write the net ionic equation for the acid-base hydrolysis equilibrium that is established when potassium cyanide is dissolved in water. Blank + H2O (l) arrow Blank + Blank
this solution is________

3. At 250°C, the equilibrium constant for the reaction is 0.111 mol/L.

4. [NO] is 25.92 M at equilibrium.

5. [NH3] is 0.45 M at equilibrium.

The balanced chemical equation for the reaction is:

H2(g) + I2(g) ⇌ 2HI(g)

At equilibrium, we have [H2] = 4 - 1 = 3 mol/L, [I2] = 4 - 1 = 3 mol/L, and [HI] = 1 mol/L.

The expression for the equilibrium constant, Keq, is:

Keq = [HI]^2 / ([H2] x [I2])

Substituting the values we obtained, we get:

Keq = (1 mol/L)^2 / ((3 mol/L) x (3 mol/L))

Keq = 0.111 mol/L

4. Using the equilibrium constant expression: Keq = [N2][O2] / [NO]^2

Since the reaction is 2 NO ⇄ N2 + O2, we can assume that at equilibrium the concentration of N2 and O2 are equal and x, and the concentration of NO is 2x (since the stoichiometry is 2:1). Therefore, we have:

Keq = x^2 / (2x)^2

42 = x^2 / 4x^2

x^2 = 42 * 4x^2

x^2 = 168x^2

x^2 - 168x^2 = 0

x^2 (1 - 168) = 0

x = 0 (not possible) or x = √(168) = 12.96 M

5. Using the equilibrium constant expression: Keq = [NH3]^2 / [H2]^3[N2]

Substitute the given values: Keq = 20, [H2] = 0.40 M, [N2] = 0.25 M, and let x be the equilibrium concentration of NH3. Then:

Keq = x^2 / (0.40)^3(0.25)

20 = x^2 / 0.010

x^2 = 0.010 x 20

x^2 = 0.2

x = √(0.2)

x = 0.45 M

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By analyzing the IR spectra for each compound and using the IR Characteristic Absorption Frequency Table as a guide, we can correctly label each spectrum and identify the functional groups present in each compound.

Based on the information provided, the student obtained IR spectra for bromobenzene (4), benzophenone (B), and triphenylmethanol (C) after performing a Grignard reaction in lab. The task at hand is to correctly label the IR spectra for each compound using the IR Characteristic Absorption Frequency Table as a guide.

To begin, it is important to note that IR spectroscopy is a technique that allows us to identify functional groups present in a molecule based on the absorption of specific frequencies of infrared radiation. The IR Characteristic Absorption Frequency Table provides a guide to the common absorption frequencies associated with different functional groups.

Using this information, we can analyze the IR spectra for each compound and identify the functional groups present. For example, bromobenzene (4) should show a peak around 3000-3100 cm^-1 corresponding to sp2 hybridized C-H stretching vibrations. It should also show a peak around 1600-1620 cm^-1 corresponding to the C-Br stretching vibration. Benzophenone (B) should show a peak around 1700 cm^-1 corresponding to the carbonyl group (C=O) stretching vibration, and a peak around 1600 cm^-1 corresponding to the aromatic C=C stretching vibration. Triphenylmethanol (C) should show a peak around 3400 cm^-1 corresponding to the O-H stretching vibration, and peaks around 1600-1620 cm^-1 corresponding to the aromatic C=C stretching vibration.

Based on these characteristic absorption frequencies, we can now correctly label the IR spectra for each compound. For example, the IR spectrum showing a peak at around 3000-3100 cm^-1 and another peak at around 1600-1620 cm^-1 should be labeled as belonging to bromobenzene (4). The IR spectrum showing a peak at around 1700 cm^-1 and another peak at around 1600 cm^-1 should be labeled as belonging to benzophenone (B). Finally, the IR spectrum showing a peak at around 3400 cm^-1 and peaks at around 1600-1620 cm^-1 should be labeled as belonging to triphenylmethanol (C).

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[ES] - D. The [S] required to reach ½ Vmax for enzymes that exhibit hyperbolic kinetic behavior; Km - B. Approximately equal to [Etotal] at saturating conditions. kcat - C. Is directly proportional to the rate of the slowest step of the enzyme. Vmax - A. Equal to the velocity (V0) at saturating conditions.

Here is the matched list:

A. Equal to the velocity (V0) at saturating conditions - Vmax
B. Approximately equal to [Etotal] at saturating conditions - Km
C. Is directly proportional to the rate of the slowest step of the enzyme - kcat
D. The [S] required to reach ½ Vmax for enzymes that exhibit hyperbolic kinetic behavior - [ES]

So, the final answer is : [ES]- D. The [S] required to reach ½ Vmax for enzymes that exhibit hyperbolic kinetic behavior ; Km- B. Approximately equal to [Etotal] at saturating conditions; kcat- C Is directly proportional to the rate of the slowest step of the enzyme; Vmax- A. Equal to the velocity (V0) at saturating conditions.

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The electron group geometry of the carboxyl carbon in a carboxylate ion is tetrahedral. The hybridization state of the carbon is sp3.sp3. Therefore, the correct answer is (b) tetrahedral, sp3.

The electron group geometry and hybridization state of the carboxyl carbon in a carboxylate ion is: Trigonal planar, sp2

The concept of hybridization is defined as the process of combining two atomic orbitals to create a new type of hybridized orbitals. This intermixing typically results in the formation of hybrid orbitals with completely different energies, shapes, and so on. Hybridization is primarily carried out by atomic orbitals of the same energy level. However, both fully filled and half-filled orbitals can participate in this process if their energies are equal. The concept of hybridization is an extension of valence bond theory that helps us understand bond formation, bond energies, and bond lengths.

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After equilibrium is established, then type of ion in solution in largest concentration, other than the k ion, is : (b) HPO₄²⁻.

In chemistry, equilibrium is a state of balance or stability achieved in chemical reaction when the rates of forward reaction and reverse reaction are equal.

The balanced chemical equation for the reaction between H₃PO₄ and KOH is:

H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O

In this reaction, one mole of H₃PO₄ reacts with three moles of KOH to produce one mole of K₃PO₄ and three moles of water.

When equal volumes of 0.10 M H₃PO₄ and 0.20 M KOH are mixed, the concentration of OH⁻ ions will be in excess because KOH is strong base and H₃PO₄ is a weak acid. The OH⁻ ions will react with H⁺ ions of H₃PO₄ to form water, according to following reactions:

H₃PO₄ + OH⁻ → H₂PO₄⁻ + H₂O

H₂PO₄⁻ + OH⁻ →  HPO₄²⁻ +H₂O

The net effect of these reactions is that H₃PO₄ reacts with OH⁻ to produce HPO₄²⁻. Therefore, the type of ion in solution in largest concentration, other than the K+ ion, is  HPO₄²⁻.

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In this case, Qc = 44.44 and Kc = 1.35x10^-4. Since Qc > Kc, the system is not at equilibrium and is favoring formation of reactants.

Based on the given values of concentrations and Kc, we can calculate the reaction quotient (Qc) using the formula:
Qc = [NH3]^2 / ([H2][N2])
Qc = (0.2)^2 / (0.03)(0.03)
Qc = 44.44
Comparing the value of Qc with Kc, we can determine the condition of the system. If Qc < Kc, the system is not at equilibrium and is favoring formation of products. If Qc > Kc, the system is not at equilibrium and is favoring formation of reactants. However, if Qc = Kc, the system is at equilibrium.
therefore In this case, Qc = 44.44 and Kc = 1.35x10^-4. Since Qc > Kc, the system is not at equilibrium and is favoring formation of reactants.

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The standard entropy change  ∆s° for the reaction c(g) + 2H₂(g) → CH₄(g) is -80.7 J/K*mol.

To calculate ∆s° for the reaction c(g) + 2H₂(g) → CH₄(g), we need to use the standard entropy values for each of the species involved in the reaction.

The standard entropy of carbon in its gaseous state (c(g)) is 5.7 J/K*mol. The standard entropy of hydrogen in its gaseous state (H₂(g)) is 130.6 J/K*mol. The standard entropy of methane in its gaseous state (CH₄(g)) is 186.3 J/K*mol.

Using these values, we can calculate the standard entropy change (∆s°) for the reaction as follows:

∆s° = ∑S°(products) - ∑S°(reactants)
∆s° = [S°(CH₄(g))] - [S°(c(g)) + 2S°(H₂(g))]
∆s° = [186.3 J/K*mol] - [5.7 J/K*mol + 2(130.6 J/K*mol)]
∆s° = [186.3 J/K*mol] - [267 J/K*mol]
∆s° = -80.7 J/K*mol

Therefore, the standard entropy change for the reaction c(g) + 2H₂(g) → CH₄(g) is -80.7 J/K*mol.

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A.)The maximum amount of iron(III) sulfide that will dissolve in a 0.278 M iron(III) nitrate solution is 0.0462 M. B.)The molar solubility of lead sulfide in a 0.238 M ammonium sulfide solution is 0.119 M.

A)  The balanced equation is:
[tex]Fe_2S_3(s) + 6HNO_3(aq)[/tex] → [tex]2Fe(NO_3)_3(aq) + 3H_2S(g)[/tex]

Assuming that the concentration of nitric acid is also 0.278 M. Therefore, the amount of nitric acid present is:
0.278 M × 0.500 L = 0.139 mol

The maximum amount of iron(III) sulfide that can dissolve is:
0.139 mol ÷ 6 = 0.0231 mol

= 0.0231 mol ÷ 0.500 L = 0.0462 M

So the maximum amount of iron(III) sulfide that will dissolve in a 0.278 M iron(III) nitrate solution is 0.0462 M.

B) The balanced equation is:

[tex]PbS(s) + (NH4)_2S(aq)[/tex] → [tex]PbS(s) + 2NH^4^+ (aq) + S^2^-(aq)[/tex]

The molar solubility of lead sulfide is,
Ksp =[tex][Pb^2^+][S^2^-][/tex]

The concentration of sulfide ions is:
0.238 M × 0.500 L = 0.119 mol

Assuming that the concentration of lead ions is negligible compared to the concentration of sulfide ions.

So, Ksp is:
Ksp = [tex][Pb^2^+][S^2^-][/tex] ≈[tex][S^2^-]^2[/tex]

Substituting the concentration of sulfide ions, we get:
Ksp = (0.119 M)2 = 0.0142

Solving for the concentration of lead ions at equilibrium.
[tex][Pb^2^+][/tex]= √Ksp = √0.0142 = 0.119 M

Therefore, the molar solubility of lead sulfide in a 0.238 M ammonium sulfide solution is 0.119 M.

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To recommend a chemical dose (mg/l) for potential effluent phosphate requirements, the following factors need to be considered: the level of phosphate in the influent wastewater, the required level of phosphate in the effluent, and the type of chemical being used.

If the influent wastewater contains a high level of phosphate, then a higher dose of the chemical would be required to reduce the phosphate to the desired level in the effluent. Additionally, the type of chemical being used can affect the dose required. For example, aluminum-based chemicals generally require a lower dose than iron-based chemicals. Assuming the influent wastewater has a phosphate concentration of 10 mg/l, the following chemical doses (mg/l) could be recommended for different effluent phosphate requirements:

- 2 mg/l: A chemical dose of 8-10 mg/l may be recommended to achieve this level of phosphate removal.
- 1 mg/l: A chemical dose of 6-8 mg/l may be recommended to achieve this level of phosphate removal.
- 0.5 mg/l: A chemical dose of 4-6 mg/l may be recommended to achieve this level of phosphate removal. These recommended doses are based on typical ranges for aluminum-based coagulants. The exact dose required may vary based on the specific wastewater characteristics and treatment process.

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The emission spectrum of multi-electron atoms, such as mercury, appears as a continuous band of color because these atoms have multiple electrons that can transition between different energy levels.

This results in the emission of photons at various wavelengths, creating a broad spectrum of colors. Unlike single-electron atoms, which produce discrete spectral lines, multi-electron atoms have many possible energy states, leading to a more complex and continuous emission spectrum. Thus, the emission spectrum of mercury (and other multi-electron atoms) appears as a band of color rather than distinct lines.

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For 2-methylcyclohexanol, two possible products can be formed due to the formation of two different carbocations: a more stable tertiary carbocation (3°) through hydride shift, and the less stable secondary carbocation (2°) without any shift. The corresponding products are 1-methylcyclohexene (major product) and 3-methylcyclohexene (minor product).

The acid-catalyzed dehydration of 2-methylcyclohexanol follows the E1 mechanism and involves three main steps: protonation, carbocation formation, and deprotonation.

1. Protonation: In the presence of an acid catalyst (usually H2SO4 or H3PO4), the hydroxyl group on 2-methylcyclohexanol gets protonated, forming a good leaving group, H2O.

2. Carbocation formation: The H2O molecule departs, generating a carbocation at the 2-position of the cyclohexane ring.

3. Deprotonation: A nearby base (usually a conjugate base of the acid catalyst) abstracts a proton from an adjacent carbon, forming a double bond and yielding the final product(s).

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The partial pressure of krypton is 415.6 mmHg and the partial pressure of carbon dioxide is 382.4 mmHg.

To find the partial pressure of each gas in the mixture, we need to use the mole fraction of each gas.

First, we need to find the moles of each gas in the mixture:

moles of Kr = 14.2 g / 83.8 g/mol = 0.1699 mol
moles of CO2 = 6.92 g / 44.01 g/mol = 0.157 mol

Next, we need to find the total moles in the mixture:

total moles = moles of Kr + moles of CO2 = 0.1699 mol + 0.157 mol = 0.3269 mol

Now we can find the mole fraction of each gas:

mole fraction of Kr = moles of Kr / total moles = 0.1699 mol / 0.3269 mol = 0.5198
mole fraction of CO2 = moles of CO2 / total moles = 0.157 mol / 0.3269 mol = 0.4802

Finally, we can use the mole fractions to calculate the partial pressure of each gas:

partial pressure of Kr = mole fraction of Kr x total pressure = 0.5198 x 798 mmHg = 415.6 mmHg

partial pressure of CO2 = mole fraction of CO2 x total pressure = 0.4802 x 798 mmHg = 382.4 mmHg

Therefore, the partial pressure of krypton is 415.6 mmHg and the partial pressure of carbon dioxide is 382.4 mmHg.

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To determine the mechanism for each reaction, we need to examine the reaction conditions and the substrate.

For example, if the reaction occurs under acidic conditions and the substrate is a tertiary alkyl halide, the mechanism is likely an E1 reaction. On the other hand, if the reaction occurs under basic conditions and the substrate is a primary alkyl halide, the mechanism is likely an SN2 reaction. It is important to consider factors such as steric hindrance and leaving group ability when determining the mechanism of a reaction. Understanding the mechanism of a reaction can provide insight into the steps involved in the reaction and can aid in predicting the products and controlling the reaction conditions.

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A transaminase can be employed to achieve the synthesis of chiral primary amines from keto acids. Option B is correct.

Transaminases, also known as aminotransferases, are enzymes found in the body that catalyze the transfer of an amino group from an amino acid to a keto acid. There are two types of transaminases that are commonly measured in blood tests: alanine aminotransferase (ALT) and aspartate aminotransferase (AST).

ALT is primarily found in the liver, while AST is found in various tissues including the liver, heart, and skeletal muscle. Elevated levels of these enzymes in the blood can indicate damage or disease in the organ or tissue where they are primarily located.

ALT and AST levels are often measured as part of a liver function test, which may be ordered by a healthcare provider if there are concerns about liver damage or disease. Elevated levels may be seen in conditions such as hepatitis, cirrhosis, or liver cancer. However, it is important to note that elevated transaminase levels can also occur in other conditions and may not necessarily indicate liver disease.

Hence, B. is the correct option.

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--The given question is incomplete, the complete question is

"Which of the following can a transaminase be employed to achieve? A) synthesis of sugars from carbohydrates B) the synthesis of chiral primary amines C) reduction of ketones to chiral alcohols D) hydroxylation of aromatic molecules."--

The Kc for this reaction of Nitrogen monoxide, a pollutant in automobile exhaust, being oxidized to nitrogen dioxide in the atmosphere is approximately 9.9 x 1[tex]0^{11}[/tex]

To find Kc for the given reaction, we first need to understand the relationship between Kp and Kc. The equation relating Kp and Kc is:

Kp = Kc × (RT)^(Δn)

Where:
- Kp is the equilibrium constant in terms of pressure
- Kc is the equilibrium constant in terms of concentration
- R is the universal gas constant (0.0821 L atm / (mol K))
- T is the temperature in Kelvin
- Δn is the change in moles of gas during the reaction (moles of products - moles of reactants)

For the given reaction:
2NO(g) + O[tex]_{2}[/tex](g) ⇌ 2NO[tex]_{2}[/tex](g)

Δn = (2 moles of NO[tex]^{2}[/tex]) - (2 moles of NO + 1 mole of O[tex]_{2}[/tex]) = 2 - 3 = -1

Now we need to convert the temperature from Celsius to Kelvin:
T = 23°C + 273.15 = 296.15 K

We are given Kp = 3.1 x 1[tex]0^{12}[/tex], and we can plug in the values into the equation:

3.1 x 1[tex]0^{12}[/tex] = Kc × (0.0821 × 296.15[tex])^{-1}[/tex]

Now, we solve for Kc:

Kc = (3.1 x 1[tex]0^{12}[/tex]) / (0.0821 × 296.15[tex])^{-1}[/tex]
Kc ≈ 9.9 x 1[tex]0^{11}[/tex]

So, the Kc for this reaction is approximately 9.9 x 1[tex]0^{11}[/tex] (to two significant figures).

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The reaction of phenylacetaldehyde with NaBH4 results in the reduction of the carbonyl group to a hydroxyl group, yielding phenylacetalcohol. The addition of H3O+ in a subsequent step results in the protonation of the alcohol group, forming the final product, phenylacetaldehyde diol.

To draw the organic product(s) of the reaction of phenylacetaldehyde with NaBH4, then H3O+, follow these steps:

1. Identify the starting material: Phenylacetaldehyde is an aldehyde with a phenyl group attached to the carbonyl carbon (C6H5CH=O).

2. Reaction with NaBH4: NaBH4 (sodium borohydride) is a reducing agent that selectively reduces aldehydes and ketones to their corresponding alcohols. In this case, phenylacetaldehyde will be reduced to phenylethanol.

3. Formation of phenylethanol: The NaBH4 donates a hydride ion (H-) to the carbonyl carbon of phenylacetaldehyde, and the oxygen atom of the carbonyl group captures a proton (H+) from water to form the alcohol group (-OH). The resulting product is phenylethanol (C6H5CH2OH).

4. Reaction with H3O+: H3O+ is a strong acid, but in this case, it does not further react with phenylethanol as the alcohol is not a good leaving group, and the reaction conditions do not favor any additional reactions.

So, the organic product of the reaction of phenylacetaldehyde with NaBH4, then H3O+ is phenylethanol (C6H5CH2OH).

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The correct answer is the ΔG for the reaction at 298 K with the given partial pressures is approximately 9.90 kJ/mol.

To calculate the ΔG (Gibbs free energy) of the reaction under non-standard conditions, we can use the equation:

ΔG = ΔG° + RT ln(Q)

where ΔG° is the standard Gibbs free energy change (-17.09 kJ/mol), R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (298 K), and Q is the reaction quotient.

First, we need to determine the reaction quotient Q. For the given reaction:

ICl(g) + Cl₂(g) → ICl₃(s)

Q = [ICl₃] / ([ICl] * [Cl₂])

Since ICl₃ is a solid, its concentration doesn't affect the reaction quotient, and we can set [ICl₃] = 1. Now, substitute the given partial pressures for ICl and Cl₂:

Q = 1 / (0.0200 * 0.00100)

Now, calculate the value of Q:

Q = 1 / (0.0000200) = 50,000

Next, plug in the values of ΔG°, R, T, and Q into the equation:

ΔG = -17.09 kJ/mol + (8.314 J/mol·K * 298 K * ln(50,000)) / 1000

Note that we divide by 1000 to convert the units of R from J/mol·K to kJ/mol·K. Now, calculate the natural logarithm of Q:

ln(50,000) ≈ 10.82

And finally, substitute the calculated value back into the equation:

ΔG ≈ -17.09 kJ/mol + (8.314 * 298 * 10.82) / 1000 ≈ -17.09 kJ/mol + 26.99 kJ/mol ≈ 9.90 kJ/mol

Therefore, the ΔG for the reaction at 298 K with the given partial pressures is approximately 9.90 kJ/mol.

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The molarity of a solution formed by dissolving 690 mg of aluminium chloride hexahydrate in enough water to yield 48.1 ml of solution is 0.0595 M or 0.0595 mol/L.

How we calculated it?

To determine the molarity of a solution formed by dissolving 690 mg of aluminium chloride hexahydrate in enough water to yield 48.1 ml of solution, follow these steps:

1. Convert the mass of aluminium chloride hexahydrate (AlCl3·6H2O) to moles. First, find the molar mass of AlCl3·6H2O:
(Al = 26.98 g/mol, Cl = 35.45 g/mol, H2O = 18.02 g/mol)
Molar mass of AlCl3·6H2O = 1(26.98) + 3(35.45) + 6(18.02) = 241.43 g/mol

2. Convert 690 mg to grams: 690 mg = 0.690 g

3. Calculate the moles of AlCl3·6H2O:
no. of moles = given mass / molar mass

no. of moles = 0.690 g / 241.43 g/mol ≈ 0.00286 moles

4. Convert the volume of the solution to litres:

48.1 ml = 48.1/1000 = 0.0481 L

5. Calculate the molarity:
Molarity =  no. of moles / volume in litres

Molarity = 0.00286 moles / 0.0481 L ≈ 0.0595 M

So, the molarity of the solution formed by dissolving 690 mg of aluminium chloride hexahydrate in enough water to yield 48.1 ml of solution is approximately 0.0595 M.

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The potential of the half-cell at 25°C for the given reaction, [tex]Fe^{3+[/tex](0.500 M) + e− → [tex]Fe^{2+[/tex](0.100 M) is approximately 0.8114 V.

To calculate the potential of the half-cell at 25°C for the given reaction: [tex]Fe^{3+[/tex](0.500 M) + e− → [tex]Fe^{2+[/tex](0.100 M), we'll need to use the Nernst equation. The Nernst equation is:
E = E° - (RT/nF) × lnQ
where:
E = the potential of the half-cell
E° = standard reduction potential of the half-cell
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin (25°C = 298.15K)
n = number of electrons transferred in the reaction (1 for this reaction)
F = Faraday's constant (96485 C/mol)
Q = reaction quotient

First, we need to find the standard reduction potential (E°) for the reaction. The standard reduction potential for [tex]Fe^{3+[/tex] + e− → [tex]Fe^{2+[/tex] is +0.77 V.

Next, calculate the reaction quotient (Q):

Q = [[tex]Fe^{2+[/tex]]/[[tex]Fe^{3+[/tex]]
Q = (0.100 M)/(0.500 M)
Q = 0.2

Now, plug all the values into the Nernst equation:

E = 0.77 - ((8.314 J/(mol·K)) × 298.15K) / (1 × 96485 C/mol) × ln(0.2)
E = 0.77 - (2.303 × (8.314 J/(mol·K) × 298.15K) / (1 × 96485 C/mol)) × log10(0.2)
E ≈ 0.77 - (0.0592 V) × (-0.69897)
E ≈ 0.77 + 0.0414
E ≈ 0.8114 V

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To determine which of the two white crystalline compounds is ionic and which is covalent, there are several methods that could be used. One way is to test their solubility in water.

Ionic compounds tend to be soluble in water, while covalent compounds tend to be insoluble or only slightly soluble. Another method is to test their conductivity in water. Ionic compounds will conduct electricity in water due to the presence of charged ions, while covalent compounds will not. Additionally, the melting and boiling points of the compounds could also provide clues. Ionic compounds tend to have higher melting and boiling points than covalent compounds due to their stronger electrostatic interactions. Finally, the chemical formula of the compounds could also give some indication. Ionic compounds tend to be made up of a metal and a non-metal, while covalent compounds are typically made up of non-metals only. By analyzing these characteristics, one can determine which of the two white crystalline compounds is ionic and which is covalent.

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Answer:

If I initially have a gas at a pressure of 10.0 atm, a volume of 54.0 liters, and a temperature of 200. K, and then I raise the pressure to 14.0 atm and increase the temperature to 300. K, what is the new volume of the gas?

Hybridization: sp2. Bonding: One sigma bond between C and each Cl and one pi bond between C and O.

In COCl2, carbon has four valence electrons and is encircled by three iotas, one oxygen and two chlorine. To decide its hybridization, we initially work out the all out number of valence electrons:

4 valence electrons for carbon

6 valence electrons for oxygen

7 valence electrons for every chlorine (absolute of 14 electrons)

The all out is 24 valence electrons. Carbon utilizes hybridization to shape four sp3 orbitals, and each orbital covers with the orbital of one of the four molecules. The two chlorine particles are single-attached to carbon, while the oxygen molecule is twofold clung to carbon.The hybridization conspire for COCl2 is:

One 2s orbital and three 2p orbitals of carbon hybridize to shape four sp3 orbitals.The two chlorine particles each structure a solitary bond with carbon utilizing their one unpaired 3p electron.The oxygen particle frames a twofold bond with carbon utilizing two of its unpaired 2p electrons.

The particle has a three-sided planar calculation, with bond points of roughly 120 degrees. The covering orbitals can be imagined as follows:

Every carbon sp3 orbital covers with a half breed orbital from one of the three encompassing molecules.

Every chlorine iota is attached to carbon utilizing its single 3p orbital, which covers with one of the sp3 orbitals on carbon.

The oxygen particle is twofold clung to carbon utilizing two of its unpaired 2p electrons, which cross-over with two of the sp3 orbitals on carbon.

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The following is a balanced reaction.

[tex]2 HgO - > 2 Hg + O_2[/tex]

Chemical reactions are interactions between reactant molecules that result in the formation of new product molecules with distinct chemical characteristics.

Atoms in the reactants rearrange themselves to form new compounds or molecules during a chemical reaction. When reactant molecules are hit with enough kinetic energy to break existing chemical bonds and form new ones, chemical reactions occur.

A chemical reaction that involves the breaking or formation of chemical bonds transforms the reactants into products. The reactants are written on the left side of the equation, and the products are written on the right, separated by an arrow showing the direction of the reaction.

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The value of δ g° at 25 °c for the decomposition of pocl3 into its constituent elements is -1117.8 kJ/mol kJ/mol.

The value of δ g° at 25 °c for the decomposition of POCl3 into its constituent elements, 2POCl3 (g) → P2 (g) +O2 (g) + 3Cl2 (g), can be calculated using the standard free energy change of formation (Δ f g°) for each of the reactants and products. The equation for δ g° is:
δ g° = Σ n Δ f g° (products) - Σ m Δ f g° (reactants)
where n and m are the stoichiometric coefficients for the products and reactants, respectively.

Using the values of Δ f g° for each species from standard tables, we can calculate δ g° for the reaction:
δ g°f (POCl₃) = -558.9 kJ/mol

δ g°f (P₂) = 0 kJ/mol

δ g°f (O₂) = 0 kJ/mol

δ g°f (Cl₂) = 0 kJ/mol

Plugging these values into the formula, we get:
δ g° = [2(0) + 0 + 3(0)] - [2(-558.9)] =0 - (- 1,117.8) =  +1,117.8 kJ/mol

Therefore, the value of δ g° at 25 °c for the decomposition of pocl3 into its constituent elements is -1117.8 kJ/mol kJ/mol.

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Approximately 21.23 mL of a 0.164 M calcium hydroxide solution is required to neutralize 19.9 mL of a 0.350 M hydrobromic acid solution.

To determine the volume of a 0.164 M calcium hydroxide solution needed to neutralize 19.9 mL of a 0.350 M hydrobromic acid solution, follow these steps:

Step 1: Write the balanced chemical equation for the reaction:
Ca(OH)2 + 2 HBr → CaBr2 + 2 H2O

Step 2: Calculate the moles of hydrobromic acid:
moles of HBr = volume x molarity
moles of HBr = 19.9 mL x 0.350 mol/L = 6.965 mmol (converting mL to L is not necessary as it will cancel out in the final calculation)

Step 3: Determine the stoichiometry ratio from the balanced equation:
1 mol Ca(OH)2 : 2 mol HBr

Step 4: Calculate the moles of calcium hydroxide needed:
moles of Ca(OH)2 = (moles of HBr / 2) = 6.965 mmol / 2 = 3.4825 mmol

Step 5: Calculate the volume of calcium hydroxide solution required to occur neutralization:
volume = moles of Ca(OH)2 / molarity
volume = 3.4825 mmol / 0.164 mol/L ≈ 21.23 mL

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The volume of 0.250 M HNO₂(aq) required to neutralize 36.0 milliliters of a 0.150 M NaOH solution is 21.6 milliliters.

To calculate the volume of HNO₂ needed for neutralization, we can use the balanced chemical equation for the reaction of HNO₂ and NaOH:

HNO₂(aq) + NaOH(aq) → NaNO₂(aq) + H₂O(l)

We can see from the equation that the stoichiometric ratio of HNO₂ to NaOH is 1:1. This means that one mole of HNO₂ will react with one mole of NaOH.

First, we need to determine the number of moles of NaOH present in the solution:

0.150 M NaOH = 0.150 moles NaOH / liter × 0.0360 liters = 0.00540 moles NaOH

Since the stoichiometric ratio of HNO₂ to NaOH is 1:1, the number of moles of HNO₂ required for neutralization is also 0.00540 moles.

We can now use the concentration of the HNO₂ solution to determine the volume needed:

0.250 M HNO₂ = 0.250 moles HNO₂ / liter

0.00540 moles HNO₂ × 1 liter / 0.250 moles HNO₂ = 0.0216 liters = 21.6 milliliters

Therefore, the volume of 0.250 M HNO₂(aq) required is 21.6 milliliters.

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The structure of the unknown compound can be classified as methyl ketone.

Based on the given conditions, it is concluded that :
1.  It has a boiling point of 98°C.

2. Iodoform test : A positive result indicates the presence of a methyl ketone group (CH3-C=O) or any other group that can be oxidized to form a methyl ketone group.
3. Tollen's test : A negative result implies that the compound does not have an aldehyde functional group.
4. Schiff's test : A negative result also suggests the absence of an aldehyde functional group.

Methyl ketone is an organic compound that contains a carbonyl group that is bonded to two hydrocarbon groups. It can be said that this compound has a methyl ketone group. The spectra is not given in the question. So, it is difficult to identify the exact answer.

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a. water :

Mass = 1g/mL × 1000mL = 1000g

Concentration = 55.56 moles / 1L = 55.56 M

b.  Acetic Acid:

Mass = 1.049g/cm³ × 1000cm³ = 1049g

Concentration = 17.48 moles / 1L = 17.48 M

a. Water:
Density = 1g/mL
To find the concentration, first, determine the mass and number of moles for a given volume. Let's assume we have 1L (1000mL) of water.

Mass = Density × Volume
Mass = 1g/mL × 1000mL = 1000g

Next, convert the mass to moles. The molar mass of water (H2O) is 18g/mol.

Moles = Mass / Molar Mass
Moles = 1000g / 18g/mol = 55.56 moles

Concentration (Molarity) = Moles / Volume in Liters
Concentration = 55.56 moles / 1L = 55.56 M

b. Acetic Acid:
Density = 1.049g/cm³
Let's assume we have 1L (1000mL) of acetic acid. First, convert the volume to cm³ (1mL = 1cm³).

Volume = 1000mL = 1000cm³

Now, determine the mass using the density.

Mass = Density × Volume
Mass = 1.049g/cm³ × 1000cm³ = 1049g

Next, convert the mass to moles. The molar mass of acetic acid (CH3COOH) is 60g/mol.

Moles = Mass / Molar Mass
Moles = 1049g / 60g/mol = 17.48 moles

Concentration (Molarity) = Moles / Volume in Liters
Concentration = 17.48 moles / 1L = 17.48 M

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Ca2+(aq) + CO32-(aq) → CaCO3(s) (total ionic equation) with spectator ions Na+ and Cl-.

The total ionic condition for the combination of watery calcium chloride (CaCl2) and fluid sodium carbonate (Na2CO3) is:

Ca2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + CO32-(aq) → 2 Na+(aq) + 2 Cl-(aq) + CaCO3(s)

In this situation, the watery calcium chloride separates into Ca2+ and 2 Cl-particles, and the fluid sodium carbonate separates into 2 Na+ and CO32-particles. At the point when these particles are combined as one, the Ca2+ and CO32-particles consolidate to frame strong calcium carbonate (CaCO3), which accelerates out of arrangement.

The 2 Na+ and 2 Cl-particles stay in arrangement and don't respond further.This condition shows the total ionic species present in the arrangement and the strong item framed. It is critical to take note of that the observer particles (Na+ and Cl-) don't partake in the response and stay unaltered in arrangement.

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Chemical bonds molecular level in substances, are broken and formed through the transfer or sharing of electrons to create new chemical substances.

During a synthetic response, the compound connections between iotas are broken and new bonds are framed to make different compound substances. The breaking and framing of synthetic bonds include the exchange or sharing of electrons between molecules.

In certain responses, the bonds break and structure in a solitary step, while in others, the cycle might happen in different advances. The bonds that are broken and shaped decide the energy change of the response.

For instance, in an ignition response, the bonds in the fuel particles (like hydrocarbons) are broken, and new bonds are framed between the fuel and oxygen particles, delivering carbon dioxide and water as the items. The energy delivered during this response is utilized to drive motors or produce heat.

In rundown, synthetic responses include the breaking and shaping of compound bonds through the exchange or sharing of electrons, bringing about the making of new substance substances.

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The complete question is:

What happens to the atoms within molecules undergoing a chemical reaction?