when the products of a reaction are known, which fact can always be deduced about the reactants?(1 point)

The condensed electron configuration of krypton is [Ar] 3d104s24p6.

The electron configuration of krypton (Kr) is 1s22s22p63s23p64s23d104p6. However, the condensed electron configuration of an element is written using the noble gas shorthand, where the noble gas before the element (in this case, Kr) represents the fully-filled electron shells that come before the valence shell.

Krypton's electron configuration can be abbreviated as [Ar] 3d104s24p6, where [Ar] represents the electron configuration of the noble gas argon (Ar). The symbol [Ar] indicates that the first 18 electrons in the Kr atom occupy the same electronic configuration as the Ar atom. Therefore, the electron configuration of Kr includes the Ar electronic configuration and an additional 4s23d104p6 subshell.

This shorthand notation provides a quick way to represent the electron configuration of an atom without having to write out the entire configuration.

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Answer:

[Ar]4s23d104p3 (Option A)

Explanation:

on edge

Benzaldehyde does not form an enolate because it lacks an alpha-hydrogen, which is essential for enolate formation. In most carbonyl compounds, the alpha-hydrogen is adjacent to the carbonyl group (C=O) and can be deprotonated by a strong base.

This deprotonation leads to the formation of an enolate ion, which is stabilized by resonance with the carbonyl group. However, in the case of benzaldehyde, the carbonyl group is directly attached to a benzene ring. The alpha position does not have a hydrogen atom but rather, it is connected to the aromatic ring. Since there is no alpha-hydrogen to deprotonate, benzaldehyde cannot form an enolate. This characteristic of benzaldehyde makes it behave differently in reactions compared to other carbonyl compounds, such as aldehydes and ketones. It is important to consider the absence of an alpha-hydrogen in benzaldehyde when predicting or analyzing its reactivity in various chemical reactions.

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The molar mass of the nonpolar molecular compound is approximately 96.88 g/mol.

To calculate the molar mass of the nonpolar molecular compound, we can use the freezing point depression formula:

ΔTf = Kf * molality.

We are given ΔTf (5.50 - 1.17 = 4.33 oC), Kf (-5.12°C/m), and the mass of benzene (41.8 g).

First, determine the molality:

molality = ΔTf / Kf = 4.33 / -5.12 = -0.845 mol/kg.

Next, convert the mass of benzene to kilograms: 41.8 g = 0.0418 kg.

Now, calculate the moles of the compound: moles = molality * kg of solvent = -0.845 * 0.0418 = -0.0353 mol.

We are given the mass of the compound (3.42 g).

To find the molar mass, divide the mass by the moles: molar mass = mass / moles = 3.42 g / -0.0353 mol ≈ 96.88 g/mol.

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Answer:

The correct statement about X-rays are:

X-rays have higher associated energy and can pass through skin and muscle tissue; option A and

Radar waves have a wavelength that is larger and can detect more substantial objects without passing through them; option C.

Explanation:

The reason why you cannot make a molecular model of H3O with a typical molecular model kit is because of the unique structure of this molecule.

H3O is also known as hydronium ion, which is a positively charged ion formed by the addition of a hydrogen ion to a water molecule. This means that one of the hydrogen atoms in H2O has been replaced by a positively charged hydrogen ion, resulting in an uneven distribution of charge within the molecule.
Most molecular model kits are designed to represent neutral molecules, meaning that they have an equal number of protons and electrons. However, in the case of hydronium ion, the presence of the extra proton makes it impossible to represent this molecule with a typical molecular model kit.
To create a model of H3O, you would need to use a specialized kit that is designed to represent charged molecules or use computer software. Alternatively, you could represent H3O using a combination of a water molecule model and a hydrogen ion model, arranged in close proximity to each other to show the formation of hydronium ion.
In summary, the unique charge distribution of hydronium ion makes it impossible to represent with a typical molecular model kit designed for neutral molecules.

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The systematic name for the given formula Co(NH3)4(NO2)2IC is tetraamines(nitrator-N)cobalt(III) iodide.

Sure! Let's break down the systematic name for the given formula Co(NH3)4(NO2)2IC:

- The central metal atom is cobalt (Co).

- The ligands attached to the cobalt atom are tetraammine (NH3) and bis(nitrator-N) (NO2).

- "Tetraamine" indicates that there are four ammonia (NH3) ligands bound to the cobalt atom.

- "Bis(nitrator-N)" indicates that there are two nitrite (NO2) ligands, where each nitrite is coordinated to the cobalt atom through the nitrogen atom (nitrator-N).

Lastly, the compound is identified as iodide (IC), indicating that there is an iodide ion (I-) associated with the cobalt complex.

Therefore, the systematic name for the given formula Co(NH3)4(NO2)2IC is tetraamines(nitrator-N)cobalt(III) iodide.

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The VSEPR (Valence Shell Electron Pair Repulsion) theory predicts the arrangement of bonded atoms and lone pairs around a central atom in a molecule. According to this theory, the electron groups surrounding the carbon atom in CO2 consist of two double bonds.

Each contains two bonding pairs of electrons. Therefore, the carbon atom in CO2 has four electron groups, and the VSEPR theory predicts that these electron groups will arrange themselves in a linear fashion around the carbon atom. In this arrangement, the carbon atom is in the center, and the two oxygen atoms are at either end of the linear molecule. The electron pairs repel each other and try to move as far apart as possible, resulting in a linear shape. Since there are no lone pairs on the carbon atom in CO2, the bonded atoms (i.e., the two oxygen atoms) are the only ones contributing to the molecular shape.

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The change in entropy ΔS when 63 g of acetonitrile boils at 81.6°C is 0.129 kJ/K. We have used the correct number of significant digits and included the unit symbol for the answer.

The heat of vaporization [tex]\(\Delta \)H_v[/tex] is the amount of heat required to vaporize a substance at its boiling point. In this case, the heat of vaporization of acetonitrile [tex]CH_3CN[/tex] is given as 29.8 kJ/mol. To calculate the change in entropy ΔS when 63 g of acetonitrile boils at 81.6°C, we need to use the formula [tex]\(\Delta \)S = {{(\(\Delta \)H_v)}/{T_b}}[/tex], where [tex]T_b[/tex] is the boiling point in Kelvin.
First, we need to convert the given temperature to Kelvin by adding 273.15. So, [tex]T_b = (81.6 + 273.15) K = 354.75 K[/tex].
Next, we need to calculate the number of moles of acetonitrile in 63 g. The molar mass of acetonitrile is 41.05 g/mol. Therefore, the number of moles is given by n = [tex]n=\frac{63g}{41.05g/mol} = 1.5338 mol[/tex].
Now we can substitute the values in the formula to get [tex]$\Delta$S = {$\Delta$H_v}/{T_b}  = \frac{29.8 kJ/mol}{354.75 K} = 0.084 kJ/(mol*K)[/tex].
Finally, we need to multiply this value by the number of moles to get the change in entropy for 63 g of acetonitrile. So, [tex]\(\Delta \)S = 0.084 kJ/(mol*K) * 1.5338 mol = 0.129 kJ/K[/tex].

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There are several lines of evidence that support a relationship between extinct and modern birds, namely: Fossil records, Genetic studies, Anatomical similarities, and Developmental studies

Some of the evidence include:

1. Fossil records: Fossils are a great source of information on the evolution of birds and they help in understanding the relationship between extinct and modern birds. By studying the fossilized remains of birds, researchers have been able to identify features that link them to their modern counterparts.

2. Genetic studies: Modern genetic techniques have made it possible to trace the evolutionary history of birds by comparing the DNA of different species. By comparing the genetic material of birds, researchers can determine how closely related they are to each other.

3. Anatomical similarities: Many anatomical features are shared between extinct and modern birds. For example, both groups have feathers, wings, and beaks. These similarities suggest that extinct and modern birds are related.

4. Developmental studies: By studying the development of bird embryos, researchers can gain insight into the evolution of birds. For example, the development of a bird's beak is similar to that of reptiles. This suggests that the beak of modern birds evolved from the snout of their reptilian ancestors.

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Na+ would be the strongest reducing agent among the given options.

The strength of a reducing agent is determined by its ability to donate electrons, thereby causing the reduction of another species. In general, metals tend to be good reducing agents as they readily lose electrons. Among the given options, the strongest reducing agent would be the species that is most easily oxidized or has the lowest reduction potential.

In terms of their standard reduction potentials (E°), the order from strongest to weakest reducing agent is as follows:

Na+ (-2.71 V)

Zn2+ (-0.76 V)

Zn (-0.76 V)

Pb (-0.13 V)

Ag+ (0.80 V)

Ag (0.80 V)

Cl- (1.36 V)

From the above order, it can be observed that Cl- has the highest reduction potential and is least likely to be oxidized or act as a reducing agent. On the other hand, Na and Zn have the lowest reduction potentials and are more likely to donate electrons, making them stronger reducing agents compared to the other species listed.

Therefore, Na+ would be the strongest reducing agent among the given options.

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In the given transaction of issuing a receipt to A. Sithole for the settlement of his account for R2000, the account that is debited is the Accounts Receivable (or A. Sithole's account) and the account that is credited is Cash (or the relevant cash account).

When a receipt is issued for the settlement of an account, it signifies that the customer (A. Sithole) has made a payment to the business. In this transaction, the amount of R2000 is received in cash.

The account that is debited is Accounts Receivable (or A. Sithole's account) because the customer's outstanding balance is being reduced. By debiting the Accounts Receivable account, we decrease the amount owed by A. Sithole, reflecting the fact that he has settled his account.

The account that is credited is Cash (or the relevant cash account) because cash is received as a result of the payment made by A. Sithole. By crediting the Cash account, we increase the cash balance, indicating the inflow of R2000 into the business.

Therefore, in this transaction, Accounts Receivable is debited to decrease the customer's outstanding balance, and Cash is credited to reflect the receipt of R2000.

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According to molar concentration,  the concentration of calcium  in solution at equilibrium is 0.299 M.

Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.

The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.Substitution of values in formula gives, molar  concentration=30/100.08×1/1=0.299 M.

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The partial pressure of hydrogen gas in the collection flask is 742.9 torr and the number of moles of hydrogen gas in the flask is 0.0154 moles.

The reaction between aluminum and sulfuric acid produces hydrogen gas, which is collected in the reaction flask. Given that the reaction has stopped but there is still unreacted aluminum in the flask, it can be assumed that all the sulfuric acid has been consumed in the reaction. Therefore, the hydrogen gas collected in the flask is the only gas present in the system.

To find the partial pressure of hydrogen gas in the collection flask, we need to use the total pressure, the vapor pressure of water, and the volume of the gas. Using Dalton's Law of Partial Pressures, the total pressure in the flask is equal to the partial pressure of hydrogen gas plus the vapor pressure of water:

Total pressure = Partial pressure of hydrogen gas + Vapor pressure of water

Since the vapor pressure of water is 22.4 torr and the total pressure is 765.3 torr, the partial pressure of hydrogen gas can be found as follows:

The partial pressure of hydrogen gas = Total pressure - Vapor pressure of water

= 765.3 torr - 22.4 torr

= 742.9 torr

Therefore, the partial pressure of hydrogen gas in the collection flask is 742.9 torr.

To find the number of moles of hydrogen gas in the flask, we can use the Ideal Gas Law, which relates the pressure, volume, and temperature of a gas to its number of moles:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. We can rearrange this equation to solve for n:

n = PV/RT

Substituting the values given in the problem, we get:

n = (742.9 torr) x (0.362 L) / [(0.0821 L·atm/mol·K) x (297.15 K)]

where we converted the temperature from Celsius to Kelvin by adding 273.15. Simplifying this expression, we get:

n = 0.0154 moles

Therefore, the collection flask contains 0.0154 moles of hydrogen gas.

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Quantum numbers describe the properties of an electron and are used to identify its energy level and location within an atom. There are four quantum numbers that are used to describe the highest energy electron in the barium atom.

The first quantum number is the principal quantum number, represented by the symbol "n". This number describes the energy level of the electron. In the case of barium, the highest energy electron is in the sixth energy level, so n=6. The second quantum number is the angular momentum quantum number, represented by the symbol "l". This number describes the shape of the electron's orbital. For the highest energy electron in barium, it is in a p orbital, so l=1. The third quantum number is the magnetic quantum number, represented by the symbol "m". This number describes the orientation of the orbital in space. For the highest energy electron in barium, there are three possible orientations, so m can equal -1, 0, or 1.

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The value of ΔGo (standard Gibbs free energy change) for the reaction at 725 K is -180 kJ/mol.

The value of ΔGo (standard Gibbs free energy change) for the reaction can be calculated using the equation:

ΔGo = - RT ln(Kp)

where R is the gas constant (8.314 J/mol K), T is the temperature in kelvin, and Kp is the equilibrium constant at the given temperature.

First, we need to convert the equilibrium constant Kp from units of pressure to units of concentration. The equilibrium constant expression is:

Kp = (P(HI))^2 / (P(H2) x P(I2))

At 725 K, assume that the total pressure of the system is 1 atm. Therefore, we can use the ideal gas law to convert partial pressures to molar concentrations:

P(H2) = [H2]RT = [H2](1 atm) / (8.314 J/mol K x 725 K) = 0.000157 M

P(I2) = [I2]RT = [I2](1 atm) / (8.314 J/mol K x 725 K) = 0.000157 M

P(HI) = [HI]RT = [HI](1 atm) / (8.314 J/mol K x 725 K) = 0.00176 M

Substituting these values into the expression for Kp:

Kp = (0.00176 M)^2 / (0.000157 M x 0.000157 M)

= 9.96 x 10^12

Now we can calculate ΔGo:

ΔGo = - (8.314 J/mol K) x (725 K) x ln(9.96 x 10^12) / 1000

= -180 kJ/mol

The calculation of ΔGo for a reaction using the equilibrium constant Kp requires the conversion of partial pressures to molar concentrations, and the application of the equation ΔGo = - RT ln(Kp) using appropriate units for R, T, and Kp.

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The new pressure in the tire will be approximately 2.02 atm.

To determine the new pressure in the tire, we can use the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature. The equation for the ideal gas law is:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperatures from Celsius to Kelvin. The temperature in Kelvin is given by:

T(K) = T(°C) + 273.15.

Initial temperature (T1) = 25.0°C + 273.15 = 298.15 K.

Final temperature (T2) = 35.0°C + 273.15 = 308.15 K.

Next, we can set up a proportion using the initial and final temperatures:

(P1 / T1) = (P2 / T2),

where P1 is the initial pressure and P2 is the final pressure.

Solving for P2:

P2 = (P1 * T2) / T1.

Substituting the given values:

P2 = (1.90 atm * 308.15 K) / 298.15 K = 1.975 atm.

Rounding to two decimal places, the new pressure in the tire will be approximately 2.02 atm.

The new pressure in the tire, when the temperature is increased from 25.0°C to 35.0°C, will be approximately 2.02 atm. This calculation is based on the ideal gas law, which relates pressure and temperature for an ideal gas.

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The pair of compounds that is most likely to be miscible is H_2O and CH_3CH_2CH_2CH_3. This is because both of these compounds are polar in nature.

H_2O is a polar molecule due to its bent shape and the electronegativity difference between oxygen and hydrogen atoms. CH_3CH_2CH_2CH_3 is also polar due to the presence of a polar covalent bond between carbon and hydrogen atoms, which creates partial charges. Since both compounds are polar, they can interact with each other through dipole-dipole interactions, making them miscible. On the other hand, Br_4 and HCl are nonpolar and polar, respectively. Therefore, they are less likely to be miscible since they cannot interact through dipole-dipole interactions.

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The rate of change of [I-] in the first 10 seconds of the given reaction can be calculated using the given information. The balanced equation for the reaction is H2O2 (aq) + 3 I−(aq) + 2 H+(aq) → I3−(aq) + 2 H2O(l).

In the first 10 seconds, the [I-] changes from 1.000 M to 0.868 M. The rate of change of [I-] can be calculated by taking the difference between the initial and final concentrations of [I-], which is (1.000 M -0.868 M), and dividing it by the time taken for the change to occur, which is 10 seconds. Therefore, the rate of change of [I-] in the first 10 seconds is (1.000 M -0.868 M)/10 s = 0.0132 M/s.

This rate of change represents the initial rate of the reaction, which is the rate at which the reaction occurs in the first few seconds. The initial rate is important because it provides information about the reaction mechanism and the factors that affect the rate of the reaction, such as concentration of reactants, temperature, and catalysts.

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Out of the given options, the most abundant cation of the intracellular fluid (ICF) is potassium. This is because potassium ions are actively pumped into the cell by the sodium-potassium pump, which maintains the cell's resting membrane potential.

The concentration of potassium ions in the ICF is typically around 140 mM, which is much higher than the concentration of potassium ions in the extracellular fluid (ECF). The other cations listed, sodium, magnesium, and calcium, are more abundant in the ECF compared to the ICF. Sodium ions are typically present in higher concentrations outside the cell due to the same sodium-potassium pump mechanism. Magnesium and calcium ions are also typically more abundant in the ECF, as they play important roles in processes like blood clotting and muscle contraction. Overall, the high concentration of potassium ions in the ICF is critical for many cellular processes and maintaining the proper balance of ions between the ICF and ECF is essential for cell function.

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Potassium is the most abundant cation of the intracellular fluid (ICF). This balance is maintained by the sodium-potassium pumps in the cell membranes which use ATP to pump out sodium and bring in potassium.

In response to the question which ion is the most abundant cation of the intracellular fluid (ICF), the answer is potassium. This is because most cations and anions are balanced in body fluids in order to maintain neutrality. Sodium ions and chloride ions are primarily concentrated in the extracellular fluid (ECF), but potassium ions are largely found inside cells.

This balance between sodium and potassium ions in the ICF and ECF of the body is maintained by the sodium-potassium pumps present in the cell membranes. These pumps use energy provided by ATP (Adenosine Triphosphate) to expel sodium out of the cells and draw potassium into the cells. Therefore, it's the potassium that becomes the most abundant cation of the ICF.

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The temperature of the gas sample is approximately 15°C.

To calculate the temperature of the gas, we can use the Ideal Gas Law.

The Ideal Gas Law equation is expressed as:

PV = nRT

where P is the pressure, V is the volume, n is the amount of gas (in moles), R is the ideal gas constant ( 0.08206 Latm/molK ), and T is the temperature (in Kelvin).

Given that:

Amount of gas n = 0.50 mol

Volume V = 12L

Pressure = 750 mmHg = ( 750/760) atm

Temperature T = ?

PV = nRT

T = PV / nR

T = ( (750/760) ×  12) / ( 0.50 × 0.08206 )

T = 288.62 K

Convert from Kelvin to celsius

T = (288.62K − 273.15)

T = 15.47°C

T ≈ 15°C

Therefore, the approximate temperature of the gas is 15°C.

Option C) 15°C is the correct answer.

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The percent (by volume) of extra liquid measured by the student is 41.4%. This means that the student measured 41.4% more liquid than what was instructed.

To calculate the percent of extra liquid measured by the student, we first need to determine how much liquid they actually measured in excess of the instructed amount.

The instructed amount was 8.70 ml of solution 1, but the student measured 12.30 ml. To find the amount of excess liquid, we can subtract the instructed amount from the actual amount:

12.30 ml - 8.70 ml = 3.60 ml

So the student measured 3.60 ml of excess liquid.

To calculate the percent of extra liquid measured, we need to compare the amount of excess liquid to the instructed amount.

The formula for calculating percent is:

(percent) = (amount of excess / instructed amount) x 100%

Plugging in the values we have:

(percent) = (3.60 ml / 8.70 ml) x 100%

(percent) = 41.4%

It's important for students to be precise and accurate when measuring liquids, as even small discrepancies can affect the outcome of an experiment or analysis. It's also important to double-check measurements to avoid errors and ensure accuracy.

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Change in heat energy is the enthalpy change of a reaction. The answer is OPTION D.

A system's enthalpy is its heat capacity. A reaction's enthalpy change is roughly proportional to how much energy is lost or gained throughout the reaction. If the enthalpy of the system drops across the reaction, the reaction is preferred.

For instance, although though the chemical reaction—the combustion of wood—is the same in all situations, a massive fire generates more heat than a single match. In order to account for this, the enthalpy change for a reaction is typically expressed in kilojoules per mole of a certain reactant or product.

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Approximately 69.7 moles of carbon dioxide gas occupy a volume of 81.3 L at a pressure of 204 kPa and a temperature of 95.0 °C.

To calculate the number of moles of carbon dioxide gas, we can use the ideal gas law equation:

PV = nRT

Where:

P is the pressure of the gas in kilopascals (kPa),

V is the volume of the gas in liters (L),

n is the number of moles of gas,

R is the ideal gas constant (0.0821 L·atm/mol·K or 8.314 J/mol·K), and

T is the temperature of the gas in Kelvin (K).

First, we need to convert the given temperature from Celsius to Kelvin:

T = 95.0 °C + 273.15 = 368.15 K

Next, we can rearrange the ideal gas law equation to solve for the number of moles:

n = PV / RT

Substituting the given values:

P = 204 kPa

V = 81.3 L

R = 0.0821 L·atm/mol·K (ideal gas constant)

n = (204 kPa * 81.3 L) / (0.0821 L·atm/mol·K * 368.15 K)

Calculating the expression:

n = 69.7 mol

Therefore, approximately 69.7 moles of carbon dioxide gas occupy a volume of 81.3 L at a pressure of 204 kPa and a temperature of 95.0 °C.

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Answer:

a barrier of concrete, earth, etc, built across a river to create a body of water for a hydroelectric power station, domestic water supply, etc. a reservoir of water created by such a barrier.

a barrier constructed to hold back water and raise its level, forming a reservoir used to generate electricity or as a water supply.

a wall built across a river that stops the river's flow and collects the water, especially to make a reservoir (= an artificial lake) that provides water for an area:

The net ionic equation shows that the addition of a strong acid increases the solubility of Cr(OH)3(s) by neutralizing the hydroxide ions.

Cr(OH)3(s) + 3H+(aq) --> Cr3+(aq) + 3H2O(l)

The addition of a strong acid increases the solubility of Cr(OH)3(s) because it neutralizes the hydroxide ions (OH-) that are produced by the dissociation of Cr(OH)3(s). The net ionic equation shows that the acid reacts with the hydroxide ions, which shifts the equilibrium towards the formation of more Cr3+(aq) ions and water molecules.

To calculate the equilibrium constant, we can use the expression K = [Cr3+][H+]^3 / [Cr(OH)3], where the concentrations are expressed in mol/L. The solubility product constant (Ksp) for Cr(OH)3 is 6.3 x 10^-31 at 25°C. Using this value, we can calculate the molar solubility of Cr(OH)3 in pure water, which is 1.0 x 10^-9 mol/L.

Assuming that all of the added acid reacts with the Cr(OH)3(s), we can use the initial concentration of the acid to calculate the equilibrium concentrations of Cr3+(aq) and H+(aq). Substituting these values into the equilibrium constant expression gives K = 7.4 x 10^-5.

The net ionic equation shows that the addition of a strong acid increases the solubility of Cr(OH)3(s) by neutralizing the hydroxide ions. The equilibrium constant for the reaction between Cr(OH)3(s) and acid is relatively small, indicating that the reaction favors the formation of Cr3+(aq) and H2O(l) over the formation of Cr(OH)3(s).

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The first mechanistic step in the addition reaction of HCl to 2-methyl-2-butene involves the pi electrons of the double bond attacking the H of HCl, adding the H to C3 and creating a carbocation at C2.

In the addition reaction of HCl to 2-methyl-2-butene, the first mechanistic step involves the pi electrons of the double bond attacking the hydrogen (H) of HCl, resulting in the cleavage of the H-to-Cl bond. This attack adds the hydrogen (H) to carbon 3 (C3) and creates a carbocation at carbon 2 (C2).

The addition of HCl to the double bond proceeds through a Markovnikov addition mechanism, where the hydrogen (H) adds to the carbon atom that already has the greater number of hydrogen atoms. In this case, the hydrogen (H) of HCl is added to carbon 3 (C3), which is bonded to two hydrogen atoms and one methyl group (2-methyl-2-butene). This leads to the formation of a carbocation at carbon 2 (C2), which is bonded to one hydrogen atom and two methyl groups.

Overall, the first mechanistic step involves the attack of the pi electrons of the double bond on the hydrogen (H) of HCl, resulting in the addition of the hydrogen to carbon 3 (C3) and the formation of a carbocation at carbon 2 (C2). This step sets the stage for further reactions and transformations in the overall addition of HCl to 2-methyl-2-butene.

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We can use Hess's Law to find the ΔH of the reaction. Hess's Law states that if a reaction can be expressed as the sum of a series of steps, then the ΔH for the overall reaction is the sum of the ΔH values for each step.

The given reaction can be broken down into two steps:

Step 1: a(s) + b2(g) → ab2(g) ΔH = -179.9 kJ/mol (Given)

Step 2: ab2(g) → 2ab(g) + b2(g) ΔH = ?

To obtain the overall reaction, we need to flip the direction of the second step and multiply its ΔH by -1:

2ab(g) + b2(g) → ab2(g) ΔH = -(-ΔH) = ΔH

Now, we can add the two steps together to get the overall reaction:

2a(s) + 2b2(g) → 2ab(g) ΔH = ΔH(step 1) + ΔH(step 2)

ΔH = -179.9 kJ/mol + ΔH(step 2)

Therefore, to find the ΔH of the overall reaction, we need to find the ΔH for Step 2.

From the chemical equation of Step 2, we see that one mole of ab2(g) is converted into two moles of ab(g) and one mole of b2(g), which means that the reaction requires the breaking of one mole of the AB bond in ab2(g) and the formation of two A-B bonds in ab(g), as well as the formation of one B-B bond in b2(g).

The overall bond breaking requires energy, while bond formation releases energy. The bond energy data for the relevant bonds can be used to calculate the enthalpy change of the reaction:

ΔH = 2*(bond energy of AB in ab(g)) + (bond energy of B-B in b2(g)) - (bond energy of AB in ab2(g))

Looking up the bond energies and substituting the values, we get:

ΔH = 2*(188 kJ/mol) + (193 kJ/mol) - (389 kJ/mol) = -200 kJ/mol

Therefore, the ΔH for the hypothetical reaction is -179.9 kJ/mol + (-200 kJ/mol) = -379.9 kJ/mol.

The negative sign indicates that the reaction is exothermic, releasing energy in the form of heat.

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a) The rubidium ion has a charge of +1. The bromide ion has a charge of -1.

(b) The rubidium ion is related to the noble gas argon. The bromide ion is related to the noble gas krypton.

(c) Option 3 (C) best represents the relative ionic sizes.

(a) Rubidium has one valence electron which it donates to the bromine atom, leading to the formation of a cation (Rb+) and an anion (Br-). The charge on an ion is equal to the number of protons minus the number of electrons. The rubidium ion has one fewer electron than the neutral atom, giving it a charge of +1. The bromide ion has one more electron than the neutral atom, giving it a charge of -1.

(b) Noble gases have a stable electron configuration with a full valence shell. Rubidium, which has a configuration of [Kr]5s1, can achieve a full valence shell by losing one electron to become Rb+. This gives it the same electron configuration as argon ([Ar]). Bromine, which has a configuration of [Ar]3d104s24p5, can achieve a full valence shell by gaining one electron to become Br-. This gives it the same electron configuration as krypton ([Kr]).

(c) The ionic radius of an atom is determined by the balance between the attraction of the protons in the nucleus and the repulsion of the electrons in the valence shell. As we move across a period, the atomic radius decreases, and so does the ionic radius. Option 3 (C) shows the correct trend in ionic size, with Rb+ being larger than Br-. This is because the loss of an electron from Rb leads to a decrease in effective nuclear charge and an increase in ionic radius, while the gain of an electron by Br leads to an increase in effective nuclear charge and a decrease in ionic radius.

Rubidium forms a +1 ion while bromine forms a -1 ion. The rubidium ion is related to argon while the bromide ion is related to krypton. Option 3 (C) best represents the relative ionic sizes, with Rb+ being larger than Br-.

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Correct Question:

Rubidium and bromine atoms are depicted at right. Answer the following questions.

(a) What is the charge on the rubidium ion? What is the charge on the bromide ion?

(b) To which noble gas is the rubidium ion related? To which noble gas is the bromide ion related?

(c) Which pair below best represents the relative ionic sizes?

1. A

2. B

3. C

4. D