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Carbon capture and storage is one experimental method being utilised to lower carbon dioxide emissions from coal (CCS).

One experimental technique being used to reduce carbon dioxide emissions from coal is carbon capture and storage (CCS). With CCS, carbon dioxide emissions from factories or power plants are captured and either stored underground in geological formations or used to improve oil recovery.

Coal and other fossil fuels have the potential to drastically cut their carbon dioxide emissions, but CCS technology currently in the experimental stage. Unfortunately, because of its expensive cost and technical implementation difficulties, the technology is not yet extensively employed.  In order to address the current climate problem, efforts to cut CO2 emissions are essential.

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What experimental technology is being used to reduce carbon dioxide emissions from coal?

Mn3+, an ion of manganese(III), can function as an acid by giving a proton (H+) to a base. Here's an illustration: Mn3+ (aq) + 3OH- (aq) Mn(OH)3 (s)

There is no need to add an indicator because MnO4's vivid purple colour serves as one enough. In the conical flask, there is Fe2+. The Fe2+ solution is added, and the Fe2+ lowers the MnO4- to Mn2+. As Mn2+ is a colourless solution, the purple colour disappears.

The divalent metal cation manganese(2+) contains manganese as the metal. It plays the part of a cofactor. It consists of a monoatomic dication, a manganese cation, and a divalent metal cation.

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The empirical formula, CH2O9(menthol) is multiplied by 5 to get the molecular formula, C10H20O.The empirical and molecular formula for menthol are CH2O and C10H20O, respectively.

The molecular formula for menthol is C5H10O.

This can be determined by dividing the molar mass of the empirical formula (156.26 g/mol) by the molar mass of CO2 (44.01 g/mol). This gives a ratio of 3.55, which is equal to the ratio of C atoms in the empirical formula, C10H20O.

Therefore, the molecular formula is C5H10O.
Given:

Menthol is composed of C, H, and O0.1005g sample of menthol is combusted and produces0.2829g of CO2 0.1159g H2O

1. Find: Empirical and molecular formula for menthol.

Let's first calculate the number of moles of CO2 produced. The balanced equation for combustion of menthol is:

C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)

From the above equation, we can see that for 10 moles of CO2 produced 1 mole of menthol is required.

2. By taking the number of moles of CO2 produced, we can calculate the number of moles of menthol burned.

Moles of CO2 = 0.2829g / 44.01g/mol= 0.00643 mol

C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)

From the balanced equation,1 mole of C10H20O requires 10 moles of CO2

Moles of C10H20O burned = 10 * 0.00643= 0.0643 mol

Next, we can calculate the number of moles of H2O produced.

Moles of H2O = 0.1159g / 18.015g/mol= 0.00643 mol

C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)

From the balanced equation,1 mole of C10H20O requires 10 moles of H2O

Moles of C10H20O burned = 10 * 0.00643= 0.0643 mol

3. Now we can calculate the empirical formula of menthol. The empirical formula can be calculated as follows:

Empirical formula = CH2O (Divide all moles by smallest moles) The molecular weight of CH2O = 30 g/mol

The empirical formula mass of the compound is:

mass = (12.011 + 2*1.008 + 15.999) = 30.026

Empirical formula mass of CH2O is 30.026g/mol, and the given sample weighs 0.1005 g.
The number of empirical formula units in the sample is 0.1005 g / 30.026 g/mol = 0.003348Units.

Empirical formula = CH2OThe empirical formula weight of menthol is CH2O, which is equal to 30.026g/mol.

4. To find the molecular formula, we need to know the molecular weight of the menthol. We can calculate it as follows:

Molecular formula mass = Empirical formula mass x n

Where n = integer Molecular formula mass of menthol is 156 g/mol, and the empirical formula mass is 30.026 g/mol.

So, n = 156 g/mol ÷ 30.026 g/mol = 5.192

Thus the empirical formula, CH2O is multiplied by 5 to get the molecular formula, C10H20O.The empirical and molecular formula for menthol are CH2O and C10H20O, respectively.

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The atom with the lowest first ionization energy is C (carbon). The order from highest to lowest is: e) Cl (chlorine) > d) P (phosphorus) > c) Ge (germanium) > b) C (carbon) > a) Ca (calcium).

The atom that would have the lowest first ionization energy is Ca (Calcium). The amount of energy that is required to remove the most loosely held electron from an isolated neutral gaseous atom to form a cation is called the first ionization energy. It is a measure of the stability of an atom. The ionization energy of an element is determined by the amount of energy required to remove an electron from its ground state. The ionization energy is a physical property of an element that varies across the periodic table. The element that has the lowest ionization energy is the most reactive and will most likely form cations.

Identify which of the following atoms would have the lowest first ionization energy. The given atoms are Ca, C, Ge, P, and Cl. Out of these atoms, Ca would have the lowest first ionization energy. The electronic configuration of Ca is 2, 8, 8, 2. Calcium belongs to group 2 and period 4 of the periodic table. It has 20 protons, 20 electrons, and 2 valence electrons. Because of its 2 valence electrons, it has a low ionization energy. The electronic configuration of Ca is most stable because of the presence of the 8 valence electrons in the outermost shell.

The electronic configurations of the other given atoms are:

C: 2, 4Ge: 2, 8, 18, 4P: 2, 8, 5Cl: 2, 8, 7

All of these elements have electrons that are either in the process of filling the valence shell or have already filled it. They have higher ionization energies because of this. Therefore, Ca would have the lowest first ionization energy.

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Answer:

Explanation:

2-bromopropane is the major product because the reaction mechanism involves the formation of the most stable carbocation intermediate. When hydrogen bromide reacts with propene, the hydrogen atom from HBr adds to the carbon atom of the double bond that has fewer hydrogen atoms attached, resulting in the formation of a carbocation intermediate. The intermediate can either form 1-bromopropane or 2-bromopropane depending on the position of the carbocation. The 2-bromopropane is the major product because the secondary carbocation formed in this case is more stable than the primary carbocation formed in the case of 1-bromopropane.

To test for an alkene, bromine water can be used. When an alkene reacts with bromine water, the bromine molecule adds across the double bond, forming a colorless dibromoalkane product. The mechanism for the reaction between hex-1-ene and bromine involves the formation of a cyclic bromonium ion intermediate, followed by the attack of water on the intermediate, resulting in the formation of the dibromoalkane product.

a) Addition polymerization is a process in which unsaturated monomers are joined together to form a polymer. The process involves breaking the double bond of the monomer and joining the monomers together to form a long-chain polymer. The process requires a catalyst to initiate the reaction.

b) The repeating unit of the polymer formed from the addition polymerization of chloroethene monomers is -CH2-CHCl-. This polymer is called polyvinyl chloride (PVC), and it has a wide range of uses, including pipes, electrical cables, and vinyl flooring.

The major organic product of an SN2 substitution reaction is an alkene, which may be either in retention or inversion of configuration relative to the original substrate.

The reaction you are asking about is an SN2 substitution reaction, in which a nucleophile (Nu) displaces a leaving group (LG) from a molecule with an alkyl halide substrate. The major organic product of this reaction will be an alkene, which has the same carbon chain as the alkyl halide substrate. Depending on the relative configuration of the substrate, the alkene product may be the same as the original substrate (retention) or have its configuration inverted (inversion). If stereochemistry is relevant to the question, then it should be specified in the answer.

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The Lewis structure of the H2SO4 has been shown in the image attached.

In H2SO4, there are a total of 32 valence electrons available for bonding. The central atom in H2SO4 is sulfur (S), which has 6 valence electrons. To achieve an octet, sulfur needs to form six covalent bonds.

The two hydrogen atoms (H) in H2SO4 each contribute one valence electron to form a single covalent bond with sulfur. This leaves sulfur with 4 valence electrons.

The four oxygen atoms (O) in H2SO4 each contribute 6 valence electrons to form a total of 24 valence electrons in four covalent bonds with sulfur. This brings the total number of valence electrons around sulfur to 28.

To complete the octet, each oxygen atom also has two lone pairs of electrons, bringing the total number of valence electrons around sulfur to 32.

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a balanced equation for the redox reaction: 2 Ca(s) + O2(g) → 2 CaO(s)

A redox reaction is a type of chemical reaction that involves the transfer of electrons between species. One species undergoes oxidation (loses electrons) while another species undergoes reduction (gains electrons).

In the reaction between calcium metal and oxygen gas, the calcium metal is being oxidized (loses electrons) and the oxygen gas is being reduced (gains electrons). This can be seen in the balanced equation where the calcium atoms go from having an oxidation state of 0 to +2, while the oxygen atoms go from having an oxidation state of 0 to -2.

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Answer:

Transcribed Image Text: Rank the following substances in order of increasing acid strength. (1 as least and 4 as most in acid strength) ✓ H₂Se ✓ HBr HI ✓ AsH3 Expert Solution

Explanation:

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Its atomic radius increases form top to bottom inside a group, then decreases from left and right across a period. As a result, francium is indeed the largest element while helium is the smallest.

Atomic radii inside the periodic table decrease across a row form left to right and increase across a column from top to bottom. Due to these two patterns, the periodic table's lower left and upper right corners, respectively, contain the largest and smallest atoms.

The atomic radius grows form top to bottom inside a group and decreases form left to right during a period, as seen in the images below. As a result, francium is indeed the largest element while helium is the smallest.

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a. The atom with the smaller atomic size is: Nitrogen

a. The atom with the smaller atomic size is: Arsenic.

Atomic size, also known as atomic radius, is the distance between the nucleus of an atom and its outermost electrons. It is typically measured in picometers (pm) or angstroms (Å). The atomic size of an element can be calculated by finding the distance between the nucleus and the outermost electron shell of an atom of that element. This distance can be determined using various methods, including X-ray diffraction and spectroscopy. The atomic size of elements generally decreases from left to right across a period and increases from top to bottom down a group in the periodic table.

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If we use the Gizmo to find the freezing, melting, and boiling points of water at 5,000 meters (16,404 feet) then,

Freezing point: 32 ºF (0ºC)

Melting point: 32 ºF (0ºC)

Boiling point: 203°F (95°C)

The freezing point is defined as the temperature at which a liquid becomes a solid. Increased pressure usually raises the freezing point with the melting point of the solid. The boiling point of a pure substance is defined as the temperature at which the substance transitions from a liquid to the gaseous phase. At the boiling point the vapor pressure of the liquid is equal to the applied pressure on the liquid. The melting point of a substance is defined as the temperature at which the substance changes from a solid to a liquid.

Melting occurs at a single temperature for the pure substances. The normal and average melting point and boiling point of water at 1 atmospheric pressure are 0°C and 100°C respectively. Decreasing the pressure under 1 atm. will lower the boiling point since the external pressure will be lower so it will become equal with the vapor pressure at a lower temperature.

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Answer:

1. Equilibrium expressions:

a. K = [HSO4-][H3O+]/[H2SO4][H2O]

b. K = [NO]^4[H2O]^6/[NH3]^4[O2]^5

c. K = [NH3][HCl]/[NH4Cl]

d. K = [NO2]^2/[N2O4]

2. Since K = 0.050, the concentrations of the reactants (N2 and H2) are larger than the concentrations of the products (NH3).

3. Since K = 6.8 x 10^3, the concentrations of the products (SO2 and O2) are larger than the concentrations of the reactant (SO3).

4. The Ksp expression for each of the reactions is:

a. Ksp = [Na+][Cl-]

b. Ksp = [Ba2+][SO42-]

5. Brønsted-Lowry acids and bases:

a. Acid: HNO3; Conjugate base: NO3-; Base: H2O; Conjugate acid: H3O+

b. Acid: HCN; Conjugate base: CN-; Base: H2O; Conjugate acid: HCN

c. Acid: H2SO4; Conjugate base: HSO4-; Base: Cl-; Conjugate acid: HCl

d. Acid: NH3; Conjugate base: NH2-; Base: H2O; Conjugate acid: NH4+

e. Acid: H2O; Conjugate base: OH-; Base: O2-; Conjugate acid: OH-

6. Conjugate acids and bases:

a. Acid: H2O; Conjugate base: OH-

b. Acid: H3O+; Conjugate base: H2O

c. Acid: H2CO3; Conjugate base: HCO3-

d. Acid: NH4+; Conjugate base: NH3

e. Acid: HSO4-; Conjugate base: SO42-

7. The strongest acid is HIO4 (highest K value), followed by HCN, HC2H3O2, and H2CO3 (lowest K value). The K values represent the degree to which the acids dissociate in solution. HIO4 is a strong acid, meaning it dissociates almost completely in solution, while H2CO3 is a weak acid, meaning it only dissociates partially.

8. pH and pOH calculations:

a. pH = -log[H3O+] = -log(0.200) = 0.699; pOH = -log[OH-] = -log(1.0 x 10^-14/0.200) = 12.301

b. pOH = -log[OH-] = -log(0.0143) = 1.844; pH = 14.000 - pOH = 12.156

c. pH = -log[H3O+] = -log(3.0) = 0.522; pOH = 13.478

d. pOH = -log[OH-] = -log(0.0062) = 2.206; pH = 14.000 - pOH = 11.794

9. Hydronium and hydroxide ion concentrations:

pH = 3.6; hydronium ion concentration = 10^-pH = 3.98 x 10^-4 M; hydro

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The completed table of maximum moles of water, limiting reactant and excess reactant is as follows:

Q: 6 moles, O₂, 1 mole H₂

R: 6 moles, O₂, 2 moles H₂

S: 5 moles, none, none

T: 5 moles, H₂, 2.5 moles O₂

U: 8 moles, H₂, 2 moles O₂

The mole ratio of the reaction of hydrogen and oxygen to form water is obtained from the equation of the reaction.

The equation of the reaction is given below:

2 H₂ + O₂ --> 2 H₂O

The mole ratio of hydrogen to oxygen is 2:1 in both the water molecule and the reactants, hydrogen gas (H2) and oxygen gas, as can be seen from the balanced equation (O2). 

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There are approximately 0.005302 moles of NOs in 3.19 × 10^16 molecules.

Mole is an SI unit used to measure the amount of any substance.

To calculate the number of moles of NOs in 3.19 × 10^16 molecules, we need to use Avogadro's number, which is 6.022 × 10^23 molecules per mole.

First, we need to convert the number of molecules to moles using the formula:

moles = molecules / Avogadro's number

moles of NOs = 3.19 × 10^16 molecules / 6.022 × 10^23 molecules per mole

moles of NOs = 0.005302 moles (rounded to 4 significant figures)

Therefore, there are approximately 0.005302 moles of NOs in 3.19 × 10^16 molecules.

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The number of moles present in 3.19×10¹⁶ molecules of nitrogen dioxide, NO₂ is 5.30×10⁻⁸ mole

The number of moles present in 3.19×10¹⁶ molecules of NO₂ can be obtained by using the Avogadro's hypothesis as illustrated below:

From Avogadro's hypothesis,

6.022×10²³ molecules = 1 mole of NO₂

Therefore,

3.19×10¹⁶ molecules = 3.19×10¹⁶  / 6.022×10²³

3.19×10¹⁶ molecules = 5.30×10⁻⁸ mole of NO₂

Thus, we can conclude that the number of mole is 5.30×10⁻⁸ mole

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The claim was correct . All elements  have same number of particles in one mole and have different number of particles  in a mole based on atomic number .

In the International System of Units, the mole (symbol mol) is the unit of substance amount (SI). The amount of substance is a measurement of how many elementary entities of a given substance are present in an object or sample. An elementary entity can be an atom, a molecule, an ion, an ion pair, or a subatomic particle such as an electron, depending on the substance. For example, despite having different volumes and masses, 10 moles of water (a chemical compound) and 10 moles of mercury (a chemical element) contain equal amounts of substance, and the mercury contains exactly one atom for each molecule of water.

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In comparison to 310.0 K, the reaction happens 1.28 times faster at 314 K.

A high K value (higher than 1) denotes an equilibrium with more products than reactants, whereas a low K value (less than 1) denotes an equilibrium with more reactants than products.

The Arrhenius equation, which connects the rate constant (k) to the activation energy (Ea) and temperature (T), can be used to determine how much faster the reaction happens at 314 K than it does at 310.0 K. k = A * exp(-Ea / (R * T))

where T is the temperature in Kelvin, R is the gas constant, and A is the preexponential factor.

For this reaction, we can assume that the pre-exponential component is fixed and that the sole variable is temperature.

exp[(Ea / R) * (1/T1 - 1/T2)] = k2 / k1

where Ea is the activation energy, R is the gas constant, k1 is the rate constant at 310.0 K, and k2 is the rate constant at 314 K.

k2 / k1 = exp[(50.0 kJ/mol / (8.314 J/mol•K)) * (1/310.0 K - 1/314 K)] is the result of substituting the provided numbers.

If we condense this phrase, we get:

k2 / k1 = 1.28

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The major organic product of the condensation of ammonia or a primary amine with the carbonyl group of an aldehyde or ketone is an imine.

A functional group or organic substance with a carbon-nitrogen double bond (C=N) is known as an imine. A hydrogen atom or an organic group may be joined to the nitrogen atom. (R). The carbon atom is connected to two more single bonds. Imines are present in numerous processes and are frequently found in manufactured and naturally occurring chemicals.

The five core atoms for ketimines and aldimines, C2C=NX and C(H)C=NX, respectively, are coplanar. The sp2-hybridization of the mutually double-bonded nitrogen and carbon atoms yields planarity. For nonconjugated imines, the C=N distance is 1.29-1.31, whereas for conjugated imines, it is 1.35. The C-N distances in amines and nitriles, on the other hand, are 1.47 and 1.16, respectively. Slow rotation occurs around the C=N bond. E- and Z-isomers were detected using NMR spectroscopy of aldimines have been detected. Owing to steric effects, the E isomer is favored.

An imine is formed when a primary amine reacts with a carbonyl group (C=O) of an aldehyde or ketone to form a new C-N bond. This reaction is known as a condensation reaction, as it involves the loss of a small molecule (e.g. water) to form the product.

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The correct questions is :

What  is the major organic product of the condensation of ammonia or a primary amine with the carbonyl group of an aldehyde or ketone?

The average value of the molar volume of hydrogen is 24.0 liters per mole (L/mol).

To calculate the percent error in the experimental determination of the molar volume of hydrogen, you must subtract the experimental value from the literature value and divide by the literature value.

One liter of hydrogen gas at STP has a mass of 0.090 grams, which is the experimental value of the density of hydrogen. This value is lower than the literature value, which is 0.089 grams per liter (g/L).
In this experiment, if a bubble of air leaked into the graduated cylinder when it was placed in the water bath, the calculated molar volume of hydrogen would be too high as a result of this error.

This is because the presence of the bubble of air would increase the measured volume of hydrogen gas.If the magnesium ribbon appeared to be oxidized, the calculated molar volume of hydrogen would be too low as a result of this error.

This is because the oxidation of the magnesium ribbon would reduce the amount of hydrogen gas produced, resulting in a lower measured volume of hydrogen gas. For demonstration purposes, the ideal gas law may be used to calculate the maximum amount or length of magnesium ribbon that may be used.

The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the amount of substance, R is the ideal gas constant, and T is the temperature. Knowing the desired volume of the gas, the amount of substance can be calculated.

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The R f values for compounds A, B, and C on a silica gel TLC plate developed in hexanes would be determined by measuring the distance each compound traveled compared to the distance the solvent traveled.

a) There is a 4 cm gap between the origin and the solvent front. The Rf value for spot A is[tex]\frac{1.5}{4}= 0.375[/tex], because it travelled 1.5 cm. Due to the 3.5 cm movement of Spot B, its Rf is[tex]\frac{3.5}{4} = 0.875[/tex]. Spot C shifted 3 cm, making its Rf [tex]\frac{3}{4} = 0.75[/tex].

b)Due to its shorter travel distance than the other two compounds, compound A is the most polar. Recall that polar substances adhere to the adsorbent more readily, move less, and have a lower Rf value.

c)Hexanes is less polar than acetone as a solvent. Each of the three compounds would move more quickly if the same method were employed to elute them.The chemicals can be removed from the polar adsorbent more effectively with a more polar eluting solvent. Each compound would have a higher Rf value if acetone were used to elute the TLC plate as opposed to hexanes because each compound travels more quickly.

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A few statements may be true regarding a biochemical oxidation-reduction (redox) reaction. The statements are as follows: A redox reaction occurs when there is a transfer of electrons between molecules or atoms.

The electron donor becomes oxidized, and the electron acceptor is reduced, causing a transfer of energy. A redox reaction produces ATP, which is the primary energy currency of the cell. Oxidation and reduction are complementary reactions that occur simultaneously in the same reaction, resulting in the release of energy. Redox reactions are vital in metabolic pathways, and the electron carriers NAD+ and FAD+ are essential in these reactions. Oxygen is frequently used as a final electron acceptor in redox reactions. Redox reactions can also occur in non-cellular environments, such as photosynthesis, respiration, and combustion. The significance of redox reactions is enormous, and they play an essential role in sustaining life on earth. They help in generating energy, breaking down complex molecules, synthesizing molecules, and many other cellular processes.

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Explanation:

The reactant contains 2N and 6H

The product contains 2N and 6H

Therefore, the chemical equation is balanced

From the equation, for every 1 mole of N2 that reacts, 3 moles of H2 are required.

We know 28.6 grams of N2 reacted, but we don’t know the mass ratio but just the mole ratio, so we have to convert 28.6 grams of N2 to the corresponding moles of N2.

From the periodic table, the molar mass of N is about 14 g/mol, so the molar mass of nitrogen gas or N2 is two times of that which is 28 g/mol.

With this, we can calculate moles of N2, but we also need to make sure the equation is setted up the right way.

Looking at the units, if we cancel out the grams, we are left with mol. We also know that in multiplication, numerator of one number cancel with the denominator of another number and vice versa

So the equation looks like this 28.6g * mol/28g = 1.021 mol N2

So the number of moles of H2 required is 1.021 mol N2 * 3 mol H2/1 mol N2 = 3.063 mol H2 (notice that mol N2 canceled out, so the equation is set up correctly)

However, the question ask for number of grams of H2 needed, so we need the molar mass of hydrogen gas or H2, which is 1*2 = 2 g/mol

3.063 mol H2 * 2 g H2/ mol H2 = 6.126 g H2

Ans: 6.126 g H2

Answer:

See below.

Explanation:

The atomic number of lead (Pb) is 82, which means it has 82 electrons. The electronic configuration of lead is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d¹⁰ 6s² 6p²

The orbital diagram for the valence electrons of lead (Pb) is

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

s s p p p p d d

2 1 6 2 6 2 10 10

|||||||||

1 2 3 4 5 6 7 8

The notation ↑↓ represents a pair of electrons with opposite spins.

To determine if lead (Pb) is diamagnetic or paramagnetic, we need to look at whether there are any unpaired electrons. Based on the orbital diagram, we can see that all the electrons in the valence shell are paired, meaning that lead (Pb) is diamagnetic.

The  atoms of lithium that  are in 18.7 g is 16 × 10²³ atoms . This is taken out by mole concept .

The mole is a unit of measurement similar to the pair, dozen, gross, and so on. It provides a precise count of the atoms or molecules in a bulk sample of matter. A mole is the amount of substance that contains the same number of discrete entities (atoms, molecules, ions, etc.)

if 7 grams of lithium contain 6 × 10²³ atoms

then 18.7 will contain 16 × 10²³ atoms

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(S)-2-butanol will undergo an SN2 reaction with HBr to produce a racemic mixture of alkyl bromides. Here option B is the correct answer.

When optically active alcohol is treated with HBr, the reaction follows an SN1 or SN2 mechanism. In the case of SN1, a carbocation intermediate is formed, and in SN2, a backside attack by the nucleophile occurs. The stereochemistry of the product depends on the configuration of the intermediate and the direction of attack.

In the case of (S)-2-butanol, the hydroxyl group is attached to the second carbon atom, which makes it a primary alcohol. When treated with HBr, it undergoes an SN2 reaction, where the hydroxyl group is replaced by the bromine atom. The nucleophile attacks from the backside of the molecule, leading to an inversion of configuration.

This results in the formation of a racemic mixture of alkyl bromides, as both enantiomers have an equal chance of being attacked from either side. On the other hand, (R)-2-butanol, being the enantiomer of (S)-2-butanol, will also undergo the same reaction and produce the same racemic mixture of alkyl bromides.

In the case of (R)-1-phenyl ethanol and (S)-1-phenyl ethanol, they are secondary alcohols and can undergo either SN1 or SN2 reactions depending on the reaction conditions. However, the reaction mechanism will lead to the formation of a mixture of diastereomers, rather than a racemic mixture of enantiomers.

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Complete question:

Which of the following optically active alcohols, when treated with HBr, results in a racemic mixture of alkyl bromides?

a) (R)-2-butanol

b) (S)-2-butanol

c) (R)-1-phenyl ethanol

d) (S)-1-phenyl ethanol

The total amount of energy necessary to heat and vaporize a 13.1 g sample of ethanol initially at a temperature of 11.1 degrees C is 7.15 kJ.

To calculate the amount of energy, in joules, necessary to heat and vaporize a 13.1 g sample of ethanol initially at a temperature of 11.1 degrees C, we must first calculate the heat necessary to heat the sample to the boiling point of ethanol, 78.3 degrees C. The formula to calculate the amount of energy is: Q = mcΔT, where m is the mass of the sample, c is the specific heat of ethanol, and ΔT is the temperature change from 11.1 degrees C to 78.3 degrees C. Thus, the amount of energy necessary to heat the sample is: Q = 13.1 g * 2.44 JK−1g−1 * (78.3-11.1) = 1,623.08 J.
Next, we must calculate the amount of energy necessary to completely vaporize the sample. To do so, we must use the heat of vaporization of ethanol, which is 38.56 kJ mol−1. To convert from moles to grams, we must use the molar mass of ethanol, which is 46 g/mol. Thus, the amount of energy necessary to vaporize the sample is: Q = (13.1 g/46 g/mol) * 38.56 kJ/mol = 7.15 kJ.
Finally, to calculate the total amount of energy necessary to heat and vaporize the sample, we must add the two values together: Q = 1,623.08 J + 7.15 kJ = 7.15 kJ. the total amount of energy necessary to heat and vaporize a 13.1 g sample of ethanol initially at a temperature of 11.1 degrees C is 7.15 kJ.

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At has a higher initial ionisation energy than Br, while Br has a bigger atomic radius. Se has a bigger atomic radius than Br, and Br has a higher initial ionisation energy than Se.

The most loosely bound electron is further from the nucleus and thus easier to remove in bigger atoms. Hence, the ionisation energy should decrease as size (atomic radius) increases.

This is because the outer electrons aren't bound as strongly because they are farther from the nucleus.

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The acid in the reaction would donate a proton and that would be HNO2.

An acid in a chemical reaction can be identified by the presence of hydrogen ions (H+): Acids are compounds that produce hydrogen ions when dissolved in water. In a chemical reaction, an acid may donate a hydrogen ion to another compound or accept a pair of electrons from a base.

When we look at the reaction, we can see that the specie that has given out the replaceable hydrogen ion is HNO2 thus it is the acid in the reaction.

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At 769 K, the rate constant equals 2.22 1/min.

The final expression is either k=A or k=A. This implies that the response rate will be equal to the value of the collision frequency rather than the temperature when the activation energy is zero.

The Arrhenius equation, which connects the rate constant (k) to the activation energy (Ea) and temperature (T), can be used to solve this issue:

[tex]A = * exp (-Ea / (R * T))[/tex]

With the rate constant (k) at 475 K, we can utilize this knowledge to calculate the pre-exponential factor (A) as follows:

0.0450 1/min = A * exp(-13.6 kJ/mol / (8.314 J/mol•K * 475 K))

[tex]A = 5.74 x 10^9 min^-1[/tex]

The rate constant (k) at 769 K can now be calculated using the Arrhenius equation once more as follows:

[tex]k = 5.74 x 10^9 min^-1[/tex] * exp(-13.6 kJ/mol / (8.314 J/mol•K * 769 K))

k = 2.22 1/min (rounded to two significant figures)

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Answer:

First, we need to calculate the change in temperature:

ΔT = final temperature - initial temperature

ΔT = 85 °C - 35 °C

ΔT = 50 °C

Next, we can use the following formula to calculate the heat energy required:

Q = m·C·ΔT

where Q is the heat energy in calories, m is the mass of the sample in grams, C is the specific heat in cal/g °C, and ΔT is the change in temperature in °C.

Plugging in the given values, we get:

Q = 35.0 g · 0.108 cal/g °C · 50 °C

Q = 189.0 calories

Therefore, 189.0 calories are required to raise the temperature of the sample from 35 °C to 85 °C