Ethylene () is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about of polyethylene are made from ethylene each year, for use in everything from household plumbing to artificial joints. Natural sources of ethylene are entirely inadequate to meet world demand, so ethane () from natural gas is "cracked" in refineries at high temperature in a kinetically complex reaction that produces ethylene gas and hydrogen gas. Suppose an engineer studying ethane cracking fills a reaction tank with of ethane gas and raises the temperature to . He believes at this temperature. Calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture. Round your answer to significant digits. Note for advanced students: the engineer may be mistaken about the correct value of , and the mass percent of ethylene you calculate may not be what he actually observes.

Answer:

0.169

Explanation:

Let's consider the following reaction.

A(g) + 2B(g) ⇄ C(g) + D(g)

We can find the pressures at equilibrium using an ICE chart.

       A(g) + 2 B(g) ⇄ C(g) + D(g)

I       1.00     1.00        0        0

C       -x        -2x        +x       +x

E    1.00-x  1.00-2x     x         x

The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

The pressures at equilibrium are:

pA = 1.00-x = 1.00-0.211 = 0.789 atm

pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm

pC = x = 0.211 atm

pD = x = 0.211 atm

The pressure equilibrium constant (Kp) is:

Kp = pC × pD / pA × pB²

Kp = 0.211 × 0.211 / 0.789 × 0.578²

Kp = 0.169

Answer:

0.108mol of Hydrogen

Explanation:

The formula for the compound is: CH3COOH

From the formula of the compound,

There are 2moles of oxygen and 4moles of Hydrogen.

If for every 2moles of oxygen, 4moles of Hydrogen is present.

Then, for 0.054 moles of oxygen = (0.054 x 4)/2 = 0.108mol of Hydrogen is present

Answer:

For 0.027 moles CH3COOH we have 0.108 moles H ≈ 1.1 *10^-1 moles H

Explanation:

Step 1: Data given

Acetic acid = CH3COOH

Number of moles oxygen in the sample = 0.054 moles

Step 2: calculate moles CH3COOH

In 1 mol CH3COOH we have 2 moles O

For 0.054 moles Oxygen we have 0.054/2 = 0.027 moles CH3COOH

Step 3: Calculate moles H

In 1 mol CH3COOH we have 4 moles H

For 0.027 moles CH3COOH we have 4*0.027 = 0.108 moles H ≈ 1.1 *10^-1 moles

Hygroscopic nature of NaOH is the main reason for the lower concentration of the sodium hydroxide in the solution.

Explanation:

The main reason for the lower concentration of the sodium hydroxide in the solution is Hygroscopic nature of NaOH.

It's obviously true that Sodium hydroxide pellets are hygroscopic in nature, which plainly implies that the sodium hydroxide pellets retains dampness from the air, so it becomes deliquescent. Whenever the NaOH precious stones are gauged, the gems assimilate dampness from the environmental factors, thus the heaviness of the gems might change, so the centralization of the arrangement was lower than the expected one. So it is absurd to expect to get ready NaOH arrangement under ordinary room temperature. In this way, while setting up the arrangement of NaOH we must be more cautious.

Find more information about Sodium hydroxide here:

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Answer:

Explanation:

Heat is a kind of energy.

The kinetic theory relates the heat with the movement of the particles: the more the particles move, the larger the kinetic energy of the system. The kinetic theory states that heat is the kinetic energy of the particles, atoms or molecules, in a substance, that is transferred from a substance at higher temperature to other substance at lower temperature.

Based on that principle, the kinetic theory explains the changes of phases of the substances in terms of the motion of the particles: the hotter an object the faster the particles move, the more energetic the particles are, and they occupy more space. Thus, when a solid is heated, the particles move faster and it can pass to liquid or gaseous state.

Explanation:

Since, the given reaction is as follows.

       [tex]2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)[/tex]

Initial:    36.1 atm                 0          0

Change:    2x                      x           x

Equilibrium: (36.1 - 2x)       x            x

Now, expression for [tex]K_{p}[/tex] of this reaction is as follows.

            [tex]K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}[/tex]

As the initial pressure of NO is 36.1 atm. Hence, partial pressure of [tex]O_{2}[/tex] at equilibrium will be calculated as follows.

              [tex]K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}[/tex]

        [tex]2.40 \times 10^{3} = \frac{x \times x}{(36.1 - 2x)^{2}}[/tex]

                 x = 18.1 atm

Thus, we can conclude that partial pressure of [tex]O_{2}[/tex] at equilibrium is 18.1 atm.

Answer:

partial pressure O2 = 17.867 atm

Explanation:

Step 1: Data given

Temperature = 200 °C

Kp = 2.40 *10^3

Pressure NO = 36.1 atm

Step 2: The balanced equation

2 NO ⇔ N2 + O2

Step 3: The initial pressure

pNO = 36.1 atm

pN2 = 0 atm

pO2 = 0 atm

Step 4: the pressure at the equilibrium

For 2 moles NO we'll have 1 mol N2 and 1 mol O2

pNO = 36.1 - 2X atm

pN2 = X atm

pO2 = X atm

Step 5: Calculate partial pressures

Kp = pN2 * pO2 / (pNO)²

2.40*10³ = x²/(36.1 - 2x)²

48.99 = x/(36.1-2x)

x = 1768.5 -97.98x

x = 17.867

x = partial pressure O2 = 17.867 atm

Explanation:

CO(g) + Cl2(g) ⇄ COCl2(g)

This question is based on Le Chatelier's principle.

Le Chatelier's principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

a) COCl2 is added to the reaction mixture

COCL2 is a product in the reaction. If we add additional product to a system, the equilibrium will shift to the left, in order to produce more reactants. The reaction would shift to the left.

b) Cl2 is added to the reaction mixture

if we add reactants to the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.

c) COCl2 is removed from the reaction mixture

if we remove products from the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.

Chemical reactions at equilibrium, such as CO(g) + Cl2(g) ⇌ COCl2(g), shift in response to changes to re-establish equilibrium. If a product or reactant is added, the reaction will respectively shift towards reactants or products. Similarly, if a product or reactant is removed, the reaction will respectively shift towards products or reactants.

In chemistry, chemical reactions at equilibrium respond to disturbances according to Le Châtelier's principle: the system shifts in a way that counters the disturbance and re-establishes equilibrium. Now, let's consider the reaction: CO(g) + Cl2(g) ⇌ COCl2(g).

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The question is incomplete, here is the complete question:

The decay constant for 14-C is [tex]0.00012yr^{-1}[/tex] In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14-C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.

Answer: The formula for the age of the charcoal is [tex]t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]

Explanation:

Carbon-14 isotope is a radioisotope and its decay process follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,

k = rate constant  = [tex]0.00012yr^{-1}=1.2\times 10^{-4}yr^{-1}[/tex]

t = time taken for decay process = ? yr

[tex][A_o][/tex] = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 20) = 80 grams

Putting values in above equation, we get:

[tex]1.2\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{20}\\\\t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]

Hence, the formula for the age of the charcoal is [tex]t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]

The rate law for the overall reaction derived from a two-step reaction mechanism is determined by the slowest (rate-determining) step. In this case, the rate law would be rate = k * K_eq * [A]^2[B], where k is the rate constant for the slow step, K_eq is the equilibrium constant for the fast step, and [A] and [B] represent the concentrations of A and B, respectively.

In the context of chemistry, one step in a multistep reaction mechanism is often significantly slower than the others. This slow step is known as the rate-determining step or the rate-limiting step. The reaction cannot proceed faster than this slowest step.

In your specific example, the given mechanism consists of two steps. The rate law for each step is generally expressed in terms of the concentration of the reactants involved in that step.

For Step 1 (A + B ⇄ C), assuming that the reaction reaches an equilibrium, the concentrations of A, B, and C would remain constant over time, and won't affect the overall rate. Therefore, we would ignore this step when deriving the rate law for the overall reaction.

On the other hand, Step 2 (C + A ⟶ D) is the slow step, and thus determines the rate of the overall reaction. The rate law for this step would be rate = k * [C][A]. But since C is also a product of Step 1, we need to express C in terms of A and B. From equilibrium of Step 1, we know [C] = K_eq*[A][B] (where K_eq is equilibrium constant).

Substituting this in rate law of step 2, we get rate = k * K_eq * [A]^2[B] for the overall reaction.

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The rate law for the overall reaction 2A + B ⟶ D, considering Step 2 is the rate-determining step, is calculated by using the equilibrium from Step 1 to express the concentration of intermediate C in terms of A and B. After substitution and simplification, the rate law for the overall reaction is rate = k [A]²[B], indicating second-order dependence on A and first-order dependence on B.

To determine the rate law for the overall reaction, we must look at the mechanism provided. Since Step 2, which involves the conversion of C and A to D, is the rate-determining step, the rate law for the overall reaction will reflect this slowest step. However, since C is an intermediate that we cannot measure directly, we must use the equilibrium established in Step 1 to express the concentration of C in terms of the concentrations of A and B.

Assuming Step 1 is at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and thus:

Rate of forward reaction (k₁ [A][B]) = Rate of reverse reaction (k₁₁ [C])

We can rearrange this to solve for [C]:

[C] = k₁/k₁₁ [A][B]

Now, since the rate-determining step is Step 2, we write the rate law based on this step:

rate = k₂ [C][A]

Substituting in the expression for [C] gives us:

rate = k₂ (k₁/k₁₁ [A][B])[A]

rate = (k₂ * k₁/k₁₁) [A]²[B]

Thus, after simplifying and combining the rate constants into a single overall rate constant (k), we have:

rate = k [A]²[B]

This shows that the reaction is second order with respect to A and first order with respect to B.

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Answer:

[tex]COD=2030\frac{mg}{L}[/tex]

Explanation:

Hello,

In this case, considering the given information the redox reaction is:

[tex]Fe^{+2}+Cr_2O_7^{-2}\rightarrow Fe^{+3}+Cr^{+3}[/tex]

Which properly balanced turns out:

[tex]Fe^{+2}\rightarrow Fe^{+3}+1e^-\\Cr^{+6}_2O_7^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow\\6Fe^{+2}+Cr_2O_7^{-2}+14H^+\rightarrow 6Fe^{+3}+2Cr^{+3}+7H_2O[/tex]

In such a way, one sees a 6 to 1 molar relationship between the standard iron (II) solution and the dichromate, therefore, by using the following equation it is possible to determine the oxygen as shown below:

[tex]n_{Fe^{+2}}=n_{Cr_2O_7^{-2}}[/tex]

Thus, the moles of iron (II) solution are:

[tex]n_{Fe^{+2}}=0.1511\frac{molFe^{+2}}{L_{sln}}*0.02684L_{sln}=0.00406molFe^{+2}[/tex]

Moreover, the moles of dichormate result:

[tex]n_{Cr_2O_7^{-2}}=0.00406molFe^{+2}*\frac{1molCr_2O_7^{-2}}{6molFe^{+2}}=6.76x10^{-4}}molCr_2O_7^{-2}[/tex]

Thereby, the volume of the sample is:

[tex]V_{sample}=\frac{n_{Cr_2O_7^{-2}}}{M_{Cr_2O_7^{-2}}}=\frac{6.76x10^{-4}molCr_2O_7^{-2}}{0.0181\frac{molCr_2O_7^{-2}}{L_{sln}} } =0.0373L_{sample}[/tex]

Finally, the chemical oxygen demand result:

[tex]COD=6.76x10^{-4}molCr_2O_7^{-2}*\frac{7molO}{1molCr_2O_7^{-2}}*\frac{16gO}{1molO}*\frac{1000mgO}{1gO}*\frac{1}{0.0373L_{sln}} \\COD=2030\frac{mg}{L}[/tex]

Best regards.

Final answer:

The estimated pKa of the weak monoprotic acid is 3.42, as this value is equal to the pH at the halfway point to the equivalence point during a titration with a strong base where the amounts of the weak acid and its conjugate base are equal.

Explanation:

The question addresses the titration of a weak monoprotic acid with a strong base and involves finding the pKa of the acid using pH measurements.

To estimate the pKa of the weak acid, we use the information that at the halfway point of the titration (when half the equivalent amount of base has been added), the pH of the solution equals the pKa of the acid. This is because, at the halfway point, the concentrations of the acid (HA) and its conjugate base (A-) are equal.

Given that 25.0 mL of 0.100 M NaOH is the halfway point since it takes 50.0 mL of NaOH to reach the equivalence point, and the pH at this stage is 3.42, we can directly say that the pKa of the weak acid is approximately 3.42.

This is because at the halfway point of the titration, the amount of acid that has been neutralized by the base is equal to the amount of acid that remains un-neutralized. In such a scenario, according to the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA]) and since [A-] = [HA] at the halfway point, we get:

Thus, the estimated pKa value of the weak acid is 3.42.

Answer : The compartment that has the higher osmotic pressure is, 10 % (m/v) starch solution.

Explanation :

Formula used for osmotic pressure :

[tex]\pi=\frac{nRT}{V}\\\\\pi=\frac{wRT}{MV}[/tex]

where,

= osmotic pressure

V = volume of solution

R = solution constant  = 0.0821 L.atm/mole.K

T= temperature of solution = [tex]25^oC=273+25=298K  [/tex]

M = molar mass of solute

w = mass of solute

Now we have to determine the osmotic pressure for the following solution.

For 1 % (m/v) starch solution :

1 % (m/v) starch solution means that 1 grams of starch present in 100 mL or 0.1 L of solution.

Molar mass of starch = 692.7 g/mol

[tex]\pi=\frac{(1g)\times (0.0821Latm/moleK)\times (298K)}{(692.7g/mol)\times (0.1L)}[/tex]

[tex]\pi=0.353atm[/tex]

For 10 % (m/v) starch solution :

10 % (m/v) starch solution means that 10 grams of starch present in 100 mL or 0.1 L of solution.

Molar mass of starch = 692.7 g/mol

[tex]\pi=\frac{(10g)\times (0.0821Latm/moleK)\times (298K)}{(692.7g/mol)\times (0.1L)}[/tex]

[tex]\pi=3.53atm[/tex]

From this we conclude that, 10 % (m/v) starch solution has the higher osmotic pressure as compared to 1 % (m/v) starch solution.

Hence, the compartment that has the higher osmotic pressure is, 10 % (m/v) starch solution.

The 10% (m/v) starch solution will have a higher osmotic pressure compared to the 1% (m/v) starch solution because osmotic pressure increases with solute concentration.

The osmotic pressure is a property that depends on the solute concentration in a solution. In comparing a 1% (m/v) starch solution to a 10% (m/v) starch solution, the solution with the higher solute concentration is the one with higher osmotic pressure. Therefore, the 10% (m/v) starch solution will have a higher osmotic pressure because it has a greater concentration of starch molecules.

The osmotic pressure of a solution can be calculated using the formula Π = MRT, where Π is the osmotic pressure, M is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin. As solute concentration increases, so does the osmotic pressure, provided that temperature and the gas constant remain the same. Consequently, hypertonic solutions will have higher osmotic pressure than hypotonic solutions.

Answer:

C-term

Explanation:

A correlation between plate height and mobile phase velocity is known as the van Deemter equation.

[tex]H = A + \frac{B}{u}+ C*u[/tex]

Where;

H is the plate height

A  is the eddy diffusion term

B is the longitudinal diffusion term

C is the resistance to mass transfer coefficient

u is the linear velocity

Here C-term describes the mass transfer of sample components between the stationary phase and the mobile phase during separation. One of the major factor affecting C-term is the particle size, changing the diameter equally changes the particle size.

The answer is C-term

E. Both A and Cux are terms of the van Deemter Equation.

The van Deemter equation describes the relationship between the height equivalent to a theoretical plate (HETP) and the linear velocity (u) of the mobile phase in chromatography. The equation is:

[tex]\[H = A + \frac{B}{u} + C \cdot u\][/tex]

When the diameter of the stationary phase particles is decreased and made more uniform in High-Performance Liquid Chromatography (HPLC), the effects on the terms of the van Deemter equation are as follows:

1. Eddy Diffusion (A term):

  - Eddy diffusion occurs due to the multiple pathways available for the analyte molecules through the packed column. Smaller and more uniform particles reduce the multiple pathways and hence minimize the A term.

2. Longitudinal Diffusion (B/u term):

  - Longitudinal diffusion is the spreading of the analyte band due to diffusion along the length of the column. It is generally more significant at low flow rates and is not directly affected by the particle size.

3. Mass Transfer (C * u term):

  - The mass transfer term relates to the time it takes for the analyte molecules to equilibrate between the stationary and mobile phases. Smaller particles decrease the distance the analytes must travel in and out of the stationary phase, thereby reducing the C.u term.

Therefore, decreasing the diameter of the stationary phase particles and making them more uniform primarily minimizes the A and [tex]\(C \cdot u\)[/tex] terms.

The correct answer is: E. Both A and Cux

Complete Question:

When you decrease the diameter of the stationary phase particles and make them more uniform, in HPLC, which term or terms of the van Deemter Equation is or are minimized?

A. A

B. B/ux

C. Cux

D. Both A and B/ux

E. Both A and Cux

F. Both B/ux and Cux

G. None of the terms

H. All of the terms

Answer: a. [tex]2NO_2(g)\rightarrow N_2O_4(g)[/tex]: [tex]\Delta S[/tex] is negative

b. [tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)[/tex] : [tex]\Delta S[/tex] is negative

c. [tex]Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)[/tex]: [tex]\Delta S[/tex] is negative

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa

a)  [tex]2NO_2(g)\rightarrow N_2O_4(g)[/tex]

In this reaction 2 moles of gaseous reactants are converting to 1 mole of gaseous products. The randomness will decrease and hence entropy will also decrease. Thus [tex]\Delta S[/tex] is negative.

b) [tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)[/tex]

In this reaction solid reactants are converting to aqueous products. The randomness will increase and hence entropy will also increase. Thus [tex]\Delta S[/tex] is positive.

c) [tex]Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)[/tex]

In this reaction aqueous reactants are converting to solid products. The randomness will decrease and hence entropy will also decrease. Thus [tex]\Delta S[/tex] is negative

The sign of ΔS° for the given reactions can be determined based on the change in the number of moles of gas. In general, an increase in the number of moles of gas results in a positive ΔS°, while a decrease in the number of moles of gas results in a negative ΔS°.

ΔS° represents the change in entropy. To predict the sign of ΔS° for a reaction, we can consider the number of moles of gas formed or consumed. In general, an increase in the number of moles of gas will result in a positive ΔS°, indicating an increase in entropy. On the other hand, a decrease in the number of moles of gas will result in a negative ΔS°, indicating a decrease in entropy.

In the reaction 2NO2(g) → N2O4(g), the number of moles of gas decreases from 2 to 1. Therefore, ΔS° for this reaction is expected to be negative.

In the reaction CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g), the number of moles of gas increases from 0 to 1. Therefore, ΔS° for this reaction is expected to be positive.

In the reaction Ag+(aq) + Cl-(aq) → AgCl(s), the number of moles of gas remains the same. Therefore, ΔS° for this reaction is expected to be close to zero.

Answer:

We need 3910.5 joules of energy

Explanation:

Step 1: Data given

Mass of aluminium = 110 grams

Initial temperature = 52.0 °C

Final temperature = 91.5 °C

Specific heat of aluminium = 0.900 J/g°C

Step 2: Calculate energy required

Q = m*c*ΔT

⇒with Q = the energy required = TO BE DETERMINED

⇒with m = the mass of aluminium = 110 grams

⇒with c = the specific heat of aluminium = 0.900 J/g°C

⇒with ΔT = the change in temperature = T2 - T1 = 91.5 °C - 52.0 °C = 39.5 °C

Q = 110 grams * 0.900 J/g°C * 39.5

Q = 3910.5 J

We need 3910.5 joules of energy

Answer:

the correct answer:

C, B, D, A.

Explanation:

The chemical compounds named are compounds that have strong chemical bonds, forming cubic structures and crystals with very high boiling and melting points to less solid structures.

some are oxides, others salts, others are even used to emit radiation.

Answer:

Mg and Zn

Explanation:

In cathodic protection, the sacrificial anode corrodes instead of the cathode which it protects. The anode is usually higher than the cathode in the electrochemical series. This also means that the reduction potential of the sacrificial anode is more negative than that of the cathode. Consider the reduction potentials of the metals listed in the question:

Mg=-1.185V

Zn= -0.7618V

Ag= +0.7996V

Au= +1.629V

Ni= -0.251V

The reduction potential of the cathode stated in the question is -0.447V hence only magnesium and zinc can function as sacrificial anode.

The metals that can be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are silver (Ag), magnesium (Mg), nickel (Ni), and zinc (Zn).

The metals that can be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are the ones with a lower standard reduction potential (E°) than steel. In this case, we need to compare the reduction potentials of steel to the standard reduction potentials of the given metals.

Therefore, the metals that could be used as sacrificial anodes in the cathodic protection of an underground steel storage tank are silver (Ag), magnesium (Mg), nickel (Ni), and zinc (Zn).

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Answer:

The value of dissociation constant of an acid [tex]5.0\times 10^{5}[/tex].

Explanation:

The [tex]pK_a[/tex] of an acid = -5.7

The dissociation constant of the reaction = [tex]K_a[/tex]

The relation between [tex]pK_a[/tex] and [tex] K_a[/tex] is given by ;

[tex]pK_a=-\log[K_a][/tex]

[tex]-5.7=-\log[K_a][/tex]

[tex]K_a=10^{-(-5.7)}=5.012\times 10^{5}\approx 5.0\times 10^{5}[/tex]

The value of dissociation constant of an acid [tex]5.0\times 10^{5}[/tex].

Answer:

[tex]9.044\times 10^{-3}[/tex] is the value of the equilibrium constant for this reaction at 756 K.

Explanation:

[tex]COCl_2\rightleftharpoons CO+Cl_2[/tex]

Equilibrium concentration of [tex]COCl_2[/tex]

[tex][COCl_2]=7.40\times 10^{-4} M[/tex]

Equilibrium concentration of [tex]CO[/tex]

[tex][CO]=3.76\times 10^{-2} M[/tex]

Equilibrium concentration of [tex]Cl_2[/tex]

[tex][Cl_2]=1.78\times 10^{-4} M[/tex]

The expression of an equilibrium constant can be written as;

[tex]K_c=\frac{[CO][Cl_2]}{[COCl_2]}[/tex]

[tex]=\frac{3.76\times 10^{-2}\times 1.78\times 10^{-4}}{7.40\times 10^{-4}}[/tex]

[tex]K_c=9.044\times 10^{-3}[/tex]

[tex]9.044\times 10^{-3}[/tex] is the value of the equilibrium constant for this reaction at 756 K.

Answer:

135 mile marker will the car reach after 2 hours.

Explanation:

Speed of the car = 55 mile/hour

Distance covered in 2 hours = d

[tex]Speed=\frac{Distance}{Time}[/tex]

[tex]55 mile/hour=\frac{d}{2 hour}[/tex]

[tex]d=55 mile/hour\times 2 hour=110 mile[/tex]

The direction of the car is in decreasing marker numbers which mienas that car had started from end where 145 mile marker was present.

So, the marker appearing after travelling 2 hours will be:

145 - 110 = 135

135 mile marker will the car reach after 2 hours.

The natural substrate of beta glycosidase enzyme is Ganglioside GMI, Lactosylceramide, lactose and glycoprotein.

Inhibitors of beta galactosidase enzyme is 1,4-dithiothreitol, beta marcaptoethanol, 4-chloromercurobenzoic acid and Acid-beta galactosidase.

Explanation:

Beta galactosidase enzyme performs the hydrolysis of beta galactosides into monosaccharides. It acts on aryl, amino, alkyl beta glycosidic linkages also.

The enzyme attacks on the bond formed between organic entity and galactose sugar.

It can act on multiple substrates. Some substrates are :

Ganglioside GMI

Lactosylceramide

Lactose : Enzyme beta galactosidase enzyme is a boon for lactose intolerant people because it breaks lactose in yoghurt, sour cream and cheese and makes it easy for consumption to such people.

glycoprotein : These have glycosidic bonds on which enzyme works to break the bonds.

The inhibitors for the β-galactosidase are :

  4-dithiothreitol

beta marcaptoethanol

4-chloromercurobenzoic acid

Acid-beta galactosidase.

When inhibitors are bind to enzyme it breaks down the inhibitor and reaction does not takes place.

β-galactosidase can act on different substrates like lactose, ONPG and IPTG, demonstrating its flexibility. Its activity can be regulated by inhibitors such as glucose and PETG, affecting enzyme function either competitively or non-competitively.

The enzyme β-galactosidase is unique as it can act on various substrates and is influenced by several inhibitors. This enzyme mainly acts on lactose but can also interact with several structurally related substrates such as o-nitrophenyl-β-D-galactoside (ONPG) and isopropyl β-D-1-thiogalactopyranoside (IPTG), exemplifying the flexibility of β-galactosidase.

The activity of β-galactosidase is subject to regulation by various inhibitors. These inhibitors could exert their effect by binding to the enzyme's active site (competitive inhibition), or noncompetitively by interacting with the enzyme's allosteric site--an alternate part where non-substrate molecules can attach. Examples of inhibitors include glucose and phenylethyl β-D-thiogalactoside (PETG).

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Answer:

Not D

Explanation:

Answer:

Before adding any KOH, the pH is 4.03

Explanation:

Step 1: Data given

Volume of a 0.220 M HClO = 50.0 mL = 0.050 L

Molarity of KOH = 0.220 M

The ionization constant for HClO is 4.0*10^–8

Step 2: The balanced equation

HClO + KOH → KClO + H2O

Step 3:  pH before any addition of KOH

When no KOH is added, we only have HClO, a weak acid.

To calculate the pH of a weak acid, we need the Ka

Ka = [H+] / [acid]  

4.0*10^-8 = [H+]² / 0.220  

[H+]² = (4.0*10^-8 ) * 0.220  

[H+]² = 8.8*10^-9  

[H+] = √( 8.8*10^-9)  

[H+] = 9.38*10^-5 M  

pH = -log [H+]  

pH = -log(9.38*10^-5)  

pH = 4.03

Before adding any KOH, the pH is 4.03

Final answer:

To calculate the pH before the addition of KOH in the titration of HClO with KOH, we can use the ionization constant for HClO (Ka). The initial concentration of HClO can be used to calculate the concentration of H+, which in turn can be used to calculate the pH using the pH formula.

Explanation:

The pH before the addition of any KOH can be calculated using the ionization constant for HClO. HClO is a weak acid, so we can use the expression for the acid dissociation constant, Ka, to calculate the pH. The expression for Ka for HClO is:

Ka = [H+][ClO-] / [HClO]

Since we know the initial concentration of HClO, we can assume that the concentration of H+ is equal to the initial concentration of HClO. Therefore, we can rewrite the expression for Ka as:

Ka = [H+]² / [HClO]

Now we can calculate the concentration of H+ using the given initial concentration of HClO:

[H+] = sqrt(Ka * [HClO])

Finally, we can use the concentration of H+ to calculate the pH using the pH formula:

pH = -log[H+]

Explanation:

Spilling room temperature water over your skin on a hot day will cook the body down. This is because water has a low heat of vapourization( water can evaporate into gas molecules easily). This evaporation causes a cooling effect on the skin surface and it requires heat energy as liquid is converted to gas. This is a typical example of an endothermic process. BUT spilling vegetable oil of the same temperature as the water over the skin will not cool the body down because oil has a high heat of vapourization and doesn't evaporate because it is a viscous liquid and the molecules are tightly bound to each other. I hope this helps.

Answer:

1. H3PO4 + 3NaOH —> Na3PO4 + 3H2O

2. 38.5mL

Explanation:

1. We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H3PO4 + 3NaOH —> Na3PO4 + 3H2O

2. H3PO4 + 3NaOH —> Na3PO4 + 3H2O

From the equation above, the following data were obtained:

nA (mole of the acid) = 1

nB (mole of the base) = 3

Data obtained from the question include:

Vb (volume of base) =?

Mb (Molarity of base) = 0.120 M

Va (volume of acid) = 35.0 mL

Ma (Molarity of acid) = 0.0440 M

Using the formula MaVa/MbVb = nA/nB, the volume of the base (i.e NaOH) can be obtained as follow:

MaVa/MbVb = nA/nB

0.0440 x 35/ 0.120 x Vb = 1/3

Cross multiply to express in linear form as shown below:

0.120 x Vb = 0.0440 x 35 x 3

Divide both side by 0.120

Vb = (0.0440 x 35 x 3) /0.120

Vb = 38.5mL

Therefore, 38.5mL of 0.120 M NaOH is needed for the complete neutralization.

Answer:

We need 38.5 mL of NaOH to neutralize the H3PO4 solution

Explanation:

Step 1: Data given

Molarity of NaOH = 0.120 M

Volume of H3PO4 = 35.0 mL = 0.035 L

Molarity of H3PO4 = 0.0440 M

Step 2: The balanced equation

H3PO4 + 3NaOH —> Na3PO4 + 3H2O

Step 3: Calculate the volume of NaOH

b*Ca*Va = a *Cb*Vb

⇒with b = the coefficient of NaOH = 3

⇒with Ca = the concentration of H3PO4 = 0.0440 M

⇒with Va = the volume of H3PO4 = 35.0 mL = 0.0350 L

⇒with a = the coefficient of H3PO4 = 1

⇒with Cb = the concentration of NaOH = 0.120 M

⇒with Vb = the volume of NaOH = TO BE DETERMINED

3*0.0440 * 0.0350 = 0.120 * Vb

Vb = 0.0385 L = 38.5 mL

We need 38.5 mL of NaOH to neutralize the H3PO4 solution

Answer:

13.5 g

Explanation:

This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.

Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.

number of moles = n = mass of Al / Atomic Weight Al

⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹

                                                         = 13.5 g

We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.

Mass of aluminum that participated in the chemical reaction is 1.35 grams.

Number of Moles:

Number of moles is defined as the ratio of given mass to the molar mass.

Given:

Moles of Aluminum = 0.050 moles

To find:

Mass of Aluminum=?

As we know, Molar mass of Aluminum = 27g/mol

On substituting the values:

[tex]\text{Number of Moles}=\frac{\text{Given mass}}{\text{Molar mass}} \\\\ \text{Given mass}= \text{Number of Moles}*\text{Molar mass}\\\\ \text{Given mass}= 0.050*27\\\\ \text{Given mass}=1.35\text{ grams}[/tex]

Thus, the mass of aluminum that participated in the chemical reaction is 1.35 grams.

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Answer:

TRUE: The equivalence point is where the amount of acid equals the amount of base during any acid-base titration.

Explanation:

The point on the titration curve where the number of base equivalents added equals the number of acid equivalents is the equivalence point or neutralization point.

Chemical indicators are substances that change color thanks to a chemical change, depending on the pH of the medium, and thus indicate the end point or point of equivalence of an acid-base volumetry.

A titration curve occurs by representing the measured pH as a function of the added volume of titrant, where the rapid change in pH for a given volume is observed. The inflection point of this curve is called the equivalence point and its volume indicates the volume of titrant consumed to fully react with the analyte.

In some cases, there are multiple equivalence points that are multiples of the first equivalence point, as in the valuation of a diprotic acid, which indicates that its pH value will not always be 7.

The sphingolipid is the lipid with this composition.

Option: C

Explanation:

The complex lipid with  one mole of fatty acid per mole of inorganic phosphate has been Sphingolipid. Thus, option C is correct.

The lipids are biomolecules with the ability to bind with the nonpolar surfaces. The lipids have been bound with other biomolecules to form complex lipids.

Thus, the complex lipid with  one mole of fatty acid per mole of inorganic phosphate has been Sphingolipid. Thus, option C is correct.

For more information about lipids, refer to the link:

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Answer:

40:1  is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration.

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=pK_a+\log(\frac{[magenta(Php)]}{[Php]})[/tex]

We are given:

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of phenolphthalein = 9.40

[tex][magenta(Php)][/tex] = concentration of magenta phenolphthalein

[tex][Php][/tex] = concentration of colorless phenolphthalein

pH = 11

Putting values in above equation, we get:

[tex]11=9.40+\log(\frac{[magenta(Php)]}{[Php]})[/tex]

[tex]\log(\frac{[magenta(Php)]}{[Php]})=11-9.40=1.6[/tex]

[tex]\frac{[magenta(Php)]}{[Php]}=10^{1.6}=39.81 :1 \approx 40:1[/tex]

40:1  is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration.

Answer:

If carbonyl oxygen of 4-hydroxypentanal is enriched with [tex]O^{18}[/tex], then the oxygen label appears in the water .

Explanation:

Calculating the final molarity of iodide anion in the solution, convert the grams of iron (II) iodide to moles, convert the volume of the silver nitrate solution to liters, use the stoichiometry of the reaction, calculate the moles of iodide anion, and divide the moles by the final volume of the solution.

To calculate the final molarity of iodide anion in the solution, we need to use the stoichiometry of the reaction between iron (II) iodide and silver nitrate. Given that there is 0.981 g of iron (II) iodide and 150 mL of a 35.0 mM aqueous solution of silver nitrate, we can determine the moles of iron (II) iodide and the moles of iodide anion produced. Finally, we can calculate the final molarity of iodide anion by dividing the moles by the final volume of the solution.

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