What are the reactions that allow the conversion of cytosolic NADHNADH into NADPHNADPH during fatty acid biosynthesis? malate+NADP+⟶pyruvate+CO2+NADPHmalate+NADP+⟶pyruvate+CO2+NADPH glucose 6-phosphate+2NADP++H2O⟶ribulose 5-phosphate+2NADPH+2H++CO2glucose 6-phosphate+2NADP++H2O⟶ribulose 5-phosphate+2NADPH+2H++CO2 oxaloacetate+NADH+H+↽−−⇀malate+NAD+oxaloacetate+NADH+H+↽−−⇀malate+NAD+ pyruvate+CO2+ATP+H2O⟶oxaloacetate+ADP+Pi+2H+pyruvate+CO2+ATP+H2O⟶oxaloacetate+ADP+Pi+2H+ What enzymes are required? malic enzyme pyruvate carboxylase glucose 6‑phosphate dehydrogenase malate dehydrogenase What is the sum of these reactions?

Answer:

C

Explanation:

Well, we can see from the equation that hydrogen gas and ethane are in a 2:1 ratio based on their mole coefficients.

You need 2 mol H2 to make 1 mol C2H6. So that means you need 2*13.78 mol H2 to make 13.78 mol C2H6!

2*13.78 = 27.56 mol

C is correct.

Answer:

CHC13

Explanation:

A scientific question may lead to a(n)

The answer is may lead to a hypothesis.

Explanation:

Answer: observations

Hypothesis

Experimentation

Explanation:in order and right on edge

Answer:

A) No , because Qsp<Ksp for calcium fluoride.

Check attachment for calculation

Without specific calculations, to determine if a precipitate forms when mixing calcium chlorate with sodium fluoride, one assesses if the product of the ion concentrations exceeds the Ksp for calcium fluoride; if Qsp > Ksp, precipitation occurs.

To determine if a precipitate will form when calcium chlorate is mixed with sodium fluoride, one must consider the ionic products formed and compare the reaction quotient (Qsp) with the solubility product constant (Ksp) of the relevant compound, in this case, calcium fluoride (CaF2). The Ksp for calcium fluoride is 3.45 × 10∑11. In a mixture of calcium chlorate and sodium fluoride, the calcium ions (Ca2+) from calcium chlorate can react with the fluoride ions (F∑) from sodium fluoride to potentially form solid calcium fluoride. However, without the exact concentrations of Ca2+ and F∑ after mixing, one cannot directly calculate Qsp. The principle, however, is that if Qsp > Ksp, a precipitate of calcium fluoride will form, whereas if Qsp < Ksp, no precipitate forms because the solution is not supersaturated with respect to CaF2.

The concentration of each ion in the final solution depends on the dilution that occurs when the two solutions are mixed. Given the initial conditions and the principle that the formation of a precipitate requires Qsp to exceed Ksp, we generally assess the likelihood of precipitation by considering the concentrations of the reactive ions and the solubility product of the insoluble compound they may form. In this scenario, without performing a specific calculation, the correct answer would be based on understanding whether the mixing leads to a situation where the product of the concentrations of calcium and fluoride ions exceeds the Ksp of calcium fluoride.

Answer:

0.0075 milliliters (7.5 microliters) will be taken from the stock solution and then diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.

Explanation:

This is a problem of dilution using the equation:

initial concentration x initial volume = final concentration x final volume.

The final volume to be prepared is 25 microliters.

The final concentration to be prepared is 3 M.

The initial volume to be taken is not known yet.

The initial concentration is 10 M.

Now, let's substitute these parameters into the the equation above.

10 x initial volume = 3 x 25

Initial volume = 3 x 25/10

     = 7.5 microliters

Note that: 1 microliter = 0.001 milliliters

Hence,

7.5 microliters = 0.0075 milliliters

This means that an initial volume of 0.0075 milliliters (7.5 microliters) will be taken from the stock solution. This amount will then be diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.

Answer:

1670 ml

Explanation:

molarity x Volume (Liters) = moles => Volume (Liters) = moles/Molarity

Volume needed = 2.50mol/1.50M = 1.67 Liters = 1670 ml.

Answer:

hes right

Explanation:

ik all

Answer: [H3O+]= 0.05 M

Explanation:

Message me for extra explanation.

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In a chemical reaction between sulfur and oxygen that produces sulfur dioxide, there is a 1:1 ratio between sulfur and sulfur dioxide. Using the molar mass of sulfur and sulfur dioxide, it is calculated that to produce 8g of sulfur dioxide, 4g of sulfur needs to be burned.

To calculate the mass of sulfur that must be burned to give 8g of sulfur dioxide, we first need to understand mole ratio and molar mass, since the question is about a chemical reaction described by a balanced chemical equation. The balanced equation is: S + O2 --> SO2.

The molar masses are: S (32.06 g/mol), O2 (32 g/mol, because there are 2 O atoms each 16 g/mol), and SO2 (64.06 g/mol, which sum the molar masses of S and O2).

From the balanced chemical equation, we see a 1:1 ratio exists between sulfur and sulfur dioxide. So, the number of moles in 8g of SO2 would be the mass divided by the molar mass: 8g SO2 * (1 mol SO2/64.06 g SO2) = 0.125 mol SO2.

Because of the 1:1 ratio between sulfur and sulfur dioxide, 0.125 mole of sulfur is needed. Convert this to grams, we have: 0.125 mol S * (32.06 g S/1 mol S) = 4 g of sulfur

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Answer:

Check the explanation

Explanation:

Kindly check the attached image below for the step by step explanation to the question above.

Answer:1.587J / g degC.

Explanation:SOLUTION

Specific heat of aluminum = 0.897 J / g degC.

Heat gained by calorimeter = 100 * 10 * 0.897 J = 897 J

Specific heat of water = 4.18 J / g degC.

Heat gained by water = 250 * 10 * 4.18 J = 10450 J

Specific heat of copper = 0.385 J / g degC.

Heat loss by copper block = 50 * (80-10) * 0.385 J =1347.5  J

Let specific heat of unknown block be x J / g degC.

Heat loss by unknown block = 70 * (100-10) * x J = 6300x J

Heat gain = Heat loss

897 + 10450 = 1347.5 + 6300x

6300x = 9999.5

x =1.58722= 1.587

Specific heat of unknown substance is 1.587J / g degC.

Answer:

4.0 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 12L

Pressure = 5.6 atm

Temperature (T) = 205K

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow

PV = nRT

5.6 x 12 = n x 0.08206 x 205

Divide both side by 0.08206 x 205

n = (5.6 x 12)/(0.08206 x 205)

n = 4.0 moles

Therefore, the number of mole of the gas is 4.0 moles

Answer:

What will the horizontal axis represent?

Temperature

What will the vertical axis represent?

number of days

What is an appropriate scale?

0-100

Which interval should be used

20

Answer:

✔ temperature

✔ number of days

✔ 0–100

✔ 20

Explanation:

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

[tex]Q = c \times m \times \Delta T[/tex]

where,

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

[tex]Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ[/tex]

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

[tex]\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol[/tex]

Absorption of 1.0 rad of gamma radiation would likely cause the greatest tissue damage due to its high energy and penetration ability.

The amount of tissue damage caused by radiation depends on various factors, including the type of radiation and its energy level.

Alpha particles are relatively heavy and highly ionizing, but they have low penetration power, so they are generally less damaging to tissues externally but can be more damaging if ingested or inhaled.

Beta particles have less mass and energy compared to alpha particles and can penetrate deeper into tissues, potentially causing more damage.

Gamma rays are electromagnetic radiation with high energy and penetration power. They can penetrate deeply into tissues, causing damage along their path.

Considering these factors, absorption of 1.0 rad of gamma radiation would likely result in the greatest tissue damage due to its high energy and penetration ability.

A slice or part of a homogeneous substance has the same density as the whole piece because the density is uniform throughout. In a heterogeneous substance, the density can vary and depends on the composition and structure of the part being measured. Density is found by dividing mass by volume and is useful for identifying substances.

The density of a substance is defined as its mass divided by its volume. In a homogeneous material, the density is consistent throughout, which means a slice or part of the substance would have the same density as the whole piece. For example, a solid iron bar is homogeneous, and therefore any part of it would have the same density as the entire bar. However, in a heterogeneous material, the density can vary from one part to another; an instance of this is Swiss cheese, which contains air pockets resulting in variable local densities.

To determine the density of a substance, you would divide the mass by the volume of the substance. This method applies to both whole objects and parts thereof. In the case of a cube, the volume is found by cubing the edge length. In practice, density is often used to help identify substances by comparison with known values, as different substances have characteristic densities.

Answer:

Explanation:

The chemical equation for the reaction is :

[tex]C_6H_{12}O_6_{(s)} \to 2C_2H_5OH _{(l)} + 2CO_{2(g)}[/tex]

The standard enthalpy of formation [tex]\Delta H ^0_f[/tex]  of the above equation is as follows:

[tex]\Delta H^0_{f, C_6H_{12}O_6}[/tex] = -1274.4 kJ/mol

[tex]\Delta H ^0_{f, C_2H_{5}OH[/tex] = -277.7 kJ/mol

[tex]\Delta H ^0_{f, CO_2}[/tex] = -393.5 kJ/mol

[tex]\Delta H ^0_{rxn }= \sum n_p \Delta H ^0_{f,p} - \sum n_r \Delta H ^0_{f,r}[/tex]

where ;

[tex]n_p[/tex] = stochiometric coefficients of products

[tex]n_r=[/tex] stochiometric coefficients of reactants

[tex]\Delta H^0_{f.p}[/tex] = formation standard enthalpy of products

[tex]\Delta H^0_{f.r}[/tex] = formation standard enthalpy of reactants

[tex]\Delta H ^0_{rxn }= (2 \ mol* -2777.7 \ kJ/mol + 2 \ mol * - 393.5 \ kJ/mol) - (1\ mol *(-1274.4 \ kJ/mol)[/tex]

[tex]\Delta H ^0_{rxn }= -68 \ kJ[/tex]

For [tex]\Delta S ^0[/tex] ;

The standard enthalpy of formation of [tex]\Delta S ^0_f[/tex] of the reactant and the products are :

[tex]\Delta S ^0 _{f \ C_6H_{12}O_6} = 212 \ \ J/Kmol[/tex]

[tex]\Delta S ^0 _{f \ C_2H_5O_H} = 160.7 \ \ J/Kmol[/tex]

[tex]\Delta S ^0 _{f \ CO_2} = 213.63 \ \ J/Kmol[/tex]

The [tex]\Delta S ^0_{rxn}[/tex] is as follows:

[tex]\Delta S ^0_{rxn} = \sum n_p \Delta S^0_{f.p} - \sum n_r \Delta S^0_{f.r}[/tex]

[tex]\Delta S ^0_{rxn} = (2 \ mol *160.7 \ \ J/Kmol + 2 \ mol *213.63 \ \ J/Kmol) -(1*212 J/Kmol)[/tex]

[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K[/tex]   (to kJ/K)

[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K * \frac{1 \ kJ}{1000 \ J}[/tex]

[tex]\Delta S ^0_{rxn} =0.5367 \frac{kJ}{K}[/tex]

Given that;

at T = 25°C = ( 25 + 273) K = 298 K

[tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex]

[tex]\Delta G^0 _{rxn} = -68 \ kJ - 298 * 0.5367 \frac{kJ}{K}[/tex]

[tex]\Delta G^0 _{rxn} = -227. 9 \ \ \ kJ[/tex]

As [tex]\Delta G^0 _{rxn}[/tex] is negative; the reaction is spontaneous

[tex]\Delta H^0 _{rxn}[/tex] = negative

[tex]\Delta S^0 _{rxn}[/tex] = positive; Therefore , the reaction is spontaneous at all temperature , We can then say that the spontaneity of this reaction [tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex]  is dependent on temperature.

Answer:

0.22 atm.

Explanation:

Acetyl bromide is a Chemical compound with molar mass of 122.95 g/mol and chemical formula of C2H3BrO. Acetyl bromide has a density of 1.66 g/cm³ and it is often classified as a volatile organic compound.

Chloroform is a Chemical compound with molar mass of 119.38 g/mol and chemical formula of CHCl₃.

So, let us delve right into the Calculations of the question above;

The number of moles of Acetyl bromide, C2H3BrO = mass/molar mass = 51.8/ 122.95 = 0.421 moles.

The number of moles of Chloroform, CHCl₃ = mass/ molar mass = 123/119.38 = 1.03 moles.

Total moles = 1.03 moles + 0.421 moles= 1.4513 moles.

Mole fraction of Acetyl bromide= 0.421 moles/1.4513 moles = 0.2901

Partial Pressure = 0.2901 × 0.75= 0.2176 atm.

= 0.22 atm.

Final answer:

To calculate the partial pressure of acetyl bromide in a solution, Raoult's Law is applied, but specific values are necessary for an exact calculation. Without these values, a general explanation of the methodology is provided.

Explanation:

To calculate the partial pressure of acetyl bromide in a solution with chloroform, we use Raoult's Law, which states the partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. The equation for Raoult's Law is Pi = Xi * P0i, where Pi is the partial pressure of the component i, Xi is the mole fraction of the component i in the liquid phase, and P0i is the vapor pressure of the pure component i. However, the question does not provide specific values for the initial vapor pressure of acetyl bromide or the quantities of acetyl bromide and chloroform, making it impossible to calculate the exact partial pressure without these inputs. Typically, you would calculate the mole fraction of acetyl bromide by dividing the moles of acetyl bromide by the total moles of solution and then use Raoult's Law to find the partial pressure of acetyl bromide vapor.

Answer:

C. The electron-withdrawing fluorine atoms pull electron density from the oxygen in trifluoroacetate. The negative charge is more stabilized in trifluoroacetate by this effect.

Explanation:

The structures of trifluoroacetate and acetic acid are both shown in the image attached.

The trifluoroacetate anion (CF3CO2-), just like the acetate anion has in the middle, two oxygen atoms.

However, in the trifluoroacetate anion, there are also three electronegative fluorine atoms attached to the nearby carbon atom attached to the carbonyl, and these pull some electron density through the sigma bonding network away from the oxygen atoms, thereby spreading out the negative charge further. This effect, called the "inductive effect" stabilizes the anion formed,the trifouoroacetate anion is thus more stabilized than the acetate anion.

Hence, trifluoroacetic acid is a stronger acid than acetic acid, having a pKa of -0.18.

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The acid dissociation constant (Ka) is used to distinguish strong acids from weak acids. Strong acids have exceptionally high Ka values.

The Ka value is found by looking at the equilibrium constant for the dissociation of the acid. The higher the Ka, the more the acid dissociates.

The structures of trifluoroacetate and acetic acid are both shown in the image attached.  It contains

However, in the trifluoroacetate anion, there are also three electronegative fluorine atoms attached to the nearby carbon atom attached to the carbonyl.

These pull some electron density through the sigma bonding network away from the oxygen atoms, thereby spreading out the negative charge further. This effect, called the "inductive effect" stabilizes the anion formed, the trifluoroacetate anion is thus more stabilized than the acetate anion.

Hence, the correct option is C that is The electron-withdrawing fluorine atoms pull electron density from the oxygen in trifluoroacetate. The negative charge is more stabilized in trifluoroacetate by this effect.

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Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

A common addiction to the common water is the elemental chlorine and chlorine ions, the phosphate ions. Recent reports of the higher levels of lead in cities have to lead to the Pb(s) as CI 2 flows in petal pipes some may be oxidized.

They balanced equation of the net ionic reactions.

Learn more about the chloride ions and phosphate ions.

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Answer: 0.225, x, 0.0788

Explanation:

Answer: This reaction is called as Chemiluminescence. A type of reaction in which light is produced due to chemical reaction. While in case of Phosphorescence and Fluorescence light is emitted due to absorption of photons.

Explanation: i looked the answer up but your welcomee ❤

Final answer:

Luminol undergoes a chemiluminescent reaction that produces light when it reacts with hydrogen peroxide and a catalyst, often used in forensic science to detect blood.

Explanation:

Luminol goes through a chemiluminescent reaction, which refers to a process where a chemical reaction produces a molecule in an excited state. In the presence of hydrogen peroxide and a catalyst, such as the iron found in hemoglobin, luminol reacts to generate 3-aminophthalate in an excited state, emitting a bluish light. This reaction is highly valuable in forensic science for the detection of blood.

Chemiluminescence is a phenomenon where the energy from a chemical reaction is released as light. This reaction is possible because certain chemicals, like peroxide and ozone used in chemiluminescent reactions, contain unstable or energetic chemical bonds. The process is a rare occurrence in chemical reactions and is a direct transduction of chemical energy into radiant energy.

Option b (O=N−F and N=O−F) represents a pair of resonance structures.

The correct choice representing a pair of resonance structures is option b. O=N−F and N=O−F.

Resonance structures are different representations of a molecule or ion that show the delocalization of electrons. In option b, the double bond can be moved between the nitrogen and oxygen atoms, resulting in two resonance structures.

Resonance occurs when there are multiple valid Lewis structures that can be drawn for a molecule or ion. It indicates that the actual structure is a combination or hybrid of all the resonance structures.

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Answer:

Yes, a hydrocarbon is any compound that contains hydrogen and oxygen.

Explanation:

Answer:

Concentration of [tex]Zn^{+2}[/tex] = 1.9 * [tex]10^{-19}[/tex]

Explanation:

Find the provided attachment for solution.

The equilibrium concentration of uncomplexed Zn²⁺ ions if the concentration of cyanide ions at equilibrium is 0.50 M, is 19.18 × 10⁻²² M.

Equilibrium constant of any reaction will be calculated as the ration of the concentration of products to the concentration of reactants with raise to to respective coefficients at the equilibrium condition.

Moles of Zn²⁺ & Zinc cyanide complex are equal, so the concentration of 5×10⁻³ moles of Zinc cyanide complex in 1 liter of the solution is 5×10⁻³ M.

Chemical reaction for the formation of given complex with ICE table is:

                       Zn²⁺  + 4(cyanide ions)  ⇄  Zinc cyanide complex

Initial:                0                  0.50                             5×10⁻³

Change:          +x                    +4x                                -x

Equilibrium:      x                 0.50+4x                         5×10⁻³-x

Equilibrium constant for the above reaction will be calculated as:

Kf = Zinc cyanide complex / [Zn²⁺][cyanide ion]⁴

Given value of Kf = 4.17 × 10¹⁹

On putting all values on the above equation we get,

4.17 × 10¹⁹ = [5×10⁻³-x] / [x].[0.50+4x]⁴

x is very small as compared to the 5×10⁻³ and 0.50, so we can neglect x and equation becomes:

4.17 × 10¹⁹ = [5×10⁻³] / [x].[0.50]⁴

x = 19.18 × 10⁻²² M

Hence, required concentration of Zn²⁺ is 19.18 × 10⁻²² M.

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Answer:

The limiting reactant is Carbon dioxide, [tex]CO_{2}[/tex]

Explanation:

The balanced reaction equation is:

[tex]2NH_{3} + CO_{2}[/tex] → [tex]CH_{4} N_{2} O + H_{2} O[/tex]

The mole ratio of ammonia to carbon dioxide is 2:1

142100/17g = 8358.8 mol of NH3

211400/44g = 4, 804.5 mole of CO2

Now:

4,804.5 mol of CO2 × [tex]\frac{2 mol NH_{3} }{1 molCO_{2} }[/tex] = 9,609 mol of NH3  present

8,358.8 mole of NH3 × [tex]\frac{1 moleCO_{2} }{2moles NH_{3} }[/tex] = 4,179.4 mol of CO2 present

NH3 needs 8, 358.8 moles but had 9, 609 moles⇒ excess reactant

CO2 needs 4, 804,5 mol but had 4, 179.4 moles⇒ limiting reactant (used up completely)

The limiting reactant is carbon dioxide, CO2.

Ammonia is the limiting reactant in the synthesis of urea from ammonia and carbon dioxide.

The reaction between ammonia (NH3) and carbon dioxide (CO2) produces urea (CH4N2O) and water (H2O) according to the balanced equation: 2NH3(aq) + CO2(aq) → CH4N2O(aq) + H2O(l).

To determine the limiting reactant, we need to compare the amount of each reactant used to the amount of urea produced. From the given information, 142.1 kg of ammonia and 211.4 kg of carbon dioxide react to produce 171.4 kg of urea.

We can use the stoichiometry of the balanced equation to find the theoretical yield of urea from both reactants:

Since the actual yield of urea is 171.4 kg, which is less than both the theoretical yields from ammonia and carbon dioxide, the limiting reactant is ammonia (NH3).

Therefore, the limiting reactant is ammonia (NH3).

Answer:

4L

Explanation:

To obtain the volume of O2 at stp, first, we need to determine the number of mole of O2.

From the question given above,

Mass of O2 = 5.72g

Molar Mass of O2 = 32g/mol

Number of mole =Mass/Molar Mass

Number of mole of O2 = 5.72/32

Number of mole of O2 = 0.179 mole

Now, we can calculate the volume of O2 at stp as follow:

1 mole of a gas occupy 22.4L at stp.

Therefore, 0.179 mole of O2 will occupy = 0.179 x 22.4 = 4L

Therefore, the volume occupied by the sample of O2 is 4L

The volume of a 5.72 gram sample of oxygen at STP is approximately 4.01 liters, after converting the mass to moles and using the molar volume of a gas at STP.

To calculate the volume of oxygen at Standard Temperature and Pressure (STP) from the given mass, first convert the mass of oxygen to moles. At STP, 1 mole of gas occupies 22.4 liters. Use the molar mass of Oxygen (32 g/mol) to convert grams to moles and then use the volume of 1 mole of gas at STP to find the volume in liters.

First, convert the given mass of oxygen (5.72 g) to moles:

5.72 g [tex]O_{2}[/tex]
x
1 mol [tex]O_{2}[/tex]/ 32 g [tex]O_{2}[/tex]
= 0.179 moles [tex]O_{2}[/tex]

Then calculate the volume at STP (using the fact that 1 mole of any gas at STP occupies 22.4 liters):

0.179 moles [tex]O_{2}[/tex]
x
22.4 liters/mol
= 4.0096 liters

Therefore, the volume of a 5.72 gram sample of [tex]O_{2}[/tex] at STP is approximately 4.01 liters.

Answer: The heat of reaction for the combustion of titanium is 15240 kJ/mol

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=C\times \Delta T[/tex]

Q = Heat absorbed by calorimeter =?

C = heat capacity of calorimeter = 9.84 kJ/K

Initial temperature of the calorimeter  = [tex]T_i[/tex] = [tex]25.00^0C=(25.00+273)=298.00K[/tex]

Final temperature of the calorimeter  = [tex]T_f[/tex]  = [tex]91.60^0C=(91.60+273)K=364.6K[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i=(364.6-298.0)K=66.60K[/tex]

Putting in the values, we get:

[tex]Q=9.84kJ/K\times 66.60K=655.3kJ[/tex]

As heat absorbed by calorimeter is equal to heat released by combustion of titanium

[tex]Q=q[/tex]

[tex]\text{Moles of titanium}=\frac{\text{given mass}}{\text{Molar Mass}}=\frac{2.060g}{47.8g/mol}=0.0430mol[/tex]  

Heat released by 0.0430 moles of titanium = 655.3 kJ

Heat released by 1 mole of titanium = [tex]\frac{655.3}{0.0430}\times 1=15240kJ[/tex]

The heat of reaction for the combustion of titanium is 15240 kJ/mol

Answer:

pH in the [tex]H_{2}/H^{+}[/tex] half cell is 0.84.

Explanation:

Oxidation: [tex]Zn(s)-2e^{-}\rightarrow Zn^{2+}(aq.)[/tex]

Reduction: [tex]2H^{+}(aq.)+2e^{-}\rightarrow H_{2}(g)[/tex]

-------------------------------------------------------

Overall: [tex]Zn(s)+2H^{+}(aq.)\rightarrow Zn^{2+}(aq.)+H_{2}(g)[/tex]

[tex]E_{cell}^{0}=E_{H^{+}\mid H_{2}}^{0}-E_{Zn^{2+}\mid Zn}^{0}[/tex]  = (0.00 V) + (0.76 V) = 0.76 V

According to Nernst equation for this cell reaction at room temperature (298 K):              

                           [tex]E_{cell}=E_{cell}^{0}-\frac{0.0592}{n}log\frac{[Zn^{2+}].P_{H_{2}}}{[H^{+}]^{2}}[/tex]

where, [tex]E_{cell}[/tex] is cell potential, n is number of electron exchanged, [tex]P_{H_{2}}[/tex] is pressure of [tex]H_{2}[/tex] in atm and species under third bracket represent molarity of the respective species.

So, [tex]0.68V=0.76V-\frac{0.0592}{2}log\frac{(1.3M)\times (8atm)}{[H^{+}]^{2}}V[/tex]

   [tex]\Rightarrow[/tex] [tex][H^{+}]=0.1436M[/tex]

pH = [tex]-log[H^{+}][/tex] = -log(0.1436) = 0.84

Final answer:

The pH in the H+/H2 half-cell of the given galvanic cell is 8.5.

Explanation:

In the given galvanic cell, the reduction half-reaction is 2H+ (aq) + 2e → H₂(g), and the overall reaction is Zn(s) + 2H+ (aq) -> Zn²+ (aq) + H₂(g). The reduction potential for the H2(g)/H+(aq) half-reaction is 0.00 V, and for the Zn(s)/Zn²+(aq) half-reaction is -0.76 V.

To find the pH in the H+/H2 half-cell, we can use the Nernst equation:

Ecell = E°cell - (0.0592 V / n) * log[H+]

Using the given cell potential of 0.68 V and plugging in the values, we can calculate the pH in the H+/H2 half-cell to be 8.5.

Answer:

1. 4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

2. 6 moles of Cl2

Explanation:

1. The balanced equation for the reaction. This is illustrated below:

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

2. Determination of the number of mole of Cl2 produce when 4 moles of FeCl3 react with 4 moles. To obtain the number of mole of Cl2 produced, we must determine which reactant is the limiting reactant.

This is illustrated below:

From the balanced equation above,

4 moles of FeCl3 reacted with 3 moles of O2.

Since lesser amount of O2 (i.e 3 moles) than what was given (i.e 4 moles) is needed to react completely with 4 moles of FeCl3, therefore FeCl3 is the limiting reactant and O2 is the excess reactant.

Finally, we can obtain the number of mole Cl2 produced from the reaction as follow:

Note: the limiting reactant is used as it will produce the maximum yield of the reaction since all of it is used up in the reaction.

From the balanced equation above,

4 moles of FeCl3 will react to produced 6 moles of Cl2.

Answer:

Zn(s) + Fe^2+(aq) ---------> Zn^2+(aq) + Fe(s)

∆G= 61760J

K= 6.5 ×10^10

Explanation:

Zn(s) + Fe^2+(aq) ---------> Zn^2+(aq) + Fe(s)

E°reaction= E°Fe - E°Zn

E°reaction= (-0.44)-(-0.76)

E°reaction= 0.32V

We know that

∆G= -nFE°

Since n= 2 from the reaction equation and F= 96500C

∆G= -(2×96500×0.32)

∆G= 61760J

From

E°= 0.0592/n log K

0.32= 0.0592/2 log K

log K= 0.32/0.0296

log K= 10.81

K= Antilog(10.81)

K= 6.5 ×10^10