The equilibrium constant, K, for the following reaction is 2.44×10-2 at 518 K: PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 15.3 L container at 518 K contains 0.300 M PCl5, 8.55×10-2 M PCl3 and 8.55×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 8.64 L?

Answers

Answer 1

Answer:

[PCl₅] = 0.5646M

[PCl₃] = 0.1174M

[Cl₂] = 0.1174M

Explanation:

In the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

K equilibrium is defined as:

K = 2.44x10⁻² = [PCl₃] [Cl₂] / [PCl₅]

The initial moles of each compound when volume is 15.3L are:

PCl₅ = 0.300mol/L×15.3L = 4.59mol

Cl₂ = 8.55x10⁻²mol/L×15.3L = 1.308mol

PCl₃ = 8.55x10⁻²mol/L×15.3L = 1.308mol

At 8.64L, the new concentrations are:

[PCl₅] = 4.59mol / 8.64L = 0.531M

[PCl₃] = 1.308mol / 8.64L = 0.151M

[Cl₂] = 1.308mol / 8.64L = 0.151M

At these conditions, reaction quotient, Q, is:

Q = [0.151M] [0.151M] / [0.531M]

Q = 4.29x10⁻²

As Q > K, the reaction will shift to the left producing more reactant, that means equilibrium concentrations are:

[PCl₅] = 0.531M + X

[PCl₃] = 0.151M - X

[Cl₂] = 0.151M - X

Where X is reaction coordinate.

Replacing in K expression:

2.44x10⁻² = [0.151M - X] [0.151M - X] / [0.531M + X]

1.296x10⁻² + 2.44x10⁻²X = 0.0228 - 0.302X + X²

0 = 9.84x10⁻³ - 0.3264X + X²

Solving for X:

X = 0.293 → False solution. Produce negative concentrations

X = 0.0336M → Right solution.

Replacing:

[PCl₅] = 0.531M + 0.0336

[PCl₃] = 0.151M - 0.0336

[Cl₂] = 0.151M - 0.0336

[PCl₅] = 0.5646M[PCl₃] = 0.1174M[Cl₂] = 0.1174M

Related Questions

What is the core charge of helium and why?

Answers

Answer:

Formula for effective nuclear charge is as follows. So, for He atom value of S = 0.30 because the electrons are present in 1s orbital. Therefore, calculate the effective nuclear charge for helium as follows. Thus, we can conclude that the effective nuclear charge for helium is 1.7

Explanation:

The effective nuclear charge experienced by a 1s electron in helium is +1.70.

A 1.00 liter solution contains 0.31 M sodium acetate and 0.40 M acetic acid. If 0.100 moles of barium hydroxide are added to this system, indicate whether the following statements are TRUE or FALSE . (Assume that the volume does not change upon the addition of barium hydroxide.)

a. The number of moles of CH3COOH will remain the same.
b. The number of moles of CH3COO- will increase.
c. The equilibrium concentration of H3O+ will decrease.
d. The pH will decrease.
e. The ratio of [CH3COOH] / [CH3COO-] will remain the same.

Answers

Answer and Explanation:

The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.

If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:

Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻

The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.

Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol

Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol

moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-

According to this, the following are the answers to the sentences:

a. The number of moles of CH₃COOH will remain the same ⇒ FALSE

The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions

b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE

Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)

c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE

The equilibrium concentration of OH⁻ is increased

d. The pH will decrease⇒ FALSE

pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:

[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]

pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15

Thus, the pH will increase.

One brand of laundry bleach is an aqueous solution containing 4.00% sodium hypochlorite (NaOCl) by mass. You may want to reference (Pages 552 - 557) Section 13.5 while completing this problem. Part A What is the molarity of this solution

Answers

Answer:

molarity of the solution = 0.548 mol/L

Note: Additional information about the question is given as follows;

One brand of laundry bleach is an aqueous solution containing 4.00% sodium hypochlorite (NaOCl) by mass

What is the molarity of this solution? (Assume a density of 1.02 g/mL .)

Explanation:

A 4.00 percentage by mass composition of sodium hypochlorite (NaOCl) solution means that 100 g of the solution contains 4.00 g NaOCl.

Thus, a 1000 g of the solution contains 40.0 g NaOCl

Density of solution = 1.02 g/mL

Therefore, the volume occupied by 1000 g solution = mass/density

volume of 1000 g solution = 1000 g/1.02 g/ml = 980.4 mL

Molar mass of NaOCl = 74.5 g/mol

Number of moles = mass/molar mass

Number of moles of NaOCl = 40.0 g/74.5 g/mol

Number of moles of NaOCl  = 0.537 mole

Therefore, molarity = number of moles / volume(L)

volume of solution in litres = 980.4/1000 = 0.9804 L

Molarity = 0.537/0.9804 = 0.548 mol/L

Therefore, molarity of the solution = 0.548 mol/L

When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose from 25.00 °C to 31.75 °C. Determine the ∆Hrxn in kJ/mol for the reaction as written below. Assume that the density of the solutions is 1.000 g/mL, and that the specific heat capacity of the solutions is that of pure water. Must show your work (scratch paper) to receive credit.

Answers

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density  * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

which resonance form would be the most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene

Answers

The question is incomplete as the options are missing, however, the correct complete question is attached.

Answer:

The correct answer is option A. ( check image)

Explanation:

The most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene in given options is option a due to the fact that this resonating form follows the octet rule is satisfied for all atoms and additional π bond is present in between C-O that makes it more stable, while in other options there are positive charge which means they do not follows octet rule completely.

Thus, the correct answer is option A. ( check image)

The following thermochemical equation is for the reaction of Fe 3 O 4 (s) with hydrogen (g) to form iron and water vapor Fe 3 O 4 (s)+4H 2 (g) 3Fe(s)+4H 2 O(g) When 64.6 grams of Fe 3 O 4 (s) react with excess hydrogenſe) how much heat is absorbed?

Answers

Answer:

41.3kJ of heat is absorbed

Explanation:

Based in the reaction:

Fe₃O₄(s) + 4H₂(g) → 3Fe(s) + 4H₂O(g) ΔH = 151kJ

1 mole of Fe3O4 reacts with 4 moles of H₂, 151kJ are absorbed.

63.4g of Fe₃O₄ (Molar mass: 231.533g/mol) are:

63.4g Fe₃O₄ × (1mol / 231.533g) = 0.274moles of Fe₃O₄

These are the moles of Fe₃O₄ that react. As 1 mole of Fe₃O₄ in reaction absorb 151kJ, 0.274moles absorb:

0.274moles of Fe₃O₄ × (151kJ / 1 mole Fe₃O₄) =

41.3kJ of heat is absorbed

What is the maximum amount of silver (in grams) that can be plated out of 4.7 L of an AgNO3 solution containing 6.3 % Ag by mass

Answers

Answer:

296.1g of Ag is the maximum amount of silver

Explanation:

A solution of 6.3% Ag by mass contains 6.3g of Ag per 100g of solution. Thus, you need to calculate the mass of the solution and then, the mass of Ag present in solution, thus:

Mass of solution:

Assuming a density of 1g/mL:

[tex]4.7L \frac{1000mL}{1L} \frac{1g}{mL} = 4700g[/tex]

If the solution contains 6.3g of Ag per 100g of solution, the mass of Ag in 4700L is:

4700L × (6.3g Ag / 100g) =

296.1g of Ag is the maximum amount of silver

The reaction rate is k[Ce4+][Mn2+] for the following reaction: 2Ce4+(aq) + Tl+(aq) + Mn2+(aq) → 2Ce3+(aq) + Tl3+(aq) + Mn2+(aq What is the catalyst?

Answers

Answer:

Manganese (II) ion, Mn²⁺

Explanation:

Hello,

In this case, given the overall reaction:

[tex]2Ce^{4+}(aq) + Tl^+(aq) + Mn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Tl^{3+}(aq) + Mn^{2+}(aq)[/tex]

Thus, since manganese (II) ion, Mn²⁺ is both at the reactant and products, we infer it is catalyst, since catalysts are firstly consumed but finally regenerated once the reaction has gone to completion. Moreover, since inner steps are needed to obtain it, we can infer that the given rate law corresponds to the slowest step that is related with the initial collisions between Ce⁴⁺ and Mn²⁺

Best regards.

The rate constant for this first‑order reaction is 0.550 s−10.550 s−1 at 400 ∘C.400 ∘C. A⟶products A⟶products How long, in seconds, would it take for the concentration of AA to decrease from 0.690 M0.690 M to 0.220 M?

Answers

Answer:

[tex]t=2.08s[/tex]

Explanation:

Hello,

In this case, for first order reactions, we can use the following integrated rate law:

[tex]ln(\frac{[A]}{[A]_0} )=kt[/tex]

Thus, we compute the time as shown below:

[tex]t=-\frac{ln(\frac{[A]}{[A]_0} )}{k}=- \frac{ln(\frac{0.220M}{0.690M} )}{0.55s^{-1}} \\\\t=-\frac{-1.14}{0.550s^{-1}}\\ \\t=2.08s[/tex]

Best regards.

an ideal gas is at a pressure 1.00 x 10^5 N/m^2 and occupies a volume 11.00 m^3. If the gass is compressed to a volume of 1.00 m^3 while the temperature remains constant, what will be the new pressure in the gas.

Answers

Answer:

[tex]P_2=1.1x10^6Pa[/tex]

Explanation:

Hello.

In this case, we can solve this problem by applying the Boyle's law which allows us to understand the pressure-volume behavior as a directly proportional relationship:

[tex]P_1V_1=P_2V_2[/tex]

In such away, knowing the both the initial pressure and volume and the final volume, we can compute the final pressure as shown below:

[tex]P_2=\frac{P_1V_1}{V_2}[/tex]

Consider that the given initial pressure is also equal to Pa:

[tex]P_2=\frac{1.00x10^5Pa*11.00m^3}{1.00m^3}\\ \\P_2=1.1x10^6Pa[/tex]

Which stands for a pressure increase when volume decreases.

Regards.

What is the value of ΔG at 25°C when the initial concentrations of A, B, and C are 1 M, 1 mM, and 1 μM, respectively?

Answers

Answer:

Explanation:

0,44

You use 10.0 mL of solution A, 10.0 mL of solution B, and 70.0 mL of water for your first mixture. What is the initial concentration of KIO3

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  initial concentration is  [tex]C_f = 0.0022 \ M[/tex]

Explanation:

From the question we are told that

    The volume of solution A is  [tex]V_i = 10.0 mL[/tex]

    The concentration of A is [tex]C_i = 0.0200 \ M[/tex]

    The volume of solution B  is  [tex]V_B = 10.0mL[/tex]

    The volume of water is  [tex]V_{w } = 70.0 mL[/tex]

Generally the law of dilution is mathematically represented as

             [tex]C_i * V_i = C_f * V_f[/tex]

Where  [tex]C_f[/tex] is the concentration of  the mixture

            [tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as

            [tex]V_f = 10 + 10 + 70[/tex]

           [tex]V_f = 90mL[/tex]

So  

      [tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]

substituting values

       [tex]C_f = \frac{0.0200 * 10 }{90}[/tex]

       [tex]C_f = 0.0022 \ M[/tex]

Note the mixture obtained is  [tex]KIO_3[/tex]

Explain why the following chemical equation represents a Lewis acid-base reaction. H+ + NH3 —>NH4+

Answers

Answer:

Due to an electron-pair acceptor and donor.

Explanations:

Lewis acid can be defined as an electron-pair acceptor. An example is Hydrogen ion(H+). This is because it is a proton and it distributes positive charge which means that it accepts electrons(negative charge).

Lewis base can be defined as an electron-pair donor. This is because it donates electrons to be accepted by the proton. An example is ammonia(NH3).

which conditions make snow

Answers

Answer:

It depends on the weather.

Explanation:

Like rain and hail, snow comes from the water vapor that forms the clouds, but obviously its formation is different: snow forms when the temperature is low in the atmosphere. In these conditions the water vapor drops freeze and when they collide form tiny crystals that join together to form snowflakes, which fall to Earth when they are heavy enough.

Consider the 1H NMR spectrum for the following compound:
CH3CH2CH3
Predict the first-order splitting patterns of the indicated proton. This question uses specific splitting patterns instead of the often used generic term "multiplet."
a. doublet of quartets
b. triplet of triplets
c. septet
d. quartet of quartets
e. quintet

Answers

Answer:

See explanation

Explanation:

In this case, we have to check the neighbors of each carbon in the molecule. In propane, we have two types of carbons (see figure 1) (blue and red ones). The red carbons are equivalent. (Both have the same neighbors). Now, we can analyze each carbon:

Blue carbon

In the blue carbon, we have 6 hydrogens neighbors (three on each methyl). If we follow the n+1 rule, we will have:

6+1= 7

For the blue carbon, we will have a Septet.

Red carbons

In the red carbon, we have only 2 neighbors (the carbon in the middle only have 2 hydrogens). If we follow the n+1 rule, we will have:

2+1=3

For the red carbon, we will have a triplet.

See figure 2

I hope it helps!

A student followed the procedure of this experiment to determine the percent NaCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of the commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-ml aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na S,O, solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g ml.
(1) Calculate the mass of commercial bleaching solution titrated. 1.1 450=559
(2) Determine the percent NaOCl in the commercial bleaching solution.
(3) Calculate the mass of one gallon of the commercial bleaching solution.

Answers

llllllllalalalalallalalalallalallalalalalalallalalalallala pls I am really sorry for this I can't take it down

Which statement is true: Mg -> Mg2+ + 2e-

(1) Mg gains 2 electrons (2) Mg2+ loses 2 electrons (3) mg loses 1 electron (4) mg loses 2 electrons

Answers

Answer:

(2) Mg²⁺ loses 2 electrons.

Explanation:

Let's consider the following half-reaction.

Mg ⇒ Mg²⁺ + 2 e⁻

Magnesium is a metal that loses 2 electrons to fulfill the octet rule. When magnesium loses electrons, the oxidation number increases from 0 to +2. This kind of half-reaction is known as oxidation, and has to be accompanied by a reduction of another species.

The substance nitrogen has the following properties: normal melting point: 63.2 K normal boiling point: 77.4 K triple point: 0.127 atm, 63.1 K critical point: 33.5 atm, 126.0 K At temperatures above 126 K and pressures above 33.5 atm, N2 is a supercritical fluid . N2 does not exist as a liquid at pressures below atm. N2 is a _________ at 16.7 atm and 56.5 K. N2 is a _________ at 1.00 atm and 73.9 K. N2 is a _________ at 0.127 atm and 84.0 K.

Answers

Answer:

- N2 does not exist as a liquid at pressures below 0.127 atm.

- N2 is a solid at 16.7 atm and 56.5 K.

- N2 is a liquid at 1.00 atm and 73.9 K

- N2 is a gas at 0.127 atm and 84.0 K.

Explanation:

Hello,

At first, we organize the information:

- Normal melting point: 63.2 K.

- Normal boiling point: 77.4 K.

- Triple point: 0.127 atm and 63.1 K.

- Critical point: 33.5 atm and 126.0 K.

In such a way:

- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).

- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.

- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.

- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.

Best regards.

How fast are the atoms moving if the temperature of a gas is cold?

A. very, very slowly

B. they are stagnant

C. very, very quickly

Answers

Answer:

i think option a is correct answer because when there is low temperature then the kinetic enegry will be very less and the atoms moves very slowly.

Answer:

A. very, very slowly

Explanation:

A is the answer because atoms will move faster in hot gas than in cold gas.

what is the electron configuration of the iodide ion?
A. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²
B. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶
C. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶
D. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰

Answers

Answer:

the answer to this question is C

The electron configuration of the iodide ion is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶. The correct option is option C.

What is electron configuration ?

The arrangement of an atom's or molecule's electrons in their respective atomic or molecular orbitals is known as its electron configuration; for instance, the electron configuration of the neon atom is 1s2 2s2 2p6.

According to electronic configurations, electrons move individually within each orbital while interacting with the average field produced by all other orbitals. The corrosion potential and reactivity of an atom may be calculated from its electron configuration. The electron configuration of the iodide ion is  1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶.

Therefore, the correct option is option C.

To learn more about electron configuration, here:

https://brainly.com/question/30161188

#SPJ6

If the charge on the chloride ion is -1, what is the charge on the magnesium ion in the compound MgCl2?

Answers

Answer:

+2

Explanation:

Because the charge of the chloride ion is negative, that means that the charge of the magnesium ion must be positive since cations and anions go together, not cation and cation nor anion and anion. Using the "reverse criss-cross method", since the subscript of Mg is 1, that means that this is the lowest whole number ratio so we don't need to worry about simplifying. Therefore, since the charge of Cl is 2, the answer is +2.

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the surroundings

Answers

Answer:

[tex]V=27992L=28.00m^3[/tex]

Explanation:

Hello,

In this case, the combustion of methane is shown below:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]

Best regards.

The melting point of sodium chloride is 801°C. The melting point of chlorine is -101°C. Explain, in terms of structure and bonding, the difference between the melting points of these two substances.

Answers

Answer:

It's harder to melt sodium chloride because of its density and stability.

Explanation:

The bond between Na+ and Cl- is an ionic bond, meaning its an ionic compound. Ionic compounds have more inter molecular forces(forces involving multiple molecules). Chlorine has much less of these forces than NaCl and isn't as dense.

Answer:

It's harder to melt sodium chloride because of its density

Explanation:

List the following compounds in order from strongest acid to weakest acid. Rank the acids from strongest to weakest.
CH2CHCH2COOH CH2CH2CH2COOH CH3CHCH2COOH CH3CH2CH2COOH
Strongest Weakest

Answers

Answer:

CH3CH2CH2COOH<CH2(F)CH2CH2COOH<CH3CH(F)CH2COOH<CH2(F)CH(F)CH2COOH

Explanation:

We know that the presence of highly electronegative elements in carboxylic acid molecules lead to -I inductive effect. This implies that electrons are withdrawn along the chain towards the electronegative element. As electrons are withdrawn towards the electronegative element, the electron cloud of the carbonyl- hydrogen bond in the acid weakens and the hydrogen can now be easily lost as a proton, that is , the molecule becomes more acidic.

The -I inductive effect increases with increase in the number of electronegative elements present in the molecule and the proximity of the electronegative element to the carbonyl group. The closer the electronegative element is to the carbonyl group, the greater the acidity of the molecule since the -I inductive effect dies out with increasing distance from the carbonyl group. Also, the more the number of electronegative elements in the molecule, the greater the - I inductive effect and the greater the acidity of the molecule, hence the answer.

Two moles of copper (II) sulfate,CuSO4,contains blank moles of O

Answers

Answer:

8 mol

Explanation:

Step 1: Given data

Moles of copper (II) sulfate: 2 mol

Chemical formula of copper (II) sulfate: CuSO₄

Step 2: Establish the molar ratio of copper (II) sulfate to oxygen

According to the chemical formula, the molar ratio of copper (II) sulfate to oxygen is 1:4.

Step 3: Calculate the moles of O in 2 mol of CuSO₄

[tex]2molCuSO_4 \times \frac{4molO}{1molCuSO_4} = 8molO[/tex]

3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 mL of pure cyclohexene product. Calculate the student’s percent yield for this reaction. Show your work.

Answers

Answer:

[tex]49.45~%[/tex]

Explanation:

In this case, we have to start with the chemical reaction:

[tex]C_6H_1_2O~->~C_6H_1_0~+~H_2O[/tex]

So, if we start with 10 mol of cyclohexanol ([tex]C_6H_1_2O[/tex]) we will obtain 10 mol of cyclohexanol ([tex]C_6H_1_0[/tex]). So, we can calculate the grams of cyclohexanol if we calculate the molar mass:

[tex](6*12)+(10*1)=82~g/mol[/tex]

With this value we can calculate the grams:

[tex]10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0[/tex]

Now, we have as a product 500 mL of [tex]C_6H_1_0[/tex]. If we use the density value (0.811 g/mL). We can calculate the grams of product:

[tex]500~mL\frac{0.811~g}{1~mL}=405.5~g[/tex]

Finally, with these values we can calculate the yield:

[tex]%~=~\frac{405.5}{820}x100~=~49.45%[/tex]%= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

Based on the data given, the percentage yield of the student's work is 49.45 %.

What is the equation of the reaction?

The equation of the clay-catalyzed dehydration of cyclohexanol is given below:

C₆H₁₂O ----> C₆H₁₀ + H₂O

From the equation of the reaction, 1 mole of cyclohexanol yields 1 mole of cyclohexene.

Therefore 10 moles of cyclohexanol should yield 10 moles of cyclohexene.

What is the moles of cyclohexene obtained?

First we determine the mass of cyclohexene obtained.

Mass = density * volume

volume of cyclohexene = 500 mL

density = 0.811 g/mL

mass of cyclohexene = 500 * 0.811

mass of cyclohexene = 405.5 g

Number of moles of cyclohexene = mass/molar mass

molar mass of cyclohexene = 82 g

moles of cyclohexene = 405.5 g/82

moles of cyclohexene = 4.945 moles

What is the percentage yield?

Percentage yield = actual yield /expected yield * 100%

Percentage yield = 4.945/10 * 100%

Percentage yield = 49.45%

Therefore, the percentage yield of the student's work is 49.45 %.

learn more about percentage yield at: https://brainly.com/question/8638404

HELPPP.
Which of the following is a property of matter?
O A. It takes up space.
OB. It is everywhere.
O C. It is constantly changing.
O D. It cannot be divided.

Answers

Answer:

a

Explanation:

Matter can be volume or density. So, this concludes that it is when it takes up space.

Answer: A.

Explanation:

it takes up space

The rate at which two methyl radicals couple to form ethane is significantly faster than the rate at which two tert-butyl radicals couple. Offer two explanations for this observation.

Answers

Answer:

1. stability factor

2. steric hindrance factor

Explanation:

stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.

ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive

Calculate the pressure exerted by 0.5600 mole of in a 1.1000-L container at 25.0°C. (The gas constant is 0.08206 L·atm/mol·K. Take absolute zero to be –273.2°C.)
Use the ideal gas law.

(Enter your answer to four significant figures.)

Pressure =
atm

Use the van der Waals equation.

(For : a = 1.39 atm L2/mol2, and b = 0.0391 L/mol. Enter your answer to four significant figures.)

Pressure =
atm

Compare the results.

(Enter your answer to two significant figures.)

The
_________
is higher by
atm, or
%.

Answers

Answer:

using ideal gas equation =12.4576atm to 4.significant figure

using vander Waals equation = 12.3504

The differences is 0.10atm

A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of moles of an unknown gas requires 6.34 minutes to effuse through the same barrier. The molar mass of the unknown gas is:________.
g/mol.

Answers

Answer:

The molar mass of the unknown gas is [tex]\mathbf{ 51.865 \ g/mol}[/tex]

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

[tex]R \ \alpha \ \dfrac{1}{\sqrt{d}}[/tex]

[tex]R = \dfrac{k}{d}[/tex]  where K = constant

If we compare the rate o diffusion of two gases;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}[/tex]

Since the density of a gas d is proportional to its relative molecular mass M. Then;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

Rate is the reciprocal of time ; i.e

[tex]R = \dfrac{1}{t}[/tex]

Thus; replacing the value of R into the above previous equation;we have:

[tex]\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}[/tex]

We can equally say:

[tex]{\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

[tex]{\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}[/tex]

[tex]M_2 = 32 \times ( \dfrac{6.34}{4.98})^2[/tex]

[tex]M_2 = 32 \times ( 1.273092369)^2[/tex]

[tex]M_2 = 32 \times 1.62076418[/tex]

[tex]\mathbf{M_2 = 51.865 \ g/mol}[/tex]

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