The composition of a compound used to make polyvinyl chloride (PVC) is 38.4% C, 4.8% H and 56.8% Cl by mass. It took 7.73 min for a given volume of the compound to effuse through a porous plug, but it took only 6.18 min for the same amount of Ar to diffuse at the same temperature and pressure. What is the molecular formula of the compound?

There are quite a bunch of typo errors in the question; here is the correct question below:

A thermos bottle (Dewar vessel) has an evacuated space between its inner and outer walls to diminish the rate of transfer of thermal energy to or from the bottle’s contents. For good insulation, the mean free path of the residual gas (air; average molecular mass = 29) should be at least 10 times the distance between the inner and outer walls, which is about 1.0 cm. What should be the maximum residual gas pressure in the evacuated space if T = 300 K? Take an average diameter of d = 3.1 × 10⁻¹⁰ m for the molecules in the air.

Answer:

9.57 × 10⁻⁷ atm

Explanation:

The mean free path ( λ ) can be illustrated by the equation:

λ =   [tex]\frac{1}{\sqrt{2} \pi d{^2}N/V }[/tex]     ----------  (1)

N/V = [tex]\frac{N_AP}{RT}[/tex]                      ------------- (2)

From the above relation, we can deduce that;

P=  [tex]\frac{RT}{\sqrt{2}\pi d{^2}N{_A}λ }[/tex]     -------------(3)

let I=  λ

From the above equations;

d= diameter of the atom

[tex]{N_A}[/tex] = avogadro's constant

P= pressure

R= universal rate constant which is given by 0.08206 L atm mol⁻¹ k⁻¹

T= temperature

From the question, we are given that the mean free path of the residual air molecules ( d = 3.1 × 10⁻¹⁰ m) is equal to 10cm = 0.1m

Therefore, we can determine the pressure using equation (3)

i.e

P=  [tex]\frac{RT}{\sqrt{2}\pi d{^2}N{_A}λ }[/tex]

=  [tex]\frac{(8.314J{K^-^1)(300K)}}{\sqrt{2}(3.14)(3.1*10^{-10}m)^2(6.022*10^{23}mol^{-1})(0.1m) }[/tex]

=97.06 × 10⁻³ Pa   ×   [tex]\frac{1atm}{1.01325*10^5Pa}[/tex]

=9.57 ×  10⁻⁷  atm

Therefore, the maximum residual gas pressure in the calculated space is; 9.57 ×  10⁻⁷  atm

Answer:

31.3 g

The answer is higher than the true answer.

Explanation:

By neglecting the heat lost by other processes, the energy conservation states that:

Qcooling + Qevaporate = 0

The cooling process happens without phase change, so the heat can be calculated by:

Qcooling = m*c*ΔT

Where m is the mass, c is the heat capacity (cwater = 4184 J/kg.K), and ΔT is the temperature variation (final - initial).

The evaporate process happen without changing of temperature (pure substance), and the heat can be calculated by:

Qevaporate = m*L

Where m is the mass evaporated and L is the heat of evaporation (2340000 J/kg).

0.350*4184*(45 - 95) + m*2340000 = 0

2340000m = 73220

m = 0.0313 kg

m = 31.3 g

Because of the assumptions made, the real mass is not that was calculated. There'll be changing mass when the coffee is cooling, and there'll be heat loses by other processes because the system is not isolated. Also, the substance is not pure. So, there'll be more factors at the energy equation, thus, the answer is higher than the true answer.

To cool 350 g of coffee in a 100-g glass cup from 95.0°C to 45.0°C, 33.2 grams of coffee must evaporate.

To solve this problem, we first need to calculate the total heat that needs to be removed from the coffee and the cup.

Steps to Calculate:

Thus, 33.2 grams of coffee must evaporate to cool the coffee and the cup from 95.0°C to 45.0°C. Neglecting other heat losses means this answer is slightly larger than the true answer.

Answer:

Keq for this reaction is 6.94x10⁻³

Explanation:

The equilibrium equation is this one:

N₂O₄ (g) ⇄  2NO₂ (g)

Initially we have 0.03 moles from the dinitrogen tetroxide and nothing from the dioxide.

In the reaction, some amount of compound (x) has reacted.

As ratio is 1:2, we have double x in products.

Finally in equilibrium we have:

       N₂O₄ (g) ⇄  2NO₂ (g)

       0.03 - x          2x

And we know [N₂O₄] in equilibrium so:

0.03 - x = 0.0236

x = 0.03 - 0.0236 → 6.4x10⁻³

As this is the amount that has reacted, in equilibrium I have produced:

6.4x10⁻³  .2 = 0.0128 moles of NO₂

This is the expression for K,

[NO₂] ² / [N₂O₄]

0.0128² / 0.0236 = 6.94x10⁻³

Answer:

6.94x10-3

Explanation:

Answer:

The concentration of the copper(II) sulfate solution is 0.99 M

Explanation:

Step 1: Data given

Mass of copper(II) sulfate = 79 grams

Volume of the flask = 500 mL

Molar mass copper(II) sulfate = 159.61 g/mol

Step 2: Calculate moles copper(II) sulfate

Moles CuSO4 = Mass CuSO4 / molar mass CuSO4

Moles CuSO4 = 79.0 grams / 159.61 g/mol

Moles CuSO4 = 0.495 moles

Step 3: Calculate concentration

Concentration CuSO4 = moles / volume

Concentration CuSO4 = 0.495 moles / 0.5 L

Concentration = 0.99 M

The concentration of the copper(II) sulfate solution is 0.99 M

Each round of Citric Acid Cycle produce 2 molecules of carbon dioxide. So one cycle of TCA cycle is enough to convert Acetyl CoA to carbon dioxide.

Citric Acid Cycle or Kreb's Cycle or the TCA cycle is the 1st dedicated step towards the aerobic respiration. The end product of glycolysis is Pyruvate which is a three carbon compound. It's acted upon by Pyruvate Decarboxylase to produce a 2 carbon compound Acetyl CoA and a molecule of carbon dioxide. This Acetyl CoA now reacts with oxaloacetate to produce Citric Acid which is the 1st step of Citric Acid Cycle. This now produce several intermediates and a lot of reduced electron carriers along with 2 molecules of carbon dioxide and ends up being oxaloacetate again. So one cycle of Citric Acid Cycle is necessary to convert Acetyl CoA to CO2.

Answer:

D. Cr

Explanation:

In order to determine which atom has at least one electron in its d orbital, we have to write their theoretical electron configurations.

Cr has 4 electrons in d orbitals. Cr belongs to the d-block in the periodic table.

Answer:

The correct answer is static, sliding and rolling friction.

Explanation:

There are three different types of friction. Static, sliding and rolling friction occur between solid surfaces. Static friction is strongest, followed by sliding friction and then rolling friction, which is the weakest of the three. The equality of these three types of friction is that they produce heat and make movement difficult. The difference is in their magnitude and in the conditions that produce each type of friction. Static is produced when a body at rest begins to move, sliding is produced when this body is already in motion and rolling is produced when a body rolls on a surface, deforming one or both of them.

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Answer:

The Answer Of this question is 0.4052 L ...

Answer:

(a) HBr;

(b) 7.61 g;

(c) 96.2 %

Explanation:

Firstly, write the balanced chemical equation:

[tex]Rb_2SO_3 (aq) + 2 HBr (aq)\rightarrow 2 RbBr (aq) + SO_2 (g) + H_2O (l)[/tex]

(a) Find moles of each reactant dividing the mass by the molar mass of rubidium sulfite, then applying the ideal gas law for HBr:

[tex]n_{Rb_2SO_3}=\frac{6.24 g}{251.00 g/mol} = 0.02486 mol[/tex]

[tex]pV_{HBr}=n_{HBr}RT[/tex]

[tex]\therefore n_{HBr} = \frac{pV_{HBr}}{RT} = \frac{0.953 atm\cdot 1.38 L}{0.08206 \frac{L atm}{mol K}\cdot 348.15 K} = 0.04603 mol[/tex]

Find the limiting reactant by dividing each moles by the stoichiometric coefficients and comparing the two numbers:

[tex]eq._{Rb_2SO_3} = \frac{0.02486 mol}{1} = 0.02486 mol[/tex]

[tex]eq._{HBr} = \frac{0.04603 mol}{2} = 0.02302 mol[/tex]

That said, the equivalent of HBr is lower, so it's the limiting reactant.

(b) According to the balanced equation, the moles of HBr are equal to the moles of RbBr, so moles of RbBr theoretically are equal to:

[tex]n_{RbBr} = 0.04603 mol[/tex]

Using the molar mass of RbBr, convert this into mass:

[tex]m_{RbBr} = 0.04603 mol\cdot 165.372 g/mol = 7.61 g[/tex]

(c) To find the percent yield, divide the actual mass produced by the theoretical mass calculated in (b) and multiply by 100 %:

[tex]\%_{yield} =\frac{7.32 g}{7.61 g}\cdot 100\% = 96.2 \%[/tex]

Answer:

BiF5

Shape- trigonal bipyramid

Hybridisation-sp3d

Ideal bond angle-120 and 90

BrO3-

Shape- tetrahedral

Hybridisation-sp3

Ideal bond angle 109o28'

IF4+

Shape- trigonal bipyramid

Hybridisation-sp3d

Ideal bond angle-120o and 90o

Explanation:

The shape adopted by a molecule according to VSEPR is such as minimizes repulsion between electron pairs. The shape of a molecule depends on electron pairs present on the outermost shell of the central atom. The bond angles are such that electrons are positioned as far apart in space as possible, given the number of electron pairs present.

BiF52- has a square pyramidal shape, sp3d2 hybridization, with ideal bond angles of 90° and 120°. BrO3- has a trigonal pyramidal shape, sp3 hybridization, with an ideal bond angle of 109.5°. IF4+ has a square planar shape, sp3d2 hybridization, with ideal bond angles of 90° and 180°.

The shape, hybridization of the central atom, ideal bond angle(s), and any expected deviations for the molecules are as follows:

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Answer:

E° = 0.29 V

Explanation:

Let's consider the following redox reaction.

Sn(s) + Sn⁴⁺(aq) → 2 Sn²⁺(aq)

We can identify both half-reactions:

Reduction (cathode): Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq)    E°red = 0.15 V

Oxidation (anode): Sn(s) → Sn²⁺(aq)  + 2 e⁻         Ered = -0.14 V

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.15 V - (-0.14 V) = 0.29 V

Answer:

Yes

Explanation:

Before dilution:

Volume of NaCl = 10 mL

Concentration of NaCl = 0.5 M

Number of moles = Molarity*Volume = 0.5*10 = 5 millimoles.

Note that number of moles of NaCl does not change on dilution as we are only adding water.

After dilution:

Volume of NaCl = 100 mL

Number of moles = 5 millimoles (no change)

New Concentration = Number of moles per volume in litres = [tex]\frac{5\times10^{-3}}{100\times10^{-3}}[/tex]= 0.05 Molar

Hence the concentration became one-tenth of the initial concentration after dilution.

Final answer:

Yes, dilution does change the concentration of a solution. By diluting a 0.50 M NaCl solution to a total volume of 100 mL, the concentration is reduced to 0.050 M, because the same amount of NaCl is spread out in a larger volume of water.

Explanation:

Concerning dilution, yes, it does change the concentration of the solution. When you add more solvent to a solution, the concentration of the solute decreases. For example, if you have a 0.50 M NaCl solution and add enough water to make the total volume 100 mL, the concentration of NaCl changes because the same amount of solute (NaCl) is now dispersed in a greater volume of solvent (water).

Step-by-step Calculation:

Determine the initial amount of NaCl in moles by multiplying the initial volume by the concentration: Moles of NaCl = 0.50 M × 0.010 L = 0.005 moles.

Keep in mind that the amount of NaCl does not change during dilution.

Calculate the final concentration by dividing the moles of NaCl by the final volume of the solution after dilution: Final concentration = 0.005 moles / 0.100 L = 0.050 M NaCl.

Answer:

B

Explanation:

The limiting reagent in a chemical reaction is the one that is consumed completely in the course of the reaction. It dictates the extent to which the reaction would proceed as the completion of the reaction is based solely on it.

There are several ways to determine the limiting reagent. The easiest way to do this that works is to divide the number of moles of each reagent by their stoichiometric coefficient in the balanced equation. The reactant with the least value is the limiting reagent.

Thus, we need to calculate the number of moles of silver and hydrogen sulphide. To do this, we simply divide the masses by the relative atomic masses or molecular masses.

The atomic mass of silver is 108.

The number of moles of silver is =

300/108 = 2.8

We now divide this by the stoichiometric coefficient: 2.8/2 = 1.4

We do same for hydrogen sulphide. The molar mass of hydrogen sulphide is 34g/mol

The number of moles is thus 175/34 = 5.15

We now divide this by stoichiometric coefficient = 5.15/1 = 5.15

We can see that silver has less number of the value and hence it is the limiting reactant.

Answer:

Sulfur hexafluoride SF6

Explanation:

In chemistry, the shape of the compounds in which six ligands (atoms, molecules or ions) are arranged around a central atom or ion, defining the vertices of an octahedron, is called octahedral molecular geometry or Oh. It is a very common structure, and it is very studied for its importance in the coordination chemistry of transition metals. From it, other important molecular geometries are derived by continuous deformation, such as the elongated octahedron, the flat octahedron, the square-based pyramid and the flat square. Indirectly, it is also related to tetrahedral molecular geometry.

The concept of octahedral coordination geometry was developed by Alfred Werner to explain the stoichiometry and isomeries in the coordination compounds. An example of a strictly octahedral compound is SF6 sulfur hexafluoride, but chemists use the term in a lax form, so that it is applied to compounds that are not mathematically octahedra, such as cobalt hexaamine (III).

Isomers are molecules that have the same molecular formula but different structure. It is classified as structural isomers and stereoisomers. Structural isomers differ in the way of joining their atoms and are classified into chain, position and function isomers.

Answer:

CH3(CH2)8CH3 > CH3CH2CH2CH3 > CH3CH3 (Option f)

Explanation:

Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a larger atom or molecule, the valence electrons are, on average, farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles.

CH3CH3 has a molar mass of 30.07 g/mol

CH3(CH2)8CH3 has a molar mass of 142.28 g/mol

CH3CH2CH2CH3 has a molar mass of 58.12 g/mol

CH3(CH2)8CH3 > CH3CH2CH2CH3 > CH3CH3 (Option f)

The compounds should be ordered by increasing strength of their London dispersion forces, which relate to their molecular size. The correct order is CH3CH3 < CH3CH2CH2CH3 < CH3(CH2)8CH3, reflecting the increasing number of electrons and molecule size from ethane, to butane, to decane. Option e) is correct.

When placing the compounds CH3CH3, CH3(CH2)8CH3, and CH3CH2CH2CH3 in order of increasing strength of intermolecular forces (IMFs), it is important to consider the types of intermolecular forces present and the size of the molecules. All three compounds are nonpolar and primarily exhibit London dispersion forces. Because dispersion forces increase with the number of electrons and, thus, with the size of the molecule, the compound with the longest carbon chain will have the strongest intermolecular forces.

Thus, the correct ordering from weakest to strongest intermolecular forces is:

Therefore, the correct answer is e. CH3CH3 < CH3CH2CH2CH3 < CH3(CH2)8CH3.

Answer: Extracellular [Ca2+]

Explanation:

The sensitivity and density of the alpha receptors serve to enhance the response to the release of norepinephrine (NE) . However, they do not exert a strong influence as the concentration of calcium ions on the amount of norepinephrine (NE) released by sympathic nerve terminals.

The release of neurotransmitters depends more on either an external or internal stimulus.This results in an action potential which on reaching a nerve terminal, results in the opening of Ca²⁺ channels in the neuronal membrane. Because the extracellular concentration of Ca²⁺ is greater than the intracellular Ca²⁺ concentration, Ca²⁺ flows into the nerve terminal. This triggers a series of events that cause the vesicles containing norepinephrine (NE) to fuse with the plasma membrane and release norepinephrine (NE) into the synapse. The higher the action potential, the higher the Ca²⁺ flow into the terminals resulting in higher amount of norepinephrine (NE) into the synapse, and vice versa.

Catechol-O-methyltransferase (COMT) is one of several enzymes that degrade catecholamines such as dopamine, epinephrine, and norepinephrine. It serves a regulatory purpose to lower the concentration of norepinephrine upon its release from nerve terminals.

Answer:

a) kc= [SO3 ]/([SO2 ][O2 ])

b) kc= 2.27*10⁶ M⁻¹

v) the reaction is product-favored

Explanation:

for the reaction, the equilibrium constant is

SO2 (g) + O2 (g) <-----> SO3 (g)

he equilibrum constant is

kc= [SO3 ]/([SO2 ]*[O2 ])

replacing values

kc= [SO3 ]/([SO2 ]*[O2 ]) = 1.01*10⁻² M/(3.61*10⁻³M*6.11 x 10⁻⁴ M) = 2.27*10⁶ M⁻¹

since kc>>1 the reaction is product-favored

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The correct options would be:

The mobile phase is more polar than the stationary phase

The stationary phase is nonpolar

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The characteristics of reversed phase chromatography are The mobile phase is more polar than the stationary phase and The stationary phase is non polar.

Hence, option II and V are correct.

RPC is Reversed Phase Chromatography. In reversed phase chromatography the stationary phase is non polar [hydrophobic] and the mobile phase is very polar [hydrophilic].  

Now lets check all the option one by one:

Option (I): The stationary phase is non polar in reverse phase chromatography.

So it is incorrect option.

Option (II): The mobile phase is more polar than the stationary phase because in polar mobile phase has high affinity towards the polar solute.

So, it is correct option.  

Option (III): Less polar mobile phase has higher eluent strength not lower eluent strength.

So, it is incorrect option.

Option (IV): More polar mobile phase has less eluent strength.

So, it is incorrect option.

Option (V): The stationary phase is non polar [hydrophobic] is reverse phase chromatography.

So, it is correct option.

Thus, from above conclusion we can say that the The mobile phase is more polar than the stationary phase and The stationary phase is non polar are the characteristics of reverse phase chromatography.

Hence option II and V are correct.

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Answer:

Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector.

Explanation:

To answer this question we need to consult the ranges in which x rays are in the electromagnetic spectrum:

The X radiation in the electromagnetic spectrum fall in the region of:

frequency: 3 x 10¹⁶ Hz  to 3 x 10¹⁹ Hz     (1Hz = 1s⁻¹)

wavelengt: 1 pm  to 10 nm

Comparing the values in our question,

0.91 nm will be detected

5.9 x 10¹¹ Hz will not be detected.

The radiation with a wavelength 0.91 or a frequency of [tex]5.9*10^{11}[/tex] will have:

A. Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector.

X-rays are both types of high energy (high frequency) electromagnetic radiation. They are packets of energy that have no charge or mass (weight).

The X radiation in the electromagnetic spectrum fall in the region of:

Frequency : 3 x 10¹⁶ Hz  to 3 x 10¹⁹ Hz     (1Hz = 1s⁻¹)

Wavelength : 1 pm  to 10 nm

On comparing to the values given in the question:

0.91 nm will be detected

5.9 x 10¹¹ Hz will not be detected.

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Answer:

Mass of hydrogen gas evolved is 0.0749 grams.

Explanation:

Total pressure of the gases = p = 758 mmHg

Vapor pressure of water = 23.78 mmHg

Pressure of hydrogen gas ,P =  p - 23.78 mmHg = 758 mmHg - 23.78 mmHg

P = 734.22 mmHg = [tex]\frac{734.22}{760} atm=0.966 atm[/tex]

Temperature  of of hydrogen gas ,T= 25°C =298.15 K

Volume of hydrogen gas = V = 0.949 L

Moles of hydrogen gas =n

PV = nRT (Ideal gas equation )

[tex]n=\frac{PV}{RT}=\frac{0.966 atm\times 0.949 L}{0.0821 atm L/mol K\times 298.15 K}[/tex]

n = 0.03745 mol

Moles of hydrogen gas = 0.03745 mol

Mass of  0.03745 moles of hydrogen gas  = 0.03745 mol × 2 g/mol = 0.0749 g

Mass of hydrogen gas evolved is 0.0749 grams.

Answer:

S°m,298K = 85.184 J/Kmol

Explanation:

∴ T = 10 K ⇒ Cp,m(Hg(s)) = 4.64 J/Kmol

∴ 10 K to 234.3 K ⇒ ΔS = 57.74 J/Kmol

∴ T = 234.3 K ⇒ ΔHf = 2322 J/mol

∴ 234.3 K to 298.0 K ⇒ ΔS = 6.85 J/Kmol

⇒ S°m,298K = S°m,0K + ∫CpdT/T(10K) + ΔS(10-234.3) + ΔHf/T(234.3K) + ΔS(234.3-298)

⇒ S°m,298K = 0 + 10.684 J/Kmol + 57.74 J/Kmol + 9.9104 J/Kmol + 6.85 J/kmol

⇒ S°m,298K = 85.184 J/Kmol

To find the Third-Law standard molar entropy of Hg(l) at 298 K, we sum up the entropy changes between 10K and 234.3 K, the entropy change due to fusion, and then the entropy change between the melting point and 298K. The sum gives us a final value of 74.51 J K−1 mol−1.

The Third-Law standard molar entropy of Hg(l) at 298 K can be calculated by summing up the entropy changes that occur from 10K to 298K.

First calculate the entropy up to the melting point from 10 K which can be determined using the equation ΔS = ∫(Cp,mdT)/T.

This integral can be approximated as a rectangle from 10K to 234.3 K, hence ΔS₁ = (234.3-10)(4.64 J K−1 mol−1)/10 = 57.74 J K−1 mol−1.

Next, calculate the entropy change associated with the fusion process using the equation, ΔS = ΔH/T, giving ΔS₂ = 2322 J mol−1 / 234.3 K = 9.92 J K−1 mol−1.

Finally, add the entropy increase from the melting point to 298.0 K, which is given as 6.85 J K−1 mol−1.

Summing these values gives the Third-Law standard molar entropy of Hg(l) at 298 K: 57.74 J K−1 mol−1 + 9.92 J K−1 mol−1 + 6.85 J K−1 mol−1 = 74.51 J K−1 mol−1.

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Answer:

The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)

Explanation:

Step 1: Data given

The temperature of a gas = 25.0°C

AT 25 °C the gas occupies a volume of 10.0L and a pressure of 667 torr.

The volume reduces to 7.88 L but the temperature stays constant.

Step 2: Boyle's law

(P1*V1)/T1 = (P2*V2)/T2

 ⇒ Since the temperature stays constant, we can simplify to:

P1*V1 = P2*V2

⇒ with P1 = the initial pressure 667 torr

⇒ with V1 = the initial volume = 10.0 L

⇒ with P2 = the final pressure = TO BE DETERMINED

⇒ with V2 = the final volume = 7.88L

P2 = (P1*V1)/V2

P2 = (667*10.0)/7.88

P2 = 846 torr

The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)

Final answer:

Using Boyle's law, the new pressure of the gas when its volume decreases from 10.0 L to 7.88 L (with initial pressure 667 torr) is calculated to be 846 torr.

Explanation:

To calculate the new pressure of a gas when its volume changes, we can use Boyle's law, which states that for a fixed amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume.

This means if the volume decreases, the pressure increases, and vice versa, as long as the temperature and the amount of gas remain unchanged.

When the volume of the gas decreases from 10.0 L to 7.88 L and the initial pressure is 667 torr, we set up Boyle's law as follows: P1 * V1 = P2 * V2

Substituting the known values gives: 667 torr * 10.0 L = P2 * 7.88 L

Assuming no typographical errors, to solve for P2 (the new pressure), we rearrange the equation:

P2 = (667 torr * 10.0 L) / 7.88 L

After performing the division, we find: P2 = 846 torr.

Therefore, the answer is (a) 846 torr, which aligns with Boyle's expectation that a decrease in volume leads to an increase in pressure.

Answer:

(a) R=k[X2][Y] (b) reaction is zero order wrt Z (c) X2 + Y --- XY +  X (slow step)

X + Z --- XZ    (fast step)

Explanation:

(a) Suppose the reaction rate R with respect to a component X2 with concentration [X2] is generally expressed as follows:

R = k [X2]^n                     (1)

Where k is the rate constant and n is the order of reaction with respect to X2.

When the reaction rate is found to double by doubling the concentration of X2, the following equation could be developed:

2R = k *(2[X2])^n           (2)

Dividing equation (2) by (1) yields

2R/R= k *(2[X2])^n / k *[X2]^n  

2=2^n , n = 1

Thus, the reaction is first order with respect to X2.

In the same manner,  

Tripling the concentration of Y triples the rate,  

R = k [Y]^m                     (3)

Where k is the rate constant and m is the order of reaction with respect to Y.

When the reaction rate is found to triple by tripling the concentration of Y, the following equation could be developed:

2R = k *(3[Y])^m           (4)

Dividing equation (4) by (3) yields

3R/R= k *(3[Y])^m / k *[Y]^m  

3=3^m  , m = 1

Thus, the reaction is first order with respect to Y.

Therefore,the rate law for this reaction is R = k [X2] [Y]

(b)

Provided that doubling the concentration of Z has no effect, we perform similar analysis as above with p representing order of reaction wrt Z:  

R = k [Z]^p                     (5)

R = k (2*[Z]^p                     (6)

R/R= k *(2[Z])^p / k *[Z]^p  

1=2^p  , p = 0

Thus, the reaction is zero order wrt Z. This explains why the change in concentration has no effect on the rate.

(c) A suggested mechanism for the reaction that is consistent with the rate law will therefore be

X2 + Y ----  XY +  X (slow step)

X + Z ---- XZ    (fast step)

The rate law is determined by the slow step, and this is very consistent with the experimentally observed data.

Besides, addition of the two step yields the overall reaction, and both steps are reasonable.

Final answer:

The rate law is Rate = k[X2][Y] indicating first order dependencies on X2 and Y. Z has no effect on the rate, suggesting it's not involved in the rate-determining step, which implies a reaction mechanism with an initial slow step not involving Z.

Explanation:

Rate Law and Reaction Mechanism

Based on the given information, we can deduce the rate law for the reaction. As the rate doubles when the concentration of X2 is doubled, and it triples when the concentration of Y is tripled, this suggests first order dependencies with respect to X2 and Y. There is no effect on the rate when the concentration of Z is doubled, which means Z is zero order in the rate law. Therefore, the rate law can be expressed as:

Rate = k[X2][Y]

The change in concentration of Z has no effect on the rate because it is not involved in the rate-determining step of the mechanism. The reaction mechanism likely involves a slow initial step that does not include Z, followed by a rapid step that includes it. Thus, the overall rate of the reaction is controlled by the first, slower step in the mechanism.

Answer:

(a) Δ[tex]S_{sys}[/tex]  = 2.881 J/K; Δ[tex]S_{sur}[/tex]  = -2.881 J/K; total change in entropy = 0

(b)Δ[tex]S_{sys}[/tex]  = 2.881 J/K; Δ[tex]S_{sur}[/tex]  = 0 ; total change in entropy = 2.881 J/K

(c) Δ[tex]S_{sys}[/tex]  = 0 ; Δ[tex]S_{sur}[/tex]  = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = [tex]V_{1}[/tex]

Final volume = [tex]V_{2}[/tex] = [tex]2V_{1}[/tex]

(a) Change in entropy of the system Δ[tex]S_{sys}[/tex] = [tex]nRIn\frac{V_{2} }{V_{1} }[/tex]

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

Δ[tex]S_{sys}[/tex] = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding Δ[tex]S_{sur}[/tex] = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = Δ[tex]S_{sys}[/tex]+Δ[tex]S_{sur}[/tex] = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ[tex]S_{sys}[/tex]  = 2.881 J/K

Since surrounding does not change in this process Δ[tex]S_{sur}[/tex] = 0.

total change in entropy = Δ[tex]S_{sys}[/tex]+Δ[tex]S_{sur}[/tex] = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

Δ[tex]S_{sys}[/tex]  = 0

Since heat energy is not transferred from the system to the surrounding

Δ[tex]S_{sur}[/tex]  = 0

total change in entropy = Δ[tex]S_{sys}[/tex]+Δ[tex]S_{sur}[/tex] = 0

Answer:

D) To minimize plate height, the optimal flow rate is the maxima for the plot of plate height versus flow rate.

Explanation:

Van Deemter equation in chromatography, relates the variance per unit length of a separation column to the linear mobile phase velocity by considering physical, kinetic, and thermodynamic properties of a separation.

The formula is:

H = A + B/u + C×u

Where:

H = HETP (plate height)

A = Eddy diffusion term .  

B = Longitudinal diffusion.

C = Resistance against mass transfer.

u = Linear velocity .

A) A takes into account multiple pathways through the column. The value of A is column specific and independent of linear flow.  TRUE. Eddy difussion term depends of column packing

B) B takes into account longitudinal diffusion of the analyte in the mobile phase. The value of B is column specific and the impact of B on the plate height is inversely proportional to the linear velocity. TRUE. Longitudinal difussion is another term that is specific to a column. Also, as the formula is H = A + B/u + C×u, the impact of B on the plate height is inversely proportional to the linear velocity.

C) C takes into account equilibrium time between the stationary and mobile phase. The value of C is column specific and the impact of C on plate height is proportional to the linear velocity.  TRUE. C is the time that system needs to equilibirum. Based on the formula, the impact of C on plate height is proportional to the linear velocity

D) To minimize plate height, the optimal flow rate is the maxima for the plot of plate height versus flow rate.  FALSE. The optimal flow is the minimum of the graph

E) Linear velocity is the flow rate of the mobile phase. TRUE. The Van Deemter equation describes u as the linear velocity of the mobile phase.

I hope it helps!

Final answer:

The correct statement that is NOT true for the van Deemter equation is option D) To minimize plate height, the optimal flow rate is the maxima for the plot of versus flow rate.

Explanation:

The correct statement that is NOT true for the van Deemter equation is option D) To minimize plate height, the optimal flow rate is the maxima for the plot of versus flow rate.

This means that in order to minimize the plate height, the optimal flow rate is not at the maximum point on the plot of plate height versus flow rate. The optimal flow rate is usually at a point of minimum plate height.

Answer:

35.3124 g  is the maximum mass of [tex]H_2O[/tex] that can be produced.

Explanation:

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

For [tex]NH_3[/tex]  :-

Mass of [tex]NH_3[/tex]  = 52.3 g

Molar mass of [tex]NH_3[/tex]  = 17.031 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{52.3\ g}{17.031\ g/mol}[/tex]

[tex]Moles\ of\ NH_3= 3.0709\ mol[/tex]

For [tex]O_2[/tex]  :-

Given mass of [tex]O_2[/tex]= 52.3 g

Molar mass of [tex]O_2[/tex] = 31.9898 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{52.3\ g}{31.9898\ g/mol}[/tex]

[tex]Moles\ of\ O_2=1.6349\ mol[/tex]

According to the given reaction:

[tex]4NH_3+5O_2\rightarrow 4NO_4+6H_2O[/tex]

4 moles of [tex]NH_3[/tex] reacts with 5 moles of [tex]O_2[/tex]

1 mole of [tex]NH_3[/tex] reacts with 5/4 moles of [tex]O_2[/tex]

Also,

3.0709 moles of [tex]NH_3[/tex] reacts with [tex]\frac{5}{4}\times 3.0709[/tex] moles of [tex]O_2[/tex]

Moles of [tex]O_2[/tex] = 3.8386 moles

Available moles of [tex]O_2[/tex] = 1.6349 moles

Limiting reagent is the one which is present in small amount. Thus, [tex]O_2[/tex] is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

5 moles of [tex]O_2[/tex] on reaction forms 6 moles of [tex]H_2O[/tex]

1 mole of [tex]O_2[/tex] on reaction forms 6/5 moles of [tex]H_2O[/tex]

Thus,

1.6349 mole of [tex]O_2[/tex] on reaction forms [tex]\frac{6}{5}\times 1.6349[/tex] moles of [tex]H_2O[/tex]

Moles of [tex]H_2O[/tex] = 1.9618 moles

Molar mass of [tex]H_2O[/tex] = 18 g/mol

Mass of sodium sulfate = Moles × Molar mass = 1.9618 × 18 g = 35.3124 g

35.3124 g  is the maximum mass of [tex]H_2O[/tex] that can be produced.

Answer:

35.3 g

Explanation:

From the balanced equation given we can say:

4 moles of NH3 reacts with 5 moles of O2 to give 6 moles of H2O.

4*17 g of NH3 reacts with 5*32 g of O2 to give 6*18 g of H2O.

68 g of NH3 reacts with 160 g of O2 to give 108 g of H2O.

Here the limiting reagent is O2 and excess reagent is NH3.

52.3 g of O2 will react with [tex]\frac{68}{160}\times52.3=22.23\ g\ of\ NH_{3}[/tex] to give :

[tex]\frac{108}{160}*52.3=35.3\ g\ of\ H_{2}O[/tex]

Hence the maximum mass of H2O that can be produced by 52.3 g of reactants is 35.3 g.

Answer:

[tex]\Delta S^{0}[/tex] for the given reaction is -99.4 J/K

Explanation:

Balanced reaction: [tex]\frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightarrow NH_{3}(g)[/tex]

[tex]\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}][/tex]

where [tex]S^{0}[/tex] represents standard entropy.

Plug in all the standard entropy values from available literature in the above equation:

[tex]\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K[/tex]

So, [tex]\Delta S^{0}[/tex] for the given reaction is -99.4 J/K

The change in entropy in the reaction forming ammonia from nitrogen and hydrogen gas can be calculated using the stoichiometry of the balanced equation and the entropy values of the reactants and products.

The question is, 'Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 1/2N2(g) + 3/2H2(g) → NH3(g)' To calculate this, we need to consider the stoichiometry of the balanced chemical equation, which tells us that one mole of N₂ will react with three moles of H₂ to form two moles of NH3. The stoichiometric factors derived from this equation can be used to determine the change in entropy (ΔS∘) for the reaction.

The change in entropy for a reaction can be determined using the equation ΔS∘ = ΣS∘(products) - ΣS∘(reactants). As per the balanced chemical equation N₂(g) + 3H₂(g) → 2NH3(g), ΔS∘ would be equal to the sum of the absolute entropies of 2 moles of NH3 (products) minus the sum of the absolute entropies of 1/2 moles of N2 and 3/2 moles of H2 (reactants). The specific entropy values would depend on data provided in a standard thermodynamic table.

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Final answer:

To electroplate 15.0 kg of copper onto the cathode using 34.5 A, first calculate the moles of copper needed then convert this to the required charge using Faraday's constant. Finally, divide the total charge by the current to find the time in seconds and convert to hours, resulting in approximately 367 hours.

Explanation:

To calculate the time required to electroplate 15.0 kg of copper using a current of 34.5 A, we need to use Faraday's laws of electrolysis. First, we need to determine the number of moles of copper to be plated. The molar mass of copper is approximately 63.55 g/mol, so:

15,000 g / 63.55 g/mol = 236.025 mol Cu

Since each copper ion (Cu2+) requires two electrons to be reduced to copper metal, the number of moles of electrons needed is twice the number of moles of copper:

2 × 236.025 mol = 472.05 mol e-

Each mole of electrons corresponds to a charge of 96,485 coulombs (Faraday's constant), so:

472.05 mol e- × 96,485 C/mol = 45,562,240.25 C

Now, we can calculate the time required to deliver this charge at a rate of 34.5 A (since 1 A = 1 C/s), using:

Time (s) = Total Charge (C) / Current (A)

Time (s) = 45,562,240.25 C / 34.5 A = 1,320,643 s

Converting seconds to hours:

1,320,643 s / (60 s/min) / (60 min/h) ≈ 367 h

Therefore, it will take approximately 367 hours to electroplate 15.0 kg of copper onto the cathode with a constant current of 34.5 A.

Answer:

i = 3,5

Explanation:

There are missing the following values:

132 g Alanine

1150g of X

4,4°C the first freezing point dercreasing

132g of Iron(III) nitrate

5,6°C the second freezing point decreasing.

The freezing point depression is a colligative property that describes the decrease of the freezing point of a solvent on the addition of a non-volatile solute.

The formula is:

ΔT = i kf mb

Where ΔT is freezing point decreasing, i is Van't Hoff factor, kf, is cryoscopic constant and mb is molality of solution.

For alanine Van't Hoff factor is 1 (Ratio between particles in dissolution and before dissolution), molality is:

132g×(1mol/89,09g) = 1,48mol / 1,150kg = 1,29 mol/kg

Replacing:

4,4K = 1 kf 1,29mol/kg

kf = 3,41 K·kg/mol

Now, for Iron(III) nitrate molality is:

132g×(1mol/241,86g) = 0,546mol / 1,150kg = 0,475 mol/kg

Replacing:

5,6K = i×3,41 K·kg/mol×0,475 mol/kg

i = 3,5

I hope it helps!

The correct expression for the equilibrium constant (K) is:

K = [PCl₃]² / [P] × [Cl₂](3/2)

The correct expression for the equilibrium constant (K) of the given reaction is:

K = [PCl₃]² / [P] × [Cl₂](3/2)

In this expression:

[PCl₃] represents the molar concentration of PCl₃ (gaseous product).

[P] represents the molar concentration of P (reactant P in its gaseous state).

[Cl₂] represents the molar concentration of Cl₂ (reactant Cl₂ in its gaseous state).

The exponents in the expression correspond to the stoichiometric coefficients in the balanced chemical equation:

P(g) + 3/2 Cl₂(g) ↔ PCl₃(g)

The stoichiometric coefficients of P and Cl₂ are both 1, and the coefficient of PCl₃ is 1. To make the equation balanced, we need to multiply Cl₂ by 3/2 (1.5).

So, the correct expression for the equilibrium constant (K) is:

K = [PCl₃]² / [P] × [Cl₂](3/2)

To know more about equilibrium:

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