Write the equilibrium constant expression for the experiment you will be studying this week. 2. If the equilibrium values of Fe3+, SCN- and FeSCN2+ are 9.5 x 10-4 M, 3.6 x 10-4 M and 5.7 x 10-5 M respectively, what is the value of Kc? 3. Write the general form of the dilution equation. 4. A solution is prepared by adding 18 mL of 0.200 M Fe(NO3)3 and 2 mL of 0.0020 M KSCN. Calculate the initial concentrations of Fe3+ and SCNin the solution.

Answer:

(d) 2

Explanation:

Lets say that there are 2 apples or 1+1 if you count them you would do 1,2 so 2 would be the final answer

Answer:

O-H stretch signal at 3300 cm-1

Explanation:

In this question, we can start with the reaction mechanism for the synthesis of Nerolin. We have to start with naphthalen-2-ol adding NaOH we can produce the alkoxide. Then this alkoxide can react by an Sn2 reaction with bromomethane to produce Nerolin (see figure 1).

In the starting molecule (naphthalen-2-ol) we have an "OH" group. Therefore we will have an O-H stretch signal around 3300 cm^-1. The alcohol signals are very broad and very intense, so this will be the main signal for the initial molecule. In the final product, we dont have the "OH" therefore this signal will disappear (see figure 2).

I hope it helps!

Answer:

d. 2500

Explanation:

In a kinetic study with 3 different reactants, you change concentrations of the reactants to see how this concentration affects rate of reaction. General law is:

v = k [A]ᵃ [B]ᵇ [C]ⁿ

If you see 1 and 3 experiments, the concentration of C change from 0.600M to 0.200M but reaction rate doesn't change, thus n=0:

v = k [A]ᵃ [B]ᵇ [C]⁰

v = k [A]ᵃ [B]ᵇ×1

Now, reaction 2 and reaction 4 change B from 0.400M to 0.200M having the other reactants constant. When B is duplicated, rate increase 4 times. That means b = 2:

v = k [A]ᵃ [B]ᵇ

v = k [A]ᵃ [B]²

Finally, if you see 3 and 4 reactions, A change from 0.200M to 0.600M and the reaction rate change from 15.0 to 5.0, That means if the concentration of A is triplicated, reaction rate will be triplicated to. Thus a=1:

v = k [A]ᵃ [B]²

v = k [A] [B]²

Relpacing this equation in any experiment (Experiment 5, for example):

20.0 = k [0.200] [0.200]²

2500 = k

That means right answer is:

Answer:

Face-centered Cubic Unit Cell (FCC) is consists of 8 atoms at the corners (one atom at each corner) and 6 atoms are present at the face-center (one atom at each face). Each eight atoms at corners contribute 1/8 and six atoms at faces contribute  1/2 in each unit cell.

Answer:

C. 100

Explanation:

Biochemical researches and studies have found out that an average health hotdog has a calorie of between 100 and 150 which is usually dependent on the additives.

Since the nutrition label on the back of a package of hotdogs (purchased within the US) indicates that one hotdog contains 100 calories then it truly contains such amount of calories. The standard number of calories present in a hotdog is independent of the amount eaten by individuals.

Answer:

a

Explanation:

its made up of carbon and hydrogen

Answer:

Mass PbCl₂ = 50.24g

Mass AgCl = 14.84g

Explanation:

The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:

Ag⁺ + Cl⁻ → AgCl(s)

Pb²⁺ + 2Cl⁻ → PbCl₂(s)

If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:

Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = 0.00719X moles of Cl⁻ from PbCl₂

And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:

(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = 0.45409 - 0.00698X moles of Cl⁻ from AgCl

Moles of Cl⁻ that were added in the KCl solution are:

0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.

Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)

0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles

0.45409 + 0.00021X = 0.46464

0.00021X = 0.01055

X = 0.01055 / 0.00021

X = 50.24g

As X = Mass PbCl₂

And mass of AgCl = 65.08 - 50.24

The masses of the compounds in the precipitate can be found my knowing

the number of moles of chloride ion contributed by each compound.

Reasons:

The given parameter are;

Volume of KCl solution added = 0.242 L

Concentration of KCl solution = 1.92 M KCl

The ions in the solution to which KCl is added = Ag⁺ and Pb²⁺ ions

Precipitates formed = AgCl and PbCl₂

The mass of the precipitate = 65.08 g

Required:

The mass of PbCl₂ and AgCl in the precipitate

Solution;

Number of moles of chloride ions in a mole of PbCl₂ = 2 moles

Number of moles of chloride ions in a mole of AgCl = 1 mole

Let X represent the mass of PbCl₂ in the precipitate, we have;

The mass of AgCl in the precipitate = 65.08 g - X

[tex]\mathrm{Number \ of \ moles \ of \ PbCl_2} = \dfrac{X \, g}{278.1 \, g} =\mathbf{ \dfrac{X }{278.1}}[/tex]

Number of moles of chloride ions from PbCl₂ is therefore;

[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ PbCl_2} =\mathbf{ 2 \times \dfrac{X }{278.1} \ moles \ of \ Cl^-}[/tex]

[tex]\mathrm{Number \ of \ moles \ of \ AgCl \ in \ the \ precipitate} = \dfrac{65.08 -X }{143.32}[/tex]

[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ AgCl} = \mathbf{ \dfrac{65.08 -X }{143.32}} \ moles \ of \ Cl^-[/tex]

The number of moles of chloride ions from one mole of KCl = 1 mole

Number of moles of chloride ions from 0.242 L of 1.92 M KCl is therefore;

0.242 L × 1.92 moles/L = 0.46464 moles

Number of moles of chloride ions from KCl = 0.46464 moles

[tex]0.46464 \ moles \ from \ KCl = \overbrace{ \dfrac{ 2 \times X }{278.1} + \dfrac{65.08 -X }{143.32}} \ moles \ in \ PbCl_2 \ and \ AgCl[/tex]

Which gives;

[tex]\displaystyle \frac{192}{896089} \cdot X + \frac{1627}{3583} = \frac{1452}{3125}[/tex]

Therefore;

[tex]\displaystyle X = \frac{\frac{1452}{3125} - \frac{1627}{3583} }{ \frac{192}{896089} } = \frac{105864850549}{2149800000} \approx \mathbf{ 49.24}[/tex]

The mass of PbCl₂ in the precipitate, X ≈ 49.24 g

The mass of AgCl in the precipitate = 65.08 g - 49.24 g ≈ 15.84 g

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Answer:

[tex]\large \boxed{\text{D. 23.34 min}}[/tex]

Explanation:

1. Find the order of reaction

Use information from the graph to find the order.

If a plot of ln[A] vs time is linear, the reaction is first order and the slope = -k.

2. Find the  half-life

[tex]k = \dfrac{\ln2}{ t_{\frac{1}{2}}}\\\\k = \text{-slope} = -(-2.97 \times 10^{-2} \text{ min}^{-1}) =2.97 \times 10^{-2} \text{ min}^{-1} \\ t_{\frac{1}{2}} =\dfrac{\ln 2}{k} = \dfrac{\ln 2}{2.97 \times 10^{-2}\text{ min}^{-1}} =\textbf{23.34 min}\\\\\text{The half-life is $\large \boxed{\textbf{23.34 min}}$}[/tex]

The half life of the reaction is 23.33 minutes.

We know that for a first order reaction;

ln[A]t = ln[A]o  - kt

A plot of ln[A]t  against time (t) will yield a straight line graph with a slope of -k.

From the question, the slope is -2.97 x 10-2 min-1.

So, -2.97 x 10-2 min-1 = - k

k = 2.97 x 10-2 min-1

The half life of a first order reaction is obtained from;

t1/2 = 0.693/k

t1/2 = 0.693/2.97 x 10-2 min-1

t1/2 = 23.33 minutes

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Answer:

[tex]M=0.727M[/tex]

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

[tex]n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-[/tex]

So the total mole of chloride ions:

[tex]N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-[/tex]

And the total volume by adding the volume of each solution in L:

[tex]V=0.500L+0.425L=0.925L[/tex]

Finally, the molarity turns out:

[tex]M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M[/tex]

Best regards.

Answer:

Group 13

Explanation:

You know X has 3 valence electrons, as oxygen has a subscript of 3. This means X has an ionic charge of +3. Group 13 consists mainly of metalloids but it also has metals such as aluminum, which has a +3 charge. If you use aluminum as an example, you know that when combined with oxygen, it forms Al2O3. Group 12 has transition metals that don't have +3 ionic charges, group 14 has metalloids, metals that don't have ionic charges of +3, and nonmetals, and group 2 has metals with ionic charges of +2. Group 13 is the answer.

Answer:

[tex]Kp=0.0386[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]

For which the equilibrium expression is:

[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]

Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:

[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]

And the total pressure:

[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]

Yet it is 748 torr, for which the extent is:

[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]

Therefore, Kp turns out:

[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]

Best regards.

Explanation:

We have to put out the balanced chemical equation before proceeding. The equation for the dissociation of sodium sulfide in water is given as;

Na₂S  --> 2Na⁺ + S²⁻

From the stoichiometry of the reaction we can tell that;

1 mol of Na₂S produces 2 moles of Na⁺ and 1 mol of S²⁻.

Sodium ion:

1 mol of Na₂S  = 2 mol of Na⁺

2,05 = x

upon solving for x;

x =  2 * 2.05 = 4.10 moles

Sulfide ion:

1 mol of Na₂S  = 1 mol of S²⁻

2.05 = x

upon solving for x;

x = 1 * 2.05 = 2.05 moles

Answer:

HEY ITS RIGHT ABOVE!!!

Explanation:

i did 1.85 instead of 2.05 and it was right... im so sorry

Answer:

The Highly acidic proton joined to one of the carbon in the ALKYNE bond.

(Kindly Check the attachment for the drawing because the solution will need us to draw).

Explanation:

So, let us start by defining some major key terms in this particular Question given above;

(1). ISOMERIZATION: isomerization can simply be defined as the kind is of chemical rearrangement whichay lead to the breaking and the formation of new bonds.

(2). NaNH2 BASE: Sodium amide is a Chemical compound which has a Molar mass of 39.01 g/mol and Heat capacity (C) of 66.15 J/mol K. It is also known as sodamide. It is a good nucleophile.

(3). ALKYNE BOND: it is a C-C joined together by three bonds.

The chemical reaction given in the Question is given in the attachment too.

Therefore, The Highly acidic proton joined to one of the carbon in the ALKYNE bond  is removed irreversibly by NaNH2 base.

Answer:

The value of Kp at this temperature is 9.0*10³

Explanation:

The equilibrium constant Kp describes the relationship that exists between the partial pressures of the reactants and products, while Kc represents the relationship that exists between the concentrations of the reactants and products that participate in the reaction.

The general relationship between the constants Kp and Kc results:

Kp=Kc*[tex](R*T)^{moles of product - moles of reagent}[/tex]

In this case:

Replacing:

Kp=2.2*10⁵*[tex](0.0821*298)^{-1}[/tex]

Solving:

Kp≅9.0*10³

The value of Kp at this temperature is 9.0*10³

Answer:

Molecular compounds are inorganic compounds that take the form of discrete (covalent) molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2).

Answer:

B. 3

Explanation:

To decrease the temperature of your coffee from 85°C to 55°C your system need to absorb energy. This energy will be absorbed from the addition of some ice.

How many energy must be absorbed? You can use:

Q = C×m×ΔT

Where Q is heat (Energy), C is specific heat of your solution (4.184J/g°C), m is its mass (mass of 355mL of coffee = 355g) and ΔT is change in temperature (85°C-55°C = 30°C)

Replacing, your ice needs to absorb:

Q = C×m×ΔT

Q = 4.184J/g°C×355g×30°C

Q = 44559.6J

The energy that is taken from an ice cube to change its temperature from -15°C to 55°C is:

Energy from -15°C to 0°C (C of ice = 2.95J/g°C):

Q = C×m×ΔT

Q = 2.95J/g°C×24g×15°C

Q = 1062J

Now the energy taken to pass the ice from solid to liquid is:

Q = ΔHf×m

Q = 334J/g×24g

Q = 8016J

And the energy to increase the temperature of 0°C to 55°C of 24g of ice:

Q = 4.184J/g°C×24g×55°C

Q = 5522.9J

And the total energy that 1 ice cube needs is:

Q = 1062J + 8016J + 5522.9J

Q = 14600.9J

But you need 44559.6J to decrease the temperature of your coffee, that is:

44600J / 14600.9J = 3.05

≈ 3 ice cubes to decrease the temperature of the coffee.

Right solution:

Answer:

3

Explanation:

I’m not positive but I think it’s correct

Answer:

[tex]m_{air}=0.947g[/tex]

[tex]n_{O_2} =0.00686molO_2[/tex]

Explanation:

Hello,

In this case, we can firstly use the ideal gas equation to compute the total moles of the gaseous mixture (air) with the temperature in Kelvins:

[tex]T=212\°F=100\°C=373.15K\\\\n=\frac{PV}{RT}=\frac{1.00atm*1.00L}{0.082\frac{atm*L}{mol*K}*373.15K}\\ \\n=0.0327mol[/tex]

In such a way, since the molar mass of air is 28.97 g/mol, we can compute the mass of air with a single mass-mole relationship:

[tex]m_{air}=0.0327mol*\frac{28.97g}{1mol} =0.947g[/tex]

Finally, knowing that the 21% of the 0.0327 moles of air is oxygen, its moles turn out:

[tex]n_{O_2}=0.0327mol*\frac{0.21molO_2}{1mol} =0.00686molO_2[/tex]

Best regards.

Answer:

[tex]\boxed{Butyne}[/tex]

Explanation:

Triple Bonds => So it is an alkyne

The suffix used will be "-yne"

4 Carbons => The prefix used will be "But-"

Combining the prefix and suffix, we get:

=> Butyne

Answer:

[tex]\boxed{\mathrm{Butyne}}[/tex]

Explanation:

Alkynes have triple bonds ≡. The molecule has one triple bond.

Suffix ⇒ yne

The molecule has 4 carbon atoms and 6 hydrogen atoms.

Prefix ⇒ But (4 carbons)

The molecule is Butyne.

[tex]\mathrm{C_4H_6}[/tex]

Answer:

An alkyl halide can undergo SN2 reaction with an amine

Explanation:

The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.

The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).

This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.

The detailed mechanism of this reaction has been attached to this answer.

Answer:

3-methylpentan-3-ol

Explanation:

In this case, we have an "Oxymercuration reaction". With this in mind, we will have to add an "OH" to the most substituted carbon of the double bond and we will obtain 3-methylpentan-3-ol. To understand  how this molecule is produced we have to check the mechanism:

The mercury(II) acetate ([tex]Hg(OAC)_2[/tex]) is an ionic substance. So, this substance can be ionized into his ions and we will have the cation [tex]HgOAc^+[/tex] and the anion [tex]AcO^-[/tex]. The cation will attack the double bond and vice-versa to produce a "cyclic intermediate". Then a water molecule will attack the most substituted carbon and the cyclic compound would be broken producing a new bond C-O with a positive charge in the oxygen. Then a deprotonation step takes place and finally, the [tex]NaBH_4[/tex] would reduce the compound to produce the final alcohol.

See figure 1

I hope it helps!

Answer:

KMnO4,H3O^+,75°C

Explanation:

The conversion of cyclohexene to trans-1,2-cylohexanediol is an oxidation reaction. Alkenes are oxidized in the presence of potassium permanganate and acids to yield the corresponding diols.

These diols may also be called glycols. They are molecules that contain two -OH(hydroxyl) groups per molecule. The reaction closely resembles the addition of the two -OH groups of hydrogen peroxide to an alkene.

The bright color of potassium permanganate disappears in this reaction so it can be used as a test for alkenes.

Answer:

Partial pressure (Ar) = 316.1mmHg

Explanation:

In the mixture of Ar and H₂ you can find the total moles of both gases using general gas law and with the mass of the sample and molar weight of each gas find the mole fraction of Argon and thus, its partial pressure.

Moles of gases:

PV = nRT

P = 910.0mmHg ₓ (1atm / 760mmHg) = 1.1974atm

V = 1.66L

n = Moles gases

R = 0.082atmL/molK

T = 54.9°C + 273.15K = 328.05K

PV = nRT

1.1974atm*1.66L = n*0.082atmL/molK*328.05K

Knowing H₂ = 2.016g/mol and Ar = 39.948g/mol you can write:

1.13g = 2.016X + 39.948Y (1)

Where X = moles of hydrogen and Y = moles of Argon.

Also we can write:

0.0739moles = X + Y (2)

Total moles of the sample are moles of hydrogen + moles Argon

Replacing 2 in 1:

1.13g = 2.016(0.0739-Y) + 39.948Y

1.13 = 0.1564 - 2.016Y + 39.948Y

0.9736 = 37.932Y

0.02567 = Y = moles of Argon

As total moles are 0.0739moles, mole fraction of Ar in the sample are:

XAr = 0.02567mol / 0.0739mol

X Ar = 0.347

Last, partial pressure of Ar = X Ar * total pressure.

Partial pressure (Ar) = 0.347*910.0mmHg

Answer

mass of aluminum metal= 7 .0497g of Al

Explanation:

current = 21 A

time = 1 hour = 60 X 60 = 3600 s

quantity of electricity passed = current X time = 21X 3600 = 75600 C

Following the electrolysis the below reaction will occur :

Al3+ + 3e- --------> Al

therefore, 3F i.e. 3 X 96500 C = 289500 C gives 1 mole of Al

so 1 C will produce 1/289500 moles of Al

so 108000 C will produce 1/289500 X 75600 = 0.2611 moles of Al

now 1 mole of aluminium weighs = 27 g/mole

so 0.2611 moles of Al = 0.2611 X 27 = 7 .0497 g

mass of aluminum metal= 7 .0497 g of Al

The mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A is 7.05 g

We'll begin by calculating the the quantity of electricity used. This can be obtained as follow:

Current (I) = 21 A

Time(t) = 1 h = 60 × 60 = 3600 s

Q = it

Q = 21 × 3600

Recall:

1 mole of Al = 27 g

1 electron (e) = 96500 C

Thus,

3 electrons = 3 × 96500 = 289500 C

From the balanced equation above,

289500 C of electricity produced 27 g of Al.

Therefore,

75600 C of electricity will produce = (75600 × 27) / 289500 = 7.05 g of Al

Thus, the mass of the aluminum metal obtained is 7.05 g

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Answer:

E. Q < K and reaction shifts right

Explanation:

Step 1: Write the balanced equation

A(s) + 3 B(l) ⇄ 2(aq) + D(aq)

Step 2: Calculate the reaction quotient (Q)

The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.

Q = [C]² × [D]

Q = 0.64² × 0.38

Q = 0.15

Step 3: Compare Q with K and determine in which direction will shift the reaction

Since Q < K, the reaction will shift to the right to attain the equilibrium.

Answer:

Compound are defined as the containing two or more different element .

(1) Ionic compound and (2) Covalent compound.

Explanation:

Covalent compound :- covalent compound are the sharing of electrons two or more atom.

Covalent compound are physical that lower points and compared to ionic .

Covalent compound that contain bond are carbon monoxide (co), and methane .

Covalent compound are share the pair of electrons.

Covalent compound are bonding a hydrogen atoms electron.

Ionic compound a large electrostatic actions between atoms.

Ionic compound are higher melting points and covalent compound.

Ionic compound are bonding a nonmetal electron.

Ionic electron can be donate and received ionic bond.

Ionic compound bonding kl.

Answer:

Final product: cyclopent-1-en-1-ylbenzene

Explanation:

In this case, we have a Wittig reaction. The addition of [tex]PPh_3[/tex] and n-Buli will produce the "Ylide compound". First, we will have an Sn2 reaction in which the iodide is replaced by triphenylphosphine. Then the base n-Buli will remove a hydrogen atom to form a double bond (Ylide compound). Then the double bond will be delocalized to produce a carbanion. This carbanion, will attack the carbon in the carbonyl group generating a negative charge in the oxygen. Then the negative charge will attack the phosphorous atom to produce a cyclic structure. Finally, the cyclic structure is broken to produce the alkene (cyclopent-1-en-1-ylbenzene).

See figure 1

I hope it helps!

Answer:

This means the the sign of q for the reaction was _NEGATIVE _____ and the reaction was _EXOTHERMIC_____.

Explanation:

In calorimetry, when heat is absorbed by the solution, the q-value of the solution will have a positive value. This means that the reaction will produce heat for the solution to absorb and thus the q-value for the reaction will be negative. This is an exothermic reaction.

Whereas, when heat is absorbed from the solution, the q-value for the solution will have a negative value. This means that the reaction will absorb heat from the solution and so the reaction is endothermic, and q value for the reaction is positive.

So, from the question, since the q-value of water is positive, it means that heat is absorbed by the solution and the reaction will produce a negative value of q and it's an exothermic reaction because the reaction produces heat for the solution.

Answer:

Rate law: [tex]r=k[A]^2[/tex]

Integrated rate law: [tex]\frac{1}{[A]}=kt+ \frac{1}{[A]_0}[/tex]

Rate constant: [tex]k=3.40x10^{-2}\frac{L}{mol*s}[/tex]

Explanation:

Hello,

In this case, since the slope is obtained by plotting 1/[A] and it has the units L/(mol*s) or 1/(M*s), we can infer the reaction is second-order, therefore, its rate law is:

[tex]r=k[A]^2[/tex]

The integrated rate law:

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

That is obtained from the integration of:

[tex]\frac{d[A]}{dt}=-k[A]^2[/tex]

And of course, since the slope equals the rate constant, its value is:

[tex]k=3.40x10^{-2}\frac{L}{mol*s}[/tex]

Regards.

Answer:

1-cyclopentylhexan-2-one

Explanation:

1-cyclopentylhexan-2-one

Answer:

DE2

Explanation: for every one D+2 you need two E-1 because +2=-2