a sample of radioactive isotope is found to have an activity of 114 bq imediately after it is pulled from the reactor that formed the isotope. its activity 3 h, 50 min later is measured to be 80.2 bq. find the decay constant

Answer:

a. To compare the concentration of 80 g/litre NaOH solution and 3 M NaOH solution, we need to convert one of the concentrations to the other unit.

One mole of NaOH weighs 40 grams. So, to convert 80 g/litre NaOH to Molarity, we can divide 80 g/litre by 40 g/mol to get:

80 g/litre NaOH = 2 M NaOH

Therefore, 3 M NaOH has a higher concentration than 80 g/litre NaOH solution.

b. To compare the concentration of 5.3 g/litre Na2CO3 and N/10 Na2CO3 solution, we need to first understand what N/10 solution means.

N/10 Na2CO3 means that the solution contains 1/10th of the normal concentration of Na2CO3. The normal concentration of Na2CO3 is the molar concentration of Na2CO3 that corresponds to the formula weight of Na2CO3, which is 106 g/mol.

So, the normal concentration of Na2CO3 is 1 mol/L or 1 M Na2CO3.

Therefore, N/10 Na2CO3 solution has a concentration of 1/10 M Na2CO3.

Now, let's compare the two concentrations:

5.3 g/litre Na2CO3 = (5.3/106) M Na2CO3 = 0.05 M Na2CO3

Since 0.05 M Na2CO3 is greater than 1/10 M Na2CO3, the concentration of 5.3 g/litre Na2CO3 solution is higher than that of N/10 Na2CO3 solution.

Explanation:

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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83.3 mL is required to provide 0.125mol KOH

Explanation:

We know that,

V=[tex]\frac{n}{c}[/tex] where n=number of moles, c= concentration and V=volume

According to the question,

c=1.50M and n=0.125 mol

Substituting the values,

V=[tex]\frac{0.125mol}{1.50M}[/tex]

=0.0833L

The volume should be in mL,

0.0833L× [tex]\frac{1000mL}{1L}[/tex]

= 83.3mL

HF is corrosive at concentration that are ≥ 1 M (1 mol/L). Therefore the correct option is option C.

HF (hydrofluoric acid) is a very poisonous and corrosive acid that, when contacted, can result in serious burns and tissue damage. The concentration of HF affects its ability to corrode.

At concentrations of less than 1 M (1 mol/L), HF is corrosive. At this concentration, HF can quickly permeate the skin, resulting in painful tissue injury.

Because of its fast skin penetration and capacity to interact with calcium ions in the body to generate insoluble calcium fluoride (CaF2), HF has a corrosive effect on tissues and cells.

Although the effects may be delayed or less severe than at higher concentrations, severe burns and tissue damage can still occur at HF concentrations of less than 0.1 M (0.1 mol/L). Therefore the correct option is option C.

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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 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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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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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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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 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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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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Fossil fuels are formed over millions of years from the remains of dead plants and animals that have been buried under layers of rock and sediment.

These fuels contain stored carbon that was originally absorbed by the plants and animals during their lifetime. Examples of fossil fuels include coal, oil, and natural gas. When these fuels are burned for energy, the carbon is released into the atmosphere in the form of carbon dioxide, which contributes to climate change. Natural gas is a combustible mixture of hydrocarbons and other organic compounds that is found beneath the Earth's surface. Coal is a non-renewable fossil fuel that is used to generate electricity and heat, and is also used in the production of steel, cement, and other industrial products.

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The bond length in carbon-carbon single bonds (C-C) is longer than that in double (C=C) and triple (C≡C) bonds, as they involve the sharing of one electron pair, while double and triple bonds share two and three electron pairs, respectively.

The bond length and bond energy of carbon-carbon bonds differ based on the type of bond they form. In a single bond, the carbon-carbon bond length is longer at 0.154 nm and has a bond energy of 348 kJ/mol. In a double bond, the carbon-carbon bond length is shorter at 0.134 nm and has a bond energy of 611 kJ/mol.

In a triple bond, the carbon-carbon bond length is even shorter at 0.120 nm and has a bond energy of 837 kJ/mol. These differences in bond length and bond energy are due to the increase in the number of shared electrons between carbon atoms in double and triple bonds.

In contrast, bond energy increases as the bond order rises; C-C single bonds have the lowest bond energy, while C≡C triple bonds possess the highest bond energy due to the stronger attractive forces between the bonded carbon atoms. Overall, carbon-carbon bonds exhibit a relationship where bond length decreases and bond energy increases as the number of shared electron pairs rises.

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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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True. A carbocation (CH3C+O) and an aluminium tetrachloride anion (AlCl4-) are produced as a result of the interaction between acetyl chloride (CH3COCl) and aluminium chloride (AlCl3).

The aluminium chloride serves as a Lewis acid catalyst in the reaction, which follows a Friedel-Crafts acylation mechanism. An acylium ion (CH3CO+) and an AlCl4- anion are created when the acetyl chloride interacts with the aluminium chloride. The carbocation (CH3C+O) is then created by a rearrangement of the acylium ion. By serving as a counter ion, the AlCl4- anion stabilises the carbocation. As a result, the chemical equation provided is accurate. True. A carbocation and an aluminium tetrachloride anion are created in this Friedel-Crafts acylation reaction between acetyl chloride and aluminium chloride.

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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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Regarding the statements about Acids & Bases and Hydrobromic Acid, here's an analysis of their validity:

1. The partial atomic charge of the anion is inversely related to the HX bond length: True. As the bond length increases, the partial atomic charge on the anion decreases, indicating weaker bonding between the hydrogen and halogen atoms.

2. The partial atomic charge measures the charge on the hydrogen atom: False. The partial atomic charge measures the charge distribution within the molecule, not just on the hydrogen atom.

3. Acid strength increases with the charge of the anion connected to the acidic hydrogen: True. A higher charge on the anion implies a stronger attraction to the hydrogen atom, resulting in a stronger acid.

4. Only HI is shown completely dissociated: I cannot confirm this statement without interacting with the Open Odyssey platform.

5. The H-X bond length increases with acid strength: False. In fact, the opposite is true. As the acid strength increases, the H-X bond length generally decreases.

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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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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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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 volume occupied by 0.50 mole of the gas at a temperature of 35 °C and at 1 atm is 12.6 L (3rd option)

We can obtain the volume of the gas by using the ideal gas equation as shown below:

PV = nRT

1 × V = 0.50 × 0.0821 × 308

V = 12.6 L

Thus, from the above calculation, it is evident that the volume of gas is 12.6 L (3rd option)

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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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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 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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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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The process by which a substance changes from a gas or vapor to a solid without first becoming a liquid is known as sublimation.

Sublimation is a phase transition process in which a solid substance is transformed directly into its gaseous form or vice versa, bypassing the liquid state. This phenomenon occurs when the pressure and temperature conditions are such that the solid substance can vaporize without melting.

The most common examples of sublimation include the freezing of dry ice (solid carbon dioxide), where it converts directly into a gas without first melting into a liquid. Another example is the process of freeze-drying or lyophilization, which is widely used in the food and pharmaceutical industries to preserve and store products for longer periods.

In addition to these industrial applications, sublimation also plays a vital role in various natural processes. For instance, the formation of snowflakes and frost on cold surfaces occurs due to sublimation of water vapor present in the atmosphere. Sublimation is also responsible for the erosion of rocks and mountains, as water vapor freezes directly onto the surface and causes physical breakdown due to expansion and contraction.

In summary, sublimation is an essential process that has many practical and natural applications, and it occurs when a substance transitions directly from a solid to a gas or vice versa without passing through a liquid phase.

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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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