PLSSSSS HELPPPPPPPPPPPP
A Model Atom
In this lab, you will examine the relationship between an element’s
location on the periodic table and its number of valence electrons.
You will use this information to draw models of three elements and
their valence shells

The number of the grains of rice that we have from the question here is  16214 grains

In this case, we know that we have to rely on the information that we have in the question so as to be able to obtain the mass of the rice that we need in this case and that is what we are going to set out to do in this question.

We know that;

Mass of the Rice = 786 grams - 332 grams

= 454 g

If 1000 grains of rice have a mass of 28 g

x grains of rice have a mass of 454 g

x = 1000 * 454/28

x = 16214 grains

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The formation of the phosphorylated intermediate (an anhydride) involves the formation of a penta-coordinate intermediate. This intermediate is formed by a nucleophilic attack of the sulfur on the phosphorus atom of the phosphate group.

In this mechanism, the sulfur atom of the sulfate group nucleophilically attacks the phosphorus atom of the phosphate group to form a penta-coordinate intermediate. This intermediate then rearranges to form a phosphorylated intermediate, which is an anhydride.

Mechanism showing the penta-coordinate intermediate and the formation of the phosphorylated intermediate are given as follows:

Step 1: Alkyl Phosphate Formation : The first step of the mechanism includes the formation of an alkyl phosphate. A proton is abstracted by OH− from the phosphate group to create the alkyl phosphate. The base catalyzes this step.

Step 2: Binding to Mg2+After the alkyl phosphate is created, the magnesium ion binds to it.

Step 3: Nucleophilic attack: Following that, the nucleophilic attack happens, with the nucleophile being the water molecule. It is coordinated with the magnesium ion. It occurs at phosphorus, causing it to be phosphorylated. It results in the creation of a pentacoordinate intermediate.

Step 4: Release of Orthophosphate: Orthophosphate is released as a result of the reaction between pentacoordinate intermediate and water. It results in the creation of a diester intermediate.

Step 5: Subsequent Hydrolysis: In the final step, the intermediate diester is hydrolyzed to form orthophosphate and the final product. This is accomplished via nucleophilic substitution.

The end result is a free phosphate group that is bound to the alcohol's oxygen. A phosphate anhydride is formed in the process.

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The volume of 0.100 M NaOH is required to precipitate all of the nickel (II) ions from 150.0 mL of a 0.321 M solution of Ni(NO₃)₂ is 483 mL.

To solve this problem, we need to first write the balanced chemical equation for the reaction between NaOH and Ni(NO₃)₂:

Ni(NO₃)₂ + 2NaOH → Ni(OH)₂ + 2NaNO₃

From this equation, we can see that 2 moles of NaOH are required to precipitate 1 mole of Ni(NO₃)₂.

Therefore, we need to calculate the number of moles of Ni(NO₃)₂ in 150.0 mL of a 0.321 M solution:

moles of Ni(NO₃)₂ = volume (L) x concentration (mol/L)

moles of Ni(NO₃)₂ = 0.150 L x 0.321 mol/L

moles of Ni(NO₃)₂ = 0.0483 mol

To precipitate all of the nickel (II) ions, we need to add an equal number of moles of NaOH.

Since the concentration of NaOH is 0.100 M, we can calculate the volume of NaOH required:

moles of NaOH = 0.0483 mol (from above)

volume of NaOH = moles of NaOH / concentration of NaOH

volume of NaOH = 0.0483 mol / 0.100 mol/L

volume of NaOH = 0.483 L or 483 mL

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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.

We can calculate the mass of NaH2PO4·H2O required mass of NaH2PO4·H2O = (0.00250 - 5.5 x 10⁻⁻⁷)

Sodium dihydrogen phosphate NaH2PO4 (monobasic) and sodium hydrogen phosphate Na2HPO4 (dibasic) are a weak acid and its conjugate base pair that are mixed to make a buffer with pH 7.2.

To calculate the amount of solid NaH2PO4·H2O required to prepare the buffer, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

First, we need to calculate the pKa of phosphoric acid (H3PO4). The dissociation reactions for phosphoric acid are:

H3PO4 ⇌ H+ + H2PO4-

[tex]Ka1 = 7.5 x 10^-3[/tex]Ka1 = 7.5 x 10^-3

H2PO4- ⇌ H+ + HPO42-

[tex]Ka2 = 6.2 x 10^-8[/tex]

HPO42- ⇌ H+ + PO43-

[tex]Ka3 = 4.8 x 10^-13[/tex]

pKa2 = -log(Ka2)

[tex]pKa2 = -log(6.2 x 10^-8) = 7.21[/tex]

[tex][A-]/[HA] = 10^(pH - pKa)[/tex]

[A-]/[HA] = 10^(2.14 - 7.21) = 1.1 x 10^-5[tex][A-]/[HA] = 10^(2.14 - 7.21) = 1.1 x 10^-5[/tex]

H3PO4 ⇌ H+ + H2PO4-

moles of H3PO4 = (0.1000 M) x (0.02500 L) = 0.00250 mol

moles of NaH2PO4·H2O = 0.00250 - x

The molar mass of NaH2PO4·H2O is 138.01 g/mol, so the mass of NaH2PO4·H2O required is:

mass of NaH2PO4·H2O = (0.00250 - x) mol x (138.01 g/mol)

Now, we can use the ratio of [A-] to [HA] to solve for x:

[A-]/[HA] = [NaH2PO4·H2O]/[H3PO4]

[tex]1.1 x 10^-5 = x / (0.05000 L x 0.100 M)[/tex]

x = 5.5 x 10^-7 mol[tex]x = 5.5 x 10^-7 mol[/tex]

Finally, we can calculate the mass of NaH2PO4·H2O required:

mass of [tex]NaH2PO4·H2O = (0.00250 - 5.5 x 10^-7)[/tex].

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Answer:

The correctly written thermochemical equation is:

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l), ΔH = -2,220 kJ/mol

This equation represents the combustion of propane (C3H8) in the presence of oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), with a heat release of -2,220 kJ/mol. The state symbols (g) for gases and (l) for liquids indicate the physical state of each substance at standard conditions.

Explanation:

ABOVE

When 100 moles of propane undergo the given combustion reaction, 500 moles of oxygen will be required.

The combustion reaction of propane is given by:C₃H₈ + 5O₂ → 3CO₂ + 4H₂OFor every mole of propane, five moles of oxygen are required to complete the combustion reaction. Therefore, for 100 moles of propane, 500 moles of oxygen will be required.Theoretical oxygen for 100 moles of propane undergoing the combustion reaction is 500 moles of oxygen. Thus, the correct option is (C) 500 moles. The theoretical oxygen required is different from the actual oxygen required, which may vary due to incomplete combustion or other factors that affect the reaction.

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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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Polymers are large molecules made up of repeating subunits called monomers, and they are essential building blocks for many cellular structures and processes.

A molecule is a group of two or more atoms that are chemically bonded together. Molecules are the fundamental units of compounds, which are substances made up of two or more different types of atoms.

Molecules can have a variety of sizes and shapes, depending on the number and arrangement of atoms. Some molecules are simple, consisting of just a few atoms, while others are much larger and more complex, such as proteins or DNA. The properties and behavior of a substance depend largely on the types of molecules it contains and how those molecules interact with each other.

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Cl2 (g) + F2 (g) ⇌ 2ClF (g)

Kc = [ClF]^2 / [Cl2][F2]

Let x be the change in concentration of ClF, Cl2, and F2 at equilibrium. Then the equilibrium concentrations can be expressed as:

[ClF] = 0.839 M + x

[Cl2] = 0.327 M - x

[F2] = 0.579 M - x

Substituting these expressions into the equilibrium constant expression and solving for x gives:

20.0 = ([0.839 + x]^2) / ([0.327 - x][0.579 - x])

Expanding the numerator and denominator and simplifying, we get:

20.0 = (0.704x^2 + 3.321x + 0.702) / (-0.189x^2 + 0.463x - 0.190)

Multiplying both sides by the denominator and rearranging, we get a quadratic equation:

0.189x^2 - 3.880x + 3.032 = 0

Using the quadratic formula, we find that:

x = 7.68 × 10^-2 M

Substituting this value back into the expressions for the equilibrium concentrations gives:

[ClF] = 0.839 M + 7.68 × 10^-2 M = 0.917 M

[Cl2] = 0.327 M - 7.68 × 10^-2 M = 0.250 M

[F2] = 0.579 M - 7.68 × 10^-2 M = 0.501 M

Therefore, the equilibrium concentrations of ClF, Cl2, and F2 at 2500 K are 0.917 M, 0.250 M, and 0.501 M, respectively.

The correct response is B. An electron EMITS a photon as it goes from a HIGHER TO LOWER energy level.

A subatomic particle with a negative electric charge is called an electron. Together with protons and neutrons, it is one of the fundamental particles that make up an atom.

When an atom is excited, its electrons have the ability to transition from one energy level to another. These stimulated electrons will eventually revert to their initial lower energy levels since they are unstable. They do this by releasing energy in the form of particular wavelength photons. Each element has a distinct set of energy levels, which leads to the emission of photons at a certain wavelength and the formation of a distinct emission spectrum.

In this instance, the existence of a line in the emission spectrum's blue section shows that an electron has released a photon with a certain wavelength that corresponds to the region's blue color. A photon with an energy equal to the difference between the two levels is released when an excited electron returns from a higher energy level to a lower energy level.

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To form each of the monochlorination products shown, you will need to draw curved arrows that demonstrate the propagation steps. The first step is when a chlorine radical combines with the double bond to form a chlorine radical cation, which then donates an electron to the double bond.

This results in the formation of two radical chlorides, one on each carbon atom. These radicals then combine with two hydrogen atoms to form the monochlorination product, completing the reaction.The curved arrows for this process should be drawn as follows:

An arrow pointing from the chlorine radical to the double bond, representing the attack of the radical all arrows have been drawn, the monochlorination product has been formed. The mechanism of propagation steps to form each monochlorination product is shown in the following reaction:To represent this reaction, you can draw a curved arrow to show the movement of electrons from the bond to the chlorine. The arrow should start from the carbon-carbon double bond and point towards the chlorine.

Then, another curved arrow can be drawn to represent the formation of the C-Cl bond. The arrow should start from the chlorine and point towards the carbon with the unpaired electron.This process can be repeated to form the second monochlorination product. The following diagram shows the mechanism of the propagation steps:Here, you can see that the curved arrows are used to represent the movement of electrons during the reaction. The arrows point towards the atom that is gaining the electrons.

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The pH of a solution that is 9.0 x 10-² M hydrogen carbonate is 1.05.

pH is a figure expressing the acidity or alkalinity of a solution on a logarithmic scale.

On the logarithmic scale, 7 is neutral, lower values are more acidic and higher values more alkaline. The pH of a solution can be calculated using the following expression;

pH = −log10 c

where;

According to this question, the concentration of a hydrogen carbonate solution is 9.0 × 10-²M. The pH can be calculated as follows:

pH = - log {9 × 10-²}

pH = 1.05

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Option (C) is correct. The pH of the solution of sodium acetate (NaCH3COO) given that the Ka of acetic acid (CH3COOH) is 2.57.

Sodium acetate is defined as the salt of a weak acid and strong base from the equation:

C2H3NaO2 ---> CH3COO−+Na+

CH3COO− + H2O  ⇌  CH3COOH + OH−

As it is a weak acid and strong base, this is a good indicator of a fairly high pH value.

Kb = [HB+] + [OH−] / [B]

where, [B] is the concentration of the base[HB+] is the concentration of base ions.[OH−] is the concentration of the hydroxide ions.

Ka Kb=1⋅10−14

So, Kb=1⋅10−14 / 1.8⋅10−5

          =5.555...⋅10−10

Putting the value of this in the expression of pH we get the value of pH .

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The pH of 0.352 M triethylammonium iodide, (C2H5)3NI, is 8.11.

This is calculated using the following equation:
pH = -log(Kb x [I-])
Where Kb is the base dissociation constant for triethylamine, (C2H5)3N, which is 5.2 x 10-4, and [I-] is the concentration of iodide ions, which is 0.352 M.
First, calculate the concentration of triethylamine, (C2H5)3N:
[(C2H5)3N] = (0.352 M) / (1 + Kb x (0.352 M))
[(C2H5)3N] = 0.351 M
Then, calculate the concentration of iodide ions, [I-], in the solution:
[I-] = 0.352 M - 0.351 M
[I-] = 0.001 M
Finally, calculate the pH of the solution:
pH = -log(Kb x [I-])
pH = -log(5.2 x 10-4 x 0.001)
pH = 8.11
Therefore, the pH of 0.352 M triethylammonium iodide, (C2H5)3NI, is 8.11.

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In general, an aldol addition reaction involves the addition of an enolate ion or enol to an aldehyde or ketone.

What is an enolate ion?

The enolate or enol acts as a nucleophile, attacking the carbonyl carbon of the aldehyde or ketone, resulting in the formation of a beta-hydroxy carbonyl compound called an aldol.

To carry out an aldol addition reaction, you would typically need an appropriate carbonyl compound, which can act as an electrophile, and a base, such as sodium hydroxide or lithium diisopropylamide (LDA), which can generate the enolate or enol. The choice of reactants would depend on the specific aldol addition reaction being carried out and the desired product to be formed.

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The mass of water released is  0.454 g

Moles of water released is 0.025 moles

Moles of CoCl2 remaining is 0.0042 moles

Value of x is 6

Mass of water released = 1. 0 g - 0.5460 g = 0.454 g

Moles of water released = 0.454 g/18 g/mol = 0.025 moles

Moles of CoCl2 remaining = 0.5460 g/130 g/mol = 0.0042 moles

We have;

Number of moles of anhydrous compound = Number of moles of hydrated compound

1/130 + 18 x = 0.0042

0.0042 (130 + 18x) = 1

0.546 + 0.0756x = 1

x = 1 - 0.546 /0.0756

x = 6

The value of x is 6

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The equilibrium constant, K is related to the amount of products and reactants at which the reaction reaches at equilibrium (a point where reaction rate is zero). So, the correct choice to fill up the blank is at equilibrium.

Equilibrium define when the rate of the forward reaction is equal to the rate of the backward reaction. The mathematical ratio showing the concentration of the product divided by the concentration of the reactant is called the equilibrium constant K. See the image above for an expression representing it. It expresses the relationship between the products and reactants of a reaction in equilibrium with a particular unit.

Equilibrium constants depend only on temperature and are independent of concentrations, pressures, and volumes of reactants and products. It is also independent of the presence of catalysts and the presence of inert materials. Thus, the right choice for filling the blank is at equilibrium.

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Answer:

Assign an oxidation state to each element in each reaction and use the change in oxidation state to determine which element is being oxidized and which element is being reduced.

1. C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(g)

2. C2H4(g)+Cl2(g)→C2H4Cl2(g)

Explanation:

1- In the reaction, C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(g), the oxidation state of each element changes as follows:

C6H12O6: C: -1 to +4; H: +1 to +1; O: -2 to -2

O2: O: 0 to -2

CO2: C: +4 to +4; O: -2 to -2

H2O: H: +1 to +1; O: -2 to -2

2- In this reaction, oxygen (O2) is being reduced, since its oxidation state changes from 0 to -2. Carbon (C6H12O6) is being oxidized, since its oxidation state changes from -1 to +4.

In the reaction, C2H4(g)+Cl2(g)→C2H4Cl2(g), the oxidation state of each element changes as follows:

C2H4: C: -3 to -2; H: +1 to +1

Cl2: Cl: 0 to 0

C2H4Cl2: C: -2 to -2; H: +1 to +1; Cl: 0 to -1

In this reaction, chlorine (Cl2) is being reduced, since its oxidation state changes from 0 to -1. Ethene (C2H4) is being oxidized, since its oxidation state changes from -3 to -2.

Explanation: the resistance that one surface or object encounters when moving over another.

or

the action of one surface or object rubbing against another.

Answer: a force that resists the motion of one object against another

(a) one radial node the Number of radial nodes = n - l - 1

And number of angular nodes = l

n = 3 and l = 1

Orbital is 3p.

(b) It has zero angular node hence s-orbital and there is 1 radial node . 1 = n - 0 - 1 ; n = 2 and l = 0

The orbital is 2s.

(c) the shape of the orbital is that of dz². There is two angular nodes and there is no radial node.

n = 3 and l = 2

Hence the orbital is 3dz².

In atomic physics, a radial node is a point in space where the probability density of finding an electron in an atom is zero. It is a type of nodal plane that occurs in atomic orbitals, which are regions of space where electrons are most likely to be found.

Radial nodes occur in the radial distribution function of an atomic orbital, which describes the probability density of finding an electron at a given distance from the nucleus. The number of radial nodes in an atomic orbital is equal to n - l - 1, where n is the principal quantum number and l is the azimuthal quantum number.

Radial nodes represent regions of space where the radial wave function of the electron changes sign.

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The enthalpy change can not be measured directly because you have to take into account how much energy was put into the reaction in the first place.

Hope this helps!!! :)

The standard enthalpy of formation of all elements in their standard states are assumed to be zero. It is not possible to determine the enthalpy of formation of calcium carbonate as it is formed from other compounds.

The standard enthalpy of formation of a compound can be defined as the enthalpy change accompanying the formation of one mole of the compound from its constituent elements, all the substances being in their standard states.

The standard enthalpy of formation is usually denoted as ΔfH⁰. For example the enthalpy of formation of CO₂ and CH₄ are -393.5 kJ mol⁻¹ and -74.8 kJ mol⁻¹ respectively.

Here CaCO₃ is formed by the reaction:

CaO + CO₂ → CaCO₃

The enthalpy change for the given reaction is not an enthalpy of formation of CaCO₃. Since CaCO₃ is not formed from its constituent elements.

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The reagents needed to carry out the halogenation of a ring are a halogen (e.g. bromine or chlorine) and ultraviolet (UV) light. The halogen is added to the ring in a reaction flask, and then the flask is exposed to UV light to activate the reaction.

The reagents required to carry out the given synthesis, which involves the halogenation of a ring with a halogen and UV light, are described below. The following are the reagents needed for the synthesis is Halogen and UV Light

The halogenation of a ring can be done with the halogen and UV light. The halogen and UV light are the two reagents needed to carry out this reaction. Halogens like fluorine, chlorine, bromine, and iodine can be used in halogenation. Halogenation of organic compounds is a common technique used in organic synthesis. Halogens' electronegativity makes them highly reactive, and their addition to a molecule usually results in the formation of new carbon-halogen bonds.

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Answer:

70.95%

Explanation:

To calculate the percent yield of water in this reaction, we need to compare the actual amount of water produced to the theoretical amount of water that could be produced based on the amount of hydrobromic acid and sodium hydroxide used.

First, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, limiting the amount of product that can be formed.

To find the limiting reactant, we can use stoichiometry to calculate the amount of water that could be produced from each reactant, assuming they react completely. The balanced chemical equation for the reaction is:

HBr + NaOH → NaBr + H2O

From the equation, we can see that the mole ratio of HBr to H2O is 1:1, and the mole ratio of NaOH to H2O is 1:1. Therefore, the amount of water produced depends on the amount of HBr and NaOH present, and the reactant that produces less water is the limiting reactant.

Using the molar masses of the compounds, we can convert the masses of HBr and NaOH to moles:

moles of HBr = 17.0 g / 80.91 g/mol = 0.210 moles

moles of NaOH = 13.9 g / 40.00 g/mol = 0.348 moles

Based on the balanced chemical equation, the theoretical amount of water that could be produced from 0.210 moles of HBr is also 0.210 moles. The theoretical amount of water that could be produced from 0.348 moles of NaOH is also 0.348 moles.

However, since the amount of water produced is given as 2.69 g, we need to convert this to moles:

moles of H2O produced = 2.69 g / 18.02 g/mol = 0.149 moles

To calculate the percent yield of water, we can use the formula:

percent yield = (actual yield / theoretical yield) x 100%

where actual yield is the amount of water produced (0.149 moles) and theoretical yield is the amount of water that could be produced based on the limiting reactant.

Since the reactant that produces less water is the limiting reactant, we need to compare the theoretical yield of water from both reactants, and the lower value will be the theoretical yield based on the limiting reactant.

The theoretical yield of water from HBr is:

0.210 moles of HBr x (1 mole of H2O / 1 mole of HBr) = 0.210 moles of H2O

The theoretical yield of water from NaOH is:

0.348 moles of NaOH x (1 mole of H2O / 1 mole of NaOH) = 0.348 moles of H2O

Since the theoretical yield of water from HBr is lower, it is the limiting reactant. Therefore, the theoretical yield of water is 0.210 moles.

Now we can calculate the percent yield of water:

percent yield = (actual yield / theoretical yield) x 100%

percent yield = (0.149 moles / 0.210 moles) x 100%

percent yield = 70.95%

Therefore, the percent yield of water is 70.95%.

The formal charge on the Z atom in the [ZO2]-1 ion is +1.

The Lewis dot structure for the [ZO2]-1 molecular ion is:

          O

          |

    Z === O

          |

          O-

1.   Determine the total number of valence electrons in the ion by adding the valence electrons of each atom and the charge of the ion.

2.  Connect the Z atom to each O atom with a single bond, which uses up 2 electrons.

3.  Add the remaining electrons in pairs as lone pairs to each atom until all valence electrons are used up.

4.  Draw the Lewis dot structure.

O

|

Z === O

|

O-

5.  Calculate the formal charge on the Z atom using the formula:

Formal charge = valence electrons - (number of lone pair electrons + 1/2 x number of bonding electrons)

Formal charge = 4 - (2 + 1/2 x 2) = 4 - 3 = +1

Therefore, the formal charge on the Z atom in the [ZO2]-1 ion is +1.

What is valence electron?

A valence electron is an electron in the outer shell associated with an atom, and that can participate in the formation of a chemical bond if the outer shell is not closed.

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The bonds would be arranged in order of decreasing polarity as follows: H-F Bond (most polar) > O-H Bond and C-H Bond (tied) > C-C Bond (least polar).

In order to arrange the bonds in order of decreasing polarity, we can look at the electronegativity difference between the two atoms of each bond. Electronegativity differences will determine whether the bond is polar, nonpolar, or ionic.

In general, the polarity of a bond is determined by the electronegativity difference between the atoms in the bond. The greater the difference in electronegativity, the more polar the bond is likely to be.

The following is a list of the bonds in order of decreasing polarity, based on the information provided in the table:

The table shows the following:

H-F Bond: Electronegativity difference = 1.9

C-H Bond: Electronegativity difference = 0.4

C-C Bond: Electronegativity difference = 0

HF is the least polar bond since the difference in electronegativity between hydrogen and fluorine is smaller than the differences between the other atoms in the list.

Therefore, the bonds would be arranged in order of decreasing polarity as follows: H-F Bond (most polar) > and C-H Bond (tied) > C-C Bond (least polar).

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The complete questions is:

Based on the information in the table, which of the following arranges the bonds in order of decreasing polarity

The table shows the following:

H-F Bond: Electronegativity difference = 1.9

C-H Bond: Electronegativity difference = 0.4

C-C Bond: Electronegativity difference = 0

The percent ionization of the nitrous acid in the 0.036 M aqueous solution is 2.1%.

The osmotic pressure (π) of a solution can be related to the molar concentration (M) of the solute and the temperature (T) of the solution by the following equation:

π = MRT

Where R is the gas constant.

We can use this equation to calculate the molar concentration of the nitrous acid solution:

M = π / RT

M = (0.93 atm) / (0.0821 L·atm/(mol·K) x 298 K)

M = 0.036 M

This is the molar concentration of the undissociated nitrous acid in the solution. To calculate the percent ionization of the acid, we need to know the concentration of the H+ and NO2- ions in the solution.

The balanced chemical equation for the dissociation of nitrous acid is:

HNO2(aq) ⇌ H+(aq) + NO2-(aq)

Let x be the extent of ionization of the nitrous acid. Then the concentration of H+ and NO2- ions can be expressed in terms of x as follows:

[H+] = x M

[NO2-] = x M

The concentration of the undissociated nitrous acid is (1-x)M.

The expression for the equilibrium constant (Ka) of the reaction can be written as:

Ka = [H+] [NO2-] / [HNO2]

Substituting the concentrations in terms of x, we get:

Ka = x^2M / (1-x)M

Simplifying the above equation, we get:

Ka = x^2 / (1-x)

The percent ionization of the acid is the fraction of the original HNO2 molecules that dissociate into H+ and NO2- ions. It can be calculated as follows:

% ionization = (concentration of H+ ions) / (initial concentration of HNO2) x 100

% ionization = (x M) / (M) x 100

% ionization = x x 100

Substituting the value of x from the above equation for Ka, we get:

Ka = x^2 / (1-x)

x = sqrt(Ka / (1+Ka))

We can calculate the value of Ka using the standard reference value of the acid dissociation constant (Ka) for nitrous acid at 25°C, which is 4.5 x 10^-4.

x = sqrt(4.5 x 10^-4 / (1+4.5 x 10^-4))

x = 0.021

% ionization = 0.021 x 100

% ionization = 2.1%

Therefore, the percent ionization of the nitrous acid in the 0.036 M aqueous solution is 2.1%.

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The correct electron configuration for Tc (technetium) is [Kr] 5s² 4d⁵. Therefore, the correct answer is: [kr]5s²4d⁵.

Technetium (Tc) is a radioactive chemical substance with the atomic number 43 and symbol Tc. It is a silvery-gray metal that belongs to the transition metals group on the periodic table. Technetium is the first element to be artificially produced, and all of its isotopes are radioactive, with no stable isotopes. It is a highly toxic and dangerous element, and therefore has no significant commercial applications. Technetium has many nuclear and medical applications due to its radioactivity, and is used in medical imaging, cancer treatment, and scientific research.

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Answer:

C

Explanation:

edg 2023