The reaction C4H10 ---> C2H6 + C2H4 has activation energy (Ea) of 450 kJ/mol, and the Ea of the reverse reaction is 250 kJ/mol. Estimate ΔH, in kJ/mol, for the reaction as written above.

Answer:

i) The pCa before initiating the titration is 2.6

ii) The pCa is 6.67

Explanation:

please look at the solution in the attached Word file

Answer:

[tex]V=22.4L[/tex]

Explanation:

Hello,

In this case, considering the ideal gas equation:

[tex]PV=nRT[/tex]

It is possible to compute the volume the gas would have for the given STP conditions as:

[tex]V=\frac{nRT}{P}[/tex]

[tex]V=\frac{1mol*0.082\frac{atm*L}{mol*K}*273K}{1atm}\\\\V=22.4L[/tex]

Which correspond to the standard volume as well.

Best regards.

Answer:

The volume of the balloon would be 22.386 L

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

Replacing:

1 atm* V= 1 mol* 0.082 [tex]\frac{atm*L}{mol*K}[/tex]*273 K

Solving:

[tex]V=\frac{1 mol*0.082\frac{atm*L}{mol*K} *273 K}{1 atm}[/tex]

V=22.386 L

The volume of the balloon would be 22.386 L

Answer:

pH = 6.82

Explanation:

To solve this problem we can use the Henderson-Hasselbach equation:

We're given all the required data to calculate the original pH of the buffer before 0.341 mol of HCl are added:

By adding HCl, we simultaneously increase the number of HOCl and decrease NaOCl:

Answer:

Explanation:

find the solution below

Answer:

[tex]p_2=347.9kPa[/tex]

Explanation:

Hello,

In this case, we use the Gay-Lussac's law which allows us to understand a gas' pressure-temperature behavior as a directly proportional relationship:

[tex]\frac{p_1}{T_1}= \frac{p_2}{T_2}[/tex]

Whereas it is convenient to use the pressure in kPa and the temperature in kelvins in order to compute the required resulting pressure, therefore:

[tex]p_1=55.0psi*\frac{6.89476kPa}{1psi} =379.2kPa\\T_1=30.0+273.15=303.15K\\T_2=5.0+273.15=278.15K[/tex]

Thus, we obtain:

[tex]p_2= \frac{p_1T_2}{T_1}=\frac{379.2kPa*278.15K}{303.15K}\\ \\p_2=347.9kPa[/tex]

Best regards.

Answer:

The pressure in the tire at 5.0 °C is 347.91 kPa

Explanation:

Step 1: Data given

The pressure in a bicycle tire is 55.0 psi

Temperature = 30.0 °C = 303 K

Temperature decreases to 5.0 °C = 278 K

Volume and Amount of moles are held constant

Step 2: Calculate the pressure at the new temperature

P1/T1 = P2 / T2

⇒with P1 = the initial pressure of the bicycle tire is 55.0 psi

⇒with T1 = the initial temeprature = 303 K

⇒with P2 = the pressure at the new temperature

⇒with T2 = the decreased temperature = 278 K

55.0 psi / 303 K = P2 / 278 K

P2 = (55.0 psi / 303 K) * 278 K

P2 = 50.46 psi

Step 3: Convert pressure from psi to kPa

50.46 psi = 50.46 * 6.895 = 347.91 kPa

The pressure in the tire at 5.0 °C is 347.91 kPa

Answer: New concentration of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] is 0.23 M.

Explanation:

The given data is as follows.

     Moles of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] = 0.06 mol

     Moles of [tex]H_{2}C_{6}H_{5}O_{7}[/tex] = 0.08 mol

Therefore, moles of [tex]OH^{-}[/tex] added are as follows.

    Moles of [tex]OH^{-}[/tex] = [tex]0.125 \times \frac{25}{1000}[/tex]

                          = 0.003125 mol

Now, new moles of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] = 0.06 + 0.003125

                     = 0.063125

Therefore, new concentration of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] will be calculated as follows.

       Concentration = [tex]\frac{0.063125}{0.275}[/tex]

                               = 0.23 M

Thus, we can conclude that new concentration of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] is 0.23 M.

Answer:

2.3 ev or 3.68 ×10^-19J

Explanation:

The spectrum is shown in the image attached

h= 4.136 × 10-15 eV⋅ s

c = 2.998 × 108 m/s

λmax= 550×10^-9 (from the spectrum attached)

E=hc/λmax

E= 4.136 × 10^-15 × 2.998 × 10^8/550×10^-9

E= 2.3 ev or 3.68 ×10^-19J

The energy change for the electronic excitation is :  3.68 * 10⁻¹⁹J

Given data :

h = 4.136 * 10⁻¹⁵ eV⋅ s

c = 2.998 * 108 m/s

λmax = 550 * 10⁻⁹  ( Obtained from image attached below )

Applying the energy and wavelength relationship equation

E = hc / λmax

  = ( 4.136 * 10⁻¹⁵  * 2.998 * 108  ) /   550 * 10⁻⁹

  =  2.3 ev   ≈ 3.68 * 10⁻¹⁹J.

Hence we can conclude that the energy change for the electronic excitation is 2.3 ev   ≈ 3.68 * 10⁻¹⁹J.

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Missing data related to your question is attached below

Answer:

The solubility of those salts increases which contains conjugate of weak acid. Conjugate of weak acid refers to strong base such as sodium hydroxide and potassium hydroxide etc.

Explanation:

The solubility of salts in strong acidic solution increases due to the presence of conjugate of weak acid which is actually a strong base. So if the salts contain strong base, it readily react with strong acid that is present in the solution.

Answer:

see explaination

Explanation:

We are given the (R)-3-bromo-2,3-dimethylpentane and asking to draw the curved arrow which is the showing the mechanism for first-order substitution and first-order elimination reactions. We know the formation of carbocation is the rate determining step in the first-order substitution and first-order elimination reactions.

So in the (R)-3-bromo-2,3-dimethylpentane there is –Br gets removed and formed the tertiary carbocation which is more stable, so the curved arrows in Box 1 to depict the flow of electrons and intermediate in Box 2.

Check attachment

Answer:

By mixing 2 Moles of Sodium Metal with 2 Moles of nitric acid, Sodium Nitrate is formed.

Explanation:

2Na+   +  HNO3   produces    2NaNO3 + H2

To find the value of ΔG for the reaction MSO4⋅3H2O(s)↽−−⇀MSO4(s)+3H2O(g) at the given vapor pressure of water, use the equation ΔG = -RTln(K), where ΔG is the change in Gibbs free energy, R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant.

The reaction can be represented as: MSO4·3H2O(s) <--> MSO4(s) + 3H2O(g)

The equilibrium vapor pressure of water above the solid is 14.7 Torr.

Since we are given the equilibrium condition, we can use the equation ΔG = -RTln(K), where ΔG is the change in Gibbs free energy, R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant.

By plugging in the given values, we can calculate the value of ΔG at the equilibrium condition.

The value of [tex]\Delta G[/tex] for the reaction when the vapor pressure of water is 14.7 Torr is approximately 2.005 kJ/mol.

The value of [tex]\Delta G[/tex] for the reaction at 298 K when the vapor pressure of water is 14.7 Torr is given by the equation:

[tex]\[ \Delta G = -RT \ln \left( \frac{P_{H_2O}^3}{P_{H_2O}^{eq}} \right) \][/tex]

where:

- [tex]\( \Delta G \)[/tex] is the change in Gibbs free energy,

- [tex]R[/tex] is the universal gas constant (8.314 J/(mol·K)),

- [tex]T[/tex] is the temperature in Kelvin (298 K in this case),

- \[tex]\( P_{H_2O}^3 \)[/tex] is the partial pressure of water raised to the power of the moles of water in the reaction (which is 3),

- [tex]\( P_{H_2O}^{eq} \)[/tex] is the equilibrium vapor pressure of water (14.7 Torr).

First, we need to convert the equilibrium vapor pressure of water from Torr to atmospheres to match the units of the gas constant [tex]R[/tex]. The conversion factor is 1 atm = 760 Torr.

[tex]\[ P_{H_2O}^{eq} = \frac{14.7 \text{ Torr}}{760 \text{ Torr/atm}} = 0.01934 \text{ atm} \][/tex]

Now we can plug in the values into the equation:

[tex]\[ \Delta G = -(8.314 \text{ J/(mol·K)}) \times (298 \text{ K}) \times \ln \left( \frac{(0.01934 \text{ atm})^3}{(0.01934 \text{ atm})} \right) \][/tex]

[tex]\[ \Delta G = -8.314 \times 298 \times \ln \left( (0.01934)^2 \right) \][/tex]

[tex]\[ \Delta G = -8.314 \times 298 \times \ln \left( 0.000373 \right) \][/tex]

[tex]\[ \Delta G = -8.314 \times 298 \times (-7.936) \][/tex]

[tex]\[ \Delta G = 8.314 \times 298 \times 7.936 \][/tex]

[tex]\[ \Delta G = 2004.8 \text{ J/mol} \][/tex]

[tex]\Delta G = 2.005 kJ/mol[/tex]

Answer:

Ca(NO2)2

Explanation:

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution.

Answer:

HF + OH- = F- + H2O

Explanation:

Since hydrofluoric acid does not ionize in aqueous solution, the fluoride ion is still present as part of the product

Answer:

H+ (aq) + OH-(aq) → H2O(l)

Explanation:

Step 1: Data given

Volume of hydrofluoric acid = 15.0 mL = 0.015 L

Molarity = 1.25 M

Molarity of KOH = 3.05 M

Step 2: The unbalanced equation

HF(aq) + KOH(aq) → KF(aq) + H2O(l)

This equation is already balanced

Step 3: The net ionic equation

The net ionic equation  shows only those elements, compounds, and ions that are directly involved in the chemical reaction.

The elements, compounds, and ions that do not take part in the chemical reaction are called spector ions.

H+ (aq) + F-(aq) + K+(aq) +  OH-(aq)→ K+(aq) +F-(aq) + H2O(l)

We'll remove all the spector ions.

H+ (aq) + OH-(aq) → H2O(l)

Answer:

a. Minimum 1.70 V

b. There is no maximum.

Explanation:

We can solve this question by remembering that the cell potential is given by the formula

ε⁰ cell = ε⁰ reduction -  ε⁰  oxidation

Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the  oxidized species  0.80 V, thus

ε⁰ reduction -  ε⁰  oxidation ≥  ε⁰ cell

Since ε⁰  oxidation is by definition the negative of ε⁰ reduction , we have

ε⁰ reduction - ( 0.80 V )  ≥  0.90 V

⇒ ε⁰ reduction  ≥ 1.70 V

Therefore,

(a) The minimum standard reduction potential is 1.70 V

(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V

Answer:

36.5 mol

Explanation:

The vapor pressure of a solution of a non volatile solute in  water is given by Raoult´s law:

P H₂O = χ H₂O x P⁰ H₂O

where  χ H₂O  is the mole fraction of water in the solution and P⁰ H₂O

In the turn the mole fraction is given by

χ H₂O = mol H₂O / total # moles = mol H₂O /ntot

Thus

P H₂O = mole H₂O / n tot  x   P⁰ H₂0

now the mol of H₂O is equal n tot - 6  mol solute

Plugging the values given in the question and  solving the resultant equation

19.8 torr = ( ntot - 6 ) x 23.7 torr / n tot

19.8 ntot  = 23.7 ntot - 142.2

ntot = 36.5 ( rounded to 3 significant figures )

If the standard reduction potential of a half-cell is positive, the reduction reaction is spontaneous when paired with a hydrogen electrode

Explanation:

Answer:

Reduction

Explanation:

Answer:

The decrease in free energy is 113.299kJ

Explanation:

K for enzyme catalyzed reaction = 500s^-1

Temperature (T) =298k

ΔG =?

ΔG = - 2.303 RT log k

ΔG = (-2.303)(8.314)(298) log 500

ΔG = - 15399.9 J

ΔG catalyzed = - 15. 399kJ

The first order reaction is given as:

t1/2= 0.693/k

or k= 0.693/t1/2

0.693/10^17

Therefore,

K= 0.693 × 10^-17

Now,

K= 0.693 × 10^-17

T= 298k

ΔG uncatalyzed =?

ΔG uncatalyzed = - 2.303 RT log k

ΔG uncatalyzed = (-2.303)(8.314)(298) log0.693 × 10^-17

= 97908.1J

ΔG uncatalyzed = 97.9081kJ

Therefore,

The decrease in free energy is:

ΔG uncatalyzed - ΔG catalyzed

97.908 - (-15.399)

= 113.299KJ

The decrease in free energy is 113.299kJ

Final answer:

The free energy of activation for the CO2 loss step can be calculated using the Arrhenius equation. The enzyme lowers the free energy of activation by comparing the activation energies of the catalyzed and uncatalyzed reactions.

Explanation:

The free energy of activation for the CO2 loss step can be calculated using the Arrhenius equation:

k = Ae^(-Ea/RT)

Where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin.

Since the rate-determining step is the loss of CO2, we can use the kcat value (500 s-1) as the rate constant for this step. To find the activation energy, we need to rearrange the Arrhenius equation:

Ea = -RT ln(k/A)

Now we can substitute the given values into the equation:

Ea = -(8.314 J/mol·K)(298 K) ln(500 s-1/A)

To calculate the value of A, we can use the half-life for the uncatalyzed reaction:

t1/2 = ln(2)/(kuncat)

Replacing kuncat with the appropriate value, we can solve for A:

A = e^(ln(2)/(kuncat) - ln(2)/(kcat))/t1/2 = e^(ln(kcat/kuncat))/t1/2

Finally, we can substitute the values of kcat, kuncat, and t1/2 into the equation to find A.

To calculate how much the enzyme lowers the free energy of activation, we can compare the activation energies of the uncatalyzed and catalyzed reactions:

∆∆G (ΔEa) = ∆Ga - ∆Ga,uncat

Where ∆Ga is the activation energy of the catalyzed reaction and ∆Ga,uncat is the activation energy of the uncatalyzed reaction.

Answer:

28 grams CaH₂(s) is required for production of 15L H₂(g) at 25°C and 825Torr.

Explanation:

CaH₂(s) + H₂O(l) => Ca(OH)₂(s) + H₂(g)

Using ideal gas law, PV = nRT

=> moles H₂(g) = PV/RT = [(825/760)Atm](15L)/(0.08206L·Atm·mol⁻¹·K⁻¹)(298K) = 0.6659 mol H₂(g)

From stoichiometry of given equation,        

=> 0.6659 mol H₂(g) requires 0.6659 mole CaH₂(s)

Converting moles to grams, multiply by formula weight,

=> 0.6659 mole CaH₂(s) = 0.6659 mole CaH₂(s) x 42g/mole = 27.966 grams CaH₂(s) ≅ 28 grams CaH₂(s)  (2 sig. figs.)

Answer:

x1= 4.5 × 10^-3, y1= 0.9

Explanation:

A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2

Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:

1. The vapor phase is ideal at pressure of 1 bar

2. Henry's law apply to dilute solution only.

3. Raoult's law apply to concentrated solution only.

Where,

Henry's constant for species 1 H= 200bar

Saturation vapor pressure of species 2, P2sat= 0.10bar

Temperature = 25°C= 298.15k

Apply Henry's law for species 1

y1P= H1x1...... equation 1

y1= mole fraction of species 1 in vapor phase.

P= Total pressure of the system

x1= mole fraction of species 1 in liquid phase.

Apply Raoult's law for species 2

y2P= P2satx2...... equation 2

From the 2 equations above

P=H1x1 + P2satx2

200bar= H1

0.10= P2sat

1 bar= P

Hence,

P=H1x1 + (1 - x1) P2sat

1bar= 200bar × x1 + (1 - x1) 0.10bar

x1= 4.5 × 10^-3

The mole fraction of species 1 in liquid phase is 4.5 × 10^-3

To get y, substitute x1=4.5 × 10^-3 in equation 1

y × 1 bar = 200bar × 4.5 × 10^-3

y1= 0.9

The mole fraction of species 1 in vapor phase is 0.9

What is Binary solution?

A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2

If we consider the pressure of the 2 phase system is 1 bar.

The assumption are as follows:

Where, these values are given:

Henry's constant for species 1 H= 200bar

Temperature = 25°C= 298.15K

P₂sat= 0.10 bar

Apply Henry's law for species 1

y₁P= H₁x₁.......... (i)

where y₁= mole fraction of species 1 in vapor phase, P= Total pressure of the system  ,x₁= mole fraction of species 1 in liquid phase.

Apply Raoult's law for species 2

y₂P= P₂sat. x₂...........(ii)

From (i) and (ii)

P=H₁x₁ + P₂sat. x₂

200bar= H₁

0.10= P₂sat

1 bar= P

Hence,

P=H₁x₁ + (1 - x₁) P₂sat

1bar= 200bar × x₁ + (1 - x₁) 0.10bar

x₁= [tex]4.5 * 10^{-3}[/tex]

The mole fraction of species 1 in liquid phase is [tex]4.5 * 10^{-3}[/tex]

To get y, substitute x₁=[tex]4.5 * 10^{-3}[/tex] in (i)

y × 1 bar = 200bar × [tex]4.5 * 10^{-3}[/tex]

y₁= 0.9

The mole fraction of species 1 in vapor phase is 0.9.

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Answer:

pC6H5CHO = 0.180 atm

Fraction dissociated = 0.756

Explanation:

Step 1: Data given

Temperature = 523 K

the value of its equilibrium constant is K = 0.558

Mass of benzyl alcohol = 1.20 grams

Molar mass of benzyl alcohol = 108.14 g/mol

Volume = 2.00 L

heated to 523 K

Step 2: The balanced equation

C6H5CH2OH(g) ⇆ C6H5CHO(g) + H2(g)

Step 3: Calculate moles benzyl alcohol

Moles benzyl alcohol = Mass / molar mass

Moles benzyl alcohol = 1.20 grams / 108.14 g/mol

Moles benzyl alcC6H5CH2OHohol = 0.0111 moles

Step 4: Initial moles

Moles C6H5CH2OH = 0.0111 moles

Moles C6H5CHO = 0 moles

Moles H2O = 0 moles

Step 5:  moles at the equilibrium

Moles C6H5CH2OH = 0.0111 - X moles

Moles C6H5CHO = X moles

Moles H2O = X moles

Step 6: Calculate the total number of moles at equilibrium

Total number of moles = (0.0111 - X moles) + X moles + X moles

Total number of moles = 0.0111 + X moles

Step 7: Calculate the total pressure at the equilibrium

p*V = n*R*T

p = (n*R*T) / V

⇒with p = the total pressure at the equilibrium = TO BE DETERMINED

⇒with n = the total number of moles = 0.0111 + X moles

⇒with R = the gas constant = 0.08206 L*atm / mol * K

⇒with T = the temperature = 523 K

⇒with V = the volume of the vessel = 2.00 L

p = (0.0111 - X moles ) * 0.08206*523 / 2.00

p = 21.46(0.0111 - X moles)

Step 8: Define the equilibrium constant K

K = 0.558 =  (pC6H5CHO)*(pH2) / (pC6H5CH2OH)

0.558 = (X / (0.0111 + X)*P)²  /  ((0.0111-X)/(0.0111+X)*P)

0.558 = (X²(21.46 * (0.0111+X))) / ((0.0111 + X) (0.0111-X))

X = 0.00839

Step 9: Calculate the equilibrium partial pressure

pC6H5CHO = X / (0.0111 + X)  * (21.46 * (0.0111 +X))

pC6H5CHO = 0.180 atm

Step 10: What fraction of benzyl alcohol is dissociated into products at equilibrium?

Fraction dissociated = Δn / n°

Fraction dissociated = X / 0.0111

Fraction dissociated = 0.00839 / 0.0111

Fraction dissociated = 0.756

Answer:

Local fishermen will lose their jobs

Answer:

See explanation

Explanation:

The central atom in the perbromate ion is bromine. The chemical symbol of bromine is Br. There are no lone pairs around the central bromine atom. The ion is tetrahedral in shape hence we expect a bond angle of 109°. 27 which is the ideal tetrahedral bond angle. The actual bond angle of the prebromate ion is 109.5°. The perbromate ion is BrO4^-

The observed bond angle is very close to the ideal value because of the absence of lone pairs of electrons from the central atom in the ion.

The perbromate anion, BrO4-, has Bromine (Br) as its central atom and two lone pairs of electrons. This configuration results in a square planar molecular structure, presenting ideal Bromine-Oxygen bond angles of 90° and 180°. These angles are expected to be virtually accurate due to minimization of lone pair-bonding pair repulsions.

The perbromate anion, represented by the chemical formula BrO4-, has bromine (Br) as its central atom. Based on the octet rule, the central bromine atom is surrounded by four oxygen atoms and has two lone pairs of electrons. Given this arrangement, the perbromate anion exhibits an octahedral electron-pair geometry, but due to the presence of the two lone pairs, its molecular structure is square planar. The ideal angle between the Bromine-Oxygen bonds in a square planar structure is 90° or 180°. Since the lone pairs occupy the positions minimizing their interactions with the bonded oxygen atoms, the actual angle in the perbromate anion is expected to closely match this ideal angle.

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Answer: Concentration or pressure of a reactant and temperature

Explanation: The two major factors that can affect the rate of a chemical change are concentration or pressure of a reactant  and temperature.

Answer:

B) The molecules gain energy, and their relative motion decreases

The change in the molecular motion should be option B.

The molecular motion refers to the movement of constituent particles or molecules in a specific direction. It should be impacted by heat and temperature. When there is the transformation from liquid to solid so the change in the molecular motion of the substance should be that the molecular gained the energy and there should be a decrease in the relative motion.

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Question:

Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.

Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

Answer:

D) 85 J/K

E) - 50 J/K

F) 62.5 J/K

G) 12.5 J/K

Explanation:

Let's make use of the entropy equation: ΔS = [tex] \frac{Q}{T} [/tex]

Part D)

Given:

T = 20°C = 20 +273 = 293K

Q = 25.0 kJ

Entropy change will be:

ΔS = [tex] \frac{25*1000}{293} [/tex]

= 85 J/K

Part E)

Given:

T = 500K

Q = -25.0 kJ

Entropy change will be:

ΔS = [tex] \frac{-25*1000}{500} [/tex]

= - 50 J/K

Part F)

Given:

T = 400K

Q = 25.0 kJ

Entropy change will be:

ΔS = [tex] \frac{25*1000}{400} [/tex]

= 62.5 J/K

Part G:

Given:

T1 = 400K

T2 = 500K

Q = 25.0 kJ

The net entropy change will be:

ΔS = [tex] (\frac{25*1000}{400}) + (\frac{-25*1000}{500}[/tex]

= 12.5 J/K

Answer:

A) The change in entropy [tex]\delta S = 0.085J/K[/tex]

B) The change in entropy [tex]\delta S = -50J/K[/tex]

C) The change in entropy [tex]\delta S = 62.5J/K[/tex]

D) The net change in entropy [tex]\delta S = 12.5J/K[/tex]

Explanation:

A)

[tex]T = 20^oC + 273k\\\\T = 293k[/tex]

expression for change in entropy,

[tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{25}{293}\\\\\delta S = 0.085J/K[/tex]

B) [tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{-25*10^3}{500}\\\\\delta S = -50J/K[/tex]

The negative sign indicates the heat lost into the surrounding

C)

[tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{25*10^3}{400}\\\\\delta S = 62.5J/K[/tex]

The entropy remains constant in the adiabatic process because no heat is given to the system in this process.

D)

[tex]\delta S_1 = \frac{\delta Q_1}{T_1}\\\\\delta S_1 = \frac{25*10^3}{400}\\\\\delta S_1 = 62.5J/K[/tex]

similarly,

[tex]\delta S_2 = \frac{\delta Q_2}{T_2}\\\\\delta S_2 = \frac{25*10^3}{500}\\\\\delta S_2 = 50J/K[/tex]

Therefore the net change in entropy is,

[tex]\delta S = \delta S_1 - \delta S_2\\\\\delta S = 62.5 - 50\\\\\delta S = 12.5J/K[/tex]

Entropy is not a conserved quantity because it can be created but cannot be destroyed.

The net change in entropy is calculated by difference of the change in entropy at two different temperatures.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  [tex]E_{cell}__{AC}} = 0.94[/tex]

Explanation:

  From the question we are told that

          the cell voltage for AD is  [tex]E_{cell}__{AD}} = 1.56V[/tex]

From the data give we can see that

               [tex]E_{cell}__{AD}} - E_{cell}__{BD}} = E_{cell}__{AB}}[/tex]

i.e           [tex]1.56 - 1.53 = 0.03[/tex]

   In the same way we can say that

              [tex]E_{cell}__{AD}}-E_{cell}__{CD}} = E_{cell}__{AC}}[/tex]

=>        [tex]E_{cell}__{AC}}=1.56- 0.62[/tex]

                       [tex]E_{cell}__{AC}} = 0.94[/tex]

Galvanic cells are the voltaic cells that generate an electric current from redox reactions. The cell potential of cells A and C will be 0.94 volts.

The cell potential is the estimation of the gained or lost electrons by the species on the electrode of the electrochemical cell.

Given, the cell voltage for cells A and D is 1.56 V.

Cell potential between A and B cells is calculated as:

[tex]\begin{aligned} \rm E_{cell}_{(AB)} &= \rm E_{cell}_{(AD)} - \rm E_{cell}_{(BD)}\\\\&= 1.56 - 1.53\\\\&= 0.03 \end{aligned}[/tex]

Similarly, cell potential between A and C cells is calculated as:

[tex]\begin{aligned} \rm E_{cell}_{(AC)} &= \rm E_{cell}_{(AD)} - \rm E_{cell}_{(CD)}\\\\&= 1.56 - 0.62\\\\&= 0.94 \end{aligned}[/tex]

Therefore, the cell potential between A and C cell is 0.94 volts.

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Answer:

Explanation:

1s²2s²2p⁶3s² = magnesium ionization energy, 0.738 MJ/mol atomic radius 160 pm

1s²2s²2p⁶3s¹ = Sodium, ionization energy, 0.496 MJ/mol, atomic radius 186 pm

1s²2s²2p⁵ = Florine ionization energy, 1.681 MJ/mol, atomic radius 64 pm

Ionization energy is the minimum amount of energy needed to remove an electron from its orbital around the atom from the influence of the atom while atomic radius is one-half the distance between the nuclei of identical atoms that are bonded together.

Answer:

"Hawaii has better weather" is an opinion.

Therefore, you would need to support it with facts in order to deem it accurate or not.

If the text suggests that Hawaii has nice weather, then the statement would be accurate.

If the text hints that Hawaii does not have ideal weather, the statement would be inaccurate.

Final answer:

The statement that 'Hawaii has better weather' is subjective; Hawaii has a tropical type A climate with wet and dry areas due to the rain shadow effect. While it's warm, Kauai receives over 460 inches of rain annually, and snow can occur on mountain peaks in winter.

Explanation:

Whether or not Hawaii has 'better weather' is subjective and depends on personal preferences. However, based on the text provided, the statement that Hawaii has better weather because it has a tropical type A climate may not be entirely accurate for everyone. While Hawaii does have a warm and tropical climate, there are variations across the islands. For example, the island of Kauai is one of the wettest places on Earth, receiving over 460 inches of rain per year. Moreover, the rain shadow effect caused by Mount Wai'ale'ale leads to heavy rainfall on the windward side and dry conditions on the leeward side, creating a semi-desert environment. Additionally, it's noteworthy that snow can be found on the tops of Hawaii's highest mountains in the winter.

The diverse climate conditions in Hawaii mean that the weather can vary significantly from one part of the island to another, which might be pleasant for some but not for others. Therefore, whether Hawaii has 'better weather' is based on an individual's weather preferences and what they consider to be better.

This is an incomplete question, here is a complete question.

Consider the reaction: [tex]NO_2(g)+CO(g)\rightleftharpoons NO(g)+CO_2(g)[/tex]

Kc = 0.30 at some temperature.

If the initial mixture has the concentrations below, the system is_______.

Chemicals   Concentration (mol/L)

- NO₂            0.024

- CO               0.360

- NO               0.180

- CO₂             0.120

Possible answers:

1) not at equilibrium and will remain in an unequilibrated state.

2) not at equilibrium and will shift to the left to achieve an equilibrium state.

3) not at equilibrium and will shift to the right to achieve an equilibrium state.

4) at equilibrium

Answer : The correct option is, (2) not at equilibrium and will shift to the left to achieve an equilibrium state.

Explanation:

Reaction quotient (Qc) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

First we have to determine the value of reaction quotient (Qc).

The given balanced chemical reaction is,

[tex]NO_2(g)+CO(g)\rightleftharpoons NO(g)+CO_2(g)[/tex]

The expression for reaction quotient will be :

[tex]Q_c=\frac{[NO][CO_2]}{[NO_2][CO]}[/tex]

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

[tex]Q_c=\frac{(0.180)\times (0.120)}{(0.024)\times (0.360)}=2.5[/tex]

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When [tex]Q>K[/tex] that means product > reactant. So, the reaction is reactant favored.

When [tex]Q<K[/tex] that means reactant > product. So, the reaction is product favored.

When [tex]Q=K[/tex] that means product = reactant. So, the reaction is in equilibrium.

The given equilibrium constant value is, [tex]K_c=0.30[/tex]

From the above we conclude that, the [tex]Q>K[/tex] that means reactant < product. So, the reaction is reactant favored that means reaction must shift to the reactant or left to be in equilibrium.

Hence, the correct option is, (2) not at equilibrium and will shift to the left to achieve an equilibrium state.