14. A piece of titanium at 100.0°C was dropped into 50.0 g of water at 20.0°C. The final temperature of the system was 22.6°C. What is the mass of the titanium? (Specific Heat of titanium = 0.54 J/g°C)

Answer:

C. CrF6, chromium(VI) hexafluoride.

Explanation:

Hello,

In this case, since we are given a hexavalent chromium we must notice it has +6 as its oxidation state. Moreover, fluorine, when forming ionic compounds works with -1, for which the chemical formula is:

[tex]Cr^{6+}F^-\\\\CrF_6[/tex]

And the stock name is indeed C. CrF6, chromium(VI) hexafluoride (looks like D. is the same) since we have six fluoride ions in the formula and we point out chrmium's oxidation state.

Regards.

Answer:

C. CrF6, chromium(VI) hexafluoride.

Explanation:

Answer:

[tex]\large \boxed{\text{0.980 L}}[/tex]

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

[tex]p_{1}V_{1} = p_{2}V_{2}[/tex]

Data:

[tex]\begin{array}{rcrrcl}p_{1}& =& \text{1.00 atm}\qquad & V_{1} &= & \text{250. mL} \\p_{2}& =& \text{2.55 atm}\qquad & V_{2} &= & ?\\\end{array}[/tex]

Calculations:  

[tex]\begin{array}{rcl}\text{1.00 atm} \times \text{250. mL} & =& \text{2.55 atm} \times V_{2}\\\text{250. mL} & = & 2.55V_{2}\\V_{2} & = &\dfrac{\text{250. mL}}{2.55}\\\\& = &\textbf{98.0 mL}\\\end{array}\\\text{The balloon's new volume is $ \large \boxed{\textbf{0.980 L}}$}[/tex]

Answer: The energy of a Br–F bond is 110 kJ/mol

Explanation:

The balanced chemical reaction is,

[tex]Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)][/tex]

[tex]\Delta H=[(n_{Br_2}\times B.E_{Br_2})+(n_{F_2}\times B.E_{F_2}) ]-[(n_{BrF_3}\times B.E_{BrF_3})][/tex]

[tex]\Delta H=[(n_{Br_2}\times B.E_{Br-Br})+(n_{F_2}\times B.E_{F_F}) ]-[(n_{BrF_3}\times 3\times B.E_{Br-F})][/tex]

where,

n = number of moles

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

[tex]\Delta H=[(1\times 193)+(3\times 155)]-[(2\times 3\times B.E_{Br-F})][/tex]

[tex]B.E_{Br-F}=110kJ/mol[/tex]

Thus the energy, in kJ/mol, of a Br–F bond is 110

Answer:

C) 5970 y

Explanation:

Given;

initial amount of wood, N₀ = 15.3 cpm/g

remaining amount of wood (charcoal), N = 7.4 cpm/g

half life of carbon 14, t 1/2 = 5700 years

The age of the ashes can be calculated using the following formula;

[tex]N = N_0(\frac{1}{2})^{\frac{t}{t_1_/_2} }\\\\(\frac{1}{2})^{\frac{t}{t_1_/_2} } = \frac{N}{N_0} \\\\(\frac{1}{2})^{\frac{t}{t_1_/_2} } = \frac{7.4}{15.3} \\\\(\frac{1}{2})^{\frac{t}{t_1_/_2} } = 0.48366\\\\t = t_{1/2} Log\frac{1}{2} (0.48366)\\\\t = \frac{t_{1/2}ln(0.48366)}{-ln(2)} \\\\t = t_{1/2}(1.0479)\\\\t = 5700(1.0479)\\\\t = 5973 \ years\\\\t = 5970 \ years(nearest \ ten)[/tex]

Therefore, the ashes are 5970 years

Answer:

131.29 kJ

Explanation:

Let's consider the following balanced equation.

H₂O(g) + C(graphite)(s)  ⇄ H₂(g) + CO(g)

Given the standard enthalpies of formation (ΔH°f), we can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(H₂(g)) + 1 mol × ΔH°f(CO(g)) - 1 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C(graphite)(s))

ΔH°r = 1 mol × (0 kJ/mol) + 1 mol × (-110.53 kJ/mol) - 1 mol × (-241.82 kJ/mol) - 1 mol × (0 kJ/mol)

ΔH°r = 131.29 kJ

Answer:

[tex]P=6.10atm[/tex]

Explanation:

Hello,

In this case, we can study the ideal gas equation that relates temperature, volume, pressure and moles as shown below:

[tex]PV=nRT[/tex]

Thus, since we are asked to compute the pressure y simply solve for it as follows:

[tex]P=\frac{nRT}{V}=\frac{11.1mol*0.082\frac{atm*L}{mol*K}*300K}{44.8L}\\ \\P=6.10atm[/tex]

Best regards.

Answer:

Trial 1: Moles acetic acid = 0.00686 moles;

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = 0.00137 mol/g

Trial 2: Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = 0.00144 mol/g

Explanation:

Equation of the reaction: 2CH₃COOH + Mg(OH)₂ ---> Mg(CH₃COO)₂ + 2H₂O

Trial 1:

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 7.8 mL = (7.8/1000) Litres = 0.0078 L

Moles acetic acid = 0.88 M * 0.0078 L

Moles acetic acid = 0.00686 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00686 moles of acetic acid will react with 0.00686/2 moles of Mg(OH)₂ = 0.00343 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00343 mole /2.5 g

Neutralization capacity = 0.00137 mol/g

Trial 2.

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 8.2 mL = (8.2/1000) Litres = 0.0082 L

Moles acetic acid = 0.88 * 0.0082

Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00722 moles of acetic acid will react with 0.00722/2 moles of Mg(OH)₂ = 0.00361 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00361 mole /2.5 g

Neutralization capacity = 0.00144 mol/g

Answer:

[tex]T_{sol}=-15.9\°C[/tex]

Explanation:

Hello,

In this case, we can analyze the colligative property of solutions - freezing point depression - for the formed solution when ethylene glycol mixes with water. Thus, since water freezes at 0 °C, we can compute the freezing point of the solution as shown below:

[tex]T_{sol}=T_{water}-i*m*Kf[/tex]

Whereas the van't Hoff factor for this solute is 1 as it is nonionizing and the molality is:

[tex]m=\frac{mol_{solute}}{kg\ of\ water}=\frac{45.0g*\frac{1mol}{62g} }{85.0g*\frac{1kg}{1000g} } =8.54m[/tex]

Thus, we obtain:

[tex]T_{sol}=0\°C+(-8.54m*1.86\frac{\°C}{m} )\\\\T_{sol}=-15.9\°C[/tex]

Best regards.

The freezing point of a solution prepared from 45.0 g ethylene glycol and 85.0 g of water is -15.9 °C.

Freezing point depression is a drop in the temperature at which a substance freezes, caused when a smaller amount of another, non-volatile substance is added.

We will use the definition of molality.

b = mass of solute / molar mass of solute × kg of solvent

b = 45.0 g / 62.07 g/mol × 0.0850 kg = 8.53 m

We will use the following expression, where Kf is the cryoscopic constant of water.

ΔT = Kf × b = 1.86 °C/m × 8.53 m = 15.9 °C

The freezing point of pure water is 0°C.

T = 0°C - 15.9 °C = -15.9 °C

The freezing point of a solution prepared from 45.0 g ethylene glycol and 85.0 g of water is -15.9 °C.

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Answer:

BaCl2

Explanation:

Barium = Ba

Chloride => Cl-

Chemical Equation:

Ba + Cl => BaCl2

Note:

The valency of barium is 2 and valency of chloride is 1 (i.e. chlorine). The formula formed by the combination of these elements is BaCl2 (there's exchange of valencies when these two elements combine).

Answer:

E. O2

Explanation:

All substances has a simple molecular structure, where between their molecules are held by van der Waals' forces. But C must be incorrect because between the H2O2 molecules, they are mainly held by hydrogen bonds on top of van der Waals' forces. Hydrogen bonds are stronger than van der Waals' forces, so more energy is required to separate the H2O2 molecules.

In structures A and D, the molecules are polar. Their van der Waals' forces are stronger than Cl2H2 and O2, which are non-polar.

Between the Cl2H2 and O2, O2 has a smaller molecular size. The van der Waals' forces between the O2 molecules are hence the weakest. Least amount of energy is required to break the intermolecular forces between the O2 molecules therefore it has the lowest boiling point.

Answer:

They are:

H2, N2, O2, F2, Cl2, Br2, and I2.

Note: whether the element At molecule is monoatomic or diatomic is incredibly arguable. While some say it exists as diatomic because it is a halogen like bromine, iodine etc, At is in fact extremely unstable and no one has ever really studied the molecules on it, so, when others say it is monoatomic, this is also based on calculations. But the other 7 elements listen above is for sure diatomic.

Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2).

Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2) are the formulas of the elements that is present as diatomic molecules under normal environmental conditions. Diatomic molecules refers to those molecules that is composed of only two atoms of the same or different elements. There are large number of diatomic molecules which is made up of two similar elements or different elements.

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Answer:B

Explanation:

Answer: i think it is c

Explanation: i checked my textbook.

Answer:

% yield of the student's experiment is

[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%

Explanation:

given

volume of acetone= 43.8 mL

molar weight of acetone = 58.08 g/mol

density of acetone = 0.791 g/mL

A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL)

43.8 mL = 43.8mL × 0.791g/mL

= 34.6458g ≈34.65g

1 mole of acetone = 58.08g

∴34.65g = 34.65g/58.08g

= 0.60mol

molecular weight of the product 1,5-diphenylpenta-1,4-dien-3-one = 234.29 g/mol

mole = mass/ molar weight

mole = 79.4g/ 234.29g/mol

mole(n) = 0.3389mol ≈ 0.34mol

1 mole of acetone will produce 1 mole of the product

∴0.60mol of acetone will produce 0.60mol of the product

but we get 0.34mol of the product

∴ % yield of the student's experiment is

[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%

Answer:

The correct answer is B. Analysis

Explanation:

Error bars are part of the statistical analysis in the scientific method. Once the scientist have collected the data, he or she proceed to the data analysis. A very common way of comparing the data variability is to use error bars in ghaphical representations. From these bars, it can be estimated the error of a determination and experimental groups are compared.

The error bars would be used during B. Analysis.

A blunders bar is a line through a factor on a graph, parallel to one of the axes, which represents the uncertainty or variant of the corresponding coordinate of the point. In IB Biology, the error bars most often represent the same old deviation of an information set.

Blunders bars can be used to examine visual quantities if various other situations preserve. This can decide whether or not differences are statistically sizable. Mistake bars can also propose the goodness of match of a given characteristic, i.e., how well the function describes the facts.

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#SPJ2

Answer:

B!

Explanation:

I got it right in class!

Answer:please see below for answers in the spaces given.

Explanation:

There are three types of cubic-unit cells of a cubic system which include Simple cubic unit cell, body-centered cubic unit cell and face-centered cubic-unit cell and Thier characteristics are completed below.

1) In the simple cubic unit cell, the centers of _______eight _____ identical particles define the _________corners___ of a cube. The particles do touch along the cube's _______edges_____ but do not touch along the cube's ____diagonal_______ or through the center. There is/are _______one_____ particle per unit cell and the coordination number is

__six______ .

2. In the body-centered cubic unit cell, the centers of _______eight _____ identical particles define the _______corners_____ of the cube plus ______one______ particle at the _______center_____ of ______the cube______ . The particles do not touch along the cube's _______edges_____ or faces but do touch along the cube's ____diagonal________ . There is/are _____two_______ particles per unit cell and the coordination number is _____eight_______ .

3. In the face-centered cubic cell, the centers of ______eight______ identical particles define the _______corner____ of the cube plus ________one____ particle in the _____center_______ of ______each face______ . The particles on the _____corners_______ do not touch each other but do touch those on the ______faces____ . There is/are ________four___ particles per unit cell and the coordination number is _____twelve_______ .

Answer:

Isomers are molecules that have the same molecular method, however have a unique association of the atoms in space. That excludes any extraordinary preparations which can be sincerely because of the molecule rotating as an entire, or rotating about precise bonds.

Answer:

28.13 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2O4(l) + 2N2H4 (l) → 3N2(g) + 4H2O(g)

Next, we shall determine the masses of N2O4 and N2H4 that reacted and the mass of H2O produced from the balanced equation.

This is illustrated below:

Molar mass of N2O4 = (14x2) + (16x4) = 92 g/mol

Mass of N2O4 from the balanced equation = 1 x 92 = 92g

Molar mass of N2H4 = (14x2) + (4x1) = 32 g/mol

Mass of N2H4 from the balanced equation = 2 x 32 = 64 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72 g

Summary:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4 to produce 72 g of H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4.

Therefore, 40 g of N2O4 will react with = (40 x 64)/92 = 27.83 g of N2H4.

From the calculations made above, we can see that it will take a higher mass i.e 27.83 g than what was given i.e 25 g of N2H4 to react completely with 40 g of N2O4.

Therefore, N2H4 is the limiting reactant and N2O4 is the excess reactant.

Finally, we shall determine the mass of H2O produced from the reaction of 40.0 g of N2O4 and 25.0 g of N2H4.

In this case the limiting reactant will be used because it will produce the maximum amount of H2O as all of it is consumed in the reaction.

The limiting reactant is N2H4 and the mass of H2O produced can be obtained as follow:

From the balanced equation above,

64 g of N2H4 reacted to produce 72 g of H2O.

Therefore, 25 g of N2H4 will react to produce = (25 x 72)/64 = 28.13 g of H2O.

Therefore, 28.13 g of H2O were obtained from the reaction.

Answer:

Ca for calcium

20 electrons

2-2s electron

Answer:

0.0826 mol (corrected to 3 sig. fig.)

Explanation:

First, balance the equation:

2NH3 →3H2 + N2

Take the atomic no. of N=14.0, and H=1.0,

no. of moles = mass / molar mass

So, no. of moles of NH3 dissociated = 2.81 / (14.0+1.0x3)

= 0.165294117mol

From the equation, the mole ratio of NH3:N2 = 2:1, meaning for every 2 moles of NH3 dissociated, one mole of N2 is formed.

So, using this ratio, the no. of moles of N2 formed will be 0.165294117 / 2

=0.0826 mol (corrected to 3 sig. fig.)

Answer:

e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)

Explanation:

All the following are oxidation–reduction reactions except:________

a. H₂(g) + F₂(g) → 2HF(g).  Redox. H is oxidized and F is reduced.

b. Ca(s) + H₂(g) → CaH₂(s).  Redox. Ca is oxidized and H is reduced.

c. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g).  Redox. K is oxidized and H is reduced.

d. 6Li(s) + N₂(g) → 2Li₃N(s).  Redox. Li is oxidized and N is reduced.

e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g). Not redox. All the elements have the same oxidation number

The reaction Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g) is not a redox reaction.

Redox reactions are those reactions in which there is a change in the oxidation number of species from left to right in the reaction. A specie is oxidized leading to increase in oxidation number while another specie is reduced leading to decrease in oxidation number.

The reaction in which there is no change in oxidation number of species from left to right is the reaction; Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g).

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The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is [tex]1.1\times 10^6[/tex] M.

[tex]HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)[/tex]

This value indicates that

A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is [tex]ONO^-[/tex]

D. The conjugate acid of CN- is HCN

Answer: A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When [tex]K_{p}>1[/tex]; the reaction is product favoured.

When [tex]K_{p};<1[/tex] ; the reaction is reactant favored.

[tex]When K_{p}=1[/tex]; the reaction is in equilibrium.

As, [tex]K_p>>1[/tex], the reaction will be product favoured and as it is a acid base reaction where [tex]HONO[/tex] acts as acid by donating [tex]H^+[/tex] ions and [tex]CN^-[/tex] acts as base by accepting [tex]H^+[/tex]

Thus [tex]HONO[/tex] is a strong acid thus [tex]ONO^-[/tex] will be a weak conjugate base and [tex]CN^-[/tex] is a strong base which has weak [tex]HCN[/tex] conjugate acid.

Thus the high value of K indicates that [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

Answer:

300.44 g

Explanation:

The balanced equation for the reaction is given below:

Mg + Cu(NO3)2 —> Mg(NO3)2 + Cu

Next, we shall determine the mass of Mg that reacted and the mass of Mg(NO3)2 produced from the balanced equation.

This is illustrated below:

Molar mass of Mg = 24 g/mol

Mass of Mg from the balanced equation = 1 x 24 = 24 g

Molar mass of Mg(NO3)2 = 24 + 2[14 + (16x3)]

= 24 + 2[ 14 + 48]

= 24 + 124 = 148 g/mol

Mass of Mg(NO3)2 from the balanced equation =

1 x 148 = 148 g

From the balanced equation above,

24 g of Mg reacted to produce 148 g of Mg(NO3)2.

Next, we shall determine the theoretical yield of Mg(NO3)2.

This can be obtained as follow:

From the balanced equation above,

24 g of Mg reacted to produce 148 g of Mg(NO3)2.

Therefore, 139.6 g of Mg will react to = (139.6 x 148)/24 = 860.87 g of Mg(NO3)2

Therefore, the theoretical yield of Mg(NO3)2 is 860.87 g

Finally, we shall determine the actual yield of Mg(NO3)2 as follow:

Theoretical of Mg(NO3)2 = 860.87 g

Percentage yield = 34.90%

Actual yield of Mg(NO3)2 =?

Percentage yield = Actual yield /Theoretical yield x 100

34.90% = Actual yield /860.87

Cross multiply

Actual yield = 34.90% x 860.87

Actual yield = 34.9/100 x 860.87

Actual yield = 300.44 g

Therefore, the actual yield of Mg(NO3)2 is 300.44 g

Answer:

The correct answer is -8.80 kJ.

Explanation:

The ΔH° can be determined by using the formula,  

ΔH°rxn = ΔH°f (products) - ΔH°f(reactants)

Based on the given information, the ΔH°f of H2S(g) is -20.6, for Ag2S (s) is -32.6 and for H2O (l) is -285.8 kJ/mole.  

Now putting the values we get,  

= [2 molΔH°f (Ag2S) + 2 molΔH°f (H2O)] - [4 molΔH°f(Ag) + 2 molΔH°f(H2S) + 1 molΔH°f(O2)]

=[2 mol (-32.6 kJ/mol) + 2 mol(-285.8 kJ/mol)] - [4 mol(0.00 kJ/mol) + 2 mol (-20.6 kJ/mol) + 1 mol (0.00 kJ/mol)

= [(-65.2 kJ) + (-571.6 kJ)] - [(-41.2 kJ)]

= -595.6 kJ

Thus, the enthalpy change of -595.6 kJ takes place when 4 mol of Ag reacts by the equation mentioned.  

The mass of Ag given is 6.38 grams, the molecular mass of Ag is 107.9 g/mol. The formula for calculating moles is,  

Moles = mass/molar mass

= 6.38 g / 107.9 g/mol

= 0.0591 mol

Now the change in enthalpy when 0.0591 mol of Ag reacts by the given reaction is (-595.6 kJ/4 mol) × 0.0591 mol = -8.80 kJ

The negative sign indicates that the heat is released in the process. Therefore, the -8.80 kJ of heat is released by 6.38 grams of Ag in the given case.  

Explanation:

The mass to charge ratio =136

Answer:

3.50 M

Explanation:

Step 1: Write the balanced equation

4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)

Step 2: Make an ICE chart

        4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)

I              0               0              3.60            3.60

C         +4x            +7x               -2x             -6x

E            4x              7x           3.60-2x      3.60-6x

Step 3: Calculate the value of x

The concentration of water at equilibrium is 0.600 M. Then,

3.60-6x = 0.600 M

x = 0.500 M

Step 4: Calculate the concentration of O₂ at equilibrium

The concentration of O₂ at equilibrium is 7x = 7(0.500M) = 3.50 M

Answer:

diffusion

Explanation:

Diffusion is the movement of particles from a region of higher concentration to a region of lower concentration in response to a concentration gradient. A concentration gradient simply means a difference in concentration.

Diffusion occurs in solids,liquids and gases. Diffusion is fastest in gases and slowest in solids. Diffusion of solid particles may take very many years while diffusion of gases takes a few milliseconds depending on the mass of the gas.

In materials, atoms and molecules also move from one part of the material to another. This is also refereed to as diffusion.

Answer:

Option A. 2096.1 K

Explanation:

The following data were obtained from the question:

Al2O3(s) + 3C(s) —> 2Al(s) + 3CO(g)? Enthalpy (H) = +1287 kJ mol¯¹

Entropy (S) = +614 JK¯¹ mol¯¹

Temperature (T) =...?

Entropy, enthalphy and temperature are related by the following equation:

Change in Entropy (ΔS) = Change in Enthalphy (ΔH) /Temperature (T)

ΔS = ΔH/T

With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:

Enthalpy (H) = +1287 kJ mol¯¹ = 1287000 Jmol¯¹

Entropy (S) = +614 JK¯¹mol¯¹

Temperature (T) =...?

ΔS = ΔH/T

614 = 1287000/ T

Cross multiply

614 x T = 1287000

Divide both side by 614

T = 1287000/614

T = 2096.1 K

Therefore, the temperature at which the reaction will be feasible is 2096.1 K.

Answer:

See attached picture.

Explanation:

Hello,

In this case, we should define each type of structural formula as shown below:

- Chain isomers: molecules with the same molecular formula, but different arrangements.

- Positional isomers are constitutional isomers that have the same carbon skeleton and the same functional groups but differ from each other in the location of the functional groups.

- Functional isomers are structural isomers that have the same molecular formula (that is, the same number of atoms of the same elements), but the atoms are connected in different ways so that the groupings are dissimilar.

Regards.