given a choice of atoms, which atom should you put in the center of a lewis structure? question 10 options: the atom that is nearer to the right and lowest down on the periodic table. the atom that is nearer to the right and highest up on the periodic table. the atom that is nearer to the left and highest up on the periodic table. the atom that is nearer to the left and lowest down on the periodic table

The mass of the excess reagent (iron) that remains after the reaction is complete is 12.1 grams.

The maximum mass of Iron(III) oxide that can be formed is 74.8 grams.

The formula for the limiting reactant is O₂, which is oxygen gas.

Calculate the number of moles of each reactant:

Number of moles of iron = 38.2 g / 55.845 g/mol = 0.683 moles

Number of moles of oxygen = 14.9 g / 32.00 g/mol = 0.466 moles

The balanced equation shows that the stoichiometric ratio between iron and oxygen is 4:3. Therefore, the mole ratio of iron to oxygen is 4:3.

Since the mole ratio of iron to oxygen is greater than 4:3, it means that there is an excess of iron and oxygen is the limiting reactant. So we need to use the number of moles of oxygen to calculate the maximum mass of Iron(III) oxide that can be formed.

The balanced equation shows that 1 mole of iron reacts with 1 mole of oxygen to produce 1 mole of Iron(III) oxide. The molar mass of Iron(III) oxide is 159.69 g/mol.

Number of moles of Iron(III) oxide = 0.466 moles (which is equal to the number of moles of oxygen)

The maximum mass of Iron(III) oxide = Number of moles of Iron(III) oxide x molar mass of Iron(III) oxide

Maximum mass of Iron(III) oxide = 0.466 moles x 159.69 g/mol = 74.8 grams

The maximum mass of Iron(III) oxide that can be formed is 74.8 grams.

The formula for the limiting reactant is O₂, which is oxygen gas.

To calculate the mass of the excess reagent (iron) that remains after the reaction is complete, we need to first calculate the mass of iron that reacted with the oxygen:

Mass of iron reacted = 0.466 moles x 55.845 g/mol = 26.1 grams

The initial mass of iron was 38.2 grams, so the mass of excess iron that remains after the reaction is complete is:

Mass of excess iron = initial mass of iron - a mass of iron reacted

Mass of excess iron = 38.2 g - 26.1 g = 12.1 grams

The mass of the excess reagent (iron) that remains after the reaction is complete is 12.1 grams.

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Answer:

The balanced chemical equation for the reaction between sulfuric acid (H₂SO4) and aqueous sodium cyanide is:

H₂SO4 + 2 NaCN → 2 HCN + Na₂SO₄

In this reaction, sulfuric acid (H₂SO4) reacts with aqueous sodium cyanide (NaCN) to produce hydrogen cyanide gas (HCN) and aqueous sodium sulfate (Na₂SO₄).

To balance the equation, two moles of sodium cyanide are required for every mole of sulfuric acid. The reaction produces two moles of hydrogen cyanide and one mole of sodium sulfate for every two moles of sodium cyanide and one mole of sulfuric acid.

It's important to note that hydrogen cyanide gas is highly toxic and dangerous, and proper safety precautions must be taken when handling this chemical.

Explanation:

Urea's molecular formula is (NH₂)2CO. The Avogadro's number, or 6.022 10²³ atoms/mol, must be used to determine the number of atoms in a given amount of urea.

Nowadays, urea molecules include a total of 8 atoms, of which 4 are in the H atom, 2 in the N atom, 1 in the C atom, and 1 in the O atom. In 5.6 g of urea, there are 2.247 1023 H atoms, 1.124 1023 N atoms, 0.562 10²³ C atoms, and 0.562 10²³ O atoms.

Carbon, nitrogen, oxygen, and hydrogen are the four elements that make up urea. It contains two electrons and a molar mass of 60.06 g/mole.

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Radial symmetry is a type of body symmetry where each plane that passes through the centre splits the body into two equally sized halves.

The capacity of an animal's body plan to be divided along a line that divides the animal's body into nearly identical right and left halves is known as bilateral symmetry. Arrangement of identical pieces in a circle around a central axis is a type of symmetry.

The organism can be divided into additional planes with identical parts in radial symmetry. In biradial symmetry, the organism can be divided solely in two planes rather than all three as in radial symmetry.

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To make a 2 M solution from an 18 M solution, you will need to dilute it by a factor of 9. (C1V1 = C2V2) denotes the relationship between the initial concentration (C1), initial volume (V1), final concentration (C2), and final volume (V2)

A concentrated solution has a high solute to solvent ratio, meaning there is a large amount of solute (such as salt or sugar) dissolved in a small amount of solvent (such as water). A diluted solution has a low solute to solvent ratio, meaning there is a small amount of solute dissolved in a large amount of solvent.

Knowing how to dilute a solution is important in many scientific and medical applications, as it allows us to create solutions with specific concentrations that are needed for experiments or treatments.

Dilution can also be used to reduce the toxicity or reactivity of a substance, making it safer to handle or use.

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According to the equation, one molecule of H₂ reacts for every two molecules of C₂H₄ that respond at the particle level. One molecule of C₂H₆ is created when these molecules join.

According to the equation, 1 mole of H₂ reacts for every 1 mole of C₂H₄ that does. To create 1 mole of C₂H₆, these reactants must be combined in a ratio of 1:1.

The balanced equation for the chemical reaction is:

[tex]C_2H_4_(g_) + H_2_(g_) -- > C_2H_6_(g_)[/tex]

This equation illustrates how ethene (C₂H₄) and hydrogen gas (H₂) combine to create ethane. (C₂H₆). The stoichiometric coefficients in the equation denote the mole ratios of the reactants and products in the balanced equation. They are represented by the coefficients in the equation.

According to the equation, one molecule of H₂ reacts for every two molecules of C₂H₄ that respond at the particle level. One molecule of C₂H₆ is created when these molecules join.

According to the equation, 1 mole of H₂ reacts for every 1 mole of C₂H₄ that does. To create 1 mole of C₂H₆, these reactants must be combined in a ratio of 1:1.

In general, the balanced equation and its coefficients offer crucial details about the proportions of reactants and products engaged in the chemical reaction.

Chemical processes are modeled by chemical equations. The proportions of the reactants and products engaged in a chemical reaction are displayed in a balanced chemical equation.

The reactant's and products' mole ratios, which show the relative amounts of each substance engaged in the reaction, are represented by the coefficients in the balanced equation.

The mass conservation principle, says that mass is neither produced nor destroyed in a chemical reaction and experimental data are used to determine these coefficients.

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The reaction rate is 12 times faster when the concentration of 1-iodopropane quadruples and the concentration of sodium hydroxide triples.

In SN₂ reaction, the rate is dependent on both the concentration of the nucleophile as well as the concentration of the substrate. The rate law for this reaction can be calculated as;

rate = k[substrate][nucleophile]

where k is the rate constant and [substrate] and [nucleophile] are the concentrations of the substrate and nucleophile, respectively.

If the concentration of 1-iodopropane quadruples and the concentration of sodium hydroxide triples, then the new rate of the reaction can be calculated using the following equation:

new rate = k[(4[substrate])×(3[nucleophile])]

new rate = k[12[substrate][nucleophile]]

new rate = 12 times the original rate

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Answer:42.4 g

Explanation:

2.35x10^24 atom / 6.02x10^23 atom

1.06 moles x 39.95 = 42.39 g Ar

Please find attached herewith the solutions for your questions.

To balance a chemical equation, you have to keep in mind these steps:

Hope it helps.

If you have any query, feel free to ask.

The weight of potassium hydrogen phthalate that will require 40.0 ml of a 0.100 m NaOH solution to reach the equivalence point is 0.8169 grams.

In order to find out the weight of potassium hydrogen phthalate that will require 40.0 ml of a 0.100 m NaOH solution to reach the equivalence point, we can use the following formula: weight of KHP = (molarity of NaOH) x (volume of NaOH) x (molar mass of KHP) / 1000

First, we need to calculate the number of moles of NaOH required to reach the equivalence point: Moles of NaOH = (molarity of NaOH) x (volume of NaOH) / 1000 Moles of NaOH = (0.100 mol/L) x (40.0 mL) / 1000 Moles of NaOH = 0.004 mol.

Now we can use the number of moles of NaOH to calculate the weight of KHP required: Weight of KHP = (0.004 mol) x (204.22 g/mol), Weight of KHP = 0.8169 g. Therefore, the weight of potassium hydrogen phthalate that will require 40.0 ml of a 0.100 m NaOH solution to reach the equivalence point is 0.8169 grams.

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Answer:

To determine if the given chemical equation is carbon balanced, we need to count the number of carbon atoms in the reactants and compare it to the number of carbon atoms in the products.

Reactants:

2 C2H6 -> 4 carbon atoms

Products:

4 CO2 -> 4 carbon atoms

Since the number of carbon atoms on the reactant side is equal to the number of carbon atoms on the product side, we can conclude that the given chemical equation is carbon balanced.

As long as the temperature and other conditions are kept constant, the rate constant will remain constant for that reaction.

Part A: Why should experimentally determined values of the rate constant, k, for each trial be fairly similar, even though the concentrations of the reactants are changed in each trial,

The experimentally determined values of the rate constant k for each trial should be fairly similar even though the concentrations of the reactants are changed in each trial because the rate constant is a measure of the intrinsic reactivity of the reaction itself, and is independent of the initial concentrations of the reactants.

As long as the temperature and other conditions are kept constant, the rate constant will remain constant for that reaction.

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The last statement is the false statement which says that A mixture can be separated into its constituent components only by chemical means.

A mixture can be separated into its constituents by both physical as well as chemical means. While homogeneous mixtures are usually separated by both chemical and physical means, heterogeneous mixtures whose components can be seen by the bare eye are usually separated by physical means.

Physical means like hand picking, distillation, filtration, and fractional distillation are widely used in the separation of mixtures. For example, a mixture of soil and water is usually separated from each other using the filtration method in which a filter paper or muslin cloth is used to filter out clear water from the mixture. Thus it is false to say that the mixtures can be separated only by chemical methods.

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The false statement about mixtures is:

"A mixture can be separated into its constituent components only by chemical means."

The correct statement is that a mixture can be separated into its constituent components by physical means such as filtration, distillation, chromatography, and so on. Chemical means are used to separate compounds into their constituent elements.

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The balanced reaction equation is;

2 HBr (aq) + Ca(OH)2 (s) → CaBr2 (aq) + 2 H2O (l)

A balanced chemical equation is an equation in which the number of atoms of each element is the same on both the reactant and product sides of the equation. This means that the law of conservation of mass is obeyed - the total mass of the reactants equals the total mass of the products.

The reactants and the products can be seen on the left and on the right hand sides of the reaction equations respectively.

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Answer:

solid this is because molecules in a sound medium are much closer together than those in a liquid or gas

The substance is nitrogen dioxide (NO₂).

Nitrogen dioxide (NO₂) is a reddish-brown irritating gas that is a major contributor to photochemical smog. In the presence of sunlight, it reacts with other pollutants such as volatile organic compounds (VOCs) to form ground-level ozone, which is a major component of smog. NO₂ is also a precursor to nitric acid, which is one of the major components of acid deposition.

NO₂ is mainly emitted by vehicles, power plants, and industrial processes. Exposure to high levels of NO2 can cause respiratory problems such as coughing, wheezing, and shortness of breath. It is therefore important to control NO₂ emissions to protect human health and the environment.

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Both XeF2 and O3 molecule has nonlinear structure. Hence option B  and C are correct.

Compounds with a geometry different than linear geometry are referred to as nonlinear molecules. This indicates that these molecules are not linear and that their atoms are not aligned in a linear fashion.

XeF2 has a linear geometry, 3 lone pairs, and 2 bond pairs. Option B is therefore incorrect.

Similar to BeCl2, which similarly has a linear shape and 3 lone pairs and 2 bond pairs. Option A is therefore unsuitable.

Ozone O3 has a triangular planar shape, two bond pairs, and one single pair. Hence, since it is a non-linear molecule, option C is valid.

A linear molecule with a bond angle of 180 is carbon dioxide. Option D is therefore incorrect.

Another linear molecule is N2O. Option E is therefore incorrect.

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1. The initial pH of the solution is 1.89.

2. At half the equivalence point (20 mL of NaOH added), the pH of the solution is 3.26.

3. At the equivalence point (30 mL of NaOH added), the pH of the solution is 10.72.

4. At some volume past the equivalence point the pH of the solution is 12.30

In the sketch (in figure), the x-axis represents the volume of NaOH added and the y-axis represents the pH of the solution. The initial pH before any NaOH has been added is 1.89, which is the pH of the 0.15 M formic acid solution.

As NaOH is added, the pH increases slowly at first, but then increases more rapidly as the solution enters the buffer region. At the half-equivalence point (20 mL of NaOH added), the pH is 3.26. At the equivalence point (30 mL of NaOH added), the pH jumps up to 10.72 due to the complete reaction of the formic acid with NaOH.

After the equivalence point, the pH continues to increase as more NaOH is added. At 50 mL past the equivalence point, the pH is 12.30, which is close to the pH of a strong base.

The titration of 0.15 M formic acid (HCOOH) with 0.25 M NaOH can be represented by the following equation:

[tex]HCOOH + NaOH[/tex] → [tex]NaCOOH + H_2O[/tex]

Before any NaOH is added, the solution consists of 0.15 M formic acid, which is a weak acid. The initial pH of the solution can be calculated using the dissociation constant (Ka) of formic acid:

[tex]HCOOH + H_2O < = > H_3O^+ + HCOO^-[/tex]

[tex]Ka = [H_3O^{+}][HCOO^{-}]/[HCOOH][/tex]

Since formic acid is a weak acid, we can assume that [tex][H_3O^+][/tex] is equal to [tex][HCOO^-][/tex]. Let x be the concentration of [tex][H_3O^+][/tex] and [[tex][HCOO^-][/tex]] at equilibrium, then:

[tex]Ka = x^2 / (0.15 - x)[/tex]

At equilibrium, the concentration of HCOOH will be (0.15 - x) M.

Let's solve for x:

[tex]Ka = x^2 / (0.15 - x)[/tex]

[tex]1.77 * 10^{-4} = x^2 / (0.15 - x)[/tex]

x = 0.0129 M

1. Therefore, the initial pH of the solution is:

[tex]pH = -log[H_3O^+][/tex]

pH = -log(0.0129)

pH = 1.89

Now let's consider the pH at different points during the titration:

Before any NaOH has been added:

The initial pH of the solution is 1.89.

2. At some fraction of the equivalence point:

At the equivalence point, all of the formic acid will have reacted with an equal amount of NaOH. Since NaOH is a strong base, the solution will be basic after the equivalence point.

At some fraction of the equivalence point, we can assume that the solution is a buffer consisting of formic acid and its conjugate base, sodium formate (NaCOOH). We can use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log([NaCOOH] / [HCOOH])

At the fraction of the equivalence point, we can assume that the concentration of HCOOH and NaCOOH are equal, and the concentration of NaOH is equal to the fraction of the equivalence point times the initial concentration of formic acid. Thus:

[HCOOH] = 0.15 M - (fraction of equivalence point) × (volume of NaOH added)

[NaCOOH] = (fraction of equivalence point) × (volume of NaOH added)

[tex][H_3O^+] = Ka * [HCOOH] / [NaCOOH][/tex]

Let's assume that the fraction of the equivalence point is 0.5, which means that half of the initial concentration of formic acid has reacted with NaOH. Let's also assume that we have added 20 mL of NaOH so far:

[tex][HCOOH] = 0.15 M - 0.5 * 0.02 L * 0.25 M\\[HCOOH] = 0.14 M\\[NaCOOH] = 0.5 * 0.02 L * 0.25 M\\[NaCOOH] = 0.0025 M[/tex]

[tex][H_3O^+] = 1.77 * 10^{-4} * (0.14 / 0.0025)[/tex]

[tex][H_3O^+] = 9.88 * 10^{-3} M[/tex]

[tex]pH = pKa + log([NaCOOH] / [HCOOH])\\\\pH = 3.75 + log([0.0025 / 0.14])\\\\pH = 3.26[/tex]

Therefore, at half the equivalence point (20 mL of NaOH added), the pH of the solution is 3.26.

3. At the equivalence point:

The pH can be calculated using the hydrolysis constant (Kb) of sodium formate:

[tex]NaCOOH +[/tex] [tex]H_2O[/tex] ⇌ [tex]NaOH + HCOOH[/tex]

[tex]Kb = [NaOH][HCOOH]/[NaCOOH][/tex]

Let's assume that we have added 30 mL of NaOH, which is the equivalent amount to the initial concentration of formic acid:

[tex][NaOH] = [HCOOH] = 0.15 M\\[NaCOOH] = 0.5 * 0.03 L * 0.25 M\\[NaCOOH] = 0.00375 M\\\\Kb = [NaOH]^2 / [NaCOOH]\\\\Kb = (0.15)^2 / 0.00375\\\\Kb = 6\\\\pOH = -log[OH-]\\pOH = -log\sqrt{(Kb * [NaCOOH])} \\pOH = -log\sqrt{6 * 0.00375} \\pOH = 3.28\\pH = 14 - pOH\\pH = 10.72[/tex]

Therefore, at the equivalence point (30 mL of NaOH added), the pH of the solution is 10.72.

4. At some volume past the equivalence point:

After the equivalence point, the solution will be basic due to the excess of NaOH. The pH can be calculated using the concentration of NaOH and the volume of NaOH added:

pOH = -log[OH-]

pOH = -log(0.25 × (volume of NaOH added - volume of NaOH at equivalence point))

pH = 14 - pOH

Let's assume that we have added 50 mL of NaOH past the equivalence point:

pOH = -log(0.25 × (0.05 L - 0.03 L))

pOH = 1.70

pH = 14 - pOH

pH = 12.30

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The answer is (A) -200 J.

The first law of thermodynamics states that the change in the internal energy of a system, ΔU, is equal to the heat added to the system, Q, minus the work done by the system, W:

ΔU = Q - W

In this case, we know that ΔU = -300 J (decrease in internal energy) and W = 100 J (work done on the system). Therefore, we can rearrange the equation to solve for Q:

Q = ΔU + W = -300 J + 100 J = -200 J

This means that the system lost 200 J of heat. However, the question asks for the change in heat, which means we need to take the negative sign into an account. Therefore, the answer is (A) -200 J.

The first law of thermodynamics is a fundamental principle of physics that states that energy can only be moved or changed from one form to another. This theory holds true for all types of energy, including mechanical, thermal, and electromagnetic energy.

The first law of thermodynamics is concerned with the conservation of energy in a system. It says that the total energy of a closed system remains constant, which includes both the system's internal energy and the work done on or by the system.

In other words, any change in a system's energy must be balanced by a change in the work or heat entering or exiting the system.

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A zinc bromide solution containing 75.0 ccs of solid sodium carbonate is gradually added until the carbonate ion concentration reaches 0.0636 m. The maximum amount of zinc ions that can remain in the solution is 5.30 x 10^-14 moles.

The addition of solid sodium carbonate to the zinc bromide solution will result in the precipitation of zinc carbonate according to the following balanced chemical equation:

Na2CO3 + ZnBr2 -> ZnCO3 + 2NaBr

The stoichiometry of the reaction shows that one mole of zinc bromide reacts with one mole of sodium carbonate to produce one mole of zinc carbonate. Therefore, the number of moles of zinc carbonate produced will be equal to the number of moles of carbonate ion in the solution, which is given as 0.0636 moles.

The initial concentration of zinc bromide is not given, so we cannot directly calculate the number of zinc ions remaining in the solution. However, we can use the solubility product constant (Ksp) of zinc carbonate to determine the maximum amount of zinc ions that can remain in the solution before precipitation occurs.

The Ksp of zinc carbonate is given as 4.5 x 10^-11. Using the balanced chemical equation, we can write the expression for the Ksp as follows:

Ksp = [Zn2+][CO32-]

Since the concentration of carbonate ion is given as 0.0636 M, we can rearrange the above equation to solve for the maximum concentration of zinc ion that can remain in the solution:

[Zn2+] = Ksp / [CO32-] = (4.5 x 10^-11) / (0.0636)

= 7.07 x 10^-13 M

Therefore, the maximum amount of zinc ion that can remain in the solution is given by multiplying the maximum concentration by the final volume of the solution:

[Zn2+] x Vfinal = (7.07 x 10^-13 M) x (75.0 mL / 1000 mL/mL)

= 5.30 x 10^-14 moles

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If we have three tuning forks of frequencies 250, 500 and 1000 Hz then the tuning fork of 250 Hz would sound the lowest and the tuning fork of 1000 Hz would have the highest pitch.

A.) The tuning fork of 250 Hz would sound the lowest, as it has the lowest frequency among the three. Frequency measures the number of oscillations per second of a wave so the 250 Hz wave will have fewer oscillations per second than the 500 Hz and 1000 Hz waves, resulting in a lower pitch. A lower frequency means that the sound waves are closer together, producing a lower, deeper sound.

B) The tuning fork of 1000 Hz would have the highest pitch, as it has the highest frequency. The 1000 Hz wave will have more oscillations per second than the 250 Hz and 500 Hz waves, resulting in a higher pitch. A higher frequency means that the sound waves are farther apart, producing a higher, more piercing sound.

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Solution d will have the lowest pH and be most acidic. The pH of a solution is inversely proportional to the strength of the acid, which means the stronger the acid, the lower the pH, and the more acidic the solution.

The strength of an acid is determined by its dissociation constant, Ka. A smaller Ka value means a weaker acid and a larger Ka value means a stronger acid. The pH of the four solutions will be calculated using the following equation: pH = pKa + log([A-]/[HA]), where [A-] and [HA] are the concentrations of the conjugate base and acid, respectively, and pKa is the dissociation constant of the acid.

Here, we have the following pKa values: acid pka ha 4.00 hb 7.00 hc 10.00 hd 11.00The strongest acid will have the smallest pKa value and the weakest acid will have the largest pKa value.

Therefore, the order of acidic strength is: d > c > b > a The lowest pH and the most acidic solution will be that which has the strongest acid. Since acid d has the lowest pKa value, it is the strongest acid, and its solution will have the lowest pH and be the most acidic.

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Answer:

C6H6, also known as benzene, is a Lewis base because it can donate a pair of electrons to form a coordinate covalent bond with a Lewis acid. A Lewis base is defined as any substance that can donate a pair of electrons to form a coordinate covalent bond.

However, benzene is not a Bronsted-Lowry base because it does not have a hydrogen ion (H+) to donate. A Bronsted-Lowry base is defined as any substance that can donate a hydrogen ion (H+).

Benzene is also not an Arrhenius base because it does not produce hydroxide ions (OH-) when dissolved in water. An Arrhenius base is defined as any substance that produces hydroxide ions (OH-) when dissolved in water.

Explanation:

There are different definitions of what a base is. Three common definitions are the Lewis, Bronsted-Lowry, and Arrhenius definitions.

According to the Lewis definition, a base is any substance that can donate a pair of electrons to form a bond. Benzene (C6H6) can do this, so it is considered a Lewis base.

The Bronsted-Lowry definition says that a base is any substance that can donate a hydrogen ion (H+). Benzene does not have a hydrogen ion to donate, so it is not considered a Bronsted-Lowry base.

The Arrhenius definition says that a base is any substance that produces hydroxide ions (OH-) when dissolved in water. Benzene does not produce hydroxide ions when dissolved in water, so it is not considered an Arrhenius base.

The mass of C₅H₅NHCl , the student should be dissolve in the solution to turn it into a buffer with ph = 5.64 is equals to the 31.45g.

We have a pyridine solution, which is a weak base solution. For a weak base,

the ionization constant, kb = 1.7 × 10⁻⁹

Buffer pH = 5.64

Volume of solution = 250 mL

Molarity of solution = 0.7 M

Using pH formula, pKb = - log ( 1.7 × 10⁻⁹)

= 8.77

For pH of equilibrium constant of water,

pKw = pKb + pKa

=> 14 = 8.77 + pKa

=> pKa = 5.23

Buffer pH formula = pKa + log( [[C₅H₅N] /[C₅H₅NHCl])

=> 5.64 = 5.23 + log( 0.7 M/[C₅H₅NHCl])

=> log(0.7 M/[C₅H₅NHCl]) = 5.64 - 5.23

=> log(0.7 M/[C₅H₅NHCl]) = 0.41

So, 0.7 M/[C₅H₅NHCl] = 10⁻⁰·⁴¹

=> [C₅H₅NHCl] = 10⁻⁰·⁴¹ × 0.7 M

Moles of [C₅H₅NHCl] = 250 mL× 10⁻⁰·⁴¹ × 0.7 mol [C₅H₅NHCl] / 1000 mL

= (0.7 ×10⁻⁰·⁴¹)/4

= 0.175 ×10⁻⁰·⁴¹ moles = 0.2723 moles.

Molar mass of C₅H₅NHCl = 115.5 g/mol

Mass of [C₅H₅NHCl] = 0.175 ×10⁻⁰·⁴¹ moles × 115.5 g [C₅H₅NHCl]/ 1 mol of [C₅H₅NHCl]

= 0.2723 × 115.5 g = 31.454 g

Hence, required mass is 31.45 g.

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Complete question:

a chemistry graduate student is given 250 ml of a 0.7 M pyridine solution C5H5N. pyridine is a weak base with Kb = 1.7 × 10-9. what mass of C5H5NHCl should the student dissolve in the solution to turn it into a buffer with ph= 5.64 ? you may assume that the volume of the solution doesn't change when the is dissolved in it. be sure your answer has a unit symbol, and round it to significant digits.

The chemical equation for the autoionization of water can be written as,

2H₂O ⇌ H₃O⁺ + OH⁻

In pure water, a small percentage of water molecules can react with each other through a process known as autoionization or self-ionization. In this equation, two water molecules (H₂O) undergo a reversible reaction to form a hydronium ion (H₃O⁺) and a hydroxide ion (OH⁻).

This process is also known as self-ionization or autoprotolysis of water. The square brackets [] are often used to indicate concentration, so the equilibrium constant for this reaction can be written as:

Kw = [H₃O⁺][OH⁻] = 1.0 x 10⁻¹⁴ (at 25°C)

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0.5 ml of heparin solution should be added to 500 ml to prepare heparin sodium 10 U/ml solution for flushing the artery intraoperatively.

Heparin sodium solution is a sterile, clear, colorless solution that contains heparin, a medication that is used as an anticoagulant or blood thinner. Heparin sodium solution is used to prevent blood clots from forming in conditions such as deep vein thrombosis, pulmonary embolism, and during certain medical procedures such as dialysis and heart surgery.

The solution is usually administered by injection or intravenous infusion, and is available in different strengths and volumes depending on the patient's condition and the intended use. Heparin sodium solution is stored in a cool and dry place, and should be handled and disposed of properly to avoid contamination and injury.

To prepare heparin sodium solution with a concentration of 10 U/ml using a vial of heparin 10,000 U/ml

Determine the total amount of heparin needed:

10 U/ml x 500 ml = 5,000 U

Calculate the volume of heparin solution needed:

5,000 U ÷ 10,000 U/ml = 0.5 ml

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The volume of oxygen made at the end of the reaction in which peroxidase was used as a catalyst is 10 cm3.

Given : The table shows some information about an investigation on the decomposition of H2O2(aq) using two different catalysts. In each experiment, 0.100g of the catalyst and 25.0 cm3 of H2O2(aq) were used. The concentration and temperature of the H2O2(aq) Use were kept constant.

Manganese(IV) oxide 25 95 Peroxidase 10.

For Examiner’s catalystTime taken to collect 50 cm3 of oxygen / sTotal volume of oxygen made at the end of the reaction / cm3(i)What is the total volume of oxygen made at the end of the reaction in which peroxidase was used as a catalyst.

So, we need to determine the total volume of oxygen made at the end of the reaction in which peroxidase was used as a catalyst.Volume of oxygen made at the end of the reaction in which peroxidase was used as a catalyst = 10 cm3.

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Using Boyle's law, we can calculate the pressure of the hydrogen gas before its volume was decreased. According to Boyle's law: PV = k where P is pressure, V is volume, and k is a constant at constant temperature. so the answer is 3.51 atm.

We can use the equation P1V1 = P2V2 to solve for the initial pressure, where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.

Substituting the given values, we get:

P1V1 = P2V2 P1(12.13 L) = (7.85 atm)(5.42 L) P1 = (7.85 atm)(5.42 L) / (12.13 L) P1 = 3.5075 atm = 3.51 atm.

Therefore, the answer is: the hydrogen gas exerted a pressure of 3.51 atm before its volume was decreased.

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The student should dissolve approximately 4.1 g of NaC₂H₃O₂ to prepare a buffer solution with pH 4.74.

To prepare a buffer solution with pH 4.74, the graduate student needs to add an appropriate amount of sodium acetate (NaC₂H₃O₂) to the given acetic acid solution.

First, we can use the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

where pH is 4.74, pKa is the acetic acid pKa (1.76), [A-] is the concentration of the acetate ion, and [HA] is the concentration of acetic acid. We need to find [A-] and [HA] in moles per liter.

4.74 = 1.76 + log ([A-]/[HA])

Rearranging the equation and solving for the ratio [A-]/[HA]:

[A-]/[HA] = 10^(4.74 - 1.76) ≈ 1

Since the ratio of acetate ion to acetic acid is approximately 1, this means that the concentrations of NaC₂H₃O₂ and acetic acid should be equal.

Now, let's calculate the moles of acetic acid present in the solution. Acetic acid's molecular weight is 60.05 g/mol. If the student has 500 mL of a 0.10 M solution:

moles of acetic acid = 0.10 mol/L * 0.5 L = 0.05 mol

Since we need an equal amount of sodium acetate, the moles of NaC₂H₃O₂ required will also be 0.05 mol. The molecular weight of NaC₂H₃O₂ is 82.03 g/mol. To find the mass of NaC₂H₃O₂ needed, we can use the formula:

mass = moles * molecular weight

mass of NaC₂H₃O₂ = 0.05 mol * 82.03 g/mol ≈ 4.1 g
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The Rack Stop is a small, movable metal stopper located at the bottom of the microscope body tube. Its purpose is to prevent the objective lenses from hitting the microscope slide or specimen, which could damage both the lenses and the sample.

Here is a list of microscope parts and their uses, including the Rack Stop:

1. Eyepiece or Ocular Lens: The lens at the top of the microscope that you look through to view the specimen.

2. Body Tube: The long, cylindrical part of the microscope that holds the eyepiece at the top and the objective lenses at the bottom

3. Arm: The curved part of the microscope that connects the body tube to the base.

4. Base: The flat, sturdy part of the microscope that supports the rest of the instrument.

5. Stage: The flat platform on which you place the specimen for viewing.

6. Stage Clips: Small metal clips that hold the microscope slide in place on the stage.

7. Coarse Focus Knob: A large knob that moves the body tube up and down to bring the specimen into rough focus.

8. Fine Focus Knob: A smaller knob that moves the body tube slightly to fine-tune the focus of the specimen.

9. Diaphragm: A rotating disc or lever that controls the amount of light entering the microscope and illuminating the specimen.

10. Light Source: The bulb or mirror that provides light for illuminating the specimen.

11. Objective Lenses: A set of lenses located at the bottom of the body tube that magnify the specimen.

12. Rack Stop: A small, movable metal stopper located at the bottom of the microscope body tube. Its purpose is to prevent the objective lenses from hitting the microscope slide or specimen.

13. Nosepiece: The rotating turret at the bottom of the body tube that holds the objective lenses.

14. High Power Objective: The objective lens with the highest magnification, typically 40x or higher. It is used for detailed examination of the specimen.

To use the microscope, first place the specimen on the stage and secure it with the stage clips. Turn on the light source and adjust the diaphragm to control the amount of light entering the microscope. Then, use the coarse focus knob to bring the specimen into rough focus. Once you have achieved this, use the fine focus knob to fine-tune the focus and bring the specimen into clear view. To change the magnification, rotate the nosepiece to select the desired objective lens. Finally, adjust the focus as needed and observe the specimen at the desired magnification.

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