A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water

Answer:

0.423 m.

Explanation:

The following data were obtained from the question:

Mass of glycine (NH2CH2COOH) = 141.5 g

Mass of water = 4.456 kg

Molality =.?

Next, we shall determine the number of mole in 141.5 g of glycine (NH2CH2COOH.

This is illustrated below:

Mass of glycine (NH2CH2COOH) = 141.5 g

Molar mass of glycine (NH2CH2COOH) = 14 + (2x1) + 12 + (2x1) + 12 + 16 + 16 + 1 = 75 g/mol

Mole of glycine (NH2CH2COOH) =.?

Mole = mass /Molar mass

Mole of glycine (NH2CH2COOH) = 141.5/75

Mole of glycine (NH2CH2COOH) = 1.887 moles

Finally, we shall determine the molality of the solution as follow:

Molality is simply defined as the mole of solute per kilogram of water. Mathematically it is expressed as:

Molality = mole / mass (kg) of water

With the above formula, we can obtain the molality of the solution as follow:

Mole of glycine (NH2CH2COOH) = 1.887 moles

Mass of water = 4.456 kg

Molality =.?

Molality = mole /mass (kg) of water

Molality =1.887/4.456

Molality = 0.423 m

Therefore, the molality of the solution is 0.423 m

Answer:

Answers are in the explanation.

Explanation:

In a chemical reaction we can determine the sign of ΔSsys based on the states of products and reactants knowing that:

Entropy of gases >>> entropy of liquid > entropy of solids.

The entropy of solids is lower than entropy of liquids that is lower than entropy of gases.

In the reactions:

1. 2H₃O⁺(aq) + CO₃²⁻(aq) → CO₂(g) +3H₂O(l)

As 1 gas is produced, entropy of products is higher than entropy of reactants. That means  ΔSsys > 0 (That because ΔSsys is ΔSProducts - ΔSReactants)

2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

3 moles of gas are converted in 1 mole of gas in products. Entropy of reactants is higher than entropy of products, ΔSsys < 0.

3. Mg(s) + Cl₂(g) → MgCǐ₂(s)

You have 1 mole of gas in reactants and 1 mole of solid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.

4. SO₃(g) + H₂O(I) → H₂SO₄(I)

1 mole of gas in reactants, a liquid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.

Determining the sign of ΔSsys for each given chemical reaction, the following can be obtained without calculations:

1. ΔSsys > 0.

2. ΔSsys < 0

3. ΔSsys < 0

4. ΔSsys < 0

Recall:

Thus:

In the first chemical reaction, 1 mole of gas is produced, therefore: ΔSsys > 0.

In the second chemical reaction, 3 moles of gasses gives a products of 1 mole of gas, therefore: ΔSsys < 0.

In the third chemical reaction, 1 mole of gas gives 1 mole of solid as product, therefore: ΔSsys < 0.

In the fourth chemical reaction, 1 mole of gas gives 1 mole of liquid as product, therefore: ΔSsys < 0.

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Answer:

The correct answer is : option D. The test was inconclusive because he forgot to add heat.

Explanation:

Benedict's test is a test that is used to confirm the presence of the simple carbohydrates (mono saccharides and some disaccharides). It is a reagent made by mixture of solution of CuSO4 with sodium citrate and Na2CO3.

Benedict's reagent is added to the substance to test and then heated if it turns yellow to orange or red the presence of simple sugar is confirmed.

Thus, the correct answer is : option D. The test was inconclusive because he forgot to add heat.

Answer:

The test was inconclusive because the student forgot to add heat.

Explanation:

If the test revealed it was not glucose, then the student could run these tests. The student, however, does not need these substances to run the glucose test properly.

Answer:

[tex]m=0.127g[/tex]

Explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}[/tex]

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

[tex]m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g[/tex]

Best regards.

Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ

Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

Enthalpy is defined as internal heat existent in the system. It is calculated as:

[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]

Using Enthalpy Formation Table:

[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]

[tex]\Delta H^{0} = 62,6 kJ[/tex]

Entropy is the degree of disorder in the system. It is found by:

[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]

Calculating:

[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]

[tex]\Delta S^{0} = -354.1J[/tex]

And so, Gibbs Free energy will be:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]

[tex]\Delta G^{0} = 168121.8 J[/tex]

Rounding to the nearest kJ:

[tex]\Delta G^{0}[/tex] = 168.12 kJ

Answer:

well, first off. the formula for carbon tetrachloride is CCl4

We need to find the molar mass of carbon tetrachloride

1(Mass of C) + 4(mass of chlorine)

1(12) + 4(35.5)

12 + 142

154 g/mol

Number of moles of CCl3 in 543.2g CCl3

n = given mass / molar mass

n = 543.2/153

n = 3.53 moles

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Answer:

3.53 moles

Explanation:

Answer:

Zeros located at the end of significant figures are significant.

Explanation:

Hope it will help :)

Answer:

3BaO + Al₂(SO₄)₃  →  Al₂O₃+ 3BaSO₄

Explanation:

Yes! A reactiin occurs between barium hydroxide and auminium sulphate.

barium sulfate (BaSO4) and aluminum hydroxide (Al(OH)3) are the products obtained in this reaction.

The reaction is given by the equation below;

3BaO + Al₂(SO₄)₃  →  Al₂O₃+ 3BaSO₄

Answer:

The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].

The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].

Explanation:

The oxidation states of atoms in a compound should add up to zero.

There are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:

[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:

[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].

Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:

[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].

The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.

Therefore:

[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].

Solve this equation for the oxidation state of [tex]\rm S[/tex] here:

[tex]\text{Oxidation state of $\rm S$} = 4[/tex].

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:

H₂O

Explanation:

The empirical formular of the compound is obtained using the following steps;

Step 1: Divide the percentage composition by the atomic mass

Hydrogen = 11.21 / 1 = 11.21

Oxygen = 88.79  / 16 = 5.55

Step 2: Divide by the lowest number

Hydrogen  = 11.21 / 5.55 = 2.02 ≈ 2

Oxygen = 5.55 / 5.55 = 1

This means the ratio of the elements is 2 : 1

The empirical formular (simplest formular of a compound) of the compound is;

H₂O

Answer:Empirical formula ======== H₂O    

Explanation:The empirical formula of a compound shows the whole number ratio for  each atom in a compound.

To find empirical formula. we follow the below steps

The total mass of the compound here  is 100 grams, that is (11.21% of hydrogen + 88.79% of oxygen) we can then  assume 11.21 grams  of hydrogen and 88.79grams of oxygen

                                          Hydrogen                   Oxygen

1.composition by mass    11.21                              88.79

molecular weight              1.007g/mol               15.990g/mol

2.Divide composition by mass  11.21/1.007            88.79/15.99    

by each molecular weight to get 11.13                            5.553

no of moles

3 Divide by the least number of moles

to get atomic ratio                       11.13/5.553          5.553/5.553

                                                         2.004                           1.00

4.Convert  to whole numbers             2                                 1

Empirical formula ======== H₂O    

Answer:

A. The atomic mass is 243 amu, and the atomic number is 95.

Answer:

A. Sharing valence electrons between atoms.

Explanation:

This is the definition of a covalent bond. Option B describes ionic bonds, Option C describes intermolecular forces, and Option D is wrong because then there wouldn't be any mention of them in our high school chemistry textbooks :).  

Answer:

[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]

Explanation:

Hello,

In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:

[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]

The equilibrium expression is:

[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]

Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:

[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]

Hence, in terms of the molar solubility [tex]x[/tex], we can write:

[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]

In such a way, solving for [tex]x[/tex], we obtain:

[tex]x=0.00238M[/tex]

Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:

[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]

Best regards.

Answer:

The total photons required for this radiation = 5.1938 × 10²⁸ photons

Explanation:

Given that:

A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat.

If the radiation wavelength is 12.5 cm,

density of water = 1g/cm³

volume of the container = 0.250 L = 250 cm³

density = mass/volume

mass of the water = density × volume

mass of the water =  1g/cm³  × 250 cm³

mass of the water = 250 g

specific heat capacity of water = 4.182 J/g°C

The change in temperature was from 20.0° C to 99° C

ΔT =( 99 -20.0)° C

ΔT = 79.0° C

The heat absorbed in the process is calculated by using the formula,

q = mcΔT

q = 250 g × 4.182 J/g°C ×  79.0° C

q = 82594.5 Joules

Recall that the radiation wavelength λ = 12.5 cm = 0.125 m

The amount of energy of one photon of the radiation wavelength is determined by using the formula:

E = hv  

since v = c/λ

E = hc/λ

where;

h = Planck's constant = 6.626 × 10⁻³⁴ J.s

c = velocity of light = 3.0 × 10⁸ m/s

∴

E = (6.626 × 10⁻³⁴ J.s × 3.0 × 10⁸ m/s)/ 0.125 m

E = 1.59024⁻²⁴ Joules

The total photons required for this radiation = total heat energy/energy of radiation

The total photons required for this radiation = 82594.5 Joules/1.59024⁻²⁴ Joules

The total photons required for this radiation = 5.1938 × 10²⁸ photons

Answer:

36.7 mg

Explanation:

The following data were obtained from the question.

Original amount (A₀) = 65.1 mg

Rate constant (K) = 2.47×10¯² years¯¹

Time (t) = 23.2 years

Amount of substance remaining (A) =?

Thus, we can obtain the amount of substance remaining after 23.2 years as follow

ln A = lnA₀ – Kt

lnA = ln(65.1) – (2.47×10¯² × 23.2)

lnA = 4.1759 – 0.57304

lnA = 3.60286

Take the inverse of ln

A = e^3.60286

A = 36.7 mg

Therefore, the amount remaining after 23.2 years is 36.7 mg.

Answer:

pH = 13.7

Explanation:

A strong acid (HCl) reacts with a strong base Sr(OH)₂ producing water and a salt, thus:

2HCl + Sr(OH)₂ → 2H₂O + SrCl₂

To solve this problem, we need to find initial moles of both reactants and, with the chemical equation find limiting reactant and moles in excess to find pH as follows:

The initial moles of HCl and Sr(OH)₂ are:

100mL = 0.1L ₓ (1.0mol / L) = 0.100 moles of both HCl and Sr(OH)₂

As 2 moles of HCl reacts per mole of Sr(OH)₂, moles of Sr(OH)₂ that reacts with 0.100 moles of HCl are:

0.100 moles HCl ₓ (1 mol Sr(OH)₂ / 2 mol HCl) = 0.050 moles Sr(OH)₂

That means HCl is limiting reactant and after reaction will remain in solution:

0.100 mol - 0.050mol =

0.050 moles of Sr(OH)₂

Find pH:

1 mole of Sr(OH)₂ contains 2 moles of OH⁻, 0.050 moles contains 0.050×2 = 0.100 moles of OH⁻. In 200mL = 0.2L:, molar concentration of OH⁻ is:

0.100 moles / 0.2L =

[OH⁻] = 0.5M

As pOH of a solution is -log[OH⁻],

pOH = -log 0.5M

pOH = 0.301

And knowing:

pH = 14 - pOH

pH = 14 - 0.301

Answer:

466 torr

Explanation:

Step 1: Given data

Step 2: Calculate the final pressure

Since we have a gas changing at a constant temperature, we can calculate the final pressure using Boyle's law.

P₁ × V₁ = P₂ × V₂

P₂ = P₁ × V₁ / V₂

P₂ = 626 torr × 5.00 L / 6.72 L

P₂ = 466 torr

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:

Explanation:

mass -  15.8 g = 0.0158 kg

volume = 32.5 - 22.5 = 10.2 ml

density = mass / volume

= 0.0158 / 10.2

= 0.00154 kg/ml

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Answer:

Following are the solution to this equation:

Explanation:

In the given-question, an attachment file of the choices was missing, which can be attached in the question and its solution can be defined as follows:

In the given question "Option (iii)" is correct, which is defined in the attachment file.

When 2,2,3-trimethylcyclohexanone reacts with hydroxylamine it will produce the 2,2,3-trimethylcyclohexanoxime.

Answer:

A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable

Explanation:

Answer:

The answer is "Tertiary carbon".

Explanation:

Accent to the results, the carbon-bromine bond is weak, whenever, the bromine is connected to tertiary carbon so, bonding energy is separation for methyl-carbon, which is connected to the bromine = 70 kcal/mol and for the primary energy to the secondary energy is=  68 kcal/mol, and for tertiary CO2 = 65 kcal/mol.

The stronger the energy dissociating connection and the weaker, its power dissociation connection and its weaker bond becomes connecting with a tertiary carbon, that's why "Tertiary carbon" is the correct answer.

Answer:

1. Zinc sulfide : about the same solubility, no common ion is found.

2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.

3. Lead iodide  : about the same solubility, no common ion is found.

4. Silver hydroxide : about the same solubility, no common ion is found.

Explanation:

Hello,

In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:

1. Zinc sulfide : about the same solubility, no common ion is found.

2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.

3. Lead iodide  : about the same solubility, no common ion is found.

4. Silver hydroxide : about the same solubility, no common ion is found.

Best regards.

CHECK COMPLETE QUESTION BELOW

At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with Xbenzene = 0.580.

Answer:

The total vapor pressure is [tex]81.3 mmHg[/tex]

Explanation:

We will be making use of Dalton and Raoults equation in order to calculate the total pressure,

Which is [tex]PT= (PA × XA) +(PB ×XB)[/tex]

PT= total vapor pressure

From the question

Benene's Mole fraction = 0.580

then to get Mole fraction of toluene we will substract the one of benzene from 1. because total mole fraction is always 1.

= (1 - 0.580) = 0.420

Vapor pressure of benzene given = 183 mmHg

Vapor pressure of toluene given= 59.2 mmHg

If we substitute those value into above equation, we have

PT=(183×0.580)+(59.2×0.420)

=81.3mmHg

Therefore,, the total vapor pressure of the solution is 81.3 mmHg

Answer:

4.22

Explanation:

We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].

If the pOH of a solution is given, one may obtain the pH of such solution from the formula;

pH + pOH =14

Hence we can write;

pH = 14-pOH

pH = 14 - 9.78 = 4.22

Hence the pH of the solution is 4.22.

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

The given question is incomplete, the complete question is:

How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures

Answer:

The correct answer is spontaneous at all the temperatures.

Explanation:

Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS

When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol

= -126000 J/mol, it is negative

ΔS = 146 J/K/mol, it is positive

Now, ΔG = ΔH-TΔS

= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.  

Answer:

It has 2

Explanation:

The significant figures are 1 and 5!

Hope this helps:)

Answer:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.

Explanation:

Hello,

In this case, for the equilibrium reaction:

[tex]NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)[/tex]

Whose equilibrium expression is:

[tex]Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}[/tex]

The proper matching is:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.

Best regards.