Nylon 88 is made from the monomers H2N(CH2)8NH2 and HOOC(CH2)6COOH. So, would you characterize nylon 88 as rather an addition or a condensation polymer? Please explain your answer.

Answers

Answer 1

Answer:

Combination of H2N(CH2)8NH2 and HOOC(CH2)6COOH leads to the loss of water molecules at each linkage position.

Explanation:

A condensation polymer is a polymer formed when two monomers combine with the elimination of a small molecule such as water. The removal of the small molecule occurs at the point where the two monomers are joined to each other.

Nylon is known to form condensation polymers. This is because it involves the linkage of an -OH group to an -NH2 group. Water is eliminated in the process.

In the case of H2N(CH2)8NH2 and HOOC(CH2)6COOH, linkage of the both monomers at the 8 position of each chain leads to the formation of nylon- 8,8 with loss of water molecules at each linkage position. This stepwise loss of water molecules at each linkage makes it a condensation polymer.


Related Questions

2) Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc. Escreva V(verdadeiro) ou F (falso) em cada afirmação.

( ) Foguetes só levam astronautas ao espaço.

( ) Satélites artificiais servem para ajudar na previsão do clima.

( ) Satélites artificiais "fotografam" o planeta para descobrir queimadas ilegais.

( ) Satélites artificiais permitem vermos jogos ao vivo até do Japão.

( ) Foguetes são movidos com pólvora e dinamite.

Answers

Answer:

F, V, V , V, F

Explanation:

1 - "Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc".

2 - Tipo Meteorologia: utilizados para monitorar o tempo e o clima no planeta Terra, por exemplo, os da série Meteosat.

3 - ...

4 - ...

5 - Usam combustivel solido, liquido, hibridos (solido e liquido), iônica:

Solido:

 São sistemas simples que unem os dois propelentes envolvidos em uma massa sólida que, quando inflamada, não para de queimar até o esgotamento completo.

Liquido:

 São muito mais complexos e envolvem o bombeamento de quantidades imensas de propelentes para as câmaras de combustão dos motores.

Hibridos:

 O propelente sólido – normalmente o combustível – é distribuído ao longo do tanque de maneira homogênea. O propelente líquido ou gasoso "normalmente o oxidante" fica armazenado em tanques.

 Podem ser desligados depois de sofrerem ignição, além de permitirem um controle de queima relativamente preciso.

Iônica:

 Usando eletricidade (captada por painéis solares ou gerada por reatores atômicos) para ionizar átomos (normalmente gases nobres, como xenônio), e expulsá-los em velocidades altíssimas.

A newly found element with the symbol J has two naturally occurring isotopes. Isotope one has an atomic mass of 139.905 amu and an abundance of 37.25%. Isotope two has an atomic mass of 141.709 amu and an abundance of 62.75%. Calculate the mass of the element.

Answers

Answer:

The mass of the element is 141.03701 amu

Explanation:

The catch here is that it notes a " newly found element. " Otherwise you could just refer to the average atomic mass of the element in the periodic table, and receive your solution in a much faster way.

The first isotope has an atomic mass of 139.905 amu, and a respective percent abundance of 37.25%. The second isotope has an atomic mass of 141.709 amu, and the remaining percent abundance, 100% - 37.25% = 62.75% ( given ). We can calculate the mass of the unknown element by associating each percentage with the mass of their respective isotope, over 100%.

Mass = ( ( 139.905 amu )( 37.25% ) + ( 141.709 amu )( 62.75% ) )/ 100,

Mass = ( ( 5211.46125 ) + ( 8892.23975 ) ) / 100,

Mass = ( 14103.701 ) / 100 = 141.03701 amu

50.0 mL each of 1.0 M HCl and 1.0 M NaOH, at room temperature (20.0 OC) are mixed. The temperature of the resulting NaCl solution increases to 27.5 OC. The density of the resulting NaCl solution is 1.02 g/mL. The specific heat of the resulting NaCl solution is 4.06 J/g OC Calculate the Heat of Neutralization of HCl(aq) and NaOH(aq) in KJ/mol NaCl produced

Answers

Answer:

-62.12kJ/mol is heat of neutralization

Explanation:

The neutralization reaction of HCl and NaOH is:

HCl + NaOH → NaCl + H₂O + HEAT

An acid that reacts with a base producing a salt and water

You can find the released heat of the reaction  -heat of neutralization- (Released heat per mole of reaction) using the formula:

Q = C×m×ΔT

Where Q is heat, C specific heat of the solution (4.06J/gºC), m its mass of the solution and ΔT change in temperature (27.5ºC-20.0ºC = 7.5ºC).

The mass of the solution can be found with the volume of the solution (50.0mL of HCl solution + 50.0mL of NaOH solution = 100.0mL) and its density (1.02g/mL), as follows:

100.0mL × (1.02g / mL) = 102g of solution.

Replacing, heat produced in the reaction was:

Q = C×m×ΔT

Q = 4.06J/gºC×102g×7.5ºC

Q = 3106J = 3.106kJ of heat are released.

There are 50.0mL ×1M = 50.0mmoles = 0.0500 moles of HCl and NaOH that reacts releasing 3.106kJ of heat. That means heat of neutralization is:

3.106kJ / 0.0500mol of reaction =

-62.12kJ/mol is heat of neutralization

The - is because heat is released, absorbed heat has a + sign

An aqueous solution is made by dissolving 29.4 grams of aluminum acetate in 433 grams of water. The molality of aluminum acetate in the solution is

Answers

Answer:

0.333 m

Explanation:

Molality (m) is moles of solute over kilograms of solvent.

Convert grams of the solute (aluminum acetate) to moles.

(29.4 g)/(204.11 g/mol) = 0.144 mol

Convert grams of the solvent (water) to kilograms.

433 g = 0.433 kg

Divide the solute by the solvent.

(0.144 mol)/(0.433 kg) = 0.333 m

The molality of the solution is 0.333 m.

The Lucas test has _______ results based on the type of alcohol present because the reaction involves a _________, which is ________ stable for tertiary alcohols compared to primary alcohols. Therefore, tertiary alcohols react ________ primary alcohols.

Answers

Answer:

1) positive

2) carbocation

3) most stable

4) faster

Explanation:

A common test for the presence of alcohols can be achieved using the Lucas reagent. Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride.

The reaction of Lucas reagent reacts with alcohols leading to the formation of an alkyl chloride. Since the reaction proceeds via a carbocation mechanism, tertiary alcohols give an immediate reaction. Once a tertiary alcohol is mixed with Lucas reagent, the solution turns cloudy almost immediately indicating an instant positive reaction.

Secondary alcohols may turn cloudy within five minutes of mixing the solutions. Primary alcohols do not significantly react with Lucas reagent obviously because they do not form stable carbocations.

Therefore we can use the Lucas reagent to distinguish between primary, secondary and tertiary alcohols.

What is/are the major organic product(s) of the following reaction, Question 2 options: A) CH3CH2CCH Br B) CH3CCCH3 Br C) CH2CH2 HCCH Br D) HCCCH2CH2Br E) HCCBr

Answers

Answer:

CH3CH2C≡CH

Explanation:

The particular reaction under study is known as the alkykation of acetylide ions. An acetylide ion can be alkykated using a suitable alkyl halide. The overall scheme of the reaction is;

CH≡C^- + RX -----> RC≡CH + X^-

This reaction is most effective when primary alkyl halides are used. It involves SN2 substitution of a halide in the alkyl halide by an acetylide ion. Secondary, tertiary or even bulky primary substrates are known to yield alkenes and alkynes owing to elimination by E2 mechanism.

The reaction, 2 NO(g) + O2(g) → 2 NO2(g), was found to be first order in each of the two reac­tants and second order overall. The rate law is therefore

Answers

Answer:

[tex]r=k[NO][O_2][/tex]

Explanation:

Hello,

In this case, rate laws allows us to compute how fast a chemical reaction is carried out by means of the change in the concentration of the species affecting the rate. In such a way, since the statement says that the reaction was found to be first order to both nitrogen monoxide and oxygen, it means that their concentrations are powered to first power by separated. It also implies that the overall order is second-order since the specific orders are added (powers properties). Therefore, the rate law is:

[tex]r=k[NO][O_2][/tex]

Whereas k is the rate constant and we find the concentration of the reactants to the first power each one.

Best regards.

We wear cotton clothes in summer.

Answers

Answer:

we wear cotton clothes because it helps to cool us down and remove the excess heat that causes us to feel hot.

Answer:

[tex]\boxed{\mathrm{view \: explanation}}[/tex]

Explanation:

We wear cotton clothes in the summer beacuse cotton absorbs and removes body moisture caused by the sweat and allows better air circulation than fabric clothes.

A 1.0 kg object absorbs 1,303 J of heat energy and experiences a temperature increase of 5.2∘C. What is the object’s specific heat, in joules per gram-degree celsius? Report your answer with the correct number of significant figures.

Answers

Answer:

c = 250.58 J/kg/[tex]^{0}C[/tex]

Explanation:

The specific heat of a substance is the required quantity of heat to increase or decrease the temperature of its unit mas by 1 kelvin.

Q = mcΔθ

where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is its specific heat and Δθ is the change in temperature of the substance.

Given that; m = 1.0 kg, Q = 1303 J and Δθ = 5.2 [tex]^{0}C[/tex], then;

c = Q ÷ (mΔθ)

  = 1303 ÷ (1.0 × 5.2)

  = 1303 ÷ 5.2

  = 250.58 J/kg/[tex]^{0}C[/tex]

The specific heat of the object is 250.58 J/kg/[tex]^{0}C[/tex].

Answer:

0.25

Explanation:

The idea that light can act as packets led to what new field of science?
A. Quantum mechanics
B. Nuclear mechanics
C. Electrical mechanics
D. Physical mechanics

Answers

The answer is A. Quantum Mechanics
Quantum mechanics has to do with light and incredibly small particles.



Nuclear mechanics has to do with protons in atoms. Electrical mechanics has to do with electricity and magnetism. Physical mechanics has to do with motion

Content attribution
QUESTION 2 • 1 POINT
Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?​

Answers

The given question is incomplete. The complete question is :

Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?​

a) [tex]O^{2-}[/tex]

b)  [tex]F^{-}[/tex]

c)  [tex]N^{3-}[/tex]

d)  [tex]S^{2-}[/tex]

Answer: b)  [tex]F^{-}[/tex]

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

Here potassium is having an oxidation state of +1 called as  cation and thus is an anion must have an oxidation state of -1 if they have to combine in 1: 1 ratio to  give neutral ionic compound.

Thus the anion has to be [tex]F^-[/tex] which combines with [tex]K^+[/tex] in 1: 1 ratio to give [tex]KF[/tex]

Sodium pentothal is a short-acting barbiturate derivative used as a general anesthetic and known in popular culture as truth serum. It is synthesized like other barbiturates, but uses thiourea, (H2N)2C=S, in place of urea. The mechanism involves the following steps:
1. Ethoxide ion deprotonates malonic ester, forming enolate anion 1;
2. Enolate anion 1 acts as a nucleophile in an SN2 reaction with ethyl bromide, forming alkylated intermediate 2;
3. Ethoxide ion deprotonates alkylated intermediate 2, forming enolate anion 3;
4. Enolate anion 3 acts as a nucleophile in an SN2 reaction with 2-bromopentane, forming alkylated intermediate 4;
5. Alkylated intermediate 4 reacts with thiourea to form tetrahedral intermediate 5;
6. Tetrahedral intermediate 5 collapses, expelling ethoxide ion and forming intermediate 6;
7. Intermediate 6 reacts with sodium hydroxide to form sodium pentothal.
Draw the mechanism out on a separate sheet of paper and then draw the structure of enolate anion 1. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading.

Answers

Answer:

Figure 1

Explanation:

In this case, we have to ester with a "malonic synthesis" in which we have to add a strong (ethoxide) to produce an enolate ion that would form a new C-C bond with an alkyl halide (ethyl bromide and bromo pentane). Then a "nucleophilic acyl substitution reaction" takes place to add thiourea, in this step two ethanol groups are eliminated to form a cyclic structure. Finally, an "elimination reaction" happen by the addition of sodium hydroxide generating a double bond and a negative charge in the sulfur atom that is neutralized with the positive charge of sodium.

See figure 1 for the total mechanism

I hope it helps!

hen adding a solute to water, the vapor pressure will __________ and the boiling point will __________.

Answers

Answer: When a solute is added to water, the vapor pressure will decrease and the boiling point will increase.

Explanation:

When a solute is added to water, a solvent's vapor pressure will decrease because of the displacement of solvent molecules by the solute. i.e. some of the solvent molecules at the surface of the water are replaced by the solute.When a solute is added to water, a solvent's boiling point will increase because water molecules need more energy to produce required pressure to escape the boundary of the liquid , so as the number of particles increase in the liquid it increase the boiling point.

Hydrogen iodide decomposes according to the equation: 2HI(g) H 2(g) + I 2(g), K c = 0.0156 at 400ºC A 0.660 mol sample of HI was injected into a 2.00 L reaction vessel held at 400ºC. Calculate the concentration of HI at equilibrium.

Answers

Answer:

[HI] = 0.264M

Explanation:

Based on the equilibrium:

2HI(g) ⇄ H₂(g) + I₂(g)

It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:

Kc = 0.0156 = [H₂] [I₂] / [HI]²

As initial concentration of HI is 0.660mol / 2.00L = 0.330M, the equlibrium concentrations will be:

[HI] = 0.330M - 2X

[H₂] = X

[I₂] = X

Where X is reaction coefficient.

Replacing in Kc:

0.0156 = [X] [X] / [0.330M - 2X]²

0.0156 = X² / [0.1089 - 1.32X + 4X² ]

0.00169884 - 0.020592 X + 0.0624 X² = X²

0.00169884 - 0.020592 X - 0.9376 X² = 0

Solving for X:

X = - 0.055 → False solution, there is no negative concentrations

X = 0.0330 → Right solution.

Replacing in HI formula:

[HI] = 0.330M - 2×0.033M

[HI] = 0.264M

A 25.0-mL sample of 0.100M Ba(OH)2(aq) is titrated with 0.125 M HCl(aq).
How many milliliters of the titrant will be needed to reach the equivalence point?

Answers

Answer:

20.0

Explanation:

NaOH = (25.0) (0.100m) \ 0.125M = 20.0mL

The volume of a sample of oxygen is 300mL when the pressure is 1 atm and the temperature is 27 C . At what temperature is the volume 1.00 L and the pressure.500 atm?

Answers

Answer:

T2 = 500K

Explanation:

Given data:

P1 = 1atm

V1 = 300ml

T1= 27 + 273 = 300K

T2 = ?

V2 = 1.00ml

P2 = 500atm

Apply combined law:

P1xV1//T1 = P2xV2/T2 ...eq1

Substituting values into eq1:

1 x 300/300 = 500 x 1/T2

Solve for T2:

300T2 = 500 x 300

300T2 = 150000

Divide both sides by the coefficient of T2:

300T2/300 = 150000/300

T2 = 500K

A meteorologist filled a weather balloon with 3.00L of the inert noble gas helium. The balloon's pressure was 765 torr. The balloon was released to an altitude with a pressure of 530 torr. What was the volume (L) of the weather balloon

Answers

Answer:

4.33 L

Explanation:

Step 1: Given data

Initial volume of the balloon (V₁): 3.00 L

Initial pressure of the balloon (P₁): 765 torr

Final  volume of the balloon (V₂): ?

Final pressure of the balloon (P₂): 530 torr

Step 2: Calculate the final volume of the balloon

If we consider Helium to behave as an ideal gas, we can calculate the final volume of the balloon using Boyle's law.

[tex]P_1 \times V_1 = P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{765torr \times 3.00L}{530torr} = 4.33 L[/tex]

At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=3.0×10−8 M?

Answers

Answer:

[tex][OH^-]=3.33x10^{-7}M[/tex]

Explanation:

Hello,

In this case, for the given concentration of hydronium, we can compute the pH as shown below:

[tex]pH=-log([H^+])=-log(3.0x10^{-8})=7.52[/tex]

Now, given the relationship between pH and pOH we can compute the pOH which is directly related with the concentration of hydroxyl in the solution:

[tex]pOH=14-pH=14-7.52=6.48[/tex]

Then, the concentration of hydroxyl turns out:

[tex][OH^-]=10^{-pOH}=10^{-6.48}[/tex]

[tex][OH^-]=3.33x10^{-7}M[/tex]

Best regards.

Assume that a compound is a cyclic, planar, completely conjugated ring. Which number of p electrons would make it aromatic?
a) 0 p electrons
b) 2 p electrons
c) 3 p electrons
d) 4 p electrons
e) 32 p electrons

Answers

Answer:

option b is correct

2 p electron makes aromatic

Explanation:

An aromatic compound which is cyclic, planar and has a complete conjugate ring must have (4n + 2)pi electrons(Huckel's rule)

Huckel's Standard (4n+2 rule): For a compound to be an aromatic, a particle must have a specific number of pi (electrons with pi bonds, or lone pairs inside p orbitals) inside a shut loop of parallel, adjoining p orbitals. The pi electron tally is characterized by the arrangement of numbers created from 4n+2 where n = zero or any positive whole number (i..e, n = 0, 1, 2, and so forth.). The most widely recognized case in six pi electrons (n = 1) which is found for example in benzene, pyrrole, furan, and pyridine.

where n is the number of pi electrons

where n = 0

(4n +2) pi electrons = 2pi electrons

attached is an example of aromatic which is cyclic, planar and a complete conjugate ring

Answer:

2 p electrons.

Explanation:

For any compound to be considered an aromatic compound it must be cyclic,flat, conjugated and it must obey Huckel's rule that states an aromatic compound must have 4n + 2 pi electrons in it's p orbitals for it to be an aromatic compound.

n can represent an integer from 2, 6,10, 14,........

The lone pair is actually in a pure 2p orbital perpendicular to the ring, which means they count as π electrons.

The following reaction is part of the electron transport chain. Complete the reaction and identify which species is reduced. The abbreviation Q represents coenzyme Q. Use the appropriate abbreviation for the product.
FADH2+Q→
The reactant that is reduced is: _____

Answers

Answer:

[tex]FADH_2+Q --> FAD + QH_2[/tex]

The reactant that is reduced is Q.

Explanation:

The complete equation for the reaction is such that:

[tex]FADH_2+Q --> FAD + QH_2[/tex]

Two molecules of H atom is lost from [tex]FADH_2[/tex] and the H atoms are gained by the coenzyme Q. Consequently,  [tex]FADH_2[/tex] becomes FAD while Q becomes [tex]QH_2[/tex].

From the definition of oxidation as loss of hydrogen and reduction as the addition of hydrogen, it can be concluded that the FADH2 that lost hydrogen is a reactant that is oxidized while the coenzyme Q that gained hydrogen is a reactant that is reduced in the reaction.

What is the percent yield for a chemical reaction if the actual yield is 36 g and the theorical yield is 45 g.

Answers

Answer:

⇒ Percent yield = 80 %

Explanation:

Given:

Actual yield = 36 g

Theoretical yield = 45 g

Find:

Percent yield

Computation:

⇒ Percent yield = [Actual yield / Theoretical yield] 100%

⇒ Percent yield = [36 / 45] 100%

⇒ Percent yield =[0.8] 100%

⇒ Percent yield = 80 %

An analytical laboratory balance typically measures mass to the nearest 0.1 mg. You may want to reference (Page) Section 21.6 while completing this problem. Part A What energy change would accompany the loss of 0.1 mg in mass

Answers

Answer:

The  energy change is  [tex]E = 9.0 *10^{9}\ J[/tex]

Explanation:

   From the question we are told that

          Mass loss  is  [tex]m_l = 0.1 \ mg = 0.1 *10^{-3} mkg = 0.1 *10^{-6} \ kg[/tex]

  Generally the energy change that  would accompany this loss  is mathematically represented as

     [tex]E = m * c^2[/tex]

Where  c is the speed of light with values [tex]c = 3.0*10^{8} \ m/s[/tex]

     [tex]E = 0.1 *10^{-6} * [3.0 *10^{8}]^2[/tex]

     [tex]E = 9.0 *10^{9}\ J[/tex]

   

1. In this experiment, the procedure instructs you to dissolve solid potassium hydrogen tartrate (KHT) in two different solvents. What are these two solvents? (2 pts)

Answers

Answer:

Water

Explanation:

Solid potassium hydrogen tartrates (KHT) is soluble in water. This is especially at room temperature.

The solvent for KHT is water.

Identify a process that is NOT reversible. A. melting of steel B. freezing water C. melting of ice D. frying an egg E. deposition of carbon dioxide (gas to solid)

Answers

I’m pretty sure it would be D. Frying an egg

A process that is not a reversible reaction is frying an egg.

What are reversible reactions?

Reversible reactions are those reactions in which product will again change into the reactant.

Melting of steel and ice are reversible reaction as after cooling again we get the original state of steel and ice.Freezing of water is also reversible reaction as at normal temperature we get the original state of water.Deposition of carbon dioxide is also a reversible reaction.Frying an egg is a non reversible reaction as after frying an egg we didn't get the original egg again.

Hence option (D) is correct.

To know more about reversible reaction, visit the below link:

https://brainly.com/question/1495850

When the optically active carboxylic acid below is decarboxylated using the conditions typical in the acetoacetate synthesis, will the ketone product also be optically active?

Answers

Answer:

ye, it will be optically active

Explanation:

a compound is said to be optically active if it can optically rotate.

the removal of carboxyl group and release of cabon dioxide from carboxylic acid in acetoacetate synthesis which will result in production of ketone as given the attachment below.

Which of the following solutions would have the highest pH? Assume that they are all 0.10 M in acid at 25°C. The acid is followed by its Ka value.

a. HCHO2, 1.8 x 10-4
b. HF, 3.5 x 10-4
c. HClO2, 1.1 x 10-2
d. HCN, 4.9 x 10-10
e. HNO2, 4.6 x 10-4

Answers

Answer:

[tex]HCN~~Ka=4.9x10^-^1^0[/tex]

Explanation:

In this case, we have to remember the relationship between the Ka value and the pH. We can use the general reaction for any acid with his Ka value expression:

[tex]HA~->~H^+~+~A^-[/tex]    [tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]

In the Ka expression, we have a proportional relationship between Ka and the concentration of [tex]H^+[/tex]. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher

So, if we have to find the higher pH value we need to search the smaller Ka value in this case [tex]HCN~~Ka=4.9x10^-^1^0[/tex].

I hope helps!

HCN has the highest pH among all the acids listed in the question.

The Ka is called the acid dissociation constant. It shows the extent to which an acid is ionized in water. The pH shows the hydrogen ion concentration of water. The higher the Ka, the higher the hydrogen ion concentration and the lower the pH.

Hence, HCN has the lowest Ka and the lowest hydrogen ion concentration. Therefore, HCN has the highest pH among all the acids listed in the question.

Learn more: https://brainly.com/question/6505878

2.50 mol NOCl was placed in a 2.50 L reaction vessel at 400ºC. After equilibrium was established, it was found that 28% of the NOCl had dissociated according to the equation 2NOCl(g) 2NO(g) + Cl 2(g). Calculate the equilibrium constant, K c, for the reaction.

Answers

Answer:

Explanation:

2NOCl(g)    ⇄      2NO(g) + Cl 2(g)

C ( 1 - .28 )            .28 C         .14 C

Kc = [ NO ]²  x [ Cl₂ ] / [ NOCl ]²

= (.28 C )² x .14 C / C² ( 1 - .28 )²

= .021173 x C

C = concentration of reactant

= 2.5 / 2/5 = 1 M

Kc = .021173 x 1

= 211.73 x 10⁻⁴ M .

The equilibrium constant will be "2.117×10⁻²".

Given:

Number of moles = 2.50 molVolume of solution = 2.5 L

At equilibrium,

Concentration of NO = 0.28 MConcentration of Cl₂ = 0.14 M

Now,

The concentration of NOCl will be:

= [tex]\frac{Number \ of \ moles}{Volume \ of \ solution}[/tex]

= [tex]\frac{2.5}{2.5}[/tex]

= [tex]1 \ M[/tex]

At equilibrium,

The concentration of NOCl will be:

= [tex]1-0.28[/tex]

= [tex]0.72 \ M[/tex]

hence,

The equilibrium constant,

→ [tex]K_c =\frac{ [NO]^2 [Cl_2]}{[NOCl]^2}[/tex]

By substituting the values, we get

        [tex]= \frac{(0.28)^2\times (0.14)}{(0.72)^2}[/tex]

        [tex]= 2.117\times 10^{-2}[/tex]

Thus the above answer is right.

Learn more about equilibrium below:

https://brainly.com/question/14954687

A vehicle travels 2345 meter in 35 second toward the evening sun in the West. What is its speed? A. 47 m/s West

Answers

Explanation:

Speed = 2345 ÷ 35 = 67m/s

What is the balanced oxidation half-reaction for the following reaction? Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq) Question 8 options: There is no reaction Cu2+(aq) + 2e– → Cu(s) Fe2+(aq) + 2e– → Fe(s) Cu(s) → Cu2+(aq) + 2e– Fe(s) → Fe2+(aq) + 2e–

Answers

Answer:

Fe(s) → Fe2+(aq) + 2e-

Explanation:

Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq)

Oxidation is the loss of electrons. When the oxidation number of an element increases, that means there is a loss of electrons and that element is being oxidized.

The oxidation half equation in this reaction is;

Fe(s) → Fe2+(aq)

The loss of electron is represented in the product side and is given by;

Fe(s) → Fe2+(aq) + 2e-

What is the molar concentration of H atoms at equilibrium if the equilibrium concentration of H2 is 0.28 M? Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

0.56M

Explanation:

Molar concentration is defined as the ratio between moles of solute and volume in liters of solution.

In a 0.28M H₂ there are 0.28moles of H₂ per liter of solution.

Now, in 1 molecule of H₂ there are 2 atoms of H. Following this idea, in 0.28 moles of H₂ there are 0.28*2 = 0.56 moles of H atoms.

Thus, molar concentration of H atoms in a 0.28M H₂ is 0.56M

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