303.15) = 7.23 ×[tex]10^25[/tex]molecules/[tex]m^3[/tex]

Now we can calculate the diffusion rate:

J = -D * A * ΔC / Δx = -1.3 × [tex]10^-13 m^2/s[/tex] * 1.54 × [tex]10^-3 m^2[/tex] * 7.23 × [tex]10^25[/tex] molecules/[tex]m^3[/tex] / 4.0 × [tex]10^-5 m[/tex] = -1.5 × [tex]10^19 molecules/s[/tex]

Note that the diffusion rate is negative because the concentration gradient is negative (oxygen molecules diffuse from high concentration at the front to low concentration at the rear).

To find the number of oxygen molecules that cross the lens in one hour, we need to multiply the diffusion rate by the number of seconds in one hour:

N = J * 3600 s = -1.5 × [tex]10^19[/tex] molecules/s * 3600 s = -5.4 × [tex]10^22[/tex]molecules

The negative sign means that the net direction of oxygen diffusion is from the rear to the front of the lens, so more oxygen molecules leave the front than enter it. However, the question only asks for the number of molecules that cross the lens, so we take the absolute value of the result:

N = 5.4 ×[tex]10^22 molecules[/tex]

Therefore, about 5.4 × [tex]10^22[/tex] oxygen molecules cross the lens in one hour.

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The three types of convergent plate boundaries are ocean-ocean, ocean-continent, and continent-continent.

A zone where two or more tectonic plates converge is referred to as a convergent border. The land inside the boundary area is altered as a result. In areas where convergent borders exist, earthquakes and volcanic eruptions are highly common.

Depending on the type of crust that is present on either side of the boundary—oceanic or continental—convergent borders, where two plates are moving toward one another, can be classified into one of three categories. Mountains and mountain ranges are created when two continental plates collide, pulling up the rock at the boundary and crumpling and folding it.

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The Ksp value for MgSO₃ is 5.5×10−21. The concentration of Mg²⁺ in solution is 8.9×10−11 M. To generate a precipitate, the concentration of SO₂⁻³ must exceed 6.2×10−11 M.

Ksp refers to the solubility product constant  that provides equilibrium constant for the dissolution of a particular solid substance into an aqueous solution. It projects the level at which a solute dissolves in solution. The greater the Ksp value of a substance.
It places  a mathematical relationship that states how the concentrations of the products differentiate with the concentration of the reactants. Furthermore, subscripts are placed to the equilibrium constant symbol K, such as K eq, K c, K p, K a, K b, and K sp.

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The formulae of the chemical compounds formed are as follows:

A chemical compound is formed when two or more elements are combined together in a definite proportion. Chemical bonds are formed when the elements interact with one another. These bonds develop as a result of atoms sharing electrons.

Examples of chemical compounds include baking soda, water, and table salt.

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Monodeuteriomethane is a type of molecule that contains one deuterium atom and three hydrogen atoms. In a disordered crystal of monodeuteriomethane, each tetrahedral CH3D molecule can be oriented randomly in one of four possible ways.

A) Using Boltzmann's formula, the entropy of the disordered state of a crystal containing 140 molecules can be calculated as S = klnW, where k is Boltzmann's constant and W is the number of microstates. The number of microstates for a crystal containing 140 molecules can be calculated as W = 4^140. Thus, S = kln(4^140) = 140kln4 + ln(140!) ≈ 1.69 × 10^25 J/K. B) To calculate the entropy of the disordered state of a crystal containing 1 mol of molecules, we need to know the Avogadro's number, which is approximately 6.022 × 10^23 molecules/mol. Thus, the number of molecules in 1 mol of monodeuteriomethane is 4 × 6.022 × 10^23 = 2.409 × 10^24 molecules. Using Boltzmann's formula, the entropy can be calculated as S = klnW, where W = 4^(2.409 × 10^24). Therefore, S ≈ 4.58 × 10^51 J/K. C) If the C-D bond of each of the CH3D molecules points in the same direction in a crystal containing 14 molecules, then there are only two possible orientations for each molecule. Thus, the number of microstates is W = 2^14, and the entropy can be calculated as S = kln(2^14) = 14kln2 ≈ 9.09 × 10^-22 J/K. D) If the C-D bond of each of the CH3D molecules points in the same direction in a crystal containing 140 molecules, then the number of microstates is W = 2^140, and the entropy can be calculated as S = kln(2^140) = 140kln2 ≈ 9.09 × 10^-20 J/K. E) To calculate the entropy of the crystal if the C-D bond of each of the CH3D molecules points in the same direction in a crystal containing 1 mol of molecules, we need to know the Avogadro's number. Thus, the number of molecules in 1 mol of monodeuteriomethane is 4 × 6.022 × 10^23 = 2.409 × 10^24 molecules. The number of microstates for 1 mol of monodeuteriomethane is W = 2^(2.409 × 10^24), and the entropy can be calculated as S = klnW. Therefore, S ≈ 4.54 × 10^50 J/K. In summary, the entropy of a crystal of monodeuteriomethane depends on the number of molecules in the crystal, the number of possible orientations for each molecule, and the direction of the C-D bond of each molecule. The more disordered the crystal, the higher the entropy, and the more ordered the crystal, the lower the entropy.

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The statement "Mass is conserved during a chemical reaction" means that the total mass of the reactants (the substances that undergo a chemical change) is equal to the total mass of the products (the new substances formed as a result of the chemical change).

A chemical reaction is a process in which one or more substances, known as reactants, are transformed into one or more new substances, known as products, through the breaking and forming of chemical bonds. Chemical reactions are a fundamental part of chemistry and occur all around us, from the food we eat to the fuel we burn.

Chemical reactions are represented by chemical equations that show the reactants on the left-hand side and the products on the right-hand side. The coefficients in front of the reactants and products indicate the relative amounts of each substance involved in the reaction. There are several types of chemical reactions, including synthesis, decomposition, single replacement, double replacement, and combustion reactions.

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The conjugate acid of NH₃ is NH₄⁺, When ammonia accepts a proton (H+), it becomes NH₄⁺. In this reaction, NH₃ is the base and NH₄⁺ is the conjugate acid, because NH₄⁺ is formed by the addition of a proton to NH₃ .

Ammonia (NH₃) is a weak base because it can accept a proton (H+) to form its conjugate acid, ammonium ion (NH₄⁺). In this reaction, ammonia (NH₃) acts as a Bronsted-Lowry base by accepting a proton (H+) to form ammonium ion (NH₄⁺), which acts as a Bronsted-Lowry acid. The key concept to understand here is the relationship between a weak base and its conjugate acid. A weak base can accept a proton to form its conjugate acid, which is always one proton (H+) more than the original weak base.

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Redox chemistry plays a crucial role in the workup process, particularly in the reaction of excess iodine with thiosulfate to form iodide and dithionate: [tex]I_2 (aq) + 2 S_2O_3^{2-} (aq)[/tex] → [tex]2 I^- (aq) + S_4O_6^{2-} (aq)[/tex]. The practical advantage of reducing excess iodine to iodide lies in the improved isolation and purification of the desired product.

In many chemical reactions, excess reactants are often used to ensure complete conversion of the limiting reactant to the product. However, the presence of excess reactants can also lead to the formation of unwanted side products or impurities. In this case, excess iodine can potentially interfere with the desired product's properties, affecting its purity and yield.

By reducing excess iodine to iodide using thiosulfate, we eliminate the possibility of it interfering with the desired product. Iodide ions are less reactive than iodine, thus minimizing unwanted side reactions. Additionally, the products of this redox reaction, iodide and dithionate, are typically more soluble in water, which simplifies their removal from the reaction mixture through aqueous washes or filtration.

In conclusion, reducing excess iodine to iodide using thiosulfate in the workup process provides a practical advantage by facilitating the isolation and purification of the desired product. This step prevents potential interference from excess iodine, minimizes side reactions, and simplifies the removal of reaction by-products, ultimately leading to a higher purity and yield of the target compound.

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The reaction for the formation of hydrogen sulfide (H2S) is given by: H2(g) + S(s) ⇌ H2S(g) Initial concentrations are 0.045 M H2, 0.070 M S, and no H2S. At equilibrium, the concentration of H2 is 0.010 M.

To determine the concentrations of S and H2S at equilibrium, we need to calculate the change in concentrations. Since the stoichiometry is 1:1, the decrease in H2 concentration (0.045 - 0.010 = 0.035 M) corresponds to an equal increase in H2S concentration. Therefore, at equilibrium, [H2S] = 0.035 M. Since S is a solid, its concentration remains unchanged (0.070 M), and it doesn't affect the equilibrium constant, K. To calculate K, use the equilibrium concentrations of H2 and H2S: K = [H2S] / [H2] K = (0.035 M) / (0.010 M) K = 3.5 Under the given reaction conditions at equilibrium, the concentrations are [H2] = 0.010 M, [S] = 0.070 M, [H2S] = 0.035 M, and K = 3.5.

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Dispersion forces are a type of intermolecular force that causes an attraction between molecules that results from a distortion of the electron cloud that causes an uneven distribution of charge.

All molecules have electrons that are in constant motion, and sometimes these electrons can accumulate on one side of the molecule, creating a temporary dipole moment.

This temporary dipole moment can then induce a dipole moment in a nearby molecule, causing an attraction between the two.

Hence, dispersion forces are a type of intermolecular force that results from temporary dipoles induced by the motion of electrons.

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If a student uses nitric acid ( HNO3) instead of sulphuric acid (H2SO4) in reaction d, it can lead to incorrect results.

Although both HNO3 and sulphuric acid are strong acids, they have different properties and react differently in certain situations. In this particular reaction, sulphuric acid is needed to remove any remaining carbonate or bicarbonate ions from the solution. If nitric acid is used instead, the reaction will not proceed as expected and the results may be inaccurate.

When students mistakenly use nitric acid (aq) instead of sulphuric acid (aq) in reaction d, the results may be different due to this error. Even though both are strong acids, their properties and reactivity are not identical, which may affect the outcome of the reaction. Therefore, it is important to use the correct acid as specified in the experiment to ensure accurate and reliable results.

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A chemical equation is balanced when E. the number of atoms of each element is the same in reactants and products because in a balanced chemical reaction the amount of reactants is equal to the amount of products.

A balanced chemical equation represents a chemical reaction with the same number of atoms of each element on both sides of the equation. In other words, the total mass and the number of atoms in the reactants are equal to those in the products.

The balance of the equation is achieved by adjusting the coefficients in front of the chemical formulas of the reactants and products to ensure that the same number of atoms of each element is present on both sides of the equation.  Hence, option E is correct.

Balancing the chemical equation is important as it helps to predict the amount of reactants and products that will be used and formed, respectively, and it also ensures that the reaction follows the law of conservation of mass.

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Mitochondria is a double membrane bound organelle which is found in most eukaryotic organisms. They are found inside the cytoplasm and essentially function as the cells digestive system. Here number 3 represents the mitochondria. The correct option is C.

Mitochondria popularly known as the power house of the cell play an important role in breaking down the nutrients and produce energy rich molecules for the cell. Its size ranges from 0.5 to 1.0 micrometer in diameter.

The mitochondria is a double membraned rod shaped structure which is found both in plants and animals. It comprises of an outer membrane, inner membrane and a gel material called matrix.

Thus the correct option is C.

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The color change of the equilibrium system of cobalt complexes can provide valuable information about changes made to the system at equilibrium. In this case, the Co(H₂0)₆²⁺ (aq) complex is pink and the CoCl₂⁻ (aq) complex is light blue.

If the solution changes from pink to light blue, it means that the concentration of CoCl₂⁻ (aq) complex has increased and the concentration of Co(H₂0)₆²⁺ (aq) complex has decreased. This could be due to the addition of more chloride ions or the removal of water molecules from the system. As a result, the equilibrium shifts towards the side of the equation with fewer chloride ions and more water molecules.

On the other hand, if the solution changes from light blue to pink, it means that the concentration of Co(H₂0)₆²⁺ (aq) complex has increased and the concentration of CoCl₂⁻ (aq) complex has decreased. This could be due to the addition of more water molecules or the removal of chloride ions from the system. As a result, the equilibrium shifts towards the side of the equation with fewer water molecules and more chloride ions.

If the solution stays light blue after adding a chemical, it means that the added chemical has no effect on the equilibrium system. This could be because the added chemical does not react with any of the species in the equilibrium system or because its effect is negligible compared to the existing concentrations of the species.

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In the liberal arts and humanities, MLA (Modern Language Association) style is most frequently used to compose papers and cite sources.

Thus, Brief parenthetical citations in the text that are keyed to an alphabetical list of the works cited at the end of the work are a feature of the MLA style.

With the publication of the most recent edition, the MLA citation style has undergone substantial alterations.

Building trust in the knowledge and ideas we share with one another may be more crucial than ever, and for almost a century, this has been the guiding principle of MLA style, a set of writing and documentation.

Thus, In the liberal arts and humanities, MLA (Modern Language Association) style is most frequently used to compose papers and cite sources.

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The balanced redox reaction equation is;

(Fe)3+ + NO2 + H2O → (Fe)2+ + NO3- + 2 H+

A chemical reaction in which electrons are moved between two species is an oxidation-reduction reaction, often known as a redox reaction. The words "reduction" and "oxidation," which describe the two half-reactions that occur in a redox reaction, are the origins of the term "redox."

In a redox reaction, one species loses electrons (becomes oxidized) and gains electrons (becomes reduced). It is possible to illustrate this electron transfer via half-reactions, in which the oxidizing agent receives electrons while the reducing agent loses them.

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The pH of a 3.90% KOH by mass solution with a density of 1.01 g/ml is 1.15.

The pH of a solution can be determined by calculating the molarity of the solute, in this case potassium hydroxide (KOH), and then using the appropriate equation to calculate the pH.

For a solution of 3.90% KOH by mass, the molarity can be found by multiplying the mass percent by the density of the solution (1.01 g/ml) and then dividing by the molar mass of KOH (56.1 g/mol).

This yields a molarity of 0.07 moles/L. The pH of a solution with this molarity can be calculated using the equation pH = -log([KOH]), where [KOH] is the molarity of KOH. Plugging in 0.07 moles/L for [KOH] yields a pH of 1.15. Therefore, the pH of a 3.90% KOH by mass solution with a density of 1.01 g/ml is 1.15.

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Among the given functional groups of an amino acid, the "coh with an o atom double-bonded to the carbon.

There is an open bond at the carbon" group would be in the ionized state at high pH. This functional group represents the carboxyl group (-COOH) of an amino acid, which acts as an acid and donates a proton to form a negatively charged carboxylate ion (-[tex]COO^-[/tex]) at high pH.

The other functional group, "[tex]ch2oh[/tex] with an open bond at the carbon," represents the hydroxyl group (-OH) of an amino acid, which does not undergo ionization at high pH. The remaining functional groups are not present in amino acids and do not undergo ionization under physiological conditions.

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NH[tex]_3[/tex] and H[tex]_2[/tex]O are the major species present in a 0. 150-M NH[tex]_3[/tex] solution. pOH is 2.79 and pH is 11.21.

pH (commonly known as acidity in chemistry, has historically stood for "the potential of hydrogen" (as well as "power of hydrogen").[1] This is a scale employed to describe how basic or how acidic an aqueous solution is. When compared to basic or alkaline solutions, acidic solutions—those with higher hydrogen (H+) ion concentrations—are measured with lower pH values.

Since NH3 is weak base . A weak base con not ionize completely to prodcue NH4+ and OH-.So the major species are NH3 & H2O only.

NH[tex]_3[/tex]+H[tex]_2[/tex]O→NH[tex]_4[/tex]⁺ +OH⁻

Kb=[NH[tex]_4[/tex]⁺][ OH⁻]/NH[tex]_3[/tex]

1.8×10⁻⁵ =X²/0. 150

X=1.64×10⁻³

pOH = -log[1.64×10⁻³]

        = 2.79

pH =14-2.79=11.21

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