The addition of KCL raises the boiling point of 1.0 kg water by 2.14 deg * C How many moles of were added ?

[tex]1-propanethiol[/tex]and [tex]2-propanethiol[/tex]are structural isomers, meaning that they have the same molecular formula (C₃H₈S) but different structural arrangements. The correct answer is option a.

In[tex]1-propanethiol[/tex], the sulfur atom is attached to the first carbon atom of the propane chain, whereas in [tex]2-propanethiol[/tex], the sulfur atom is attached to the second carbon atom of the propane chain. This structural difference results in different chemical and physical properties for each compound.

On the other hand, ethylmethylsulfide has a different molecular formula (C₃H₈S), which means it is not an isomer of[tex]1-propanethiol[/tex]. Ethylmethylsulfide has an ethyl group (-CH₂CH₃) and a methyl group (-CH₃) attached to the sulfur atom, which makes it a different compound altogether.

Therefore, the answer to the question is that [tex]2-propanethiol[/tex] is an isomer of [tex]1-propanethiol[/tex], but ethylmethylsulfide is not an isomer of [tex]1-propanethiol.[/tex]

The correct answer is option a.

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Complete question

Which of these compounds is an isomer of 1-propanethiol?

a. 2-propanethiol

b. ethylmethylsulfide

c. both of these

d. either of these

Therefore, a neutral sulfur atom has a total of 16 protons and 16 electrons. So, the correct option is 16.

The atomic number of sulfur is 16, which means it has 16 protons and 16 electrons in a neutral atom (since the number of protons and electrons are equal in a neutral atom). The atomic mass of sulfur is 32, which is the sum of the number of protons and neutrons in the nucleus. Since the number of protons in sulfur is 16, we can deduce that the number of neutrons in sulfur is 16 as well (atomic mass - atomic number = number of neutrons).

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The equilibrium constant for the given reaction is approximately 0.973.

The equilibrium constant (K) for the given reaction can be calculated using the Nernst equation:

Ecell = E°cell - (RT/nF)lnQ

where E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred in the reaction, F is the Faraday constant, and Q is the reaction quotient.

The standard cell potential, E°cell, can be calculated using the standard reduction potentials of Cu2+/Cu and Ag+/Ag:

E°cell = E°(Cu2+/Cu) - E°(Ag+/Ag)

      = 0.34 V - 0.80 V

      = -0.46 V

where the reduction potential values are taken from standard reduction potential tables.

At equilibrium, the reaction quotient Q is equal to the equilibrium constant K. The equilibrium concentration of Cu2+ is not given, but we can assume that it is much smaller than the concentration of Ag+ (since copper is a less noble metal than silver, and therefore less likely to oxidize in solution). Therefore, we can approximate the Q expression as:

Q ≈ [Ag+]^2/[Cu2+]

Substituting the given values and constants into the Nernst equation, we get:

0.47 V = -0.46 V - (8.314 J/K·mol)(298 K)/(2 mol)(96,500 C/mol) ln [Ag+]^2/[Cu2+]

Simplifying the equation:

ln [Ag+]^2/[Cu2+] = -0.0274

[Ag+]^2/[Cu2+] = e^-0.0274

[Ag+]^2/[Cu2+] = 0.973

K = [Ag+]^2/[Cu2+] = 0.973

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Table MM.1: Bond Angles

Compound C-C-C Angle (straight chain or cyclic)

Propane 109.5 degrees

Butane 109.5 degrees

Pentane 109.5 degrees

Cyclopropane 60 degrees

Cyclobutane 90 degrees

Cyclopentane 108 degrees

The bond angles in organic compounds are primarily determined by the hybridization of the central atom and the number of electron pairs around it. In straight chain alkanes such as propane, butane, and pentane, the carbon atoms are sp³ hybridized and have tetrahedral geometry with bond angles of approximately 109.5°.

In cyclic compounds such as cyclopropane, cyclobutane, and cyclopentane, the bond angles are different from those in straight chain alkanes due to ring strain caused by the bond angles deviating from the ideal tetrahedral angle.

In cyclopropane, the bond angles are approximately 60°, while in cyclobutane they are approximately 90°, and in cyclopentane they are approximately 108°.

Overall, the bond angles in organic compounds are important in determining the overall shape and properties of the molecule.

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The mass of solid aluminum produced at the cathode is approximately 7.43 grams after six hours.

Using Faraday's equation:

moles of substance = (electric charge passed) / (Faraday's constant x number of electrons transferred)

In this case, the substance being produced is aluminum, Al, and the number of electrons transferred is 3, since Al3+ ions each gain 3 electrons to form Al atoms.

The electric charge passed can be calculated from the current and time using the equation:

electric charge = current x time

                         = 3.7 A x 6 hours x 3600 seconds/hour = 80,064 C

The Faraday constant is the charge on one mole of electrons, which is 96,485 C/mol. So we can calculate the moles of aluminum produced as:

Moles of Al = 80,064 / (96,485 x 3) = 0.276 mol

Mass of Al = 0.276 mol x 26.98 g/mol = 7.43 g

Therefore, the mass of solid aluminum produced at the cathode is approximately 7.43 grams after six hours.

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We need to make a lucite sleeve with a thickness of at least 1.29 mm to prevent beta particles from passing through.

To calculate the required thickness of the lucite sleeve, we need to consider the stopping power of the material. Beta particles are electrons that can be stopped by materials with high electron density, such as lucite.

First, we need to calculate the range of the beta particles in the glass wall of the syringe. We can use a range-energy relationship to estimate the range:

R = 0.412 * Emax^1.65

Where R is the range in g/cm2, and Emax is the maximum energy of the beta particles in MeV. For 90Sr, the maximum energy is 2.28 MeV.

R = 0.412 * 2.28^1.65
R = 0.412 * 7.07
R = 2.91 g/cm2

Next, we need to calculate the thickness of lucite required to stop the beta particles. We can use the concept of range straggling to estimate the thickness:

d = 1.96 * Rho * R / (Z * Z * dE/dx)

Where d is the thickness in cm, Rho is the density of the material in g/cm3, Z is the atomic number of the material, and dE/dx is the stopping power of the material in MeV/(g/cm2).

For lucite, Z is approximately 6, and dE/dx is 2.27 MeV/(g/cm2) for beta particles with energies between 0.5 and 3 MeV.

d = 1.96 * 1.2 * 2.91 / (6 * 6 * 2.27)
d = 0.129 cm = 1.29 mm

Therefore, we need to make a lucite sleeve with a thickness of at least 1.29 mm to prevent beta particles from passing through.

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The answer is option (b) C17H35COOH.

The melting point of a fatty acid depends on its molecular weight and the degree of saturation.

The longer the carbon chain length and the fewer the double bonds in the fatty acid, the higher its melting point.

Among the given options, (b) C17H35COOH has the highest molecular weight and is the longest chain fatty acid, so it will have the highest melting point.

Therefore, the answer is option (b) C17H35COOH.

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The molar mass of the 1120 mL sample of a pure gaseous compound is found to be 48 g/mol.

To find the molar mass of the compound, we can use the ideal gas law:

PV = nRT, the pressure is P, volume is V, number of moles is n, gas constant is R, temperature. is T At STP (standard temperature and pressure), we have, P = 1 atm, V = 1.120 L, T = 273.15 K and R = 0.08206 L·atm/mol·K. From the mass of the sample, we can calculate the number of moles of the compound using its density at STP,

density = mass/volume

= 2.86 g/1.120 L

= 2.55 g/L

The molar mass of the compound is then,

molar mass = mass/number of moles

= 2.86 g/(2.55 g/L × 0.08206 L·atm/mol·K × 273.15 K) ≈ 48 g/mol

Therefore, the molar mass of the compound is approximately 48 g/mol.

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when 1.00 mol of both H2 and I2 are placed in a 1.00 L flask, both H2 and I2 are the limiting reactants, and the total amount of HI produced is 4.00 mol.

The balanced chemical equation for the reaction is:

H2 + I2 → 2HI

The experimenter has placed 1.00 mol of both H2 and I2 in a 1.00 L flask, which means the initial concentrations of both substances are 1.00 M. According to the balanced chemical equation, the two substances react to produce 2 moles of HI. Therefore, the limiting reactant in this reaction is the one that will be completely consumed first, which can be determined by calculating the theoretical yield of HI from each reactant. Since both reactants have equal amounts in moles, either H2 or I2 can be the limiting reactant.

To determine which is the limiting reactant, we need to compare the theoretical yields of HI from each reactant. The theoretical yield of HI from 1.00 mol of H2 is 2.00 mol, while the theoretical yield of HI from 1.00 mol of I2 is also 2.00 mol. Therefore, neither H2 nor I2 is in excess and both are limiting reactants.

The amount of HI produced can be calculated by using the stoichiometry of the balanced equation:

1.00 mol H2 x (2 mol HI / 1 mol H2) = 2.00 mol HI
1.00 mol I2 x (2 mol HI / 1 mol I2) = 2.00 mol HI

Therefore, the total amount of HI produced is 4.00 mol.

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I understand you'd like to know about the solubility of compounds containing CO32- and PO43- ions.

In general, most compounds containing CO32- (carbonate) and PO43- (phosphate) ions are insoluble in water. However, there are exceptions when these ions are combined with cations from Group 1 of the periodic table, such as Li+, Na+, K+, Rb+, and Cs+. Compounds with these cations and CO32- or PO43- are generally soluble in water.

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Answer:

Explanation:

The atoms combine to attain a noble or inert gas electronic configuration

The percent yield when 8.00 g of NaI are produced from a mixture of 10.0 g I₂ and 10.0 g NaOH is 67.7%.

To calculate the percent yield, first determine the theoretical yield and then compare it to the actual yield. In this case, 8.00 g of NaI are produced (actual yield).

1. Write the balanced equation for the reaction:
2NaOH + I₂ → 2NaI + H₂O

2. Calculate the moles of each reactant:
I₂: (10.0 g) / (253.8 g/mol) = 0.0394 mol
NaOH: (10.0 g) / (40.0 g/mol) = 0.250 mol

3. Determine the limiting reactant:
For I₂: (0.0394 mol) / (1 mol I₂) = 0.0394
For NaOH: (0.250 mol) / (2 mol NaOH) = 0.125

Since 0.0394 < 0.125, I₂ is the limiting reactant.

4. Calculate the theoretical yield of NaI:
(0.0394 mol I₂) x (2 mol NaI / 1 mol I₂) x (149.9 g/mol NaI) = 11.81 g NaI

5. Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) x 100
Percent yield = (8.00 g / 11.81 g) x 100 = 67.7%

The percent yield for this reaction is 67.7%.

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The reaction tBuCl + ROH → tBuOR + HCl follows an SN1 mechanism, where the rate-determining step is the formation of the carbocation intermediate.

The solvent can have a significant effect on the rate of the reaction by stabilizing or destabilizing the intermediate.

In general, polar protic solvents stabilize the carbocation intermediate by solvating the positive charge, while polar aprotic solvents destabilize the carbocation intermediate by not solvating the positive charge.

Solvent 1:  Water is a polar protic solvent that can stabilize the carbocation intermediate by solvating the positive charge.

Therefore, the reaction rate is expected to be relatively slow in water due to increased stabilization of the intermediate.

Solvent 2:  Acetone is a polar aprotic solvent that can destabilize the carbocation intermediate by not solvating the positive charge.

Therefore, the reaction rate is expected to be relatively fast in acetone due to decreased stabilization of the intermediate.

Solvent 3:  Dichloromethane is a non-polar solvent that cannot stabilize or destabilize the carbocation intermediate by solvating the positive charge.

Therefore, the reaction rate is expected to be intermediate in dichloromethane.

In summary, the relative rates of the reaction in these solvents can be ordered as follows:

Solvent 2 (acetone) > Solvent 3 (dichloromethane) > Solvent 1 (water).

This is because acetone is a polar aprotic solvent that destabilizes the intermediate, dichloromethane is a non-polar solvent that does not affect the intermediate, and water is a polar protic solvent that stabilizes the intermediate.

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The mass of the graduated cylinder is 17228.05 units.

The volume of a cylinder is obtained using the formula;

V = πr²h

Now, we have the following information;

Volume of the cylinder = 100 mL or 100 cm³

Inner Diameter (I.D.) = 23 mm

Outer Diameter (O.D.) = 25 mm

Height of the cylinder  = h

Radius of the cylinder = 11.5 + 2 = 13.5 mm

Volume = 3.14 × 13.5 × 13.5 × 13.5 = 7725.58

Density = Mass / Volume

2.23 = Mass / 7725.58

Mass = 7725.58 × 2.23 = 17228.05 units

Hence, the mass of the cylinder is 17228.05 units.

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The atom with the smallest atomic radius is At (Astatine) and in second par the atom with the smallest atomic radius is Be (Beryllium).As we move down a group in the periodic table, the atomic radius generally increases due to the addition of new electron shells.

Astatine is an exception to this trend. Being in the last group of the periodic table, Astatine has a higher atomic number and more electrons than any other element in Group I. This results in strong electron-electron repulsions that cause its electron cloud to shrink and its atomic radius to decrease.

As we move across a period in the periodic table, the atomic radius generally decreases due to the increase in effective nuclear charge, which pulls the electrons closer to the nucleus. Beryllium, being the first element in Group II, has the smallest atomic radius because it has a higher effective nuclear charge than the other elements in the group.

To summarize, the atoms with the smallest atomic radii in each group are Astatine (At) in Group I and Beryllium (Be) in Group II.

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The use of a computational program can help to accurately determine thermodynamic properties of a reaction and provide insights into its behavior over different temperature ranges.

The conversion of fumarate ion to malate ion is catalyzed by the enzyme fumarase, and the average standard reaction enthalpy ΔrHo over the given temperature range can be calculated using the provided data and an appropriate computational program. To do this, one can plot the given data points and use linear regression to obtain the slope, which corresponds to ΔrHo.

Once ΔrHo is determined, it is possible to calculate ΔrCP and A,So for each temperature using appropriate thermodynamic equations. These calculations can be used to comment on the validity of the approximation that ΔrHo is independent of temperature over the given range.

Overall, the use of a computational program can help to accurately determine thermodynamic properties of a reaction and provide insights into its behavior over different temperature ranges.

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To create a 0.1M solution of Na₂S₂O₃, you would need 16.98 g of Na₂S₂O₃·5H₂O in 1 liter of solution.

The yellow color reappears after the endpoint is reached due to the formation of the iodine-starch complex. In the titration of iodine with thiosulfate, the endpoint is reached when all the iodine has reacted and the solution becomes colorless.

However, when excess thiosulfate is added, it can react with the iodine-starch complex that was formed during the titration, resulting in the reformation of iodine and the reappearance of the yellow color. This is known as the "iodine clock reaction" and is commonly used in chemistry experiments to demonstrate the concept of reaction kinetics.

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The magnitude of the velocity and acceleration of points A and B on the vertical-axis windmill's parabolic blade when t = 4 s are as follows:

Velocity of A: 2 rad/s

Acceleration of A: 0.5 rad/s²

Velocity of B: 4 rad/s

Acceleration of B: 0.5 rad/s²

1. Calculate angular velocity (ω) using the equation ω = α_c * t, where α_c is the constant angular acceleration (0.5 rad/s²) and t is the time (4 s).

2. For point A, ω = 0.5 * 4 = 2 rad/s.

3. For point B, ω = 2 * 2 = 4 rad/s.

4. Since the angular acceleration is constant, the acceleration of points A and B remains 0.5 rad/s².

5. The velocity of point A is 2 rad/s, and the velocity of point B is 4 rad/s.

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The number of atoms of each element present are 1.4 × 10²³ atoms of water, 9.27 × 10²² atoms of carbon dioxide, 1.62 × 10¹⁸ atoms of benzene and 1.36 × 10²⁴ atoms of sucrose.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.

A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.

a. Mass of water = 4.21g

Moles = 4.21 / 18

= 0.233 moles

number of atoms = 6.023 × 10²³ × 0.233

= 1.4 × 10²³ atoms.

b. mass of carbon dioxide = 6.81g

moles = 6.81 / 44

= 0.154 moles

number of atoms = 6.023 × 10²³ × 0.154

= 9.27 × 10²² atoms.

c. mass of benzene = 0.00021g

moles = 0.00021 / 78

= 2.69 × 10⁻⁶ moles

number of atoms = 6.023 × 10²³ × 2.69 × 10⁻⁶

= 1.62 × 10¹⁸ atoms.

d. 2.26 moles

number of atoms = 6.023 × 10²³ × 2.26

= 1.36 × 10²⁴ atoms.

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The relative polarity order of the molecules will be in the order shown as above:

Methanol > Ethanol > 1-butanol > 1-hexanol

The polarity principle is that the greater the electronegativity difference between atoms in a bond, the more polar the bond. The most electronegative atom carries the partial negative charge whereas the partially positive charge is found on the electropositive atoms. The hydrocarbons part containing carbons and hydrogens is non-polar and provides a non-polar surface area to the molecule.

Methanol is a polar molecule as the alcohol group dominates the molecule makes it polar. The oxygen is partially negative whereas the carbons and hydrogens are partially positive. Therefore the methanol has a larger polar surface area and has a greater electronegativity difference.

Ethanol is comparatively less polar than methanol but the alcohol group still gives the polar effect. When compared with the methanol the two carbons and multiple hydrogens attached to it will show a bit more non-polar properties with a larger surface area of non-polar.

1-butanol has now four carbons and multiple hydrogens which will contribute non-polarity to it and has larger non-polar surface areas than methanol and ethanol.

Hexanol is mostly non-polar with some non-polar properties as the alcohol group still gives a small polar effect but when compared to the rest of the molecules the total of six carbons and the multiple hydrogens attached show the dominance of non-polar properties and will have larger non-polar surface areas.

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The standard free energy change for the autoionization of water at 25°C is 75.3 kJ/mol.

The standard free energy change (ΔG°) for the autoionization of water at 25°C can be calculated using the following equation:

ΔG° = -RTln(Kw)

where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (25°C = 298 K), and Kw is the ion product constant of water (1.0 × 10^-14 at 25°C).

Substute the values in the above equation, we get:

ΔG° = - (8.314 J/mol·K) × 298 K × ln(1.0 × 10^-14)

ΔG° = 75.3 kJ/mol

Therefore, the standard free energy change for the autoionization of water at 25°C is 75.3 kJ/mol.

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The change in enthalpy when 138.03 g of NO₂ is produced is 342.6 kJ.

The thermochemical equation is shown below.

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -114.2 kJ

This means that 114.2 kJ of energy is released when 2 moles of NO(g) and 1 mole of O₂(g) react to form 2 moles of NO₂(g). We can use this information to calculate the change in enthalpy when a certain amount of NO₂(g) is produced.

The molar mass of NO₂ is 46.01 g/mol

The number of moles of NO₂ can be calculated as shown below.

n(NO₂) = mass / molar mass

= 138.03 g / 46.01 g/mol

= 3.00 mol

According to the balanced equation, 2 moles of NO₂ react to produce 2 moles of NO₂.

Therefore, the number of moles of NO needed to produce 3.00 moles of NO₂ is 3.00 mol.

Calculate the change in enthalpy for the production of 138.03 g of NO₂ is shown below.

ΔH = (n(NO₂)) x ΔH

= (3.00 mol) x (-114.2 kJ/mol)

= -342.6 kJ

Therefore, the change in enthalpy when 138.03 g of NO₂ is produced is -342.6 kJ.

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The statement "Alkenes can be converted into alcohols by acid-catalyzed addition of water" is true because Acid-catalyzed addition of water, also known as hydration, is a common way to convert alkenes into alcohols.

The reaction involves the addition of a water molecule across the carbon-carbon double bond of the alkene, forming an alcohol.

The reaction is typically carried out in the presence of a strong acid catalyst, such as sulfuric acid or phosphoric acid, which protonates the alkene to make it more reactive towards nucleophilic attack by the water molecule.

This reaction is a well-known addition reaction and is used extensively in organic chemistry for the synthesis of alcohols.

It is an important reaction for the industrial production of alcohols from alkenes, especially in the case of simple alkenes like ethene, which can be hydrated to produce ethanol. This reaction is also used in the production of various chemicals, including plastics, solvents, and detergents.

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When anion undergoes aerobic decomposition, a total mass of 2.28 kg O2 is required to biodegrade 1.4 g of the substance.

To determine the mass of O2 required to biodegrade 1.4 g of the given substance, we need to first determine the molar mass of the substance. From the given formula, we can see that the anion has a molar mass of 502 g/mol.

Using the given stoichiometric coefficients, we can write the balanced equation for the aerobic decomposition of the anion:

2C18H29SO−3(aq) + 51O2(aq) → 36CO2(aq) + 28H2O(l) + 2H+(aq) + 2SO2−4(aq)

From the balanced equation, we can see that 51 moles of O2 are required to biodegrade 2 moles of the anion. Therefore, the number of moles of O2 required to biodegrade 1 mole of the anion is:

51/2 = 25.5 moles of O2

The mass of O2 required to biodegrade 1 mole of the anion is:

25.5 moles × 32 g/mol = 816 g

Therefore, the mass of O2 required to biodegrade 502 g (1.4 g × 1000/502 = 2.79 moles) of the anion is:

2.79 moles × 816 g/mol = 2279 g or 2.28 kg (rounded to two decimal places)

In summary, 2.28 kg of O2 is required to biodegrade 1.4 g of the given substance.

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The volume of oxygen gas required to completely react with 51.6 g of aluminum at 770 mmHg and 34 ∘C is 87.9 mL.

we need to use the balanced chemical equation for the reaction between aluminum (Al) and oxygen (O2), which is:
4 Al + 3 O2 → 2 Al2O3

This means that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.

Next, we need to use the given conditions to calculate the number of moles of aluminum we have:
molar mass of Al = 26.98 g/mol
moles of Al = mass of Al / molar mass of Al
moles of Al = 51.6 g / 26.98 g/mol
moles of Al = 1.912 mol

Now we can use the mole ratio from the balanced chemical equation to find the number of moles of oxygen needed:
moles of O2 = (3/4) x moles of Al
moles of O2 = (3/4) x 1.912 mol
moles of O2 = 1.434 mol

Finally, we can use the ideal gas law to calculate the volume of oxygen gas at the given conditions:
PV = nRT

where:
P = pressure = 770 mmHg
V = volume (unknown)
n = number of moles = 1.434 mol
R = gas constant = 0.08206 L·atm/mol·K
T = temperature = 34 + 273.15 = 307.15 K

Rearranging the equation, we get:
V = nRT / P
V = (1.434 mol) x (0.08206 L·atm/mol·K) x (307.15 K) / (770 mmHg)
V = 0.0879 L or 87.9 mL

Therefore, the volume of oxygen gas required to completely react with 51.6 g of aluminum at 770 mmHg and 34 ∘C is 87.9 mL.

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3 equivalent resonance structures can be drawn for the BrO₃- ion. Each structure has one double bond with one of the Oxygen atoms while the other two have single bonds. Option (d) 3.

Let's analyze the structure and follow these steps:

1. Identify the central atom: In this case, the central atom is Bromine (Br).
2. Determine the number of valence electrons: Bromine has 7 valence electrons, and each Oxygen atom has 6 valence electrons. Since it is a negative ion, we also need to add 1 electron. So, in total, there are (7 + 3 * 6 + 1) = 26 valence electrons.

3. Distribute the valence electrons and create the basic structure: Bromine is surrounded by 3 Oxygen atoms, and each Oxygen forms a single bond with Bromine. The remaining 20 electrons are distributed as lone pairs (2 pairs for each Oxygen).

4. Check for the possibility of creating double bonds to satisfy the octet rule: Since Br has only used 6 of its valence electrons, we can create double bonds with Oxygen to fulfill the octet rule.


So, the answer is d. 3.

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The angle between one of the carbon-hydrogen bonds and one of the carbon-chlorine bonds in the methylene chloride (CH2Cl2) molecule is approximately 109.5 degrees.

This angle is a result of the tetrahedral geometry of the carbon atom in the molecule, which has four electron pairs around it. Two of these electron pairs are involved in the carbon-hydrogen bonds, and the other two are involved in the carbon-chlorine bonds. The shape of the molecule is determined by the repulsion between these electron pairs, which results in a tetrahedral arrangement. The angle between any two bonds in a tetrahedral molecule is approximately 109.5 degrees. This geometry is important for understanding the properties and behavior of molecules, as it affects how they interact with other molecules and their reactivity in chemical reactions.

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The number of vacancies comes out to be 3.52 *10¹⁸/cm³ that can be shown in the below expalnation.

The number of vacancies can be calculated using the below formula-

Nv = N exp (-Qv / kT)

     = (Na x D / A) exp (-Qv / kT)

It is given that,

D = 18.63 g/cm³

Na is known which is 6.022*10²³ atoms/mol also called Avogadro's number.

Qv = 0.98 eV/atom

T = 1173 K

K = 8.65*10⁻⁵ eV/atom-K

Substituting these values in the above equation as follows-

Nv = (6.02*10²³ atoms/mol) (18.63 g/cm3) / (1969.9 g/mol) exp (0.98 ev/atom / (86.2*10⁻⁵ ev atom -K)(1173 K)

    = 3.52 *10¹⁸/cm³

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In conducting an experiment that involves transferring heat between two substances, it is essential to ensure that the process is carried out accurately to obtain accurate results. Two common mistakes that are often made during this process are the loss of heat when the hot substance is transferred to the cold water and the failure to attain a constant final temperature. However, there are steps that can be taken to avoid these mistakes and ensure accurate results.

The first mistake of heat loss can be avoided by using an insulated container to hold the cold water. This container will prevent the loss of heat and ensure that the cold water maintains its temperature. Additionally, it is essential to ensure that the hot substance is transferred as quickly as possible to the cold water to prevent any significant loss of heat.
The second mistake of failing to attain a constant final temperature can be avoided by ensuring that the two substances are thoroughly mixed together. This mixing will ensure that the hot substance is evenly distributed within the cold water, allowing for an accurate measurement of the final temperature. Additionally, it is essential to monitor the temperature regularly to ensure that the temperature remains constant and that the measurements are accurate.
In conclusion, accurate measurements are essential in any scientific experiment, and steps should be taken to avoid common mistakes such as heat loss and failure to attain a constant final temperature. Using insulated containers and ensuring that the two substances are mixed thoroughly are effective ways of preventing these mistakes and ensuring accurate results.

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The conversion of an alkyne to an alkene is reduction because it involves the addition of hydrogen atoms, which results in a decrease in the number of bonds to more electronegative atoms.

This reduction reaction is typically carried out using a catalyst such as Lindlar's catalyst or sodium in liquid ammonia.

True, the conversion of an alkyne to an alkene is indeed a reduction. This process involves the addition of hydrogen atoms to the alkyne, reducing the number of carbon-carbon triple bonds and forming a carbon-carbon double bond in the alkene.

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