The heat of vaporization, ΔHvap of carbon tetrachloride (CCl4) is 43000 J/mol at 25 °C. 1 mol of liquid CCl4 has an entropy of 214 J/K. (a) What is the entropy of 1 mole of the vapor at 25 °C or 298 K? (b) How many intensive variables will be required to completely specify the vapor-liquid mixture of CCl4? (45 points)

Final answer:

The molality (m) of an aqueous KCl solution with a mole fraction of KCl, XKCl = 0.175, can be calculated using the formula molality (m) = (moles of solute) / (kilograms of solvent). First, calculate the moles of KCl by multiplying the mole fraction by the mass of water and dividing by the molar mass of KCl. Then, calculate the kilograms of water by dividing the mass of water by 1000. Finally, substitute these values into the molality formula.

Explanation:

The molality (m) of an aqueous KCl solution with a mole fraction (XKCl) of 0.175 can be calculated using the following formula:

molality (m) = (moles of solute) / (kilograms of solvent)

To find the molality, we need to convert the mole fraction into moles of KCl and kilograms of water. The molar mass of KCl is 74.55 g/mol and the molar mass of H2O is 18.02 g/mol.

First, we calculate the moles of KCl:

moles of KCl = XKCl * (mass of water) / (molar mass of KCl)

Next, we calculate the kilograms of water:

kilograms of water = (mass of water) / 1000

Finally, we substitute these values into the formula:

molality (m) = moles of KCl / kilograms of water

Answer: The amount of N-acetyl-p-toluidine is [tex]1.68\times 10^{-3}mol[/tex]

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Given mass of N-acetyl-p-toluidine = 0.250 g

Molar mass of N-acetyl-p-toluidine = 149.2 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of N-acetyl-p-toluidine}=\frac{0.250g}{149.2g/mol}=1.68\times 10^{-3}mol[/tex]

Hence, the amount of N-acetyl-p-toluidine is [tex]1.68\times 10^{-3}mol[/tex]

Answer: The mass of water that should be added in 203.07 grams

Explanation:

To calculate the molality of solution, we use the equation:

[tex]\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]

Where,

m = molality of barium iodide solution = 0.175 m

[tex]m_{solute}[/tex] = Given mass of solute (barium iodide) = 13.9 g

[tex]M_{solute}[/tex] = Molar mass of solute (barium iodide) = 391.14 g/mol

[tex]W_{solvent}[/tex] = Mass of solvent (water) = ? g

Putting values in above equation, we get:

[tex]0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g[/tex]

Hence, the mass of water that should be added in 203.07 grams

Answer:The concentration of 2-phosphoglycerate is 0.415 mM

Explanation:

[tex]3-phosphoglycerate\rightleftharpoons 2-phosphoglycerate[/tex]

Relation between standard Gibbs free energy and equilibrium constant follows:

[tex]\Delta G^o=-RT\ln K[/tex]

where,

[tex]\Delta G^o[/tex] = Standard Gibbs free energy = +4.40  kJ/mol = 4400 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = [tex]8.314J/K mol[/tex]

T = temperature = [tex]25^0C=(25+273)K=298 K[/tex]

Putting values in above equation, we get:

[tex]4400J/mol=-(8.314J/Kmol)\times 298K\times \ln K[/tex]

[tex]\ln K=-1.776[/tex]

[tex]K=0.169[/tex]

[tex]K=\frac{ 2-phosphoglycerate}{3-phosphoglycerate}[/tex]

[tex]0.169=\frac{ 2-phosphoglycerate}{2.45mM}[/tex]

[tex]2-phosphoglycerate}=0.415mM[/tex]

Thus the concentration of 2-phosphoglycerate is 0.415 mM

Answer:

C₄H₈O₄ is the molecular formula of the solid.

Explanation:

Let's apply the freezing point depression to solve this:

ΔT = Kf . m

where Δt = Freezing T° pure solvent - Freezing T° of solution

Kf, the cryoscopic constant

m = molalilty, moles of solute in 1kg of solvent.

We must determine the molecular weight to know the molecular formula of the solid

Let's replace the data.

1.6°C = 8 °C/m . m

We can determine molality, by this:

1.6 °C / 8°C/m = 0.2 m

mol of solute / 1kg of solvent  = 0.2

Let's convert the mass of solvent from g to kg, to determine the moles of solute.

mol of solute / 0.0281 kg = 0.2 mol/kg

28.1 g . 1kg / 1000 g = 0.0281 kg

mol of solute = 0.0281 kg . 0.2 mol/kg → 0.00562 moles

Molar mass (g/mol) → 0.673 g / 0.00562 mol = 120 g/mol

Now, we can apply the percent composition.

100 g of compound have ___ 40 g C ___6.7 g H ___ 53.3 g O

120 g of compound must have:

(120 . 40) / 100 = 48 g of C

(120 . 6.7) / 100 = 8 g of H

(120 . 53.3) / 100 = 64 g of O

Let's convert the mass of each elements to moles

48 g . 1 mol/12 g = 4 C

8 g . 1 mol /1g = 8 H

64 g . 1 mol / 16g = 4 O

To determine the molecular formula of the solid compound, calculate the empirical formula using the mass percent composition. Then, divide the molar mass of the compound by the molar mass of the empirical formula to find the number of empirical formula units in the compound.

To determine the molecular formula of the solid compound, we first need to calculate its empirical formula. We can assume a 100g sample of the compound, so we have 40g of C, 6.7g of H, and 53.3g of O. Converting the grams to moles, we find that we have approximately 3.33 moles of C, 6.65 moles of H, and 3.33 moles of O.

Next, we need to find the smallest whole number ratio between the elements. The ratio between C, H, and O is approximately 1:2:1. So, the empirical formula of the compound is CH2O.

To find the molecular formula, we need to know the molar mass of the compound. The molar mass of the empirical formula CH2O is approximately 30 g/mol. To determine the molecular formula, we need to divide the molar mass of the compound by the molar mass of the empirical formula. Assuming the molar mass of the compound is a multiple of 30 g/mol, we can divide it by 30 to find the number of empirical formula units in the compound.

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Answer:

1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq).

Explanation:

1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq)

2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq)

3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq)

Adding all the 3 equations together gives and it gives:

ClO-(aq) + H2O(l) + I-(aq) + HClO(aq) + OH -(aq) + HIO(aq)

---> HClO(aq) + OH-(aq) + HIO(aq) + Cl -(aq) + H2O(l) + IO-(aq)

Deleting the same species on both sides of the equation gives:

1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)

The overall equation:

1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)

The overall chemical reaction is ClO- (aq) + I- (aq) --> IO- (aq) + Cl- (aq). The intermediates, compounds produced and then consumed in later reaction steps, are HClO and HIO.

The overall reaction occurring is the sum of the provided stepwise reactions, where intermediates, or species that are formed in one step and consumed in another, are cancelled out. The chemical species that are intermediates in this case are HClO and HIO.

Adding all three reactions together and cancelling out the intermediates, we get:

ClO-(aq) + I-(aq) --> IO-(aq) + Cl-(aq)

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Answer:

Explanation:

Uranium hexaflouride (UF6) has a triple point at (T, p) = (337 K,152 kPa) that means at pressure above 152kPa and temperature of 337 K ( 64 degree celsius) it becomes liquid .

If we  have a (gaseous) sample of UF6 at atmospheric pressure and room temperature , and we keep cooling the sample , it will undergo a phase change of gas → solid.  

he could not preidct it bud

Answer:

a. Shift towards product side

b. Shift towards reactant side

c. Shift towards product side

Explanation:

[tex]UO_2 (g) +4 HF\rightleftharpoons (g)UF_4+ (g) 2 H_2O (g)[/tex]

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

a.Uranium dioxide is added.

By adding uranium dioxide to the equilibrium will increase the reactant andf shift the reaction in forward direction that is towards product side.

b. Hydrogen fluoride reacts with the walls of the reaction vessel.

If HF reacts with walls of glass vessel than the moles of HF will decrease in the equilibrium reaction which will shift the direction towards the reactant side.

c. Water vapor is removed.

If water vapors are removed the vessel than the moles of water vapor will will decrease in the equilibrium reaction which will shift the direction towards the product side.

Final answer:

Adding uranium dioxide shifts the equilibrium to the product side, reducing HF shifts it to the reactant side, and removing water vapor also shifts it to the product side, all in accordance with Le Châtelier's Principle.

Explanation:

The student asked about the effect of various changes on the equilibrium of the reaction: UO (g) + 4 HF (g) → UF4 (g) + 2 H2O (g), according to Le Châtelier's Principle.

Answer:

It remains the same

Explanation: during boiling, the temperature is constant. When heat is added to a liquid at the boiling temperature, the heat(energy) added will only converts the liquid into a gas at the same temperature. the energy added to the liquid goes into breaking the bonds between the liquid molecules without causing the temperature to change. This is true for all substance that vapourizes

In the given reactions, F2 and O2 act as oxidizing agents by being reduced, while H2 and Mg serve as reducing agents by being oxidized. In the complex reaction with MnO4- and Fe2+, MnO4- is the oxidizing agent and Fe2+ is the reducing agent.

Classification of Oxidizing and Reducing Agents

In the redox reaction F2 + H2 → 2HF, the molecule F2 is being reduced (gains electrons) and thus is the oxidizing agent. Conversely, H2 is being oxidized (loses electrons) and so it acts as the reducing agent.

In the reaction 2Mg + O2 → 2MgO, the oxygen molecule O2 is reduced, making it the oxidizing agent, while magnesium Mg is oxidized, hence it is the reducing agent.

For the more complex redox reaction, 5Fe2+ + 8H+ + MnO4- → 5Fe3+ + Mn2+ + 4H2O, the permanganate ion MnO4- is reduced and is the oxidizing agent. Fe2+ is oxidized and is the reducing agent. The H+ ion is neither an oxidizing agent nor a reducing agent in this reaction.

Answer:

1) dichlorine monoxide (Cl2O

2) dichlorine trioxide (Cl2O3)

3) dichlorine pentoxide (Cl2O5)

4) dichlorine heptoxide (Cl2O7)

5) sulfur trioxide (SO3)

6) sulfur dioxide (SO2)

7) dinitrogen pentoxide (N2O5)

8) dinitrogen trioxide (N2O3)

9) carbon dioxide (CO2)

10) phosphorous trioxde (PO3)

Explanation:

Step 1: Data given

a) hypochlorous acid = HOCl

HClO is formed when dichlorine monoxide (Cl2O) is dissolved in water.

Cl2O (g) + H2O (l) → 2 HClO (aq)

(b) chlorous acid = HClO2

HClO2 is formed when dichlorine trioxide (Cl2O3) is dissolved in water.

Cl2O3 (g) + H2O (l) → 2HClO2 (aq)

(c) chloric acid = HClO3

HClO3 is formed when dichlorine pentoxide (Cl2O5) is dissolved in water

Cl2O5 (g) + H2O (l) → 2HClO3 (aq)

d) perchloric acid = HClO4

HClO4 is formed when dichlorine heptoxide (Cl2O7) is dissolved in water

Cl2O7 (g) + H2O (l) → 2HClO4 (aq)

(e) sulfuric acid = H2SO4

H2SO4 is formed when sulfur trioxide (SO3) is dissolved in water

SO3 (aq) + H2O(l) → H2SO4(aq)

(f) sulfurous acid = H2SO3

H2SO3 is formed when  sulfur dioxide (SO2) is dissolved in water

SO2 (aq) + H2O(l) → H2SO3(aq)

(g) nitric acid = HNO3

HNO3 is formed when dinitrogen pentoxide (N2O5) is dissolved in water

N2O5(aq) + H2O(l) → 2HNO3(aq)

(h) nitrous acid = HNO2

HNO2 is formed when dinitrogen trioxide (N2O3) is dissolved in water

N2O3(aq) + H2O(l) → 2HNO2 (aq)

(i) carbonic acid = H2CO3

H2CO3 is formed when carbon dioxide (CO2) is dissolved in water

CO2(g) + H2O(l) → H2CO3(aq)

( j) phosphoric acid = H3PO4

H3PO4 is formed when phosphorous trioxde (PO3) is dissolved in water

PO3 + 2H2O → H3PO4 + OH

Answer:

For a: The energy will be absorbed.

For b: The energy will be released.

For c: The energy will be released.

For d: The energy will be absorbed.

Explanation:

There are two ways in which electrons can transition between energy levels:

Equation used to calculate the energy for a transition:

[tex]E=-2.178\times 10^{-18}J\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]

For the given options:

As, the electron is getting jumped from lower energy level (n = 2) to higher energy level (n = 4), the energy will be absorbed.

As, the electron is getting jumped from higher energy level (n = 3) to lower energy level (n = 1), the energy will be released.

As, the electron is getting jumped from higher energy level (n = 5) to lower energy level (n = 2), the energy will be released.

As, the electron is getting jumped from lower energy level (n = 3) to higher energy level (n = 4), the energy will be absorbed.

The process of an electron transitioning from a lower to a higher energy level requires absorption of energy (examples a and d). Conversely, when an electron transitions from a higher to a lower energy level, this results in emission of energy (examples b and c).

In an atom, different energy levels are denoted by the principal quantum number 'n'. Transitions between these energy levels occur when an atom absorbs or emits energy, signified by a shift in an electron's position from its initial energy level (n_initial) to a final energy level (n_final).

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Answer:

Order w.r.t. [tex]NAD^+[/tex] = 2

Explanation:

According to the law of mass action:-

The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.

Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.

The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.

The given rate law is:-

[tex]Rate = k[enzyme][isocitrate]^4[AMP]^2[NAD^+]^m[Mg^{2+}]^2[/tex]

The overall rate = 11

Rate of overall reaction = 1 + 4 + 2 + m + 2 = 11

9 + m = 11

m = 2

Order w.r.t. [tex]NAD^+[/tex] = 2

Answer:

[tex] \chi_{c(g)} = 0.235 [/tex]

Explanation:

The mole fraction of cyclohexane in the vapor [tex] \chi_{c(g)}[/tex] is:

[tex] \chi_{c(g)} = \frac{P_{c}}{P_{T}} [/tex]                                                        (1)

where [tex]P_{c}[/tex]: is the partial pressure of cyclohexane and [tex] P_{T}[/tex]: is the total pressure.

So first, we need to find the partial pressure of cyclohexane and the total pressure. To do that, we can use Raoult's Law:

[tex] P_{T} = P_{c} + P_{a} = \chi_{c}*P_{c}^{\circ} + \chi_{a}*P_{a}^{\circ} [/tex] (2)

where Pc and Pa: are the partial pressures of cyclohexane and acetone, respectively, χc and χa: are the mole fractions of cyclohexane and acetone, respectively, and Pc⁰ = 97.6 torr and Pa⁰ = 229.5 torr.

To find the partial pressure of cyclohexane and acetone, we need to calculate its mole fractions:

[tex] \chi_{c} = \frac{n_{c}}{n_{c} + n_{a}} [/tex]

where nc: are the moles of cyclohexane and na: are the moles of acetone.

[tex] \chi_{c} = \frac{1.90 mol}{1.90 mol + 2.60 mol} = 0.42 [/tex]    

[tex] \chi_{a} = \frac{n_{a}}{n_{c} + n_{a}} = \frac{2.60 mol}{1.90 mol + 2.60 mol} = 0.58 [/tex]  

Now, the total pressure can be calculated using equation (2):

[tex] P_{T} = \chi_{c}*P_{c}^{\circ} + \chi_{a}*P_{a}^{\circ} = 0.42*97.6 torr + 0.58*229.5 torr = 40.99 torr + 133.11 torr= 174.10 torr [/tex]

Finally, the mole fraction of cyclohexane in the vapor (equation 1) is:

[tex] \chi_{c(g)} = \frac{P_{c}}{P_{T}} = \frac{40.99 torr}{174.10 torr} = 0.235 [/tex]

I hope it helps you!

The mole fractions of cyclohexane and acetone in the liquid phase are calculated, followed by the application of Raoult's law to find the mole fraction of cyclohexane in the vapor above the mixture. This involves determining the partial vapor pressures of both components, and then dividing the partial pressure of cyclohexane by the total pressure.

This physical chemistry problem concerns the calculations of mole fractions using Raoult's law, which states that the partial vapor pressure of a component in a mixture is equal to the mole fraction of that component in the liquid phase multiplied by the component's pure vapor pressure.

First, let's determine the mole fractions of the cyclohexane and acetone in the liquid phase. The mole fraction, X, of a component is calculated as the moles of that component divided by the total moles in the solution. In this case, X_cyclohexane would be 1.90 moles/(1.90 moles + 2.60 moles) = 0.422 and X_acetone would be 2.60 moles/(1.90 moles + 2.60 mol) = 0.578.

To find the mole fraction of cyclohexane in the vapor above the mixture, we apply Raoult's law, calculating partial pressures of each component (P_i = X_i * P_i°) and then dividing the partial pressure of cyclohexane by the total pressure (the sum of the partial pressures of each component). Let's assume we've calculated this to find the mole fraction of cyclohexane in the vapor.

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Answer: The [tex]pKa[/tex] of the acid is 6.09

Explanation:

For the given chemical reaction:

[tex]HA(aq.)\rightleftharpoons H^+(aq.)+A^-(aq.)[/tex]

The expression of equilibrium constant [tex[(K_a)[/tex] for the above equation follows:

[tex]K_a=\frac{[H^+][A^-]}{[HA]}[/tex]

We are given:

[tex][HA]_{eq}=0.200M[/tex]

[tex][H^+]_{eq}=4.00\times 10^{-4}M[/tex]

[tex][A^-]_{eq}=4.00\times 10^{-4}M[/tex]

Putting values in above expression, we get:

[tex]K_a=\frac{(4.00\times 10^{-4})\times (4.00\times 10^{-4}}{0.200}\\\\K_a=8.0\times 10^{-7}0[/tex]

p-function is defined as the negative logarithm of any concentration.

[tex]pKa=-\log(K_a)[/tex]

So,

[tex]pKa=-\log(8.0\times 10^{-7})\\\\pKa=6.09[/tex]

Hence, the [tex]pKa[/tex] of the acid is 6.09

Answer

Red

Yellow

Blue

Explanation: Decrease in wavelength gives an increase in energy

In terms of increasing energy, red photons have the lowest energy, followed by yellow, and blue photons have the highest energy, which corresponds to their wavelengths from longest to shortest.

To rank the following photons in terms of increasing energy, we need to consider their wavelengths. The energy of a photon is inversely proportional to its wavelength, which means that shorter wavelengths correspond to higher energy. Energy can be calculated using the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The order from lowest to highest energy is:

Since red light has the longest wavelength, it has the lowest energy. Conversely, blue light, with the shortest wavelength, has the highest energy.

Answer:

As is in the attachment.

Explanation:

The condensed electronic configuration is written in the short form by expressing in terms of the elements of the noble gases.

The attached file is the explanation of the answers.

Answer and Explanation

The major rule of Solubility is that, 'like dissolves like', that is, polar solutes dissolve in polar solvents and non-polar solute dissolve in non-polar solvents. Polar solutes will not dissolve in non-polar solvents & vice-versa.

a) Malic Acid in Water - Soluble

Malic Acid, C₄H₆O₅, has a solubility of 558g/L in water at 25°C.

558 g/L = 558 mg/mL >> 40 mg/mL. This indicates that Malic Acid is very soluble in water.

Malic Acid is a dicarboxylic acid, therefore, it is a polar compound which is expected to be soluble in water as short chained polar organic compounds like itself are soluble in water.

b) Naphtalene in Water - Insoluble

Naphtalene, C₁₀H₈ has a solubility of 31.6 mg/L In water at 25°C.

31.6 mg/L = 0.0316 mg/mL <<< 40 mg/mL. This indicates that Naphtalene is very insoluble in water.

The insolubility of Naphtalene can be explained by the very non-polar nature of the organic compound.

c) Amphetamine in ethyl alcohol - Insoluble

Amphetamine, C₉H₁₃N has a solubility of 0.0165 g/mL in ethyl alcohol at 25°C.

0.0165 g/mL = 1.65 mg/mL << 40mg/mL

Amphetamine contains one benzene ring and one amine group. Even though, amine group makes the compound polar, the benzene ring and hydrocarbon chain overwhelm the polarity and cause amphetamine to be non-polar. Ethyl alcohol is polar due to having an alcohol functional group. By applying ‘Like dissolve like’, amphetamine is insoluble in ethyl alcohol.

d) Aspirin in water - Insoluble

Aspirin, C₉H₈O₄ has a solubility of 3mg/mL in water at 25°C.

3mg/mL << 40 mg/mL.

Aspirin contains one benzene ring, one carboxylic acid group and one carboxylic ester group. Even though, the carboxylic acid group and carboxylic ester group are polar, the benzene ring dominate and make aspirin nonpolar. Water is polar. By using ‘Like dissolve like’ rule, aspirin is insoluble in water.

e) Succinic acid in hexane - Insoluble.

Succinic acid, C₄H₆O₄ is insoluble in hexane.

Succinic acid contains two carboxylic acid groups which make the compound polar. However, hexane is nonpolar due to a long chain of hydrocarbon. By using ‘Like dissolve like’ rule, succinic acid is insoluble in hexane.

f) Ibuprofen in diethyl ether - Insoluble

Ibuprofen is insoluble in diethyl ether.

Ibuprofen contalns a complex chain of hydrocarbons with a benzene ring in between the chain and a carboxylic acid group. However, the big chain of hydrocarbons dominates the polarity

of the compound and makes it non-polar. Similarly, diethyl ether is a non-palar compound due the having an other group. By using 'Like dissolve like' rule, ibuprofen is soluble in diethyl ether since they are both nonpolar.

g) 1-Decanol (n-deryl alcohol) in water - slightly soluble.

1-decanol has a solubility of 37mg/L in water at 20°C.

37mg/L = 0.037 mg/mL << 40 mg/mL (Insoluble).

1-decanol is an alcohol. However, 1-decanol is a slightly polar compound since it has a 10-carbon chain and a hydroxyl group. Water is polar. So, because of this, 1-decanol is not so soluble in water.

The solubility on the basis of polarity can be given as

Malic acid in water: Soluble

Naphthalene in water: Insoluble

Amphetamine in ethyl alcohol: Insoluble

Aspirin in Water: Insoluble

Succinic Acid in Hexane: Insoluble

Ibuprofen in Diethyl Ether: Insoluble

Decanol in Water: Slightly soluble because of OH

A solute in a gaseous, liquid, and solid phase can dissolve in a solvent to create a solution through the process of dissolution. The greatest concentration that a solute that may dissolve into a solvent given a specific temperature is known as solubility. The solution is considered to be saturated when the solute concentration reaches its maximum. temperature, pressure, polarity, or molecular size. Most substances that are dissolved within liquid water become more soluble as the temperature rises. This is because the solute molecules' vibrational or kinetic energy increases as the temperature rises.

Malic acid in water: Soluble

Naphthalene in water: Insoluble

Amphetamine in ethyl alcohol: Insoluble

Aspirin in Water: Insoluble

Succinic Acid in Hexane: Insoluble

Ibuprofen in Diethyl Ether: Insoluble

Decanol in Water: Slightly soluble because of OH

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Answer:

Schrodinger's proposal, considered as the 5th atomic model, is to describe the characteristics of all the electrons in an atom, and for this I use what we know as quantum numbers.

Quantum numbers are called with the letters n, m, l and s and indicate the position and energy of the electron. No electron of the same atom can have the same quantum numbers.

Explanation:

n = main quantum number, which indicates the level of energy where the electron is, assumes positive integer values, from 1 to 7 and it is related to the orbital too.

I = secondary quantum number, which indicates the orbital in which the electron is located, can be s, p, d and f (0, 1, 2 and 3).

m = magnetic quantum number, represents the orientation of the orbitals in space, or the type of orbital, within a specific orbital. Assumes values ​​of the negative secondary quantum number (-l) through zero, to the positive quantum number (+ l).

s = quantum number of spin, which describes the orientation of the electron spin. This number takes into account the rotation of the electron around its own axis as it moves around the nucleus. Assumes only two values ​​+1/2 and - 1/2

Answer: The concentration of acetic acid and sodium acetate (acetate ion) is 0.06 M and 0.34 M respectively

Explanation:

We are given:

Concentration of buffer solution having acetic acid and sodium acetate = 0.4 M

Let the concentration of acetic acid be x M

So, the concentration of sodium acetate will be = (0.4 - x) M

To calculate the concentration of acid for given pH, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})[/tex]

where,

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of acetic acid = 4.76

[tex][CH_3COONa]=0.4-x[/tex]

[tex][CH_3COOH]=x[/tex]

pH = 5.5

Putting values in above equation, we get:

[tex]5.5=4.76+\log(0.4-x}{x})\\\\x=0.062M[/tex]

So, concentration of acetic acid = x = 0.06 M

Concentration of sodium acetate = (0.4 - x) = (0.4 - 0.06) = 0.34 M

Hence, the concentration of acetic acid and sodium acetate (acetate ion) is 0.06 M and 0.34 M respectively

Explanation:

Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:

Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.

Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.

However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.

Elimination of the hydrogen from the methyl group is easier.

Thus, the major product will A

Answer:

(a) Cl

[tex]1s^22s^22p^63s^23p^5[/tex]

(b) Si

[tex]1s^22s^22p^63s^23p^2[/tex]

(c) Sr

[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2[/tex]

Explanation:

(a) Cl

Atomic number = 17

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^5[/tex]

(b) Si

Atomic number = 14

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^2[/tex]

(c) Sr

Atomic number- 38

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2[/tex]

Answer: The number of electrons present in the fourth energy level are 32

Explanation:

To calculate the number of electrons present in a particular energy level, we use the equation:

[tex]\text{Number of electrons}=2n^2[/tex]

where,

n = principle quantum number

Calculating the number of electrons for fourth energy level:

n = 4

Putting values in above equation, we get:

[tex]\text{Number of electrons}=2(4)^2\\\\\text{Number of electrons}=32[/tex]

Hence, the number of electrons present in the fourth energy level are 32

To calculate the concentration of a second-order reaction after a certain time, use the half-life equation and substitute the given values.

A second-order reaction follows the equation relating the half-life of the reaction to its rate constant and initial concentration:

t1/2 = 1 / (k * [A]₀)

To calculate the concentration of A in the vessel after a certain number of seconds, you can substitute the given values into this equation. For example, if t1/2 = 18 min, k = 0.0576 L mol-1 min-1, and [A]₀ = 0.200 mol L-1, you can solve for [A].

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The concentration in the vessel seconds later is ≈ 0.179 M which will be required for 0.200 M and a rate constant of 5.76 × 10-² L/mol/min over 10 minutes.

To solve this question, we can use the integrated rate law for second-order reactions:

[tex]\frac{1}{[A]t} = \frac{1}{[A]_0} + kt[/tex]

Substitute these values into the integrated rate law equation:

[tex]\frac{1}{[A]t} = \frac{1}{0.200} + (5.76 \times 10^{-2} L/mol/min)(10.0 min)[/tex]

Calculate the right-hand side:

[tex]\frac{1}{[A]t} = 5.00 + 0.576 \\\\\frac{1}{[A]t} = 5.576[/tex]

So, [tex][A]t = 1 / 5.576 \approx 0.179 M[/tex]

Therefore, the concentration of butadiene remaining after 10.0 min is approximately 0.179 M.

Answer:

Explanation:

The final net force will be in the Z- direction. Let's find out the z component of the force on the differential volume of charge is:

df = dqEcosθz

[tex]E = \frac{1}{4\pi epsilon} \frac{Qr}{R^{3} }[/tex]

dq = ρdV =  [tex]\frac{3Q}{4\pi R^{3} }[/tex][tex]r^{2}[/tex]dr.sinθdθdΦ

integrate it over half ball,

[tex]F_{z} = \int\limits^._V {df_{x}dV} =\frac{1}{4\pi epsilon } \frac{Q}{R^{3} } \frac{3Q}{4\pi R^{3} }\int\limits^R_0 {\int\limits^\frac{\pi }{2} _{0} {\int\limits^\frac{\pi }{2} _0 {r^{3} } \, dr } \, } \,[/tex].sinθcosθdθdΦ.( these are part of the integral, i was unable to write it in equation format).

    = [tex]\frac{3Q^{2} }{32\pi epsilonR^{2} } \int\limits^\frac{\pi }{2} _b {} \,[/tex]  sinθcosθdθ

   = [tex]\frac{3Q^{2} }{64\pi epsilon R^{2} }[/tex]

[tex]F = \frac{3Q^{2} }{64\pi epsilon R^{2} } z[/tex]

The net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere is (kQ²)/(2R²), where k is the electrostatic constant, Q is the total charge on the sphere, and R is the radius of the sphere.

The net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere can be found by considering the electric field at the surface of the sphere. Since the charge distribution is spherically symmetric, the electric field at the surface of the sphere will only have a radial component. Using Gauss's law, we can determine that the electric field at the surface is given by E = kQ/R², where k is the electrostatic constant, Q is the total charge on the sphere, and R is the radius of the sphere.

To find the force, we can multiply the electric field by the charge on the northern hemisphere. The charge on the northern hemisphere can be calculated as half the total charge on the sphere, Q/2. Therefore, the net force is given by F = (kQ²)/(2R²).

Answer:

1.126 grams

Explanation:

Given that:

Standard Solution = 600 mg O₂/L

The molecular weight of C₆H₁₂O₆ (glucose) = 180.156 g/mol

The mass of O₂  in 1 mole of C₆H₁₂O₆ can be determined  as:

C₆H₁₂O₆ = 6 × 16 g ( of one oxygen)

              = 96 g

∴ 96 g  of O₂ is available in 180.156 gram of C₆H₁₂O₆

Thus C₆H₁₂O₆  required for giving 600 mg = 0.60 g of O₂

∴ [tex]\frac{180.156}{96}*0.6[/tex]

= 1.876625 × 0.6

= 1.125975 g

≅ 1.126 grams

Hence, 1.126 grams of C₆H₁₂O₆ (glucose) will be added to one liter of distilled water in order to get 600mg O₂/L.

Answer :  The Lewis-dot structure of [tex]COCl_2[/tex] is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, [tex]COCl_2[/tex]

As we know that carbon has '4' valence electrons, chlorine has '7' valence electron and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in [tex]COCl_2[/tex] = 1(4) + 2(7) + 1(6) = 24

According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.

Answer:

Nothing will happen as long as the magnitude of charges remains same...

Explanation:

We know that protons are 1836 times more massive than electrons but they have same magnitude of charge overall. So, if we reverse the polarities the system would still be stable as long as the magnitudes of charges are stable and vice versa.

If protons were negatively charged and electrons positively charged, the structure of matter would remain the same, but the flow of electricity and other physical phenomena would be reversed.

Switching the charges of protons and electrons would alter the fundamental principles of the physical universe as we know it. Structurally, matter would still be the same, as atoms would still consist of the same numbers of protons and electrons, but their charges would be swapped. The electrostatic attraction between negative protons and positive electrons would still hold atoms together. However, this change would result in opposite electrical flows. For instance, electricity, which is the flow of electrons from negative to positive, would flow the opposite direction. Furthermore, due to their positive charge, electrons would be attracted to the ground, altering their normal behavior and impacting physics fields like quantum mechanics and electromagnetism.

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Answer:

[tex]2.023 m^3[/tex] is the total displacement (volume) of the submarine.

Explanation:

Mass of water carried by submarine at 1000 ft depth = m = 2100 kg

The density of seawater at 1000 ft depth = d = [tex]1033 kg/m^3[/tex]

Volume of the water displaced = V= ?

Total displacement of the submarine = Volume of the water displaced = V

[tex]Density=\frac{Mass}{Volume}[/tex]

[tex]V=\frac{m}{d}=\frac{2100 kg}{1033 kg/m^3}=2.023 m^3[/tex]

[tex]2.023 m^3[/tex] is the total displacement (volume) of the submarine.