The combination of a carbonyl group and a hydroxyl group on the same carbon atom is called a ________ group.

a. carbamate group
b. carbonate
c. carboxlate
d. carboxyl

Answer:

²³⁴₉₀Th --> ²³⁴₉₁Pa + ⁰₋₁e

Explanation:

thorium-234 = ²³⁴₉₀Th

beta decay = ⁰₋₁e

protactinium-234 = ²³⁴₉₁Pa

The balanced nuclear equation is given as;

²³⁴₉₀Th --> ²³⁴₉₁Pa + ⁰₋₁e

Answer:

The reaction is given as:

[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]

Explanation:

Bases are defined as those chemical substances which give hydroxide ions in their aqueous solutions.

[tex]BOH(s)\rightarrow B^+(aq)+OH^-(aq)[/tex]

When sodium hydroxide is added to water it gets dissociated into two ions that are sodium ions and hydroxide ions. Along with this heat energy also releases during this reaction.

The reaction is given as:

[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]

The equation to show how NaOH behaves in water is NaOH → Na⁺ + (OH)⁻

The compound that produce negative hydroxide (OH−) ions when dissolved

in water are called bases .

This compounds NaOH (sodium hydroxide) is an example of a base.

When it dissolves in water it dissociate to form  negative hydroxide (OH−)

ions  and positive sodium (Na+) ions.

It can be represented by the following equation:

NaOH → Na⁺ + (OH)⁻

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Answer:

pH = 8.72

Explanation:

This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

As this compound acts like a base, we propose this equilibrium:

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰

Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M

So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

We can avoid the quadractic equation because Kb is so small

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH = 8.72

The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.

Since the following equation should be used.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

Now

(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

So,

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

Now

Kw = Ka. Kb

Kb = Kw/Ka

And,

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵

= 5.55×10⁻¹⁰

Now

[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

Now

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH

= 8.72

Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.

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Answer:

See explanation

Explanation:

So, I'm gonna take a shot at this one and say this:

With a strongly acidic/basic solution, you'll get a high conductivity when preforming a conductivity test.

The more acidic or basic a substance is, the higher the electrical conductivity.

Based on how high or low the conductivity is, it will give you an idea of the substance's pH.

Hope that made since or gave you an idea of what you're looking for. Good luck :)

Answer:

We can then determine that a reaction will shift to the right if Q<K

Explanation:

Comparing Q with K allows to find out the status and evolution of the system:

In this case, we can then determine that a reaction will shift to the right if Q<K

Answer:

a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe

b. Glu-Ala-Phe + Gly-Ala-Tyr

Explanation:

In this case, we have to remember which peptidic bonds can break each protease:

-) Trypsin

It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.

-) Chymotrypsin

It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.

With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the "Lis" and "Arg" (See figure 1).

In "peptide b", the peptidic bond that would be broken is the one in the "Phe" (See figure 2). The second amino acid that can be broken is tyrosine, but this amino acid is placed in the C terminal spot, therefore will not be involved in the hydrolysis.

Answer:

[tex]6.4knips[/tex]

Explanation:

Hello,

In this case, given the stated equivalences, we can use the following proportional factor in order to compute the required knips:

[tex]knips=1.00knop*\frac{6bippy}{1knop} *\frac{8pringle}{10bippy}* \frac{3blip}{18pringle} *\frac{4clips}{5blips} *\frac{10knip}{1clip} \\\\=6.4knips[/tex]

Regards.

1 knop=6.4 knips

First convert knop to bippy:-

[tex]1\ knop\times\frac{6\ bippy}{1\ knop} =6\ bippy[/tex]

Now, Convert 6 bippy to pringle:-

[tex]6\ bippy\times\frac{8\ pringle}{10\ bippy} =4.8\ pringle[/tex]

Now, convert 4.8 pringle to blip:-

[tex]4.8\ pringle\times\frac{3\ blip}{18\ priangle} =0.8\ blip[/tex]

Now, convert 0.8 blip to clips as follows:-

[tex]0.8\ blip\times\frac{4\ clips}{5\ blip} =0.64\ clip[/tex]

Now, convert 0.64 clip to knips:-

[tex]0.64\ clip\times\frac{10\ knip}{1\ clip} =6.4\ knip[/tex]

Hence, the following conversion is as follows:-

1.00 knop=6.4 knips

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Answer:

The number of electrons transferred in the reaction

Explanation:

Answer:

A

Explanation:

Answer:

Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1

Explanation:

The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:  

Vrms = [tex]\sqrt{3RT/M}[/tex]

where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol

For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R

Vrms = √(3 * R *4T)/0.04 = √300RT

For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R

Vrms = √(3 * R *T)/0.001 = √3000RT

Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316

Ratio of Vrms of argon to that of hydrogen = 0.316 : 1

Answer:

Cu2 + 2Mg-> 2Cu+ Mg2

Explanation:

Balance the equation and make sure both the reactant and the products are the same

Hope it will be helpful

[tex]Cu^{+2} + 2Mg[/tex]  -> [tex]2Cu + Mg^+2[/tex]  is the correct half-reactions.

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.

[tex]Cu^{+2} + 2Mg[/tex]  -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.

Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.

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Answer:

CO2 will diffuse more rapidly.

Explanation:

From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:

Rate (R) & 1/√Density (d)

R & 1/√d

But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.

Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:

Rate (R) & 1/√Molar mass (M)

R & 1/√M

From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.

Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.

This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71 g/mol

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Summary

Gas >>>>>> Molar mass

Cl2 >>>>>> 71 g/mol

CO2 >>>>> 44 g/mol

From the illustration above, we can see that CO2 is lighter than Cl2.

Therefore, CO2 will diffuse more rapidly.

Answer: CO2

Explanation:

Answer:

For the mentioned reaction, the balanced chemical equation is:  

KBr (aq) + AgNO3 (s) ⇒ KNO3 (aq) + AgBr (s)

The number written in front of the ion, atoms, and molecules in a chemical reaction so that each of the elements on both the sides of reactants and products of the equation gets balanced is known as the stoichiometric coefficient.  

From the mentioned balanced equation, the stoichiometric coefficient before KBr is 1, AgNO3 is 1, KNO3 is 1, as well as before AgBr is also 1. Thus, it is clear that 1 mole of potassium bromide reacts with 1 mole of silver nitrate to produce 1 mole of potassium nitrate and 1 mole of silver bromide.  

Answer:

The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0.  It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it

Thats all i know

Answer:

Explanation:

A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.

Based on the question, the only tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain is

2-hydroxy-2,3-dimethylbutane

     H  OH   H    H

      |     |       |      |

H - C - C -   C  - C - H

      |     |       |      |

     H  CH₃  CH₃ H

From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain

Answer:

The correct answer is - 1.02 V

Explanation:

From the reduction-oxidation reaction:

Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)

Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:

Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s)                    Eº= ‑0.25 V

Oxidation (anode) :  2 x (Fe²⁺ → Fe³⁺ + e-)(aq)                Eº= -0.77 V

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

                     Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)

In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):

Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V

Answer:

The correct answer is CaO > LiBr > KI.

Explanation:

Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.  

With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.  

The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.  

Arranging the chemical compounds in order of decreasing magnitude of lattice energy, we have:

c. CaO.

a. LiBr

b. KI

Lattice energy can be defined as a measure of the energy required to dissociate one (1) mole of an ionic compound into its constituent anions and cations, in the gaseous state.

Hence, it is typically used to measure the bond strength of ionic compounds.

Generally, lattice energy is inversely proportional to the size of the ions and directly proportional to their electric charges.

Lithium bromide (LiBr) comprises the following ions:

Potassium iodide (KI) comprises the following ions:

Calcium oxide (CaO) comprises the following ions:

From the above, we can deduce that there is an increase in the charge possessed by the ionic chemical compounds and as such this would result in an increase in the lattice energy.

In order of decreasing magnitude of lattice energy, the chemical compounds are arranged as:

I. CaO.

II. KI.

III. LiBr.

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Explanation:

To calculate [H3O+] in the solution we must first find the pH from the [ OH-]

That's

pH + pOH = 14

pH = 14 - pOH

To calculate the pOH we use the formula

pOH = - log [OH-]

And [OH-] = 5.5 × 10^-5 M

So we have

pOH = - log 5.5 × 10^ - 5

pOH = 4.26

Since we've found the pOH we can now find the pH

That's

pH = 14 - 4.26

pH = 9.74

Now we can find the concentration of H3O+ in the solution using the formula

pH = - log H3O+

9.74 = - log H3O+

Find the antilog of both sides

The solution is basic since it's pH lies in the basic region.

Hope this helps you

Answer:

[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]

Explanation:

1. Density from mass and volume

[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]

2. Volume from density and mass

[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]

3. Mass from density and volume

[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                = 12.8 mL

Volume of object               = 11.8 mL

[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

Answer:

Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.

Explanation:

I hope this helps bro

Answer:

d) NH4F

Explanation:

Hello,

In this case, the base resulting from mixing a weak acid and a weak base is d) NH4F since ammonium hydroxide is a wear base and hydrofluoric acid is a weak acid.

Ammonium hydroxide is a weak base since it is not completely ionized in ammonium and hydroxyl ions:

[tex]NH_4OH\rightarrow NH_4^++OH^-[/tex]

Moreover, hydrofluoric acid is a weak acid since it is not completely ionized in hydrogen and fluoride ions:

[tex]HF\rightleftharpoons H^++F^-[/tex]

For the both of the substances, the limit is established by the basic and the acid dissociation constant respectively.

Regards.

Answer:

B) geminal diol

Explanation:

Hello,

In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.

Regards.

Answer:

While balancing a chemical equation, we change the coefficient  to balance the number of atoms on each side of the equation

Explanation:

While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.

To summarize in chemistry terms, a chemical equation depicts the initial chemicals, or reactants, on the left-hand side and the final compounds, or products, just on right-hand side, divided by an arrow. In the chemical equation, the number of atoms in each element as well as the total charge are the same on opposite of the equation's sides.

Chemical equations are used in chemistry to depict chemical processes by writing the reactants and products in terms of their corresponding chemical formulas. While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.

Therefore, while balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.

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Answer:

406 K.

Explanation:

The following data were obtained from the question:

Number of mole (n) = 0.750 mole

Volume (V) = 10.0 L

Pressure (P) = 2.50 atm

Temperature (T) =.?

Note: Gas constant (R) = 0.0821 atm.L/Kmol

The temperature, T can be obtained by using the ideal gas equation as follow:

PV = nRT

2.5 x 10 = 0.75 x 0.0821 x T

Divide both side by 0.75 x 0.0821

T = (2.5 x 10) /(0.75 x 0.0821 )

T = 406 K.

Therefore, the temperature is 406 K.

Answer: 406 K

Explanation:

We can rewrite the ideal gas law to solve for T:

PV = nRT

T=PV / nR

We are given the following from the problem:

n=0.750 mol P=2.50 atm V=10.0 L

Plugging in our values and using R=0.08206 L⋅atm / K⋅mol we get:

T=(2.50 atm)(10.0 L) / (0.750 mole)(0.08206L ⋅ atm ⋅ mole K) = 406 K

Answer:

D

Explanation:

Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.

The action that destroys the buffer is option c. adding 0.050 moles of HCl.

It is a solution of a weak acid and salt.

Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.

The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this,  there will be only acid in the solution.

Since

moles of HC2H3O2 = 1*0.250 = 0.250

moles of NaC2H3O2 = 1*0.050 = 0.050.

moles of HCl is added = 0.050

Now

The reaction between HCl and NaC2H3O2

[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]

Now

BCA table is

            NaC2H3O2  HCl       HC2H3O2

Before 0.050 0.050 0.250

Change -0.050 -0.050 +0.050

After 0 0 0.300

Now, the solution contains the acid (HC2H3O2 ) only.

Therefore addition of 0.050 moles of HCl will destroy the buffer.

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Answer:

pH of solution B is 5

Explanation:

A weak acid, HA, is in equilibrium with water as follows:

HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)

Where Ka (10^-pKa = 1x10⁻⁹) is:

Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]

Where concentrations of this species are equilibrium concentrations

As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:

[HA] = 0.1M - X

[A⁻] = X

[H₃O⁺] = X

Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.

Replacing in Ka expression:

1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]

1x10⁻⁹ = [X] [X] / [0.1 - X]

1x10⁻¹⁰ - 1x10⁻⁹X = X²

1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0

Solving for X:

X = -0.00001 → False solution, there is no negative concentrations.

X = 1x10⁻⁵ → Right solution.

As [H₃O⁺] = X

[H₃O⁺] = 1x10⁻⁵M

And pH = -log[H₃O⁺]

pH = 5

Answer:

A. Ka = [CO2] / [C] [O2]^1/2

B. Kb = [CO2] / [CO] [O2]^1/2

Explanation:

Equilibrium constant is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Now, we shall obtain the expression for the equilibrium constant for the reaction as follow:

A. Determination of the expression for equilibrium constant Ka.

This is illustrated below:

C(s) + 1/2 O2(g) <==> CO(g)

Ka = [CO2] / [C] [O2]^1/2

B. Determination of the expression for equilibrium constant Kb.

This is illustrated below:

CO(g) + 1/2 O2(g) <==> CO2(g)

Kb = [CO2] / [CO] [O2]^1/2

Answer:

-69 kJ

Explanation:

Step 1: Given data

Standard cell potential (E°cell): +0.24 V

Electrons involved (n): 3 mol

Step 2: Calculate the standard Gibbs free energy change (ΔG°) for the voltaic cell

We will use the following expression.

ΔG° = -n × F × E°cell

where,

F is Faraday's constant (96,485 C/mol e⁻)

ΔG° = -n × F × E°cell

ΔG° = -3 mol e⁻ × 96,485 C/mol e⁻ × 0.24 V

ΔG° = -69 kJ

Explanation:

An atom undergoes alpha decay by losing a helium atom.

So when bismuth undergoes alpha decay, we have;

²¹⁰₈₃Bi --> ⁴₂He + X

Mass number;

210 = 4 + x

x = 206

Atomic number;

83 = 2 + x

x = 81

The element is Thallium. The symbol is Ti.

For the second part;

X --> ⁴₂He + ²³⁴₉₀Th

Mass number;

x = 4 + 234 = 238

Atomic Number;

x = 2 + 90 = 92

The balanced nuclear equation is;

²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate [tex]SO_4~^-^2[/tex] and the chloride [tex]Cl^-[/tex]:

Sulfate

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.([tex]Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2[/tex]), which are NOT soluble.

Chloride

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ([tex]Pb^+^2~,Ag^+~,Hg_2~^+^2[/tex]), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I),  lead" and our possibilities are:

"Calcium, strontium, or barium".

I hope it helps!

Answer:

(a) when a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants

Determining whether the statements about equilibrium is True or False

A) The concentration of the products is equal to the concentration of the reactants at equilibrium : TRUE

B) When a system is at equilibrium, Keq = 1 : TRUE

C) The rates of the forward reaction and the reverse reaction are equal at equilibrium :  TRUE

D) Adding a catalyst to a reaction system will shift the position of equilibrium to the right : FALSE

In a chemical reaction at equilibrium the value of Keq will be equal to 1 because the concentration of the products is equal to the concentration of the reactants in the chemica reaction. Also at equilibrium the rate of forward reaction is same as the rate of reverse reaction.

A catalyst can only affect the rate of reaction and not the amount of product ( yield of reaction).

Hence we can conclude that the answers to your questions are as listed above.

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