Enthalpy change of atomization is always endothermic/exothermic

The initial concentration of NaOH in the beaker is approximately 0.0007289 M.

- Volume of NaOH solution: 225.0 mL
- Concentration of HCl: 0.0100 M
- Volume of HCl needed to reach equivalence point: 16.4 mL

Step 1: Convert the volumes from mL to L.
- 225.0 mL NaOH = 0.225 L NaOH
- 16.4 mL HCl = 0.0164 L HCl

Step 2: Calculate the moles of HCl using its concentration and volume.
Moles of HCl = Concentration × Volume
Moles of HCl = 0.0100 M × 0.0164 L = 0.000164 mol

Step 3: At the equivalence point, the moles of HCl and NaOH are equal.
Moles of NaOH = Moles of HCl = 0.000164 mol

Step 4: Calculate the initial concentration of NaOH using its moles and volume.
Concentration of NaOH = Moles of NaOH / Volume of NaOH
Concentration of NaOH = 0.000164 mol / 0.225 L = 0.0007289 M

So, the initial concentration of NaOH in the beaker is approximately 0.0007289 M.

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A mole of any substance contains the same number of entities as exactly 12 g of carbon-12 contains, which is known as Avogadro's number (6.022 x 10²³ entities).

Avogadro's number is a fundamental constant in chemistry that represents the number of entities (such as atoms, molecules, or ions) in one mole of a substance. It is based on the observation that one mole of any substance always contains the same number of entities as exactly 12 grams of carbon-12, which has six protons, six neutrons, and six electrons.

This means that if we have one mole of an element, it contains 6.022 x 10²³ atoms of that element. The mass of one mole of an element in grams is equal to its atomic mass in atomic mass units (amu). Therefore, the concept of a mole allows us to easily convert between the number of atoms/molecules and the mass of a substance.

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The hydronium ion concentration, [H₃O⁺] =5.3 x 10⁻⁷ M, which is calculated in the below section.

The value of Kw = 2.8 x 10⁻¹³

In the autoionization of water, a proton is transferred from one water molecule to another to produce a hydronium ion (H₃O⁺) and a hydroxide ion (OH⁻). The equilibrium expression for this reaction is Kw = [H₃O⁺][OH⁻],

The concentration of hydronium ion and hydroxyl ion when a water molecule dissociates is the same which is 1 mol.

Kw = [H₃O] [OH⁻]

2.8 x 10⁻¹³ = [H₃O⁺]²

[H₃O⁺] = √(2.8 x 10⁻¹³)

[H₃O⁺] =5.3 x 10⁻⁷ M

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Before assembling the fractional distillation apparatus, it is important to ensure that all the components are clean and free of any residue from previous use.

This can be done by washing them with soap and water, followed by rinsing with distilled water and drying with a clean cloth.

It is also important to ensure that all the joints are tightly connected to prevent any leaks during the distillation process.

Before assembling the fractional distillation apparatus, you must ensure that all components are clean and properly functioning.

Additionally, gather the required materials and set up the apparatus in a safe, well-ventilated area according to your experimental procedure.

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an electron in the nth energy level of a hydrogen atom can make approximately 2.49 x 10^15 revolutions per second.

The time it takes for an electron in a hydrogen atom to complete one orbit in the nth energy level is given by the formula:

t = (n^3)/(Z^2) * t0

where t0 is the time it takes for an electron to complete one orbit in the ground state (n=1) and Z is the atomic number (which is 1 for hydrogen).

t0 = 2.42 x 10^-17 s

If the hydrogen atom is in an excited state for 10^-8 s, we can calculate the maximum value of n for which the electron can complete at least one orbit during this time by solving the above equation for n:

n^3 = Z^2 * t * (1/t0)

n^3 = 1^2 * 10^-8 s * (1/2.42 x 10^-17 s)

n^3 = 4.13 x 10^8

n ≈ 690

So the maximum value of n for which the electron can complete at least one orbit during 10^-8 s is n = 690.

The number of revolutions that an electron in the nth energy level makes in one second is given by:

rev/s = (2πn)/(t0 * n^2/Z^2)

Substituting n=690 and Z=1 for hydrogen, we get:

rev/s = (2π * 690)/(2.42 x 10^-17 s * 690^2)

rev/s ≈ 2.49 x 10^15 revolutions per second

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The compound BaCl₂ will conduct electricity. Therefore, the correct options are C.

Substances that conduct electricity dissolve in solution to give ions. These ions are the charge carriers in solution. Only ionic substances can dissolve in water to give ions that conduct electricity. MgSO₄ and BaCl₂ are ionic substances. They yield ions in solutions which conduct electrical current.

Ionic compounds have high points of melting and boiling and appear to be strong and brittle. Ions may be single atoms, such as sodium and chlorine in common table salt (sodium chloride) or more complex groups such as calcium carbonate.

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The ionization of HOCl (hypochlorous acid) in solution is dependent on the concentration of H+ ions. Therefore, a salt that could decrease the concentration of H+ ions in solution would also decrease the ionization of HOCl.

Option (D) NaOH is a strong base that could neutralize H+ ions and hence decrease their concentration in solution. Therefore, adding NaOH to a solution of HOCl would decrease the ionization of HOCl in solution.

Option (A) NaCl, option (B) NaOCl, option (C) Na2O, and option (E) BaCl2 do not directly affect the concentration of H+ ions in solution, and hence would not have a significant effect on the ionization of HOCl.

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To determine the pH of a 0.30 M FeCl2 solution, we need to consider the hydrolysis of the hydrated ferrous ion [Fe(OH2)6]2+ in water. This hydrolysis reaction can be represented as follows:

[Fe(OH2)6]2+ + H2O ⇌ [Fe(OH)(OH2)5]+ + H3O+

The equilibrium constant for this reaction is given by the expression:

Kw/Ksp[Fe2+] = [H3O+][Fe(OH)(OH2)5]+]/[Fe(OH2)6]2+

Where Kw is the ion product constant for water, Ksp[Fe2+] is the solubility product constant for Fe(OH)2, and [Fe2+] is the concentration of ferrous ions in solution.

We can use this equation to calculate the concentration of H3O+ ions in the solution, which will give us the pH of the solution. Plugging in the given values, we get:

Kw/Ksp[Fe2+] = [H3O+][Fe(OH)(OH2)5]+]/[Fe(OH2)6]2+
1.0 x 10^-14/8.7 x 10^-17 = [H3O+][Fe(OH)(OH2)5]+]/(0.30)^2
[H3O+] = 3.22 x 10^-3 M
pH = -log[H3O+] = 2.49

Therefore, the pH of a 0.30 M FeCl2 solution is approximately 2.49.

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Glycerol-3-phosphate is derived from two primary sources that are glycolysis and the glycerol phosphate shuttle.

During glycolysis, glucose is metabolized into pyruvate, which can then be converted into acetyl-CoA, the precursor for fatty acid synthesis. Acetyl-CoA is used to produce glycerol-3-phosphate, which is then used as a backbone for triacylglycerol biosynthesis. The glycerol phosphate shuttle, which occurs in some tissues, such as adipose tissue and liver, converts dihydroxyacetone phosphate into glycerol-3-phosphate.

This conversion allows for the incorporation of fatty acids into triacylglycerol, which is then stored as a form of energy in adipose tissue. Overall, glycerol-3-phosphate serves as a crucial precursor for the biosynthesis of triacylglycerol, an important energy storage molecule in the body.

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The entropy change of the ideal gas in the rigid tank is 0.485 kJ/kg-K. The entropy change of the ideal gas in the rigid tank undergoing a process.

where its temperature doubles can be determined using the equation:

ΔS = C ln(T2/T1)

where ΔS is the entropy change, C is the specific heat capacity of the gas, and T2 and T1 are the final and initial temperatures, respectively.

Using the given values of C = 0.7 kJ/kg-K and doubling of temperature, T2/T1 = 2, we can calculate the entropy change:

ΔS = 0.7 kJ/kg-K * ln(2) = 0.485 kJ/kg-K

Therefore, the explanation is that the entropy change of the ideal gas in the rigid tank is 0.485 kJ/kg-K. It is important to note that entropy is a measure of the disorder or randomness of a system, and it tends to increase in irreversible processes. In this case, the increase in temperature results in an increase in the randomness of the gas molecules, leading to an increase in entropy.

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The amount of the radioactive isotope remaining after 16 days is 2.5 mg

The amount of the radioactive isotope remaining after 16 days can be calculated using the formula:

A = a*(1/2)^(t/h)

where A is the amount remaining, a is the initial amount, t is the time elapsed, and h is the half-life of the substance.

In this case, we have an initial amount of 10 mg and a half-life of 8 days. Therefore, we can plug in these values into the formula to get:

A = 10*(1/2)^(16/8)
A = 10*(1/2)^2
A = 10*(1/4)
A = 2.5 mg

So, after 16 days, only 2.5 mg of the radioactive isotope remains.

Thus, the amount of the radioactive isotope remaining after 16 days is 2.5 mg.

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The concentration of glucose in a solution can be estimated using colorimetry.

How can colour comparisons be used to measure the content of glucose in a solution?

The concentration of glucose in a solution can be estimated using colorimetry, which involves comparing the color of a sample with that of a standard solution of known concentration. A common method for determining the concentration of glucose is the use of Benedict's reagent, which consists of copper sulfate, sodium citrate, and sodium carbonate.

To perform the test, a sample of the solution containing glucose is mixed with Benedict's reagent and heated in a water bath. The heat causes the glucose to reduce the copper ions in the reagent, forming a brick-red precipitate of copper(I) oxide.

The intensity of the red color of the precipitate is proportional to the concentration of glucose in the sample. This can be compared to a series of standard solutions of known glucose concentrations, which have been similarly treated with Benedict's reagent and heated to produce a range of colors.

By matching the color of the sample to the closest standard solution, the concentration of glucose in the sample can be estimated. For example, if the sample produces a color similar to the standard solution with a glucose concentration of 50 mg/dL, then the concentration of glucose in the sample can be estimated to be around 50 mg/dL as well.

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The acidic aqueous solution among the given options is NH4CN. This compound dissociates in water to form NH4+ (ammonium ion) and CN- (cyanide ion). The NH4+ ion can act as a Bronsted-Lowry acid, donating a proton (H+) to the surrounding water molecules, resulting in the formation of NH3 (ammonia) and H3O+ (hydronium ion). The presence of H3O+ ions increases the acidity of the solution.

On the other hand, NH4OCl does not form an acidic solution. It dissociates into NH4+ and OCl- (hypochlorite ion) in water. Although NH4+ can still donate a proton, the OCl- ion acts as a weak base, accepting protons and neutralizing the solution. As a result, the overall solution remains nearly neutral or slightly basic.

In summary, the acidic solution among the given options is NH4CN, as it dissociates in water to form NH4+ ions that increase the acidity by donating protons, leading to the formation of hydronium ions (H3O+).

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Group 2 carbonates become more thermally stable as you descend the group because the size of the metal cation increases, leading to a decrease in the lattice energy and an increase in the polarizability of the carbonate ion.

As you descend Group 2 of the periodic table, the size of the metal cation increases. This increase in size leads to a decrease in the lattice energy of the carbonate, as the larger cation is less effective in attracting and holding onto the carbonate ion. At the same time, the carbonate ion becomes more polarizable, meaning that it is better able to distort its electron cloud in response to the electric field of the metal cation. This combination of factors leads to an increase in the stability of the carbonate as you move down the group. As a result, the carbonates of the heavier Group 2 elements, such as barium carbonate, are more thermally stable than those of the lighter elements, such as magnesium carbonate.

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If 1.00 mole of [tex]C_{2}H_{6}[/tex] is combusted, then the amount of [tex]CO_{2}[/tex] produced would be 2 times that amount, or 2.00 moles of [tex]CO_{2}[/tex].

The balanced chemical equation for the combustion of ethane is [tex]C_{2}H_{6}[/tex] + 3.5[tex]O_{2}[/tex] -> 2[tex]CO_{2}[/tex] + 3[tex]H_{2}O[/tex]. This means that for every 1 mole of [tex]C_{2}H_{6}[/tex] that is combusted, 2 moles of [tex]CO_{2}[/tex] are produced.

Therefore, if 1.00 mole of [tex]C_{2}H_{6}[/tex] is combusted, then the amount of [tex]CO_{2}[/tex] produced would be 2 times that amount, or 2.00 moles of [tex]CO_{2}[/tex].

It is important to note that this calculation assumes that the reaction proceeds to completion, meaning that all of the reactants are consumed and all of the products are formed.

In reality, reactions may not always go to completion due to factors such as incomplete mixing or side reactions.

Additionally, there may be other factors that affect the stoichiometry of the reaction, such as the temperature, pressure, and presence of catalysts.

Therefore, the actual amount of [tex]CO_{2}[/tex] produced in a real-world scenario may differ from the theoretical calculation based on the balanced equation.

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Among all the given option, Phenolphthalein is the Most Commonly Used Indicator in Chemistry.

What is the most commonly used indicator in chemistry and why?

Among the indicators listed, phenolphthalein is the most commonly used indicator in chemistry. It is a weak acid that displays different colors depending on the pH of the solution it is in. In acidic solutions, it appears colorless, while in basic solutions, it turns pink or magenta.

Its sensitivity to changes in pH and ease of use make it a popular choice in acid-base titrations and other experiments where pH is critical. Additionally, its low cost and availability contribute to its widespread use. Overall, phenolphthalein is a versatile and reliable indicator that plays an essential role in many laboratory applications.

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50 mL of 0.090 M NaOH is required to titrate 50.0 mL of 0.090 M benzoic acid to the equivalence point.

We need to use the balanced chemical equation for reaction:

[tex]HC_7H_5O_2 + NaOH\ - > NaC_7H_5O_2 + H_2O[/tex]

From the equation, we can see that the molar ratio of benzoic acid to NaOH is 1:1.

Therefore, the number of moles of NaOH required to reach the equivalence point is equal to the number of moles of benzoic acid:

moles of NaOH = moles of benzoic acid = [tex](0.090 M)(0.050 L)[/tex]

= 0.0045 moles

To calculate the volume of 0.090 M NaOH needed to reach the equivalence point, we can use the molarity and moles of NaOH:

The volume of NaOH = moles of NaOH / molarity of NaOH

The volume of NaOH = [tex]0.0045 moles / 0.090 M = 0.050 L[/tex]

(50 mL).

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The 250 liters of gas that is collected in the expandable, sealed container. The sample is then heated from the 15.0 °C to the 45.0 °C at the constant pressure. The new volume of the container is 226.4 L.

The temperature and the volume at constant pressure is as :

V₁ / T₁ = V₂ / T₂

V₂ = V₁ T₂ / T₁

The initial volume of the gas, V₁ = 250 L

The final volume of the gas, V₂ = ?

The initial temperature of the gas, T₁ = 15 + 273

The initial temperature of the gas, T₁ = 288 K

The final temperature of the gas, T₂ = 45 + 273

The final temperature of the gas, T₂ = 318

V₂ = V₁ T₂ / T₁

V₂= ( 250 × 288 ) / 318

V₂ = 226.4 L

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H₂O have a normal meniscus while Hg has an inverted meniscus because the attraction between mercury molecules is greater than that between a molecule

And the container's walls, mercury forms a convex meniscus. The water molecules are drawn to the molecules in the glass beaker's wall.

A sunken meniscus, which is what you ordinarily will see, happens when the particles of the fluid are drawn to those of the holder. With water and a glass tube, this happens. A curved meniscus happens when the particles have a more grounded fascination with one another than to the holder, similarly as with mercury and glass.

The upper meniscus is the reverse U bend on the highest point of the outer layer of a fluid while the lower meniscus is the U bend on the highest point of the fluid's surface. By looking at the liquid's surface, the lower and upper meniscus are typically visible to the eye.

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The pH of an aqueous solution of 0.184M carbonic acid, H2CO3, is approximately 2.98. The correct answer is (c) 2.97.

To find the pH of an aqueous solution of 0.184M carbonic acid, H2CO3, we need to consider the dissociation of H2CO3 into its respective ions.

H2CO3 ⇌ H+ + HCO3- (Ka1 = 4.2 x 10-7)
HCO3- ⇌ H+ + CO32- (Ka2 = 4.8 x 10-11)

Ka1 and Ka2 represent the acid dissociation constants for the two protonation steps of carbonic acid. We can use these values to calculate the equilibrium concentrations of H+, HCO3-, and CO32- ions in solution.

First, we need to calculate the equilibrium concentration of HCO3- ions. Since carbonic acid is a weak acid, we can assume that most of it remains undissociated in solution. Therefore, the initial concentration of H2CO3 is equal to the concentration of HCO3- ions at equilibrium.

[HCO3-] = [H2CO3] = 0.184 M

Using the equilibrium constant expression for the first dissociation step, we can solve for the concentration of H+ ions.

Ka1 = [H+][HCO3-] / [H2CO3]
4.2 x 10-7 = [H+] x 0.184 / 0.184
[H+] = 4.2 x 10-7 M

Now that we know the concentration of H+ ions, we can use the equilibrium constant expression for the second dissociation step to calculate the concentration of CO32- ions.

Ka2 = [H+][CO32-] / [HCO3-]
4.8 x 10-11 = [4.2 x 10-7] x [CO32-] / 0.184
[CO32-] = 5.5 x 10-10 M

Finally, we can use the equation for the conservation of charge to calculate the concentration of OH- ions in solution.

[H+] x [OH-] = Kw = 1.0 x 10-14
[OH-] = Kw / [H+]
[OH-] = 1.0 x 10-14 / 4.2 x 10-7
[OH-] = 2.4 x 10-8 M

Now that we know the concentration of OH- ions, we can calculate the pH of the solution using the equation:

pH = -log[H+]
pH = -log(4.2 x 10-7)
pH = 6.38

However, we need to take into account that the solution contains a weak acid and its conjugate base, so we need to calculate the pH using the Henderson-Hasselbalch equation.

pH = pKa + log([A-]/[HA])
pH = 6.37 + log(5.5 x 10-10 / 0.184)
pH = 2.98

Therefore, the pH of an aqueous solution of 0.184M carbonic acid, H2CO3, is approximately 2.98. The correct answer is (c) 2.97.

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Though not explicitly covered in this lab, gradient elution solves general elution by increasing the solvent B amount throughout the run.

D is the correct answer.

A separation technique called gradient elution distributes the components between two phases, one of which is stationary and the other of which flows in a specific direction. The elution solvent strength of the mobile phase is gradually increased during the separation in gradient-elution chromatography.

Gradient elution primarily serves the following three goals: (1) decreasing the overall run time of separations; (2) changing retention times in a chromatographic run that does not effectively separate specific compounds; and (3) cleaning and/or regeneration of the chromatographic column.

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The graph of atomic size, also known as atomic radius, shows the size of the atoms of the elements as a function of their atomic number.

While it is true that the atomic radius generally increases as we move down a group and from right to left across a period in the periodic table, there are some notable deviations from this trend that result in peaks and valleys in the graph.One of the main factors that contributes to the observed peaks and valleys is the effect of the electron configuration on the size of the atom. In the case of the alkali metals, for example, each successive element in the group has an additional electron shell, which results in a significant increase in atomic size. This is because the additional electron shell increases the average distance between the outermost electrons and the nucleus, thereby increasing the size of the atom.

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Osmosis is the movement of a solvent through a semipermeable membrane from an area of low solute concentration to an area of high solute concentration.

A membrane is a thin layer of tissue or material that serves as a barrier between two environments, such as a biological tissue or a cell wall. It is typically composed of a single or multiple layers of molecules, and is often semi-permeable, meaning that it allows certain molecules to pass through while blocking others. The most common type of membrane found in the body is the cell membrane, which is composed of a lipid bilayer with embedded proteins. These proteins facilitate the movement of substances across the membrane, and can be either passive or active. Other types of membranes include the nuclear membrane, the mitochondrial membrane, and the endoplasmic reticulum.

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The average bond energy of oxygen is directly related to its enthalpy of atomisation. As the average bond energy increases, the enthalpy of atomisation also increases.

In more detail, the enthalpy of atomisation is the energy required to break one mole of a substance into its individual atoms in the gas phase. For oxygen, this means breaking the O2 molecule into two separate O atoms. The energy required to break this bond is the bond energy of oxygen.

The bond energy of oxygen is the amount of energy required to break one mole of O2 molecules into individual oxygen atoms in the gas phase. This bond energy is related to the strength of the bond between the two oxygen atoms in the molecule. As the bond energy increases, the bond between the two oxygen atoms becomes stronger, which makes it more difficult to break the bond and requires more energy to do so. This increased energy requirement results in a higher enthalpy of atomisation for oxygen.

In summary, the average bond energy of oxygen and its enthalpy of atomisation are directly related, with an increase in bond energy resulting in a higher enthalpy of atomisation.

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The rate constant of the reaction at 25°C is approximately 17.25 s⁻¹. To calculate the rate constant of the reaction at 25°C, we can use the Arrhenius equation.

The Arrhenius equation is given by: k =[tex]Ae^{-E_{a}/RT }[/tex]

Where:
- k is the rate constant
- A is the frequency factor (1.5 x 10¹¹ s⁻¹)
- Ea is the activation energy (56.8 kJ/mol)
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin (25°C + 273.15 = 298.15 K)

First, convert the activation energy to Joules per mole:
56.8 kJ/mol × 1,000 J/kJ = 56,800 J/mol

Now, plug the values into the Arrhenius equation:
k = (1.5 x 10¹¹) * [tex]e^{-56800/(8.314 * 298.15)}[/tex]

Calculate the exponent:
-56,800 / (8.314 * 298.15) = -22.966

Next, find the value of [tex]e^{-22.966}[/tex]:
which will be approximately 1.15 x 10⁻¹⁰

Finally, multiply the frequency factor by the exponent result:
k = (1.5 x 10¹¹) * (1.15 x 10⁻¹⁰)
k ≈ 1.725 x 10

So, the rate constant of the reaction at 25°C is approximately 17.25 s⁻¹.

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The concentration of PCl5 is 0.9815 M , When 1.31 g of Cl2 is added.

it reacts with PCl3 to form more PCl5 until equilibrium is reestablished. The equation for the reaction is:

PCl3 + Cl2 ⇌ PCl5

We can use the initial concentration of PCl3 and the amount of Cl2 added to calculate the change in concentration of PCl3. From there, we can use the stoichiometry of the reaction to determine the change in concentration of PCl5, and ultimately, the concentration of PCl5 at equilibrium.

Assuming the initial concentration of PCl3 is 1.0 M, we can calculate the moles of Cl2 added:

1.31 g Cl2 × (1 mol Cl2/71 g Cl2) = 0.0185 mol Cl2

Since the stoichiometry of the reaction is 1:1 for PCl3 and Cl2, this means that 0.0185 mol of PCl3 will be consumed. The new concentration of PCl3 will be:

[PCl3] = (1.0 mol - 0.0185 mol) / 1.0 L = 0.9815 M

Using the stoichiometry of the reaction, we can see that for every 1 mol of PCl3 consumed, 1 mol of PCl5 is produced. Therefore, the change in concentration of PCl5 will also be 0.0185 M. The new concentration of PCl5 will be:

[PCl5] = (0 + 0.0185 mol) / 1.0 L = 0.0185 M

So, the concentration of PCl5 at equilibrium after the addition of 1.31 g Cl2 will be 0.0185 M.
To find the concentration of PCl5 when equilibrium is reestablished after the addition of 1.31 g Cl2, follow these steps:

Step 1: Write the balanced chemical equation for the reaction:
PCl5 ⇌ PCl3 + Cl2

Step 2: Convert the mass of Cl2 to moles using its molar mass (70.9 g/mol):
(1.31 g Cl2) / (70.9 g/mol) = 0.0185 mol Cl2

Step 3: Set up an ICE (Initial, Change, Equilibrium) table:

        PCl5       PCl3      Cl2
Initial   x           0         0
Change   -y          +y        +y
Final    x-y          y        0.0185+y

Step 4: Write the equilibrium expression (Kc) for the reaction:
Kc = [PCl3][Cl2] / [PCl5]

Step 5: Plug the equilibrium concentrations from the ICE table into the equilibrium expression:
Kc = (y)(0.0185+y) / (x-y)

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The choice of recrystallization method depends on several factors such as the solubility of the impurities in the solvent, the melting point of the sample, and the amount of sample available. In this case, a 100-mg sample size suggests a microscale recrystallization procedure would be most appropriate.

Microscale recrystallization involves dissolving the sample in a small amount of solvent and then cooling the solution to allow for crystal formation. This method is ideal for small sample sizes and is relatively easy to perform. Macroscale recrystallization, on the other hand, involves dissolving larger amounts of sample in a larger volume of solvent, which is then allowed to slowly cool to promote crystal formation.

This method is best suited for larger sample sizes and is more complicated to perform than microscale recrystallization. Ultimately, the choice of recrystallization method will depend on the specific characteristics of the unknown sample and the resources available to the experimenter.

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We leave an opening at the top of the separatory funnel during the addition of NaOCl to the flask because it allows for the release of any gases or pressure that may build up during the reaction.

NaOCl can react with other compounds in the flask and release gases, so the opening prevents the pressure from building up and potentially causing an explosion.

Additionally, the opening allows for the controlled addition of NaOCl and prevents any splashing or spillage.

We leave an opening at the top of the separatory funnel during the addition of NaOCl to the flask to allow for the release of any gas or vapor that may form during the reaction. This prevents pressure buildup inside the funnel, which could lead to dangerous situations and inaccurate results.

The opening ensures safe and efficient mixing of the reactants.

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A 1.00 L flask is charged with 0.0400 mol of N2O4. At equilibrium at 373K, 0.0055 mol of N2O4 remains. Keq for this reaction is 0.8656.

Chemical equilibrium refers to the situation in a chemical process where both the reactants and products are present in concentrations that have no further tendency to vary over time, preventing any discernible change in the system's characteristics. When the forward reaction and the reverse reaction go forward at the same speed, this condition arises. The forward and backward reactions often have equal, if not zero, reaction rates. The concentrations of the reactants and products do not change on a net basis as a result. Dynamic equilibrium is the name given to such a situation.

N₂O₄(g) = 2NO₂(g)

Initially, [N₂O₄] = 0.04 M  & [NO₂] = 0 M

Let at eqb, [N₂O₄] = (0.04 - x) M  & [NO₂] = 2x M

But given that at equilibrium, [N₂O₄] = 0.0055 M = 0.04 - x

or, x = 0.0345 M

Thus, at equilibrium, [NO₂] = 2x = 0.069 M

Hence Kc = [NO₂]₂/[N₂O₄] = (0.069)₂/(0.0055) = 0.8656.

Therefore, Keq is 0.8656.

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The answer to question is that metal sulfides can be prepared by bubbling hydrogen sulfide gas through an aqueous solution containing the metal ion, and then adjusting the pH. The process involves saturating the solution with H2S, which occurs when the concentration of H2S reaches a certain level.

At this point, the H2S gas reacts with the metal ions in the solution to form metal sulfides, which are insoluble and can be filtered and collected.

An explanation for why this method works is that metal sulfides are generally insoluble in water, which means that they can be precipitated out of solution. By adding H2S gas, which is a reducing agent, to the solution, the metal ions are reduced to their sulfide form. This reaction is favored at low pH, which means that adjusting the pH of the solution can help to promote the formation of metal sulfides. Overall, this method is a relatively simple and efficient way to prepare metal sulfides, and can be used for a wide range of metals and applications.

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