What is the molarity (M) of chloride ions in a solution prepared by mixing 194.4 ml of 0.439 M calcium chloride with 363 ml of 0.497 M aluminum chloride? Enter to 3 decimal places.

Answer:

The product is cyclohexanol

Explanation:

Firstly,

A ketone undergo a borohydride reduction reaction to form an alcohol as below,

R-CO-R'  ⇒ R-CO(OH)-R'

        1.56 δ (4H, triplet) - (-CH₂-CH₂-CH-OH) ; triplet as coupling with 2 H,

        1.78 δ (4H, multiplet)  - (-CH₂-CH₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH

         3.24 δ (1H, quintet); - (-CH₂-CH₂-CH(CH₂-)-OH), coupling with4 H of 2 group of CH₂

         3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-OH), hydrogen of alcohol group, not tend to coupling with other hydrogen

Answer: 6.7*10^-7 m

Explanation:

The full explanation is shown in the image attached. The energy of the photon is obtained by dividing the bond energy by the Avogadro's number. Using the Plank's equation, we cash obtain the frequency or wavelength of radiation required by substituting into the given equation appropriately.

Final answer:

To calculate the shortest wavelength of light capable of dissociating the Br-I bond, the bond energy given (179 kJ/mol) is first converted to Joules per photon. Then, applying the equation E = hc / λ with Planck's constant and the speed of light gives the shortest wavelength needed to break the bond.

Explanation:

The question asks to calculate the shortest wavelength of light capable of dissociating the Br–I bond in iodine monobromide, given that the bond energy is 179 kJ/mol. To solve this, we apply the equation relating the energy of a photon (E) to its wavelength (λ), using Planck's constant (h) and the speed of light (c).

The energy of the photon needed can be calculated using the equation E = hc / λ, where h = 6.626 x 10-34 J·s (Planck's constant) and c = 3.00 x 108 m/s (speed of light). First, convert the bond dissociation energy from kJ/mol to Joules (J) by multiplying by 1000 and dividing by Avogadro's number (6.022 x 1023 mol-1), giving the energy required per molecule.

Finally, rearrange the equation to solve for λ: λ = hc / E. Plugging in the values, we find the shortest wavelength of light capable of breaking the Br–I bond. Remember, since the question gives the bond dissociation energy in kJ/mol, conversion to Joules is necessary for the calculation to proceed correctly.

This is a incomplete question. The complete question is:

It takes 348 kJ/mol to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon. Round your answer to correct number of significant digits

Answer: 344 nm

Explanation:

[tex]E=\frac{Nhc}{\lambda}[/tex]

E= energy  = 348kJ= 348000 J  (1kJ=1000J)

N = avogadro's number = [tex]6.023\times 10^{23}[/tex]

h = Planck's constant = [tex]6.626\times 10^{-34}Js [/tex]

c = speed of light = [tex]3\times 10^8ms^{-1}[/tex]

[tex]348000=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]

[tex]\lambda=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{348000}[/tex]

[tex]\lambda=3.44\times 10^{-7}m=344nm[/tex]    [tex]1nm=10^{-9}m[/tex]

Thus the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon is 344 nm

Answer:

Solution with 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol will have lowest freezing point

Explanation:

[tex]\Delta T_f=K_f\times m[/tex]

where,

[tex]\Delta T_f[/tex] =depression in freezing point =  

[tex]K_f[/tex] = freezing point constant  

m = molality

As we can see that higher the molality of the solution more will depression in freezing point of the solution and hence lower will the freezing point of solution.

[tex]Molality=\frac{moles}{\text{mass of solvent in kg}}[/tex]

A. 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol.

Moles of [tex]C_3H_8O[/tex]=[tex]\frac{35.0 g}{60 g/mol}=0.5833 mol[/tex]

Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)

[tex]m=\frac{0.5833 mol}{0.25 kg}=2.33 m[/tex]

B. 35.0 g of [tex]C4H_{10}O[/tex] in 250.0 g of ethanol

Moles of [tex]C_4H_{10}O[/tex][tex]=[tex]\frac{35.0 g}{74 g/mol}=0.4730 mol[/tex]

Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)

[tex]m'=\frac{0.4730 mol}{0.25 kg}=1.89 m[/tex]

C. 35.0 g of [tex]C_2H_{6}O_2[/tex] in 250.0 g of ethanol

Moles of [tex]C_2H_{6}O_2[/tex]=[tex]\frac{35.0 g}{62g/mol}=0.5645 mol[/tex]

Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)

[tex]m''=\frac{0.5645 mol}{0.25 kg}=2.26 m[/tex]

[tex]m>m'''>m''[/tex]

Solution with 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol will have lowest freezing point

Answer:

8.6 mL

Explanation:

Glacial acetic acid (or anhydrous acetic acid) can be thought of as pure acetic acid, and has a density of 1.05 g/cm³ (or g/mL).

To answer this problem first we convert moles of acetic acid (CH₃CO₂H) to grams, using its molar mass (60 g/mol):

Now we convert grams of acetic acid to mL, using its density:

Answer:

a relief valve is inserted in the downstream line from the pressure regulator.

Explanation:

Whenever dry nitrogen from a portable cylinder is used in service and installation practice the item of most importance in consideration of safety is a relief valve is inserted in the downstream line from the pressure regulator.

The most important item to consider for safety when using dry nitrogen from a portable cylinder in service and installation practice is a pressure regulator, which controls the flow and release of the gas. It is also crucial to inspect the system for leaks regularly.

When using dry nitrogen from a portable cylinder in service and installation practice, the most important item to consider for safety is a pressure regulator. A pressure regulator is used to control the flow and release of nitrogen gas from the cylinder. It ensures that the pressure does not exceed safe limits and reduces the risk of an explosion or other accidents.

Additionally, it is crucial to always check for leaks in the system before using dry nitrogen. Leaks can result in the accumulation of nitrogen gas, which can displace oxygen in the air and lead to asphyxiation. Therefore, regularly inspecting the system and using appropriate leak detection methods, such as soapy water or a leak detection solution, is essential for safety.

In summary, when working with dry nitrogen from a portable cylinder, the two most important safety considerations are the use of a pressure regulator and regular inspection for leaks.

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When 5 moles of silver sulfide react with 8 moles of aluminum metal, there is an excess of aluminum. After the reaction is complete, 3.3 moles of the excess aluminum remain unreacted.

The balanced chemical equation for the reaction between solid silver sulfide (Ag2S) and aluminum metal (Al) is:

3Ag2S + 2Al → 6Ag + Al2S3

Based on this equation, we can see that every 3 moles of Ag2S react with 2 moles of Al to produce 6 moles of Ag and 1 mole of Al2S3.

In the given question, 5 moles of Ag2S react with 8 moles of Al. Therefore, we have an excess of Al. To determine the moles of excess Al remaining unreacted, we can set up a ratio:

(8 moles Al reacted) / (2 moles Al required to react with 3 moles Ag2S) = x moles Ag2S / 5 moles Ag2S

Simplifying this ratio, we find:

x = (8 moles Al / 2) × (5 moles Ag2S / 3 moles Al)

x = 20/6 = 3.3 moles

Therefore, 3.3 moles of the excess reactant (Al) remain unreacted when the reaction is over.

Final answer:

After writing a balanced chemical equation for the reaction between silver sulfide and aluminum, we determine that 4.7 moles of aluminum remain unreacted when 5 moles of silver sulfide react with 8 moles of aluminum.

Explanation:

To determine the number of moles of the excess reactant that remain unreacted, we first need to write a balanced chemical equation for the reaction between silver sulfide and aluminum metal. Here is the balanced equation:

3 Ag₂S (s) + 2 Al (s) → 6 Ag (s) + Al₂S₃(s)

Using the balanced equation, we see that 3 moles of silver sulfide react with 2 moles of aluminum. Therefore, if we had 5 moles of silver sulfide, we would need 2/3 × 5 = 10/3 moles of aluminum to react completely with the silver sulfide.

Since 8 moles of aluminum were originally present, we subtract the amount of aluminum that reacted to find the excess:

8 moles Al - 10/3 moles Al = 14/3 moles Al

Thus, 14/3 moles or 4.7 moles of aluminum remain unreacted.

Answer:

1100 millimeters

Explanation:

1 mole of isopentyl alcohol = 22.4L = 22.4×1000 mL = 22,400L

0.050 moles of isopentyl alcohol = 0.050 × 22,400mL = 1120mL = 1100mL (to two significant figures)

Answer:

Option b. 0.048 M

Explanation:

We have the molecular weight and the mass, from sulcralfate.

Let's convert the mass in g, to moles

1 g . 1 mol / 2087 g = 4.79×10⁻⁴ moles.

Molarity is mol /L

Let's convert the volume of solution in L

10 mL . 1L/1000 mL = 0.01 L

4.79×10⁻⁴ mol / 0.01 L = 0.048 mol/L

Answer:

Well in comparison of Ni (II) and Ni (IV), Ni (IV) is a stronger vation with the +4 charge so it will attract the more oxygen ions in the aqueous solution. That is why Ni (IV) will be more strongly hydrated.

Explanation:

Final answer:

In an aqueous solution, the Ni (IV) cation is more strongly hydrated than the Ni (II) cation due to its higher charge density, which attracts more water molecules into its hydration sphere.

Explanation:

In an aqueous solution containing Ni (II) and Ni (IV) salts, the Ni (IV) cation is expected to be more strongly hydrated. This outcome is attributed primarily to the charge density. The Ni (IV) has a higher charge (+4) compared to Ni (II) which has a +2 charge. In terms of hydration, water molecules, which act as dipoles, are more strongly attracted to ions with a higher charge density. This means that the Ni (IV) ion, with its higher charge, attracts and binds water molecules more strongly than the Ni (II) ion.

This strong attraction results in a greater degree of hydration for the Ni (IV) ion as it pulls more water molecules into its hydration sphere. This process is crucial in understanding the properties of solutions, especially in predicting the behavior of ions in biological and chemical systems.

Here is the complete question

An aqueous solution is saturated with both a solid and a gas at 5 °C. What is likely to happen if the solution is heated to 85 °C ?

View Available Hint(s)

a.) Some gas will bubble out of solution and more solid will dissolve.

b.) Some gas will bubble out of the solution and some solid will precipitate out of the solution.

c.) Some solid will precipitate out of solution.

d.) More gas will dissolve and more of the solid will dissolve.

Answer:

a.) Some gas will bubble out of solution and more solid will dissolve.

Explanation:

Temperature increase usually increases the dissolution of solids in liquids. From the question; Some gas will be bubble out of the solution as the temperature is being increased to 85 °C because that aqueous solution is saturated(i.e equal amount of solute and solvent in the solution) with both  solid and gas at 5 °C, but when the solution is heated to 85 °C, the solution becomes supersaturated( i.e the solute is now more at the given temperature than the solvent).  

Answer:

As the temperature increases, the solubility of the solid increases and the solubility of the gas decreases. When the solution that is saturated between a solid and a gas at 5 ° C and heated to 85 ° C, the gas comes out of the solution first.

Explanation:

A saturated solution is one that has the maximum amount of solute that is dissolved. An unsaturated solution is one that has a low amount of solute compared to the saturated solution. According to Henry's law, the solubility of a gas at a specific temperature is directly proportional to its partial pressure, that is

C ∝ p

C = kp

Where

p is the partial pressure

k ia a proportionality constant

C is the concentration of the gas

Answer:

As shown in the attachment

Explanation:

The four possible isomers are as shown in the attachment.

Answer:

1. Group 1 — 3

2. Transition metals

Explanation:

d-block elements

These elements are also known as transition elements as their positioning and transition of properties lies between s and p block elements.

d-block elements have number of valence electrons equal to their group number, which is equal to the number of electrons in the "valence shell".

For example, Consider a transition metal or d block element Scandium. It's atomic number is 21.

Electronic configuration of Scandium(Sc)- [Ar] 3d¹ 4s²,  it has three electrons in its outermost shell and has a valency of three.

The electron configuration of scandium indicates that the ultimate shell(orbit) of scandium has a complete of electrons. But the electron configuration of scandium within side the Aufbau approach indicates that its ultimate electron([tex]3d^1\\[/tex]) has entered the d-orbital. Thus we can say that scandium has 3 valence electrons.

Therefore, we can say that d block elements have same number of outer electrons and valence electrons.

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Answer:

48.5 kJ

Explanation:

Let's consider the sublimation of carbon dioxide at its sublimation temperature, that is, its change from the solid to the gaseous state.

CO₂(s) → CO₂(g)

The molar mass of carbon dioxide is 44.01 g/mol. The moles corresponding to  66.0 g are:

66.0 g × (1 mol/44.01 g) = 1.50 mol

The heat of sublimation for carbon dioxide is 32.3 kJ/mol. The heat required to sublimate 1.50 moles of carbon dioxide is:

1.50 mol × (32.3 kJ/mol) = 48.5 kJ

To calculate the radius of an atom in a body-centered cubic (BCC) structure, we can use the formula: radius (r) = √(3/4) * a/2, where a is the lattice constant.

In a body-centered cubic (BCC) unit cell, the atoms in the corners do not touch each other but contact the atom in the center. The unit cell contains two atoms: one-eighth of an atom at each of the eight corners and one atom in the center. An atom in a BCC structure has a coordination number of eight. To calculate the radius of an atom in a BCC structure, we can use the formula:

Radius (r) = √(3/4) * a/2

where a is the lattice constant. Plugging in the given value of the lattice constant as 409, we can calculate the radius of the atom of metal M in a BCC structure using this formula.

Answer:

Molarity for solution is 1.44 M

Explanation:

Molarity = Mol of solute / 1L of solution

We would need the volume of solution (To be calculated with density)

We would need the moles of solute (To be calculated with  mass and molar mass of solute)

7.41 % by mass means 7.41 g of solute in 100 g of solution

So, moles of solute → 7.41 g / 58.45 g/mol = 0.127 mol

Let's determine the volume by density

Density = Mass / volume

1.14 g/mL = 100 g / Volume

Volume = 100 g / 1.14 g/mL → 87.7 mL

To reach molarity we must have the volume in L

87.7 mL . 1L / 1000 mL = 0.0877 L

Molarity → mol /L = 0.127 mol / 0.0877L → 1.44 M

Answer:

8.08 × 10⁻⁴

Explanation:

Let's consider the following reaction.

COCl₂(g) ⇄ CO (g) + Cl₂(g)

The initial concentration of phosgene is:

M = 2.00 mol / 1.00 L = 2.00 M

We can find the final concentrations using an ICE chart.

     COCl₂(g) ⇄ CO (g) + Cl₂(g)

I       2.00            0            0

C        -x             +x           +x

E    2.00 -x          x             x

The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.

The concentrations at equilibrium are:

[COCl₂] = 2.00 -x = 1.96 M

[CO] = [Cl₂] = 0.0398 M

The equilibrium constant (Keq) is:

Keq = [CO].[Cl₂]/[COCl₂]

Keq = (0.0398)²/1.96

Keq = 8.08 × 10⁻⁴

Answer: a) The [tex]K_a[/tex] of acetic acid at [tex]25^0C[/tex] is [tex]1.82\times 10^{-5}[/tex]

b) The percent dissociation for the solution is [tex]4.27\times 10^{-3}[/tex]

Explanation:

[tex]CH_3COOH\rightarrow CH_3COO^-H^+[/tex]

 cM              0             0

[tex]c-c\alpha[/tex]        [tex]c\alpha[/tex]          [tex]c\alpha[/tex]

So dissociation constant will be:

[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]

Give c= 0.10 M and [tex]\alpha[/tex] = ?

Also [tex]pH=-log[H^+][/tex]

[tex]2.87=-log[H^+][/tex]  

[tex][H^+]=1.35\times 10^{-3}M[/tex]

[tex][CH_3COO^-]=1.35\times 10^{-3}M[/tex]

[tex][CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M[/tex]

Putting in the values we get:

[tex]K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}[/tex]

[tex]K_a=1.82\times 10^{-5}[/tex]

b)  [tex]\alpha=\sqrt\frac{K_a}{c}[/tex]

[tex]\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}[/tex]

[tex]\alpha=4.27\times 10^{-5}[/tex]

[tex]\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}[/tex]

The Ka of acetic acid is calculated using the pH and the definition of Ka using an ICE table, then percent dissociation is calculated using the initial concentration and the concentration of H+ found. We first calculate [H+] using the pH, then input those values into the Ka expression to find Ka, then use the formula for percent dissociation to find the percent dissociation.

To solve this problem, we will first calculate the Ka using the known pH and the equilibrium relationship between the pH, Ka and [H+] concentration. Then we will calculate the percent dissociation of the acetic acid.

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Explanation:

As per the measurements that are made by the student are as follows.

        21.6 cm is equivalent to 8.5 inch

Therefore, 1 cm for the given situation will be equivalent to inch as follows.

              1 cm = [tex]\frac{8.5}{21.6}[/tex] inch

                      = 0.393 inch

Hence, we will calculate the ratio of cm : inch as follows.

               Ratio of cm : inch = [tex]\frac{8.5}{21.6}[/tex]

                                             = 0.39

Thus, we can conclude that the ratio of cm/inch using the students data is 0.39.

Answer: The value of [tex][OH^-][/tex] for the solution is [tex]10^{-6}M[/tex]

Explanation:

To calculate the concentration of hydroxide ion for the solution, we use the equation:

[tex][H^+]\times [OH^-]=10^{-14}[/tex]

We are given:

[tex][H^+]=100\times [OH^-][/tex]

Putting values in above equation, we get:

[tex]100\times [OH^-]\times [OH^-]=10^{-14}[/tex]

[tex][OH^-]^2=\frac{10^{-14}}{100}[/tex]

[tex][OH^-]=\sqrt{10^{-12}}[/tex]

[tex][OH^-]=10^{-6}M[/tex]

Hence, the value of [tex][OH^-][/tex] for the solution is [tex]10^{-6}M[/tex]

Answer: The number of moles of hydrogen gas generated is [tex]2.33\times 10^{-3}mol[/tex]

Explanation:

To calculate the number of moles of hydrogen gas, we use the equation given by ideal gas which follows:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 0.781 atm

V = Volume of the gas = 71.85 mL = 0.07185 L     (Conversion factor:  1 L = 1000 mL)

T = Temperature of the gas = [tex]20^oC=[20+273]K=293K[/tex]

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

[tex]0.781atm\times 0.07185L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 293K\\\\n=\frac{0.781\times 0.07185}{0.0821\times 293}=2.33\times 10^{-3}mol[/tex]

Hence, the number of moles of hydrogen gas generated is [tex]2.33\times 10^{-3}mol[/tex]

Explanation:

It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.

This means that volume occupied by 2 atoms is equal to volume of the unit cell.

So, according to the volume density

        [tex]5 \times 10^{26} atoms = 1 [tex]m^{3}[/tex]

        2 atoms = [tex]\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms[/tex]

                     = [tex]4 \times 10^{-27} m^{3}[/tex]

Formula for volume of a cube is [tex]a^{3}[/tex]. Therefore,

           Volume of the cube = [tex]4 \times 10^{-27} m^{3}[/tex]

As lattice constant (a) = [tex](4 \times 10^{-27} m^{3})^{\frac{1}{3}}[/tex]

                                   = [tex]1.59 \times 10^{-9} m[/tex]

Therefore, the value of lattice constant is [tex]1.59 \times 10^{-9} m[/tex].

And, for bcc unit cell the value of radius is as follows.

                 r = [tex]\frac{\sqrt{3}}{4}a[/tex]

Hence, effective radius of the atom is calculated as follows.

                 r = [tex]\frac{\sqrt{3}}{4}a[/tex]

                   = [tex]\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m[/tex]

                   = [tex]6.9 \times 10^{-10} m[/tex]

Hence, the value of effective radius of the atom is [tex]6.9 \times 10^{-10} m[/tex].

To calculate the concentration of Cl2O5 in the vessel 5.70 seconds later, use the first-order rate law equation and the given rate constant and initial concentration.

To calculate the concentration of Cl2O5 in the vessel 5.70 seconds later, we can use the first-order rate law equation.

The rate law equation for a first-order reaction is: ln([A]t/[A]0) = -kt

Where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.

In this case, we know the rate constant (k) is 0.184 s^-1 and the initial concentration ([A]0) is 1.16 M. We need to find the concentration at 5.70 seconds ([A]t).

Plugging in the values: ln([A]t/1.16) = -0.184 * 5.70

Solving for [A]t, we find that the concentration of Cl2O5 in the vessel 5.70 seconds later is approximately 0.64 M.

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Given data: Rate constant (k) = 0.184 s⁻¹

Initial concentration (Cl₂O₅) = 1.16 M

Time (t) = 5.70 s

To find the concentration of Cl₂O₅ after 5.70 seconds  

We know that the first-order integrated rate law equation is:

ln([A]t/[A]0) = -where [A]t and [A]0 are the concentrations of reactant at time 't' and initial time '0' respectively. We can rearrange this equation to find the concentration of the reactant at any time 't':[A]t = [A]0 × e^(-kt)Putting the values in the equation, we get:

[Cl₂O₅]5.70 = 1.16 M × e^(-0.184 s⁻¹ × 5.70 s)

[Cl₂O₅]5.70 = 0.501

Therefore, the concentration of Cl₂O₅ in the vessel 5.70 seconds later is 0.501 M.

Answer:

a) The atomic mass of the potassium is 54.23 amu.

b) The melting point of bromine gas is 6.3°C.

Explanation:

a) Atomic mass of Na =22.99 amu

Atomic mass of Rb = 85.47 amu

Döbereiner triad = Na , K ,Rb

Taking average of  atomic masses of  Na and Rb

Atomic mass of the K = [tex]\frac{22.99 amu+85.47 amu}{2}=54.23 amu[/tex]

The atomic mass of the potassium is 54.23 amu.

b) Melting point of chlorine gas =-101.0°C

Melting point of iodine gas =113.6°C

Döbereiner triad = Cl, Br , I

Melting point of bromine gas :

=[tex]\frac{-101.0^oC +113.6^oC}{2}=6.3^oC[/tex]

The melting point of chlorine gas is 6.3°C.

The full set of quantum numbers varies depending on the electron being considered in each element or ion, with the numbers consisting of the principal quantum number (n), angular momentum quantum number (l), magnetic quantum number (m_l), and spin quantum number (m_s).

Quantum numbers describe values of conserved quantities in the dynamics of a quantum system, which in this case are the electrons in various elements or ions.

Answer:

(a) F⁻(g) ⇒ F⁺(g)         ΔE=2009 KJ/mol

(b) Na⁺(g) ⇒ Na⁻(g)   ΔE=-548.6 KJ/mol

Explanation:

Hess's Law

"Energy changes for a reaction is independent of steps or route involved."

Now the conversion F⁻ to F⁺ is a two step reaction. In first step F⁻ loses an electron e⁻ to become neutral atom.

F⁻ ⇒ F + e⁻   ΔH=328 KJ/mol (i)

Kindly note this energy is derived from electron affinity, which is the energy changes while adding an electron to neutral atom and its same while removing an electron from ion.

In second step,

Another electron is removed from F.

F ⇒ F⁺ + e⁻    ΔH= 1681 KJ/mol (ii)

This energy is 1st ionization energy for F, which is the energy required to remove first electron from outer most shell of neutral atom.

Now the total energy change can be found by applying Hess's law and adding equations (i) and (ii);

F⁻ ⇒ F + e⁻   ΔH=328 KJ/mol

F ⇒ F⁺ + e⁻    ΔH= 1681 KJ/mol

F⁻ ⇒ F⁺ + 2e⁻  ΔH= 2009 KJ/mol

This reaction involves two steps in first Na⁺(g) gains electron and become neutral atom while in second step Na(g) accepts another electron to become Na⁻(g).

Na⁺(g) + e⁻⇒ Na(g)    ΔH=-495.8 KJ/mol   (i)

Na(g) + e⁻ ⇒ Na⁻(g)   ΔH=-52.8 KJ/mol     (ii)   (By Applying Hess's Law)

Na⁺(g) + 2e⁻ ⇒ Na⁻(g)  ΔH=-548.6 KJ/mol

Kindly note as there is no molar change in above mentioned reactions therefore enthalpy changes(ΔH) and internal energy changes(ΔE) are same.

Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰  

Explanation :

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom are determined by the electronic configuration.

Paramagnetic compounds : They have unpaired electrons.

Diamagnetic compounds : They have no unpaired electrons that means all are paired.

The given electron configurations of Palladium are:

(a) [Kr] 5s²4d⁸

In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.

(b) [Kr] 4d¹⁰

In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.

(c) [Kr] 5s¹4d⁹

In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.

The correct electron configuration for a diamagnetic substance like Palladium (Pd) is [Kr] 4d¹⁰. This is because diamagnetic substances like Palladium need to have all their electron orbitals fully filled, thereby having no unpaired electrons.

The subject of this question is about the electron configuration of Palladium (Pd; Z 46), which is a diamagnetic element. Diamagnetic substances have no unpaired electrons and are not attracted to a magnetic field. Thus, to be consistent with this fact, the electron configuration of Palladium must not have any unpaired electrons.

(a) [Kr] 5s²4d⁸: This configuration would imply there are unpaired electrons in the 4d orbital, which contradicts the fact that Palladium is diamagnetic.

(b) [Kr] 4d¹⁰: This configuration correctly states that all the orbitals are filled, including the 5s orbital before the 4d orbital. Therefore, the correct electron configuration for Palladium, a diamagnetic element, is [Kr] 4d¹⁰.

(c) [Kr] 5s¹4d⁹: The proposed configuration would also suggest an unpaired electron exists in the 4d orbital, which contradicts Palladium being diamagnetic.

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Answer:

We need 21.0 moles of O2

Explanation:

Step 1: Data given

Moles of propanoic acid = 6.0 moles

CH3CH2COOH = propanoic acid

Step 2: The balanced equation

2CH3CH2COOH + 7O2 → 6CO2 + 6H2O

Step 3 :Calculate moles O2

For 2 moles propanoic acid we need 7 moles O2 to produce 6 moles CO2 and 6 moles H2O

For 6.0 moles propanoic acid we need 6.0 * 3.5 = 21 moles O2

We need 21.0 moles of O2

To combust 6 moles of propanoic acid completely, 15 moles of [tex]O_{2}[/tex] are required, based on the balanced chemical equation for the combustion of propanoic acid. 2 [tex]CH_{3} CH_{2} COOH[/tex] + 5 [tex]O_{2}[/tex] → 6 [tex]CO_{2}[/tex] + 4 [tex]H_{2}O[/tex] is the balanced chemical equation.

The question asks for the amount of Oxygen ([tex]O_{2}[/tex]) needed for the complete combustion of propanoic acid ([tex]CH_{3} CH_{2} COOH[/tex]). The combustion of propanoic acid can be represented by a balanced chemical equation:

2 [tex]CH_{3} CH_{2} COOH[/tex] + 5 [tex]O_{2}[/tex] → 6 [tex]CO_{2}[/tex] + 4 [tex]H_{2}O[/tex]

This means that 2 moles of propanoic acid require 5 moles of [tex]O_{2}[/tex] for complete combustion. If we have 6 moles of propanoic acid, a simple stoichiometric calculation would be to multiply the amount of [tex]O_{2}[/tex] required for 2 moles of propanoic acid by 3 (since 6 moles is three times larger than 2 moles), resulting in:

5 moles [tex]O_{2}[/tex]/2 moles [tex]CH_{3} CH_{2} COOH[/tex] × 6 moles [tex]CH_{3} CH_{2} COOH[/tex] = 15 moles [tex]O_{2}[/tex]

Therefore, to combust 6 moles of propanoic acid completely, 15 moles of [tex]O_{2}[/tex] are required.

Answer:

Lithium and sodium are the most similar because they are both alkali elements located in the same group, and therefore have similar properties.

Nitrogen and oxygen are not the most similar because although they are both non metals elements, are each located in a different group

Explanation:

Li and Na are both alkali elements from group 1 that shares some similities. The both can be obtained by the  water hydrolysis. These are common reactions:

Metal from group 1 + H₂O → Base + H₂

Metal from group 1 + O₂ → oxides

Metal from group 1 + group 17 →  ionic halides

Both form cations with 1+ charge, they can release only 1 e-

N is an element from group 15 and O, from group 16. They are both non metal.

Nitrogen can make a variety of oxides.

They react in water to produce nitric acid:

N₂O₃ + H₂O → 2HNO₃

N₂O₅ + H₂O → 2HNO₃

It has an anion with -3, as oxidation state. (Nitride)

The N with H, makes a well known hidride → ammonia

N₂ + 3H₂ → 2NH₃

The Oxygen also makes a well known hidride → water

2H₂ + O₂ → 2H₂O

Both are covalent hidrides.

N can have many oxidation's states. O always acts with -2 except for the peroxydes, with -1. O can have a great power of oxidation, that N does not have.

O₂ always acts as a reactant, at combustion reactions.

Lithium and sodium are similar as they are both alkali elements, found in the same group of the periodic table, leading to similar properties. Nitrogen and oxygen, while both nonmetals, are in separate groups, yielding different properties.

a. Lithium and sodium are the most similar because they are both alkali elements located in the same group, and therefore have similar properties. Lithium and sodium both belong to the alkali metals group in the periodic table. They share similar chemical behaviors, as they have only one electron in a valence s subshell outside a filled set of inner shells. This fact influences their reactivity and the compounds they can form.

b. Nitrogen and oxygen are not the most similar because although they are both nonmetal elements, they are each located in a different group. Even though nitrogen and oxygen are close to each other in the periodic table, they do not share the same group, hence they have different properties and different numbers of valence electrons.

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Answer:

b. Pepsin

Explanation:

In the stomach, when dygestion takes place,pepsinogen is secreted  in the stomach for the dygestion of proteins, which are catalyzed to aminoacids for the use of the body.

When the pepsinogen gets in contact with the HCl it converts to pepsin, which is the actived form of the pepsinogen.

Pepsinogen is converted into its active form, called Pepsin, under acidic conditions rendered by Hydrochloric acid (HCl). Pepsin, once activated, aids in protein digestion.

Pepsinogen, an inactive protease, is secreted by the chief cells located in the stomach. Under the acidic condition produced by the presence of Hydrochloric acid (HCl), pepsinogen is converted into its active form, known as Pepsin. Pepsin plays an integral role in the process of digestion, specifically assisting in the breakdown of proteins in the stomach.

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