Give a term for each description.Organic chemicals that can vaporize into the air.Secondary chemicals whose formation is facilitated by sunlight.Amount of gases and small particles in the atmosphere that influence ecosystem or human well-being.Gases or particles that are present in high enough concentrations to harm humans, other organisms, and buildings and other structures.Very small solid and liquid particles suspended in the air.

Answer:

I think that depends on the type of gas and the volume of the container.

Answer:

The molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.

Explanation:

Consider a slightly soluble solid with formula M₃X₂. Its solubility product expression is

[tex]\begin{array}{rcccc}M_{3}X_{2}(s) & \rightleftharpoons&3M^{2+}(aq) & + & 2X^{3-}(aq)\\& & 3s & &2s\\\\K_\text{sp}& = & [3s]^{3}[2s]^{2}&= & 108s^{5}\\\\\end{array}[/tex]

Thus, the molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.

A is wrong. The solubility product constant is a constant. It does not change in the presence of a common ion.

B is wrong. It is correct only for compounds with formula MX.

C is wrong. Ksp does not equal the concentration of the compound in solution.

The value of Ksp is related to the molar solubility in that molar solubility helps calculate the ion concentrations in a solution, which are used to calculate Ksp. The presence of a common ion affects both solubility and Ksp through the common ion effect.

The value of Ksp, or the solubility product constant, is closely related to the molar solubility of a compound, which is a measure of how much of the compound can dissolve in a given amount of solvent to form a saturated solution. The connection between Ksp and molar solubility can be understood through a series of steps:

Furthermore, if a common ion is present in the solution, it affects the solubility of the compound and thus the Ksp value. This is a demonstration of the common ion effect, where the presence of a common ion decreases the solubility and Ksp of the compound according to Le Chatelier's principle.

Therefore, understanding the relationship between Ksp and molar solubility is essential for predicting the solubility of compounds and determining the possible concentrations of ions in solution.

First we need to calculate the number of moles of FeS[tex]_{2}[/tex]:

number of moles = mass (grams) / molecular mass (g/mol)

number of moles of FeS[tex]_{2}[/tex] = 198.2/120 = 1.65 moles

From the chemical reaction we deduce that:

if            4 moles of FeS[tex]_{2}[/tex] produces 8 moles of SO[tex]_{2}[/tex]

then  1.65 moles of FeS[tex]_{2}[/tex] produces X moles of SO[tex]_{2}[/tex]

X = (1.65×8)/4 = 3.3 moles of SO[tex]_{2}[/tex]

Now we can calculate the mass of SO[tex]_{2}[/tex]:

mass (grams) = number of moles × molecular mass (grams/mole)

mass of SO[tex]_{2}[/tex] = 3.3×64 = 211.2 g

Answer : The enthalpy of the following reaction is, -390.3 KJ

Explanation :

The given balanced chemical reactions are,

(1) [tex]C(s)+2H_2(g)\rightarrow CH_4(g)[/tex]     [tex]\Delta H_1=-74.6KJ/mole[/tex]  

(2) [tex]C(s)+2Cl_2(g)\rightarrow CCl_4(g)[/tex]     [tex]\Delta H_2=-95.7KJ/mole[/tex]

(3) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]     [tex]\Delta H_3=-184.6KJ/mole[/tex]

The final reaction of is,

[tex]CH_4(g)+4Cl_2(g)\rightarrow CCl_4(g)+4HCl(g)[/tex]       [tex]\Delta H_{rxn}=?[/tex]

Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of of the reaction.

The expression for enthalpy for the following reaction will be,

[tex]\Delta H_{rxn}=[2\times \Delta H_3]+[-1\times \Delta H_1]+[1\times \Delta H_2][/tex]

where,

n = number of moles

Now put all the given values in the above expression, we get:

[tex]\Delta H_{rxn}=[2mole\times (-184.6KJ/mole)]+[-1mole\times (-74.6KJ/mole)]+[1\times (-95.7KJ/mole)]=-390.3KJ[/tex]

Therefore, the enthalpy of the following reaction is, -390.3 KJ

Answer:

-390.3 KJ

Explanation:

For Hess's Law, we need to get the corresponding equation below using the sequence of reactions given

By manipulating the reaction, either reversing them or multiplying/dividing them to a certain factor, we can get to the target equation as well as the total enthalpy

CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

C(s) + 2H2(g) → CH4(g)     ΔH = −74.6kJ (needs to reverse)

C(s) + 2Cl2(g) → CCl4(g)    ΔH = −95.7kJ (retain)

H2(g) + Cl2(g) → 2HCl(g)   ΔH = −184.6kJ (multiply by 2 to get 4Cl2 and cancel out 4 HCl and 4 H2)

Therefore, it is -390.3 KJ

Answer:

2

Explanation:

In balancing nuclear reactions the mass number and atomic numbers are usually conserved. This implies that from the given equation, the sum of the number of the subscript on the right hand side must be equal to that on the left hand side. This also applies to the superscript:

   For the mass numbers(superscript):

   235 + 1 = 1 + 139 + 95

       236 = 235

This is not balanced

  For the atomic number:

  92 + 0 = 0 + 53 + 39

      92 = 92

This is balanced.

We simply inspect to see how to balance the mass number.

By putting a coefficient of 2 behind the neutron atom, the equation becomes balanced.

Answer:

B. Pulverizing the zinc metal into a fine powder

Answer: B. Pulverizing the zinc metal into a fine powder

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]

A. Decreasing the concentration of [tex]HCl[/tex] and [tex]Zn[/tex]: Thus rate of the reaction would decrease on decreasing the concentration of hydrochloric acid and zinc.

B.  Pulverizing the zinc metal into a fine powder : If the solute particles is present in smaller size, more it can take part in the chemical reaction due to large surface area, hence increasing the rate of reaction.

C. Performing the reaction at a lower temperature. Decreasing the temperature means the energy of the particles is less and thus lesser reactants would cross the energy barrier and thus lesser will be the rate.

First you need to calculate the number of moles of aluminium and copper chloride.

number of moles = mass / molecular weight

moles of Al = 512 / 27 = 19 moles

moles of CuCl = 1147 / 99 = 11.6 moles

From the reaction you see that:

if        2 moles of Al will react with 3 moles of CuCl

then  19 moles of Al will react with X moles of CuCl

X = (19 × 3) / 2 = 28.5 moles of CuCl, way more that 11.6 moles of CuCl wich is the quantity you have. So the copper chloride is the limiting reagent.

Answer: T2= 962.2 K

Explanation:

The ideal gases is often written like PV=nRT, where P is pressure, V is volume, n is moles, R is the universal constant of the gases and T is Temperature.

So, in this problem there is a container that is a closed system, therefore n is constant and volume too. The initial point is 1 and the final point is 2, so

V1=V2  ⇒

[tex]\frac{n1RT1}{P1} = \frac{n2RT2}{P2} \\\\\frac{T1}{P1} = \frac{T2}{P2} \\\\T2= \frac{T1P2}{P1} =\frac{170 K 283KPa}{50 KPa} =962.2 K[/tex]

Answer:

The heat of the reaction is 105.308 kJ/mol.

Explanation:

Let the heat released during reaction be q.

Heat gained by water: Q

Mass of water ,m= 1kg = 1000 g

Heat capacity of water ,c= 4.184 J/g°C

Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

Q=mcΔT

Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C

Q'=CΔT'=CΔT

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

[tex]q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ[/tex]

Moles of propane =[tex]\frac{1.860 g}{44 g/mol}=0.0422 mol[/tex]

0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.

The heat of the reaction will be:

[tex]\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol[/tex]

Answer:

0.006 48 km/s

Explanation:

1. Convert miles to kilometres

14.5 mi × (1.609 km/1 mi) = 23.33 km

2. Convert hours to seconds

1 h × (60 min/1h) × (60 s/1 min) = 3600 s

3. Divide the distance by the time

14.5 mi/1 h = 23.3 km/3600 s = 0.006 48 km/s

Answer:

A. Arrhenius acid    

B. Bronsted-Lowry acid

Explanation:

An acid base reaction invoves an acid and a base and yields salt and water as products.

A neutralization is a reaction between an acid and a base to yield salt and water only. The highlighted compound is not shown here however we we shall tell what is compound is in this reaction.

Hence, this an acid base reaction in the Arrhenius sense.

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Answer : The rate constant at 785.0 K is, [tex]1.45\times 10^{-2}s^{-1}[/tex]

Explanation :

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_1[/tex] = rate constant at [tex]600.0K[/tex] = [tex]6.1\times 10^{-8}s^{-1}[/tex]

[tex]K_2[/tex] = rate constant at [tex]785.0K[/tex] = ?

[tex]Ea[/tex] = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]600.0K[/tex]

[tex]T_2[/tex] = final temperature = [tex]785.0K[/tex]

Now put all the given values in this formula, we get:

[tex]\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}][/tex]

[tex]K_2=1.45\times 10^{-2}s^{-1}[/tex]

Therefore, the rate constant at 785.0 K is, [tex]1.45\times 10^{-2}s^{-1}[/tex]

The rate constant at the new temperature is 1.81 ×10^−2 s−1.

We have the reaction written as follows; C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g).

We have to use the formula;

ln(k2/k1) =Ea/R(1/T1 - 1/T2)

Now;

k2 = ?

k1 = 6.1×10−8 s−1

Ea = 262 ×10 ^3 J/mol

T1= 600.0 K

T2 = 785.0 K

R = 8.314 J/K. mol

ln (k2/6.1×10−8) = 262 ×10 ^3 /8.314 (1/600.0 - 1/785.0)

ln (k2/6.1×10−8) = 12.6

k2/6.1×10−8 = e^12.6

k2 = 6.1×10−8 (e^12.6)

k2 = 1.81 ×10^−2 s−1.

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Answer: The unknown gas is B) [tex]H_2[/tex] .

Explanation: The problem is based on Graham's law of effusion rates of gases.

Carbon dioxide takes 4.69 times as long to escape as the unknown gas. It means the unknown gas is lighter than carbon dioxide since the lighter gas takes less time to escape.

From Graham's law, "Rate of effusion of a gas is inversely proportional to the molar mass of the gas."

Molar mass of carbon dioxide is 44.0 gram per mol. Let's say the molar mass of the unknown gas is M.

[tex]\frac{rate_A}{rate_C_O_2}=\sqrt{\frac{44.0}{M}}[/tex]

[tex]4.69=\sqrt{\frac{44.0}{M}}[/tex]

Do the square to both sides:

[tex]21.9961={44.0}{M}}[/tex]

[tex]M=\frac{44.0}{21.9961}[/tex]

M = 2.00

2.00 gram per mol is the molar mass of hydrogen gas. So, the correct option is B) [tex]H_2[/tex] .

Final answer:

Using Graham's Law of Effusion, the unknown gas effusing 4.69 times faster than carbon dioxide is identified as hydrogen (H2), which has a much lower molar mass, making answer B) H2 the correct choice.

Explanation:

The question involves the application of Graham's Law of Effusion, which relates the rates of effusion of two gases to their molar masses. Given that carbon dioxide takes 4.69 times as long to escape as the unknown gas, we can use the inverse relationship between the rate of effusion and the square root of the molar mass (rate ≈ 1/sqrt(molar mass)) to identify the unknown gas.

The molar mass of carbon dioxide (CO₂) is approximately 44 g/mol.

By setting up a ratio using Graham's Law, (rate of unknown gas/rate of CO₂) = sqrt(molar mass of CO₂/molar mass of unknown gas), and substituting the given rates, we can solve for the molar mass of the unknown gas:

(1/4.69)² = (44 g/mol) / (molar mass of unknown gas)

(1/22.0361) = (44 g/mol) / (molar mass of unknown gas)

Molar mass of unknown gas = 44 g/mol * 22.0361 = 969.8684 g/mol

Answer:

The correct answer is option A) generally free of technical jargon

Explanation:

Hello!

Let's solve this!

When we read popular magazines or newspapers, we see scientific findings. They generally do not use scientific or technical vocabulary so that more people can understand it. They are also not the most reliable sources, since to look for reliable information, we have to look for scientific journals.

After the analysis we conclude that the correct answer is option A) generally free of technical jargon

Answer:

a) 52.8

Explanation:

M1V1 = M2V2

(7.91 M)(x ml) = (2.13 M) (196.1 ml)

(7.91M) (xml) = 417.693 M.ml

x ml = 417.693/ 7.91

x   =  52.8

Answer:

A. 52.8

Explanation:

We make use of the dilution formula to solve this

[tex]M_1\times V_1 =M_2\times V_2[/tex]

Where ([tex]M_1[/tex] and [tex]M_2[/tex] are the initial molarity and final molarity and [tex]V_1[/tex] and [tex]V_2[/tex] are  the initial volume and final volume)

Plugging into the values in the formula  

[tex]M_1\times V_1 =M_2\times V_2[/tex]

[tex]7.91M \times V_1=2.13M \times 196.1mL[/tex]

[tex]V_1=\frac {(2.13M \times 196.1mL)}{7.91M}[/tex]

= 52.8 mL is the Answer

Answer:

Explanation:

A reversible reaction is indicated by using a double arrow ⇄

The upper arrow (→), from left to right,  indicates the direct or forward reaction, which goes from left to right.

In the direct reaction, the reactants are the substances shown on the left side of the equation, and the products are the substances shown on the right side.

The lower arrow (←), from right to left, indicates the reverse reaction, which goes from right to left.

In the reverse reaction, the reactants are the substances shown of the right side and the products are the substances shown of the left side.

Summarizing:

In the moment that the rates of both forward and reverse reactions are equal it is said that the equilibrium has been reached.

Answer:

just did edge test:

c. NH4CI(s)<—–>NH3( g)+HCI(g)

1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?

Using the graph form the picture you count 8 times the halving of C¹⁴ and you arrive at 45600 years.

2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the half-life.

3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.

What the problem is telling you is that at 276 days only 25% original sample remains. If you divide the number of days by two the quantity of original sample will be multiplied by two, and you will have 138 days and 50% of original sample. This is the answer because the the half-life of a isotope is the time in which 50% of original quantity of radioactive atoms will disintegrate.

The order of increasing dipole moment is II < III < I. BCl3 has no dipole moment, BIF2 has a dipole moment, and BClF2 has a greater dipole moment compared to BIF2.

The order of increasing dipole moment is II < III < I.

The dipole moment depends on the electronegativity difference of the bonded atoms and the bond length. In this case, BCl3 has no dipole moment since the molecule is symmetrical and the individual dipole moments cancel out. BIF2 has a dipole moment due to the electronegativity difference between B and I atoms, and BClF2 has a greater dipole moment compared to BIF2 due to the additional electronegativity difference between B and Cl atoms.

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The correct answer is Option A. The order of increasing dipole moment is I < II < III.

BCl₃ is a symmetrical molecule with no net dipole moment because its individual B-Cl bonds cancel out. Hence, BCl₃ has the smallest dipole moment.

For BIF₂, the asymmetry is introduced as iodine and fluorine have different electronegativities, creating a net dipole moment. However, with only one I atom, the effect is moderate.

BClF₂ is more polar than BIF₂ because the different electronegativities of Cl and F combined in an asymmetrical structure increase the net dipole moment.

The value of Q for the reaction A (g) → 2 B (g) at 298 K when ΔG = -20.5 kJ/mol is approximately 0.00069.

To find the value of the reaction quotient Q when ΔG is known, we use the relationship between Gibbs free energy change, the equilibrium constant K, and Q.

The equation is: ΔG = ΔG° + RT ln(Q)

Where:

First, we need to calculate ΔG°. This can be done using the equilibrium constant (K): ΔG° = -RT ln(K)

Given K = 14.7:

Now we substitute ΔG and ΔG° back into the equation: ΔG = ΔG° + RT ln(Q)

-20.5 kJ/mol = -2.475 kJ/mol + (0.008314 kJ/mol·K) x (298 K) x ln(Q)

Solving for ln(Q):

-20.5 + 2.475 = (0.008314 x 298) x ln(Q)

-18.025 = 2.477 x ln(Q)

ln(Q) = -18.025 / 2.477

ln(Q) ≈ -7.28

Finally, taking the exponent to find Q: Q = [tex]e^(^-^7^.^2^8^)[/tex]

   Q ≈ 0.00069

To find the reaction quotient (Q) when ΔG = -20.5 kJ/mol, we calculate ΔG° using the given value of K and the equation ΔG° = -RT ln K. Then, we use the relationship ΔG = ΔG° + RT ln Q to solve for Q, resulting in Q =0.0037.

To determine the reaction quotient (Q) for the reaction A(g) → 2 B(g) at 298 K when ΔG = -20.5 kJ/mol, we use the relationship between ΔG, Q, and K. The equation is ΔG = ΔG° + RT ln Q. We rearrange to find Q: ln Q = (ΔG - ΔG°) / RT, where ΔG° = -RT ln K

First, calculate ΔG°:

Therefore, Q = 0.0037

In a bond, Sulfur (S) is most likely to attract electrons due to its higher electronegativity. This is based on the trend in the periodic table where electronegativity increases across a group from left to right and decreases down a group.

In a bond, elements that have high electronegativity are more likely to attract electrons. Electronegativity is a measure of the ability of an atom to attract the electrons in a bond. It increases across a period from left to right, and decreases down a group in the periodic table. Based on this premise, among the atoms listed, Sulfur (S) and Selenium (Se) have higher electronegativities than Potassium (K) and Sodium (Na).

However, between Sulfur and Selenium, Sulfur (S) is more likely to attract electrons in a bond because it is higher up in the table, given that electronegativity decreases down a group. Therefore, the answer to your question is C) Sulfur (S).

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The isotope symbol for the barium (Ba) nucleus in the given reaction is 13756Ba. The isotope symbol for the strontium (Sr) nucleus in the second reaction is 8838Sr.

In the given nuclear fission reaction, the isotope symbol for the barium (Ba) nucleus is 13756Ba. This can be determined by balancing the atomic numbers and mass numbers on both sides of the reaction.

In the second reaction, the isotope symbol for the strontium (Sr) nucleus is 8838Sr. Again, this can be obtained by balancing the atomic numbers and mass numbers.

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Answer: A. The reaction takes place in one step.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

[tex]aA=bB\rightarrow cC+dD[/tex]

[tex]Rate=k[A]^a[B]^b[/tex]

k= rate constant

a= order with respect to A

b = order with respect to B

Answer:

anwer is a

Explanation:

Answer:

Boiling water breaks intermolecular attractions and electrolysis breaks covalent bonds.

Explanation:

When water boils, hydrogen bonds are broken between adjacent water molecules. The hydrogen bond is an intermolecular bond between adjacent oxygen and hydrogen atoms of water molecules.  

During electrolysis, water dissociates in the presence of electric current. Here, ions are formed in the process. Therefore, covalent bonds are broken here.

Boiling water breaks the intermolecular hydrogen bonds between water molecules, causing it to change from liquid to gas. Electrolysis on the other hand, breaks the covalent bonds within individual water molecules, splitting them into hydrogen and oxygen.

There is a distinct difference between the type of interactions broken when water is boiled and when it's electrolyzed. Boiling water primarily breaks the intermolecular attractions, specifically hydrogen bonds that exist between water molecules, thereby allowing water molecules to escape into the air as steam or water vapor. On the other hand, electrolysis of water involves breaking of the covalent bonds within individual water molecules, thereby splitting water into its constituent hydrogen and oxygen atoms. This process requires more energy compared to boiling water due to the strength of covalent bonds when compared to hydrogen bonds.

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Answer:

a. 52.8

Explanation:

To find the number of moles of HCl we use the relation M₁V₁=M₂V₂

where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.

M₁=7.91 M

M₂=2.13 M

V₁=?

V₂=196.1 mL

Replacing these values in the relationship.

M₁V₁=M₂V₂

7.91 M× V₁=2.13 M×196.1 mL

V₁=(2.13 M×196.1 mL)/7.91 M

=52.8 mL

Answer:

acetyl CoA

Explanation:

The starting molecule for the krebs cycle is acetyl CoA.

Answer:2500-3300[tex]cm^{-1}[/tex]

Explanation:We can distinguish between 2-butanone and butanoic acid by analyzing the spectra .                                          

The structure of the two compounds are slightly different as butanoic acid has an OH group whereas for 2-butanone there is no OH group present.

Please refer the attachments for structure of compounds.

Both the compounds will show carbonyl stretching frequency at around 1700-1750cm^{-1}[/tex] due to the presence of carbonyl group in both the compounds.

A broad intense peak at around 3000[tex]cm^{-1}[/tex] will be observed only for butanoic acid due to the presence of OH group and 2-butanone would not show any broad intense peak at around 3000[tex]cm^{-1}[/tex].

So by analyzing the IR spectra and identifying the intense broad peak of OH we can easily distinguish between butanoic acid and 2-butanone.

Due to hydrogen bonding in between two carboxylic acid molecules ,peak broadening for OH group in butanoic acid takes place.

Generally the ketones show IR stretching in around 1700-1730cm^{-1}[/tex] due to the presence of carbonyl group.

Generally carboxylic acids show IR stretching  for carbonyl group at around 1700-1760cm^{-1}[/tex] and a broad intense peak for OH at around 2500-3000cm^{-1}[/tex].

The C=O stretching region in the IR spectrum can be used to distinguish between butanoic acid and 2-butanone.

In the IR spectrum, the region that could be used to distinguish between butanoic acid and 2-butanone is the C=O stretching region. The C=O stretching vibration band of saturated aliphatic ketones, such as 2-butanone, appears at around 1715 cm-1. However, butanoic acid contains a carboxyl group (COOH), which exhibits a broad and intense peak in the 1680-1750 cm-1 range.

The equilibrium concentration of H2O(g) in the given reaction is approximately 12.91 × 10^-6 M.

To determine the equilibrium concentration of H2O(g) in the given reaction, we need to use the equilibrium constant expression and the equilibrium concentrations of the other species. The equilibrium constant expression for the reaction C2H4(g) + H2O(g) ⇌ C2H5OH(g) is Kc = [C2H5OH]/([C2H4][H2O]). Given that [C2H4]eq = 0.015 M and [C2H5OH]eq = 1.69 M, we can substitute these values into the expression to solve for the equilibrium concentration of H2O(g).

Kc = [C2H5OH]/([C2H4][H2O])

9.0 × 10^3 = 1.69/((0.015)([H2O]))

[H2O] = 1.69/((0.015)(9.0 × 10^3))

[H2O] ≈ 12.91 × 10^-6 M

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The equilibrium concentration of H2O(g) is approximately 0.0125 M.

The equilibrium concentration of H2O(g) given the reaction C2H4(g) + H2O(g) ⇌ C2H5OH(g), the equilibrium constant (Kc), and the equilibrium concentrations of C2H4 and C2H5OH at a particular temperature. We start by writing the expression for the equilibrium constant:

Kc = [C2H5OH]/[C2H4][H2O]

Then, we plug in the known values:

9.0 × 10^3 = 1.69/[0.015][H2O]

Now, we solve for [H2O]:

[H2O] = 1.69/(9.0 × 10^3 × 0.015)

[H2O] ≈ 1.69/(135) ≈ 0.0125 M

Answer: 2.796 grams

Explanation:

[tex]Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4[/tex]

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

[tex]\text{Number of moles of sodium sulphate}=\frac{2.54g}{142g/mol}=0.018moles[/tex]

[tex]\text{Number of moles of barium chloride}=\frac{2.54g}{208g/mol}=0.012moles[/tex]

According to stoichiometry:

1 mole of [tex]BaCl_2[/tex] reacts with 1 mole of [tex]Na_2SO_4[/tex]

0.012 moles of [tex]BaCl_2[/tex] will react with=[tex]\frac{1}{1}\times 0.012=0.012moles[/tex] of [tex]Na_2SO_4[/tex]

Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product. [tex]Na_2SO_4[/tex]  is the excess reagent as (0.018-0.012)=0.006 moles are left unused.

1 mole of [tex]BaCl_2[/tex] produces 1 mole of [tex]BaSO_4[/tex]

0.012 moles of [tex]BaCl_2[/tex] will produce=[tex]\frac{1}{1}\times 0.012=0.012moles[/tex] of [tex]BaSO_4[/tex]

Mass of [tex]BaSO_4=moles\times {\text {Molar mass}}=0.012\times 233=2.796g[/tex]

Thus 2.796 grams of [tex]BaSO_4[/tex]  are produced.