The degradation of CF3CH2F (an HFC) by OH radicals in the troposphere is first order in each reactant and has a rate constant of k = 1.6 x 10^8 M^-1s^-1 at 4°C.
Part A) If the tropospheric concentrations of OH and CF3CH2F are 8.1 x 10^5 and 6.3 x 10^8 molecules/cm^3, respectively, what is the rate of reaction at this temperature in M/s?

Answer:

Methane

Explanation:

The gas that you could keep in an outdoor storage tank in winter in Alaska is Methane.

The reason is the extreme low temperature during the winter. The boiling point of butane is 44 ºF ( -1ºC) and that of propane is a higher -43.6 º F but still within the range of average minimum winter temperature in Alaska (-50 ªF). Therefore we will have condensation in the tanks and not enough gas pressure.

Methane having  a boling point  of  -259 ºF will not condense at the low wintertime temperatures in Alaska.

Answer:

D) 5.36 hours

Explanation:

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.

We know that:

Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]

Also, copper will produce 2 electrons. So, out of 5 moles of copper, 10 moles of electrons will be produced.

So,

Charge on 10 mole of electrons = [tex]10\times 1.6\times 10^{-19}\times 6.022\times 10^{23}=9.6352\times 10^5C[/tex]

To calculate the time required, we use the equation:

[tex]I=\frac{q}{t}[/tex]

where,

I = current passed = 50.0 A

q = total charge = [tex]9.6352\times 10^5C[/tex]

t = time required = ?

Putting values in above equation, we get:

[tex]50.0A=\frac{9.6352\times 10^5C}{t}\\\\t=\frac{9.6352\times 10^5C}{50.0A}=19270.4s[/tex]

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, [tex]19270.4s\times \frac{1hr}{3600s}=5.36hr[/tex]

Hence, the amount of time needed is 5.36 hrs.

Final answer:

HCl is a polar molecule that readily dissolves in water, while O2 and N2 are nonpolar and insoluble. Ionic compounds like CaCl2 and KNO3 are highly soluble in water. BeCl2 is soluble in water, while BCl3 is insoluble.

Explanation:

When classifying materials as polar, ionic, or nonpolar, it's important to consider the types of chemical bonds present and the distribution of charge.

HCl is a polar molecule due to the unequal sharing of electrons between hydrogen and chlorine. It will readily dissolve in water, forming an acidic solution.

O2 is a nonpolar molecule because oxygen atoms share the electrons equally. It is insoluble in water due to the differences in polarity.

CaCl2 is an ionic compound, where calcium ions (positive) and chloride ions (negative) are attracted to each other due to the strong electrostatic forces. It is highly soluble in water.

N2 is a nonpolar molecule with equal sharing of electrons. It is insoluble in water.

C2H6 is a nonpolar molecule, and it is insoluble in water.

KNO3 is an ionic compound, made up of potassium ions and nitrate ions, and it is highly soluble in water.

BeCl2 is a polar molecule due to the unequal sharing of electrons. It is soluble in water, forming acidic solutions.

BCl3 is a nonpolar molecule, and it is insoluble in water.

To determine the equilibrium concentration of O2 in the given reaction, we need to use the equilibrium constant expression and the initial concentration of O2. The equilibrium constant (Kc) for the reaction is 1.5. By substituting the known values into the equilibrium constant expression, we can solve for the equilibrium concentration of O2.

In order to determine the equilibrium concentration of O2 in the given reaction, we need to use the equilibrium constant expression and the initial concentration of O2. The equilibrium constant (Kc) for the reaction is 1.5. The reaction is in the form of a reversible reaction, so the concentrations of the reactants and products play a role in determining the equilibrium concentration of O2.

Using the equilibrium constant expression, we have:

Plug in the values of [Cu] and [SO2] from the given reaction equation to calculate the equilibrium concentration of O2.

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For the given reaction, the equilibrium constant Kc is the ratio of the concentrations of the products to the reactants. Without a change or shift in the equilibrium, the equilibrium concentration of O2 remains the same as its initial concentration, 2.9 M.

This problem involves understanding the concept of equilibrium in chemical reactions. From the reaction,

CuS(s) + O2(g) ⇌ Cu(s) + SO2(g)

We know that at equilibrium, the rate of the forward reaction equals to the rate of the reverse reaction. Considering the equilibrium constant, Kc = [Products]/[Reactants], being given as 1.5, we also know that [Cu][SO2]/[CuS][O2] equals 1.5.

Since the concentration of a solid like CuS or Cu in the reaction does not affect the equilibrium, the equilibrium expression is simply [SO2]/[O2] = Kc

In the context of this question, we are provided that the initial concentration of O2 is 2.9 M and are asked to determine the equilibrium concentration of O2. Since there's no shift mentioned in the equilibrium, it indicates that no reaction has taken place. This means the equilibrium concentration of O2 remains the same as its initial concentration, that is, 2.9 M.

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To approach this problem, let's start by understanding what we have and what we need to find. We know:

1. The fuselage is in the shape of a cylindrical shell with approximate dimensions: length of 20 meters, diameter of 4 meters, and thickness of 0.01 meters.
2. The densities of aluminum and steel are 2700 kg/m³ and 7800 kg/m³ respectively.
3. The average mass of a passenger is 70 kg.

Now, we must find: the number of passengers that a steel-made Airbus A380 could carry, given that the total mass that the engines can lift stays the same.

First, we have to calculate the volume of the aluminum fuselage. Since the fuselage is a cylindrical shell, the volume can found using the formula for the volume of a cylinder: `π*r²*h` where r is the radius and h is the height or length of the cylinder in this context. The radius is the width divided by 2. To account for thickness, we subtract the volume of the inner cylinder from the volume of the outer cylinder.

To find the mass of the aluminum fuselage, we multiply the volume just calculated with the given density of aluminum.

Next, we calculate the mass of the equivalent steel fuselage. Given that the volume remains the same, we can multiply the volume of the fuselage with the density of steel to find the mass, as mass is the product of volume and density.

To find the difference in mass between the steel and aluminum fuselages, we subtract the mass of the aluminum fuselage from the steel fuselage.

This gives us 12785.653 kg, which is the additional weight when we switch from an aluminum fuselage to a steel fuselage.

Finally, we calculate how many passengers could be carried with this additional weight. Given the average mass of a passenger, knowing that this mass difference will directly affect how many passengers can be carried, we divide the weight difference by the average passenger mass of 70 kg to get the number of lost seats.

The mass of the aluminum fuselage is approximate 6768.88 kg, the mass of the steel fuselage is approximate 19554.53 kg. The difference in mass is about 12785.65 kg. This extra weight when changing to a steel fuselage would reduce the number of passengers by around 183.

So the aircraft would lose around 183 seats if the fuselage was made from steel instead of aluminum.

Learn more about Aluminum vs Steel Aircraft here:

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Final answer:

If the Airbus A380's fuselage were made from steel instead of aluminium, it would significantly increase the aircraft's weight, thereby reducing the maximum number of passengers it could carry, given the fixed capacity of the engines.

Explanation:

The question revolves around the impact of using steel instead of aluminium to construct the fuselage of an Airbus A380 and how that would affect the number of passengers it could carry. To address this, one must understand the properties that make aluminum preferable for such applications. Aluminum is a lightweight metal with a density significantly lower than that of steel. This attribute allows airplanes to carry more passengers and cargo without exceeding the maximum lift capacity of the aircraft's engines.

Steel, on the other hand, is much denser than aluminum. If the Airbus A380's fuselage were made from steel, its overall mass would substantially increase, thus reducing the number of passengers it could carry. As the engines are designed with specific thrust capacities that cannot be exceeded, the added weight from using steel would necessitate a reduction in the payload (e.g., passengers). Since the student hasn't provided the exact densities or other needed numerical values, a numeric calculation cannot be performed. However, the general principle is clear: using steel would mean transporting fewer passengers due to the increased weight of the aircraft's structure.

The evolution of aircraft construction materials from wood and fabric to metal alloys and composites has been instrumental in the advancement of modern aviation, leading to more efficient and larger aircraft, as exemplified by the Airbus A380, the world's largest commercial jetliner capable of carrying up to 850 passengers.

Answer:

109° 27'

Explanation:

The ammonium ion is tetrahedral in shape, all the HNH bonds are exactly at the tetrahedral bond angle since there are only bond pairs in the structure and no lone pairs. Recall that lone pairs decrease the bond angke from the ideal value in a tetrahedron due to higher repulsion.

Answer:

(D.) No, because polar covalent molecules have stronger intermolecular forces than nonpolar covalent molecules.

Explanation:

In water molecules intermolecular hydrogen bonding occurs between hydrogen and highly electronegative elemnt oxygen that increases the intermolecular forces. Hence water has high boiling point.

Answer:

D. No, because polar covalent molecules have stronger intermolecular forces than nonpolar covalent molecules.

Explanation:

Hello,

Intermolecular forces are those which attract or repel molecules to, or from each other. In this case, since water is the involved molecule, it is known that it is a polar covalent one, as the difference between the electronegativities of oxygen and hydrogen is 1.24, in this manner, it is also known that the more polar the molecule is, the stronger its intermolecular forces are are they are tend to stick the molecules. In such a way, the answer is: D. No, because polar covalent molecules have stronger intermolecular forces than nonpolar covalent molecules.

Best regards.

Answer: The cell potential of the cell is +0.118 V

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):  [tex]Ag(s)+Cl^-(aq.)\rightarrow AgCl(s)+e^-[/tex]

Reduction half reaction (cathode):  [tex]AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq.)[/tex]

In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}]_{diluted}}{[Cl^{-}]_{concentrated}}[/tex]

where,

n = number of electrons in oxidation-reduction reaction = 1

[tex]E_{cell}[/tex] = ?

[tex][Cl^{-}]_{diluted}[/tex] = 0.0222 M

[tex][Cl^{-}]_{concentrated}[/tex] = 2.22 M

Putting values in above equation, we get:

[tex]E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}[/tex]

[tex]E_{cell}=0.118V[/tex]

Hence, the cell potential of the cell is +0.118 V

The given statement, some type of path is necessary to join both half-cells in order for electron flow to occur, is true.

Explanation:

Flow of electrons is possible with the help of a conducting medium like metal wire.

A laboratory device which helps in completion of oxidation and reduction-half reactions of a galvanic or voltaic cell is known as salt bridge. Basically, this salt bridge helps in the flow of electrons from anode to cathode and vice-versa.

If salt bridge is not present in an electrochemical cell, the electron neutrality will not be maintained and hence, flow of electrons will not take place.

Thus, we can conclude that the statement some type of path is necessary to join both half-cells in order for electron flow to occur, is true.

Answer:

D. Carboxylate ion

Explanation:

The nitriles are hydrolyzed wiyh aqueous soda, under heating, to form carboxilates and ammonia:

H3C-C≡CN + NaOH(aq) → H3C-C=OONa+  + NH3

Answer:

C₂H₃Cl

Explanation:

We can calculate the compound's molar mass using the data given by the problem and Graham's law:

Rate₁/Rate₂ = [tex]\sqrt{\frac{M_{2}}{M_{1}} }[/tex]

In this case the subscript 1 refers to the compound and 2 refers to Ar.

Keeping in mind that Rate = volume/time, and that the volume is the same for both compounds, we can rewrite the equation as:

Time₂/Time₁ = [tex]\sqrt{\frac{M_{2}}{M_{1}} }[/tex]

6.18/7.73 =  [tex]\sqrt{\frac{39.95}{M_{1}} }[/tex]

M₁ = 62.5 g/mol

Now we determine the molecular formula by using the elemental % analysis:

Assuming we have 1 mol of the compound:

C ⇒ 62.5 g * 38.4/100 = 24 g C = 2 mol C

H ⇒ 62.5 g * 4.8/100 = 3 g H = 3 mol H

C ⇒ 62.5 g * 56.8/100 = 35.5 g Cl = 1 mol Cl

Thus the molecular formula is C₂H₃Cl

Answer:

ΔHv = 17.04 KJ/mol

Explanation:

T(°C)   T(K)      Pv(atm)         1/T(K)                LnPv

34      307      0.236       0.00325733      - 1.4439

44      317       0.292       0.0031545         - 1.2310

54      327      0.355       0.003058          - 1.0356

Clausius-Clapeyron:

⇒ δLnP = ΔH/R (δT/T²)

∴ δT/T² = δ/δT(- 1/T )

⇒ δLnP/δT = - ΔH/R

Graphing: LnP vs 1/T

we get an ecuation that corresponds to a straight line:

y = - 2049.6x + 5.2331 ...... R² = 1

where the slope of this line is:

y = mx + b

⇒ m = - 2049.6 = - ΔH/R.....Clausius-Clapeyron

⇒ ΔH = (2049.6)(R)

∴ R = 8.314 E-3 KJ/mol.K

⇒ ΔHv = 17.04 KJ/mol

Answer:

i: answer is in attachment.

ii: 24 valance electron in overall structure

iii: carbon having positive charge while oxygen having negative

iv: a bond angle of approximately 104.5 degrees.

v:  CH3 having sp3 hybridization  and both CH having sp2 hybridization

vi: yes. resonance occur between CH=CH and oxygen.

Explanation:

iii: oxygen is electronegative so attraction of electrons occur at oxygen results in negative charge.

iv: a bond angle is about 104.5 degrees because the oxygen has 2 lone pairs and 2 bond pairs of electrons so lone pairs repulsion occur more than bonding pairs that's why the angle is less than that of an atom with tetrahedral or pyramidal geometry.

The Lewis structure for CH3CHCHOH comprises a chain of three carbon atoms with associated hydrogens and an alcohol group, having 22 valence electrons in total. The formal charges for the carbon and oxygen atoms are zero, the ideal C-O-H bond angle is approximately 109.5 degrees, the first and third carbon atoms are sp3 hybridized, and the second carbon is sp2 hybridized. There are no resonance structures for CH3CHCHOH.

The Lewis structure for CH3CHCHOH can be drawn by considering that carbon typically forms four bonds, hydrogen forms one bond, and oxygen forms two bonds. Here's a step-by-step process: Place the carbons in a row since they are connected to each other, attach the three hydrogens to the first carbon, add a double bond between the second and third carbon, and complete the octets for each carbon. Attach the hydroxyl group (-OH) to the third carbon. After drawing this, complete the structure by adding lone pairs to the oxygen atom.

Total number of valence electrons: Carbon has 4 valence electrons, Hydrogen has 1, and Oxygen has 6. The compound has three carbons (3x4=12), four hydrogens (4x1=4), and one oxygen (1x6=6), totaling 22 valence electrons.

Formal charges on each atom are calculated by using the formula: Valence electrons - (Nonbonding electrons + 1/2 Bonding electrons). After calculating, you will find that the formal charges for the carbons and the oxygen in CH3CHCHOH are all zero.

The ideal bond angle for the C-O-H atoms in an alcohol group is approximately 109.5 degrees, based on a tetrahedral arrangement.

The hybridization for the carbon atoms depends on the number of sigma bonds and lone pairs around each carbon. For CH3CHCHOH, we have: first carbon (C1) is sp3 hybridized, second carbon (C2) is sp2 hybridized, and the third carbon (C3, which is part of the alcohol group) is sp3 hybridized.

Regarding the presence of resonance structures, there are no resonance structures for CH3CHCHOH since there are no delocalizable pi electrons or lone pairs adjacent to the pi bonds that could result in resonance.

The equilibrium concentration of H2 in the given reaction is approximately 0.614 mol/L.

Given that the equilibrium constant (K) is 1.00×10^2 for the reaction H2(g) + F2(g) ⇌ 2HF(g), we can use the stoichiometry of the reaction to determine the equilibrium concentration of H2. Since 1 mol of H2 reacts with 1 mol of F2 to form 2 mol of HF, the concentration of H2 at equilibrium will be half of its initial concentration. Therefore, the equilibrium concentration of H2 in the 1.14-L flask would be 1.40 mol divided by 2 and then divided by 1.14 L, which gives us approximately 0.614 mol/L.

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The equilibrium concentration of H₂ is calculated to be 0.205 M.

To find the equilibrium concentration of H₂, we first need to set up an ICE (Initial, Change, Equilibrium) table:

Initial concentrations:

Changes in concentration:

Equilibrium concentrations:

According to the equilibrium constant expression (K):

[tex]K = [HF]^2/ [H_2][F_2][/tex]

Substitute the equilibrium concentrations into the expression:

[tex]1.00 \times 10^2 = (2x)^2 / (1.23 - x)(1.23 - x)[/tex]

This simplifies to:

[tex]100 = 4x^2 / (1.23 - x)^2[/tex]

Taking the square root of both sides:

10 = 2x / (1.23 - x)

Solve for x:

10(1.23 - x) = 2x

12.3 - 10x = 2x

12.3 = 12x

x = 1.025

The equilibrium concentration of H₂  is:

[H₂ ] = 1.23 - 1.025 = 0.205 M

Answer:

0

Explanation:

The relation between Kp and Kc is given below:

[tex]K_p= K_c\times (RT)^{\Delta n}[/tex]

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

[tex]SO_3_{(g)}+NO_{(g)}\rightleftharpoons SO_2_{(g)}+NO_2_{(g)}[/tex]

Δn = (2)-(2) = 0

Thus, Kp is:

[tex]K_p= Kc\times \times (RT)^{0}[/tex]

[tex]K_p= Kc[/tex]

Final answer:

The Δn represents the difference in moles of gaseous products and reactants for a reaction when relating Kc to Kp. For the reaction SO3(g) + NO(g) ↔ SO2(g) + NO2(g), Δn would be 0. Kp is related to Kc by the equation Kp = Kc(RT)Δn.

Explanation:

The question relates to the concept of the reaction quotient (Δn) when relating the equilibrium constants Kc (equilibrium constant in terms of concentration) to Kp (equilibrium constant in terms of partial pressure) for a given chemical reaction involving gases. The value of Δn is the difference in the sum of the moles of gaseous products and the sum of the moles of gaseous reactants in a balanced chemical equation. In the example given:

[tex]SO_{3}[/tex] (g) + NO(g) ↔ [tex]SO_{2}[/tex] (g) + [tex]NO_{2}[/tex] (g), Δn would be (1 + 1) - (1 + 1) = 0.

1[tex]N_{2}[/tex](g) + [tex]2H_{2} O[/tex](g) ↔ 2NO(g) + [tex]2H_{2}[/tex] (g), Δn would be (2 + 2) - (1 + 2) = 1.

To relate Kc and Kp, we use the equation Kp = Kc(RT)Δn, where R is the gas constant and T is the temperature in Kelvins. In cases where Δn is zero, as in the first equation, Kp will be equal to Kc because (RT)0 equals 1.

Answer:D

Explanation:

Both thionyl chloride and phosphorus trichloride are used in the synthesis of acid chlorides from carboxylic acids. The thionyl chloride leads to a chlorosulphite intermediate before the acid chloride is formed

Answer:

monochlorinated products: 4

dichlorinated products: 12

Explanation:

Chlorination of alkanes is a reaction that takes place when the chlorine is in presence of light. This actually decomposes the chlorine, and one atom of Chlorine substracts an hydrogen from the alkane. Now, this hydrogen substracted comes usually from the most substitued carbon, because it's more stable (A tertiary carbon is more stable than a secondary carbon, and this more stable than primary).

When this happens, the other chlorine atom, goes as electrophyle in that carbon and formed the chlorinated product. Now, although a tertiary carbon is more stable, we can still have (in minor quantities) chlorinated products that comes from a secondary and primary carbon. The first picture shows the general mechanism of the chlorination, and the possible products for a monochlorinated.

The second picture shows the possible dichlorinated products, which are in higher quantities than the monochlorinated basicallu because of the variety of positions the chlorine can be. So, second picture shows all the products.

The chlorination of 3-methylpentane can produce varying products, depending on the number of hydrogen atoms being substituted by chlorine. Monochlorination can produce 4 unique products, while dichlorination can yield 9 unique products.

The chlorination process can result in multiple products due to the number of hydrogen atoms in 3-methylpentane that can be substituted by chlorine. The original compound, 3-methylpentane, has 12 hydrogen atoms. Therefore, monochlorination can produce various constitutional isomers. The unique monochlorinated constitutional isomers that can be obtained when we replace one hydrogen by one chlorine atom are 1-chloro-3-methylpentane, 2-chloro-3-methylpentane, and a pair of 3-chloro-3-methylpentanes, giving a total of four.

For dichlorination (two chlorine substituents), we consider each of the monochlorinated products and determine where the second chlorine atom can be placed. This will give 1,1-dichloro-3-methylpentane, 1,2-dichloro-3-methylpentane and others, totaling up to 9 unique dichlorinated constitutional isomers. Therefore,monochlorination of 3-methylpentane can produce 4 products and dichlorination can produce 9 products.

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Answer:

a. NaHCO₃ + HCl → NaCl + H₂O + CO₂

b. 39.14 g is the mass of NaHCO₃ required to produce 20.5 moles of CO₂

Explanation:

A possible reaction for NaHCO₃ to make dioxide is this one, when it reacts with hydrochloric to produce the mentioned gas.

NaHCO₃ + HCl → NaCl + H₂O + CO₂

Ratio in this reaction is 1:1

So 1 mol of baking soda, produce 1 mol of CO₂

Let's calculate the moles

20.5 g CO₂ / 44 g/m = 0.466 moles

This moles of gas came from the same moles of salt.

Molar mass baking soda = 84 g/m

Molar mass . moles = mass

84 g/m .  0.466 moles = 39.14 g

Answer:

Using valence bond theory, HgCl2 is sp hybridized while PCl3 is sp3 hybridized.

Explanation:

sp hybridized molecules are largely covalent and cannot conduct electricity hence the nonconducting nature of HgCl2. PCl3 contains a lone pair and three unpaired electrons in sp3 hybridized orbitals.

Answer:

The molar concentration of hydrogen peroxide is 0.01164 M

Explanation:

Step 1: Data given

Volume of gas yielded = 75.3 mL = 0.0753 L

Temperature = 25.0 °C

Atmosphere = 0.976 atm

(Water = 0.032 atm at 25°C)

The original volume of H2O2 is 500 mL

Molar mass of O2 =32 g/mol

Step 2: The balanced equation

2H2O2(aq) → 2 H2O(l) + O2(g)

Step 3: Calculate pressure of O2

P = P(02) + P(water)

P(O2) = P - P(water)

P(O2) = 0.976 atm - 0.032 atm

P(O2) = 0.944 atm

Step 4: Calculate moles O2

p*V=n*R*T

⇒ with p = the pressure of O2 gas = 0.944 atm

⇒ with V= the volume of O2 = 0.0753 L

⇒ with n = the number of moles of O2

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 25°C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.944 * 0.0753)/(0.08206*298K)

n = 0.00291 moles O2

Step 5: Calculate moles of H2O2

For 1 mole of O2 produced, we need 2 moles of H2O2

For 0.00291 moles O2 we need 2*0.00291 = 0.00582 moles H2O2

Step 6: Calculate molar concentration of H2O2

Molar concentration = moles / volume

Molar concentration = 0.00582 moles / 0.500L

Molar concentration = 0.01164 M

The molar concentration of hydrogen peroxide is 0.01164 M

The percentage of I2 converted = 78.6 %

Explanation:

Write down the given values

Kp= 54.4

The percentage of I2 will be converted to HI is 0.200 moled each of H2 and I2.

It should come to an equilibrium in 1.00 lit container .

Change in I2 (iodine) and H2 (hydrogen)  = x in each

Change in HI = 2x

total ni (nickel)

number of moles = 0.2 -x + 0.2 -x + 2x

=0.4 moles

Mole fractions :

I2 = 0.2-x / 0.4 H2

=0.2-x  / 0.4 HI

= 2x /0.4

Kp = HI ^2 / H2* I2

= (2x) ^2 / (0.2-x) ^2 = 54.4

by  taking square root:

2x / 0.2-x = 7.375

= x=0.157

percentage of I2 converted = 78.6 %

Adding sulfuric acid to the half-cell with the lead electrode will decrease the [tex]Pb^{2+}[/tex] ion concentration as [tex]PbSO_{4}[/tex] (s)forms, leading to an increase in cell potential due to a shift in equilibrium to release more [tex]Pb^{2+}[/tex] ions from the electrode. (option C)

The question deals with the behavior of [tex]Pb^{2+}[/tex] ions in a galvanic cell. In general, the addition of sulfuric acid to the half-cell with the lead electrode, causing the formation of [tex]PbSO_{4}[/tex] (s), would lead to a decrease in the concentration of [tex]Pb^{2+}[/tex]ions in the solution.

This is because the [tex]Pb^{2+}[/tex]ions participate in forming the insoluble [tex]PbSO_{4}[/tex] , and the removal of ions from the solution would shift the equilibrium of the half-reaction to replace the ions used in the precipitation.

According to Le Chatelier's principle, this shift in equilibrium would cause more lead ions to be released into the solution from the lead electrode, essentially increasing the cell potential.

Regarding the half-reactions:

The electrode potentials indicate the tendency of each half-reaction to occur. A negative electrode potential suggests that the substance loses electrons more easily and is therefore prone to oxidation. Conversely, a positive electrode potential indicates a substance that gains electrons readily, suggesting a reduction is favorable.

Answer:

potassium, nitrogen and phosphorous cycle

Explanation:

A fertilizer is a substance which is applied on the plants by farmers to increase the supply of nutrients for the plants. Fertilizers have known to be toxic in many ways such as they alter the potassium, nitrogen and phosphorus cycles. Nitrogen, potassium and phosphorus are present in abundant amounts in the fertilizers. Draining of these fertilizers into rivers and ponds is toxic for the aquatic life. Hence, the use of fertilizers disrupts the natural cycles and is toxic for many aquatic plants and animals.

Answer:

T = 3006.976 K

Explanation:

∴ ΔH° = 125 KJ/mol ( 800K - 1500K)

∴ ΔG° = 25 KJ/mol (1150 K)

⇒ T = ? ∴ K > 1

In the equilibrium:

∴ K > 1

If ΔG°/RT = 1

⇒ e∧(1) = 2.72 > 1

∴ ΔG°/RT = 1

⇒ T = ΔG°/R = (25 KJ/mol)/(8.314 E-3 KJ/mol.K)

⇒ T = 3006.976 K

verifying:

⇒ K = e∧(25/((8.314 E-3)(3006.976)))

⇒ K = 1.000000061 > 1

Final answer:

The student is asking about estimating the temperature at which the equilibrium constant becomes greater than 1 based on the given standard enthalpy and Gibbs energy values.

Explanation:

The given question is related to standard enthalpy and Gibbs energy of a reaction, and the estimation of temperature at which the equilibrium constant becomes greater than 1.

Firstly, it is stated that the standard enthalpy of the reaction is approximately constant at +125 kJmol-1 from 800K to 1500K. This information helps us understand the change in enthalpy with respect to temperature.

Secondly, the standard Gibbs energy is given as +25 kJmol-1 at 1150K. This value allows us to calculate the equilibrium constant using the equation: ΔG = -RT lnK.

To estimate the temperature at which the equilibrium constant becomes greater than 1, we need to solve for temperature in the equation K = e-ΔG/T and find the value of T that results in K > 1.

The problem is solved by applying the principles of thermodynamics and fluid dynamics for an ideal gas under isentropic conditions in a diffuser, and involving calculations for temperature, exit area and entropy production.

This problem involves the principles of thermodynamics and fluid dynamics, specifically in relation to the behavior of nitrogen gas in a diffuser. To solve it, we need to apply the ideal gas law and the principles of energy and entropy conservation.

For part (a), we can start by using the formula for isentropic relations in an ideal gas, which states that T2/T1 = (P2/P1)^((k-1)/k), where k is the specific heat ratio of the gas. By substituting the given values, and also considering the conservation of mass (continuity equation), we can calculate the exit temperature (T2) and the exit area.

Similarly, for part (b), using the formula for rate of entropy production (ΔS_gen = m_dot*[s2 - s1 + R*ln(P2/P1)]), where m_dot is the mass flow rate and R is the specific gas constant, we can find the rate of entropy production.

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Answer:

t = 55 min ⇒ m = 1497.692 g

Explanation:

polyaramide fiber:

∴ L = 1 m

∴ d = 710 μm

∴ m = 0.059 g

∴ t = 0.13 s

mass flow:

∴ mf = 0.059 g / 0.13 s = (0.454 g/s)×(60s/min) = 27.23 g/min

If t = 55 min:

⇒ m = mf×t = (27.23 g/min)×(55 min) = 1497.692 g

Answer:

[tex]m_{fiber}=1497.7g[/tex]

Explanation:

Hello,

In this case, as both the diameter and the length of the fiber is the same for the second case, one applies a simple rule of three to compute the mass of fiber that can be spun in 55 min as shown below:

[tex]m_{fiber}=\frac{55min*0.059g}{0.13s*\frac{1min}{60s} }\\m_{fiber}=1497.7g[/tex]

Best regards.

Answer: moles of Ne = 0.149 moles

moles of F₂ = 0.221 moles

Explanation:

The process occurs at costant volume.

Neon is a monoatomic gas, Cv =3/2R, Flourine is a diatomic gas, Cv = 5/2R

n = PV/RT ; where n is total = number of moles, P is total pressure = 3.32atm, V is volume = 2.5L, R is molar gas constant = 0.08206 L-atm/mol/K = 8.314 J/mol/K, T is temperature = 0.0°C = 273.15K

n(total) = (3.32 atm)(2.5 L)/(0.08206 L-atm/mol/K)(273.15) = 0.3703 mol

For one mole heated at constant volume,

Change in entropy, ∆S = ∫dq/T = ∫(Cv/T)dT

From T1 to T2, Cvln(T2/T2) = Cvln(288.15/273.15) = 0.05346•Cv

So, for 0.3703 moles,

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

⇒ Cv = 17.43 J/mol/K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R = 1.5 • 8.314 J/mol/K = 12.471 J/mol/K

For pure F₂, Cv = (5/2)R = 2.5 • 8.314 J/mol/K = 20.785 J/mol/K

If Y is the mole fraction of Ne, Y can be determined by setting the observed entropy change (∆S) to the weighted average of the entropy changes expected for the two different gases in the mixture:

17.43 J/mol/K = Y * 12.471 J/mol/K + (1 – Y) * 20.785 J/mol/K

⇒ 20.785 J/mol/K – 8.314 J/mol/K * Y = 17.43J/mol/K

Y = 0.403 ; 1 – Y = 0.597

moles of Ne = (0.403)(0.3703 mol) = 0.149 moles

moles of F₂ = (0.597)(0.3703 mol) = 0.221 moles

Answer:

a. The specific heat capacity of the gaseous ethanol is less than the specific heat capacity of liquid ethanol.

Explanation:

The heating curve is a curve that represents temperature (T) in the y-axis vs. added heat (Q) in the x-axis. The slope is T/Q = 1/C, where C is the heat capacity. Then, the higher the slope, the lower the heat capacity. For a constant mass, it can also represent the specific heat capacity (c).

Heats of vaporization and fusion cannot be calculated from these sections of the heating curve.

Which statement below explains that?

a. The specific heat capacity of the gaseous ethanol is less than the specific heat capacity of liquid ethanol. YES.

b. The specific heat capacity of the gaseous ethanol is greater than the specific heat capacity of liquid ethanol. NO.

c. The heat of vaporization of ethanol is less than the heat of fusion of ethanol. NO.

d. The heat of vaporization of ethanol is greater than the heat of fusion of ethanol. NO.

The correct option is b. The specific heat capacity of the gaseous ethanol is greater than the specific heat capacity of liquid ethanol.

To understand why option b is correct, let's consider the heating curve for ethanol and what the slope of the line represents. The slope of the line on a heating curve indicates the rate at which the temperature of the substance increases per unit of heat added. A steeper slope means that the temperature rises more quickly for a given amount of heat.

 In the context of the heating curve for ethanol:

 - The slope of the line for the gaseous state is steeper than the slope for the liquid state. This means that for the same amount of heat added, the temperature of gaseous ethanol increases more than the temperature of liquid ethanol.

- The specific heat capacity (c) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. A higher specific heat capacity means that more heat is required to achieve the same temperature change.

Now, let's analyze the options:

 a. The specific heat capacity of the gaseous ethanol is less than the specific heat capacity of liquid ethanol. - This would mean that gaseous ethanol requires less heat to increase its temperature by a certain amount compared to liquid ethanol. However, the steeper slope for the gaseous state indicates that it takes more heat to increase the temperature of the gas by a certain amount, not less. Therefore, this option is incorrect.

b. The specific heat capacity of the gaseous ethanol is greater than the specific heat capacity of liquid ethanol. - This is consistent with the observation that the temperature of the gas increases more slowly than that of the liquid for the same amount of heat added (as indicated by the steeper slope for the gas). Since the gas requires more heat to increase its temperature, it has a higher specific heat capacity. This option is correct.

 c. The heat of vaporization of ethanol is less than the heat of fusion of ethanol. - The heat of vaporization is the amount of heat required to change a substance from the liquid to the gaseous state, while the heat of fusion is the amount of heat required to change a substance from the solid to the liquid state. This option is incorrect because the heat of vaporization is typically much greater than the heat of fusion for most substances, including ethanol.

 d. The heat of vaporization of ethanol is greater than the heat of fusion of ethanol. - This statement is generally true for most substances, including ethanol. However, it does not explain the difference in slopes between the gaseous and liquid states on the heating curve. The heat of vaporization is related to the amount of heat required to change the phase of the substance, not the rate at which the temperature changes within a given phase. Therefore, this option is not relevant to the explanation of the slopes on the heating curve.

 In conclusion, the correct explanation for the observation that the slope of the line for gaseous ethanol is greater than the slope for liquid ethanol is that the specific heat capacity of gaseous ethanol is greater than that of liquid ethanol. This means that more heat is required to raise the temperature of the gas by a certain amount compared to the liquid, which is consistent with the steeper slope on the heating curve for the gaseous state.

All listed are common goals about energy in chemistry, including producing, conserving, storing, and using energy. However, these aren't the inherent goals of chemistry, which is more focused on the study of substances and their transformations.

In the context of energy within the field of chemistry, all of the options you've listed are typical goals: finding ways to produce energy, conserve energy, store energy, and use energy. Chemistry plays a crucial role in these areas, such as in the development of new energy sources or improving energy efficiency. However, these aren't necessarily inherent goals of chemistry. The inherent goal of chemistry is to study the properties, composition, and structure of substances, and the transformations they undergo.

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The volume of the hollow gold sphere can be calculated by subtracting the volume of the inner void (hollow space) from the volume of the entire sphere.

To get the same density as real iron ore (5.15 g/cm³), the hollow gold spheres must be designed in a way that the total volume of gold is equal to the volume of the iron ore the same mass. The density formula can be rearranged to solve for volume:

Density = Mass / Volume

Volume = Mass / Density

Density of fake "iron ore" = Mass of fake "iron ore" / Volume of fake "iron ore"

The equation:

Density of fake "iron ore" = Mass of fake "iron ore" / Volume of fake "iron ore"Volume of hollow sphere = (4/3) * π * [(radius of outer sphere)³ - (radius of inner sphere)³]

5.15 g/cm³ = Mass of fake "iron ore" / [(4/3) * π * (1³ - (1 - t)³)]

So one then Solve for t:

t = 1 - ( (3 * Mass of fake "iron ore") / (4 * π * 5.15) )^(1/3)

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To achieve the same density as real iron ore, the required thickness of the walls of each hollow lump of 'iron ore' should be approximately 4.02 cm.

To calculate the required thickness of the walls of each hollow lump of 'iron ore,' we can use the formula for density: density = mass/volume.

Since the density of real iron ore is 5.15 g/cm^3 and we want the fake 'iron ore' to have the same density, we can set up the following equation:

5.15 g/cm^3 = (mass of gold)/(volume of hollow sphere)

We know that the mass of gold is the same as the volume of the hollow sphere, since the density of gold is 19.3 g/cm^3. Therefore, we can rewrite the equation as:

5.15 g/cm^3 = 19.3 g/cm^3 / (4/3 * π * (outer radius^3 - inner radius^3))

Solving for the outer radius, we get:

outer radius = (3 * (19.3 g/cm^3) / (5.15 g/cm^3 * 4 * π) )^(1/3) ≈ 4.02 cm

And since the inner radius is 0 (since it's hollow), the required thickness of the walls is the difference between the outer radius and the inner radius:

required thickness = outer radius - inner radius = 4.02 cm - 0 cm = 4.02 cm

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