Which of these polymers are most resistant to attack by chemicals, and, as such, are often used as coatings? O Fluorocarbons O Polystyrene O Polyethylene O Rubber

If a nucleus (radioactive) decays by successive alpha, beta, and beta particle emissions, its atomic number will decrease by 2 for each alpha particle emitted and increase by 1 for each beta particle emitted.

Alpha particles are helium nuclei and have a mass number of 4 and an atomic number of 2. When an alpha particle is emitted, the original nucleus loses 2 protons and 2 neutrons, resulting in a decrease of 2 in its atomic number.

Beta particles are electrons or positrons emitted during radioactive decay. When a beta particle is emitted, a neutron in the nucleus is converted into a proton, increasing the atomic number by 1.

Therefore, if a nucleus undergoes successive alpha, beta, and beta particle emissions, its atomic number will decrease by 4 for each alpha particle emitted and increase by 2 for each beta particle emitted. The resulting nucleus will have a lower atomic number than the original nucleus.

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Lattice energy is a measure of the strength of the electrostatic attraction between ions in an ionic compound. The higher the lattice energy, the stronger the ionic bond is between the ions.The lattice energy is dependent on several factors, including the charge of the ions, the size of the ions, and the distance between the ions.

(a) In the case of KCl and MgO, both are ionic compounds with one metal ion (K and Mg) and one non-metal ion (Cl and O). Both K+ and Mg2+ have the same charge, but the size of the Mg2+ ion is smaller than the K+ ion. Similarly, both Cl- and O2- have the same charge, but the size of the O2- ion is smaller than the Cl- ion.
Smaller ions have a stronger electrostatic attraction between them than larger ions, as the distance between them is smaller. Therefore, MgO should have a higher lattice energy than KCl.

(b) In the case of LiF and LiBr, both are ionic compounds with one metal ion (Li) and one non-metal ion (F and Br). Both Li+ and F- have a smaller size than Li+ and Br-. However, since both Li+ and F- have the same charge as Li+ and Br-, the distance between the ions will be the deciding factor in determining the lattice energy.
Since Br- is a larger ion than F-, the distance between Li+ and Br- will be greater than the distance between Li+ and F-. Therefore, LiF should have a higher lattice energy than LiBr.

(c) In the case of Mg3N2 and NaCl, both are ionic compounds with one metal ion (Mg and Na) and one non-metal ion (N and Cl). Mg2+ and Na+ have the same charge, but the size of the Mg2+ ion is smaller than the Na+ ion. Similarly, both N3- and Cl- have the same charge, but the size of the N3- ion is larger than the Cl- ion.

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The osmotic pressure of the solution is 0.893 atm.

To calculate the osmotic pressure of the solution, we can use the equation:

π = MRT

Where:

π = osmotic pressure (in atm)

M = molarity of the solution (in mol/L)

R = ideal gas constant = 0.08206 L·atm/(mol·K)

T = temperature (in K)

First, we need to calculate the molarity of the solution:

Number of moles of Ca(NO3)2 = 1.64 g / (164.1 g/mol) = 0.01 mol

Volume of solution = 275 mL = 0.275 L

Molarity of solution = 0.01 mol / 0.275 L = 0.036 M

Now we can calculate the osmotic pressure:

π = (0.036 mol/L) x (0.08206 L·atm/(mol·K)) x (298.15 K) = 0.893 atm

Therefore, the osmotic pressure of the solution is 0.893 atm.

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The osmotic pressure of an ideal solution of 1.64 g Ca(NO3)2 in 275 mL of water at 25 °C is 0.89 atm.

First, we need to find the molarity of the solution. Given the formula weight of Ca(NO3)2 is approximately 164.087 g/mol, the number of moles of Ca(NO3)2 in 1.64 g is 1.64 g/164.087 g/mol = 0.01 mol. As it is dissolved in a solution with a volume of 275 mL (or 0.275 L), the molarity (M) is the number of moles/volume in L, or 0.01 mol/0.275 L = 0.03636 mol/L. We use the osmotic pressure formula, Π = MRT, where R is the ideal gas constant 0.0821 L·atm/mol·K and T is the temperature in Kelvin. The temperature in Kelvin is 25 °C + 273.15 = 298.15 K. Therefore, the osmotic pressure (Π) is 0.03636 mol/L × 0.0821 L·atm/mol·K × 298.15 K = 0.89 atm.

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An unstable atomic nucleus loses energy during radioactive decay and changes into a more stable state, frequently by producing radiation in the form of particles or electromagnetic waves.

a. Th-234 alpha decay:

Th-234 -> He-4 + Ra-230

In this process, Th-234 releases an alpha particle, which is a helium-4 nucleus, and transforms into Ra-230.

b. Fe-59 beta decay:

Fe-59 -> Co-59 + e- + anti-neutrino

In this process, Fe-59 releases a beta particle, which is an electron, and transforms into Co-59. At the same time, an anti-neutrino is also released.

c. Tc-99 gamma decay:

Tc-99m -> Tc-99 + gamma

In this process, Tc-99m transitions from a higher energy state to a lower energy state and releases a gamma ray.

d. C-111 electron capture:

C-111 + e- -> B-11 + gamma

In this process, C-111 captures an electron and transforms into B-11. At the same time, a gamma ray is also emitted.

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The complete question:

Write a balanced nuclear equation for each decay process indicated.

a. The isotope Th-234 decays by an alpha emission.

b. The isotope Fe-59 decays by a beta emission.

c. The isotope Tc-99 decays by a gamma emission.

d. The isotope C-1ll decays by a electron capture.

The Molarity of the H₂SO4 solution is 0.025 M.

The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H₂SO4) is:

2 NaOH + H₂SO4 → 2 H₂O + Na₂SO4

From the equation, we can see that 2 moles of NaOH react with 1 mole of H₂SO4 to produce 2 moles of water and 1 mole of Na₂SO4.

To find the molarity of the H₂SO4, we can use the formula:

Molarity = moles of solute / volume of solution in liters

First, we need to determine the number of moles of NaOH used in the reaction.

moles of NaOH = M x V = 0.15 M x 0.01 L = 0.0015 moles

Since 2 moles of NaOH react with 1 mole of H₂SO4, the number of moles of H₂SO4 used in the reaction is:

moles of H₂SO4 = 0.0015 moles / 2 = 0.00075 moles

The volume of the H₂SO4 solution is 30.0 mL or 0.03 L. Using the molarity formula, we can calculate the molarity of H₂SO4:

Molarity = moles of solute / volume of solution in liters = 0.00075 moles / 0.03 L = 0.025 M

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If the spots are made too small when preparing a TLC plate for development, it can affect the accuracy and reliability of the results obtained from the TLC experiment.

When the spots are too small, it can be difficult to accurately apply the sample to the TLC plate. This can lead to uneven distribution of the sample and inaccurate results. In addition, small spots may not provide enough material for detection by the TLC system.

It can be challenging to identify and distinguish them from one another. This can lead to difficulties in analyzing the results and interpreting the data obtained from the experiment. Therefore, it is essential to ensure that the spots are of appropriate size and are applied uniformly to the TLC plate to obtain accurate and reliable results from the TLC experiment.
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The main answer to your question is that during nitrogen fixation, the chemical transformation that occurs is the reduction of N2 (nitrogen gas) to NH3 (ammonia).

This is accomplished through the use of nitrogenase enzymes by certain bacteria and archaea. The explanation for this process involves the breaking of the triple bond between the two nitrogen atoms in N2, which requires a large input of energy.

Once the bond is broken, the nitrogen atoms can be combined with hydrogen atoms to form NH3. This process is essential for the creation of biologically available nitrogen that can be used by plants and other organisms.

In summary, nitrogen fixation involves the reduction of N2 to NH3 through the use of nitrogenase enzymes, and is a crucial step in the global nitrogen cycle.

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The number of moles of calcium chloride 6-hydrate produced is 7.20 moles. Please note that this calculation assumes excess calcium carbonate and water, meaning that all the hydrochloric acid is consumed in the reaction.

The balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl) to produce calcium chloride 6-hydrate (CaCl2H12O6) and carbon dioxide (CO2) is:

CaCO3 (s) + 2HCl (aq) + 5H2O (l) → CaCl2H12O6 (s) + CO2 (g)

According to the equation, 1 mole of CaCO3 reacts with 2 moles of HCl to produce 1 mole of CaCl2H12O6. Given that 7.20 moles of HCl are added, we can conclude that 7.20 moles of CaCl2H12O6 will be produced. Therefore, the number of moles of calcium chloride 6-hydrate produced is 7.20 moles. Please note that this calculation assumes excess calcium carbonate and water, meaning that all the hydrochloric acid is consumed in the reaction.

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Gravity is the attractive force between all objects. The correct answer is A.

Gravity is the fundamental force of attraction that exists between all objects with mass. It is responsible for the formation and behavior of planets, stars, galaxies, and the entire universe. The force of gravity depends on the masses of the objects and the distance between them, and it acts in all directions. Gravity is what keeps us grounded on Earth, and it is also responsible for the motion of objects in space. The laws of gravity were first described by Sir Isaac Newton in the 17th century and later refined by Albert Einstein's theory of general relativity in the 20th century. The correct answer is option A.

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In a Beryllium oxide (BeO) crystal structure with an HCP arrangement of O²⁻ ions, the Be²⁺ ions will occupy the tetrahedral interstitial sites.

In this crystal structure, the O²⁻ ions form a hexagonal close-packed (HCP) arrangement. The available interstitial sites in an HCP lattice are tetrahedral and octahedral. To determine which site the Be²⁺ ions will occupy, we can consider the size of the ions. The ionic radii of Be²⁺ and O²⁻ ions are, respectively, 0.035 nm and 0.140 nm. Since the Be²⁺ ions are smaller, they can easily fit into the smaller tetrahedral interstitial sites.

In a Beryllium oxide (BeO) crystal structure with an HCP arrangement of O²⁻ ions, the Be²⁺ ions will occupy the tetrahedral interstitial sites due to their smaller ionic radii.

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Answer:

Type1 Decay Mode Half-Life

Tc-99m γ decay 8.01 hours

I-131 β decay 8.02 days

Tl-201 electron capture 73 hours

When the mole fraction of glucose in (c6h12o6) and 162 g of water is calculated, the solution is 0.10.

First, we need to calculate the total number of moles of the glucose and water.

Moles of glucose = Mass of glucose / Molar mass of glucose

Molar mass of glucose (C6H12O6) = 6*(12.01) + 12*(1.01) + 6*(16.00) = 180.16 g/mol

Moles of glucose = 180 g / 180.16 g/mol = 1 mol

Moles of water = Mass of water / Molar mass of water

Molar mass of water (H2O) = 2*(1.01) + 16.00 = 18.02 g/mol

Moles of water = 162 g / 18.02 g/mol = 9 mol

The total number of moles in the solution is the sum of the moles of glucose and water:

Total moles = 1 mol + 9 mol = 10 mol

The mole fraction of glucose can then be calculated as the ratio of the moles of glucose to the total number of moles:

Mole fraction of glucose = Moles of glucose / Total moles

Mole fraction of glucose = 1 mol / 10 mol = 0.10

The mole fraction of glucose in the given solution, containing 180 g of glucose (C6H12O6) and 162 g of water, is 0.10.

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Polar molecules are more likely to be found in sugar than paraffin. This is due to the fact that sugar molecules are made up of polar molecules like carbon, hydrogen, and oxygen.

Due to the existence of lone pairs of electrons, the oxygen atoms in sugar molecules are particularly polar because they have a partial negative charge. Due to their lone electron, hydrogen atoms also have a little positive charge. These interactions between these polar molecules result in the formation of hydrogen bonds, which give sugar molecules their shape and structure.

Contrarily, the only elements found in paraffin molecules are carbon and hydrogen, both of which are non-polar molecules. As a result, the molecules are unable to interact with one another and create hydrogen bonds. The outcome is Due to their inability to take on the same forms and structures as sugar molecules, paraffin molecules are unlikely to include polar molecules.

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The scientist needs to use 1.23 L of the 0.930 m solution for the experiment.

moles = concentration x volume

0.930 mol/L = 0.930 M

moles = concentration x volume

1 mol = 0.930 M x volume

volume = 1 mol / 0.930 M

volume = 1.075 L

So 1 L of the 0.930 m solution contains 1.075 mol of potassium chlorite.

To find the volume of the 0.930 m solution that contains 1.32 mol of potassium chlorite, we can use the following proportion:

1.075 mol / 1 L = 1.32 mol / x

where x is the volume of the solution we need to use.

Solving for x, we get:

x = 1.32 mol / (1.075 mol / 1 L) = 1.23 L

A solution is a homogeneous mixture of two or more substances. The substance that is present in the largest quantity is called the solvent, while the substance that is present in smaller quantities is called the solute. Solutions can exist in all three states of matter, namely solid, liquid, and gas.

The properties of a solution depend on the concentration of the solute in the solvent. The concentration of a solution can be expressed in several ways, such as molarity, molality, mole fraction, and weight percent. Solutions play a crucial role in many chemical reactions, as they allow the reactants to come into close contact with each other, increasing the likelihood of a reaction taking place. Solutions are also used in many industries, such as pharmaceuticals, food and beverage, and chemical manufacturing.

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The correct statement regarding the delta of the option presented in Exhibit 3 is: "Delta is -0.4024 for the first step and is different for the second step." Option B is Correct.

This indicates that the delta value changes over time and is not constant for both steps. Hess' law states that when the primary reaction is conducted at the same temperature, all intermediate reactions that can be divided into the main reaction have standard enthalpies that add up to the same value.

The enthalpy change for a reaction is independent of the number of possible ways a product might be created if the starting and finishing conditions are the same. A reaction's negative enthalpy change denotes an exothermic process, whereas a reaction's positive enthalpy change denotes an endothermic activity.

Because the energy required for each stage of the process is the same, a reaction that occurs in just one step will have the same enthalpy as a reaction that occurs in several phases. The enthalpy of a reaction does not rely on the reaction route, according to Hess's law.

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Which of the following is true regarding the delta of the option presented in exhibit 3?

A. delta is -0.7357 for the first step and it changes over time.

B. delta is -0.4024 for the first step and is different for the second step

C. delta may be 1.00 in the second step delta will be the same for both steps

When given polyatomic ions such as nitrate, phosphate, sulfate, acetate, ammonium, chromate, carbonate, dichromate, and permanganate with a charge, valency can be used to predict the formulae for the compounds formed. The formula for ammonium and sulfite is NH₄SO₃. The correct option is D. NH₄SO₃.

Valency is the measure of an atom's combining power with other atoms when it comes to forming chemical compounds or molecules. A compound's valency is determined by the number of electrons required by an atom to reach the noble gas electronic configuration. Therefore, the valency of an element is either positive or negative. The valency of polyatomic ions is the charge present on the ion.

The formula of a compound formed between a metal and a polyatomic ion is determined by the valency of the polyatomic ion and the valency of the metal. When forming a compound between a metal and a polyatomic ion, it is vital to remember that the net charge of the compound should always be zero. For example, NH₄⁺¹ and SO₃⁻² combine to form NH₄SO₃. Hence, D is the correct option.

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The following chemical compounds are;

V₂O₅ Vanadium (V) oxide,

SF₆ Sulfur hexafluoride

HClO₂ Chlorous acid

(NH₄)₂SO₄·6H₂O Ammonium sulfate hexahydrate

Chlorous acid is a weak and unstable acid. This acid exists mainly  in aqueous solutions. They are used in disinfectants, and bleaching products. They are good sanitizers and helps to reduce bacteria.

Apart from being in products we commonly use in our house, Chlorous acid is also used in the paper and textile industries for bleaching purposes.

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The formal charges on each of the nitrogen atoms in the N3- ion shown are:

- Middle nitrogen atom: 0

- End nitrogen atoms: -1 (x²)

To calculate the formal charges on each of the nitrogen atoms in the N3- ion shown, we need to first determine the valence electrons of nitrogen. Nitrogen has five valence electrons, so in the N3- ion, there are a total of 15 valence electrons (5 valence electrons per nitrogen atom).

To calculate the formal charge, we need to subtract the number of non-bonded electrons (lone pairs) and half of the bonded electrons from the valence electrons of each nitrogen atom.

For the middle nitrogen atom, it has four non-bonded electrons and two bonded electrons, giving it a formal charge of 0.

For the two end nitrogen atoms, they each have two non-bonded electrons and four bonded electrons, giving them a formal charge of -1.

Overall, the N3- ion has a charge of -3, which is the sum of the formal charges on each nitrogen atom.

In summary, the formal charges on each of the nitrogen atoms in the N3- ion shown are:

- Middle nitrogen atom: 0

- End nitrogen atoms: -1 (x²)

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The charge on the central metal ion (Fe) in Ca3[Fe(CN)6]2 is 0. The charge of the central metal ion can be calculated using the charges of the other ions present in the compound and the overall charge of the compound.

In Ca3[Fe(CN)6]2, the overall charge of the compound is 0 since it is neutral. The charge of the cyanide ion (CN-) is -1 and there are six of them, so the total charge contributed by the cyanide ions is -6. The charge of the iron ion (Fe) can be calculated using the fact that the compound has a 2- charge overall:

Charge on Ca3[Fe(CN)6]2 = 3(+2) + 2x(charge on Fe) + 6(-1) = 0

Simplifying this expression, we get:

6 + 2x(charge on Fe) - 6 = 0

2x(charge on Fe) = 0

Charge on Fe = 0/2 = 0

Therefore, the charge on the central metal ion (Fe) in Ca3[Fe(CN)6]2 is 0.

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To determine the vapor pressure of acetone at 25 °C using the given information, we can utilize the Clausius-Clapeyron equation: the vapor pressure of acetone at 25 °C is approximately 1.0114 atm.

ln(P₂/P₁) = (ΔHvap/R) × (1/T₁ - 1/T₂)

First, we need to convert the heat of vaporization from kilojoules to joules:

ΔHvap = 31.3 kJ/mol = 31.3 × 1000 J/mol

Now, we can plug in the values into the equation and solve for P₂:

ln(P₂/1 atm) = (31.3 × 1000 J/mol / (8.314 J/(mol*K))) * (1/329 K - 1/298 K)

ln(P₂/1 atm) = 3.755 × (0.0030 K^-1)

Taking the exponential of both sides to eliminate the natural logarithm:

P₂/1 atm = e^(3.755 * 0.0030 K^-1)

Finally, solving for P₂:

P₂ = 1 atm * e^(3.755 * 0.0030 K^-1)

Calculating P₂:

P₂ ≈ 1 atm * e^(0.0113 K^-1)

P₂ ≈ 1 atm * 1.0114

Therefore, the vapor pressure of acetone at 25 °C is approximately 1.0114 atm.

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When petroleum is distilled to separate the components by boiling point, the component with the highest boiling point is called residuum.

When petroleum is distilled to separate its components by boiling point, a process called fractional distillation is used.

In this process, the crude oil is heated, and different hydrocarbon components are separated based on their boiling points.

The component with the highest boiling point is called the residuum, also known as residual fuel oil or heavy fuel oil.

Residuum is the heaviest and most viscous component obtained from the fractional distillation of petroleum. It is commonly used in industrial applications, such as marine engines and power plants, due to its high energy content and low cost.

Keep in mind that the residuum may require further processing or blending with lighter fuels to meet specific requirements for its intended use.

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The molar enthalpy of vaporization for the substance, given that 3.21 mole of the substance absorbs 28.4 KJ of heat energy is 8.85 KJ/mol

From the question given above, the following data were obtained:

Heat absorbed is related to heat of vaporization according to the following formula:

Q = n × ΔHv

Inputting the given parameters from the question, we can obtain the molar enthalpy of vaporization of substance as follow:

Q = n × ΔHv

28.4 = 3.21 × ΔHv

Divide both sides by 3.21

ΔHv = 28.4 / 3.21

ΔHv = 8.85 KJ/mol

Thus, we can conclude that the molar enthalpy of vaporization of substance is 8.85 KJ/mol

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Answer:

a. An element contains only one type of particle

Explanation:

When 68.00 J of energy is added to a sample of gallium initially at 25.0 °C, causing the temperature to rise to 38.0 °C, the volume of the gallium sample is approximately 0.84 cm³. it can be calculated using the specific heat capacity of gallium and the equation relating heat, specific heat capacity, mass, and temperature change.

To calculate the volume of the sample, we need to use the equation q = mcΔT, where q represents the heat energy added, m is the mass of the sample, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

First, we need to know the specific heat capacity of gallium. Assuming the specific heat capacity of gallium is 0.371 J/g°C, we can proceed with the calculation. Given that the temperature change (ΔT) is (38.0 °C - 25.0 °C) = 13.0 °C, and the energy added (q) is 68.00 J, we can rearrange the equation q = mcΔT to solve for the mass (m).

m = q / (cΔT)

= 68.00 J / (0.371 J/g°C * 13.0 °C)

= 4.98 g

Assuming the density of gallium is approximately 5.91 g/cm³, we can calculate the volume (V) of the sample.

V = m / density

= 4.98 g / 5.91 g/cm³

≈ 0.84 cm³

Therefore, the volume of the gallium sample is approximately 0.84 cm³.

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The free energy change for the reaction at 298 K is +81.8 kJ/mol.

We can calculate the free energy change for this reaction using the following equation:

ΔG = ΔH - TΔS

Where ΔH is the enthalpy change, ΔS is the entropy change, T is the temperature in Kelvin, and ΔG is the free energy change.

For the reaction NaHCO₃(s) ⇌ NaOH(s) + CO₂(g), the enthalpy change and entropy change can be determined from the balanced chemical equation:

NaHCO₃(s) → NaOH(s) + CO₂(g)

ΔH = ΔH(products) - ΔH(reactants) = [ΔHf°(NaOH) + ΔHf°(CO₂)] - ΔHf°(NaHCO₃)

ΔH = [( -425.9 kJ/mol + (-393.5 kJ/mol))] - (-950.7 kJ/mol) = +131.3 kJ/mol

ΔS = ΔS(products) - ΔS(reactants) = [ΔSf°(NaOH) + ΔSf°(CO₂)] - ΔSf°(NaHCO₃)

ΔS = [(+51.5 J/(mol·K) + 213.7 J/(mol·K))] - (+100.4 J/(mol·K)) = +165.8 J/(mol·K)

Substituting these values into the equation for ΔG gives:

ΔG = ΔH - TΔS = +131.3 kJ/mol - (298 K)(0.1658 kJ/(mol·K))

ΔG = +131.3 kJ/mol - 49.5 kJ/mol = +81.8 kJ/mol

Therefore, the free energy change for the reaction at 298 K is +81.8 kJ/mol.

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Count the number of valence electrons used in the bonds and lone pairs of each atom. In NBrl2, each Br atom has 8 valence electrons (6 lone pairs and 1 bond pair) and the N atom has 8 valence electrons (3 lone pairs and 1 bond pair). Therefore, all atoms have a complete octet.

To draw the Lewis dot structure for NBrl2, follow these steps:

Step 1: Determine the total number of valence electrons.

N (nitrogen) has 5 valence electrons, Br (bromine) has 7 valence electrons each, so the total number of valence electrons in NBrl2 is:

5 + 2(7) = 19 valence electrons

Step 2: Determine the central atom.

Nitrogen (N) is the least electronegative element and can be the central atom in this molecule.

Step 3: Connect the outer atoms to the central atom.

Each Br atom will form a single bond with the N atom.

Step 4: Place the remaining electrons around the atoms.

Distribute the remaining valence electrons as lone pairs on each Br atom.

Step 5: Check if all atoms have a complete octet.

The Lewis dot structure for NBrl2 is:

Br

|

Br-N-Br

|

Br

Each Br atom is bonded to the central N atom with a single bond, and each Br atom has six lone pairs around it. The N atom has three lone pairs and one bond pair with each Br atom.

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A polymerization acrylic resin is formed through a chemical reaction known as polymerization, which occurs when a liquid monomer, as methyl methacrylate, is mixed powder of small polymer beads, polymethyl methacrylate.

A polymerization acrylic resin is formed when a liquid monomer is mixed with a powder of small polymer beads. This process creates a strong and durable material commonly used in various applications, such as dental prosthetics and acrylic nails. The polymerization reaction occurs when the monomer and polymer beads chemically bond, resulting in a solid acrylic resin with enhanced properties.

This mixture, when activated by a catalyst or initiator, undergoes a chain reaction that links the individual monomer molecules together to form a large, three-dimensional network of polymers, resulting in a solid and durable material. This polymerized acrylic resin can be used in various applications, including dentistry, orthopedics, and as a coating or adhesive.

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The Complete question is

A polymerization acrylic resin is formed when a liquid monomer is mixed with a powder of small polymer beads. What is the polymerise acrylic resin is mixed with powder?

Option D depicts the initial atoms when sodium and oxygen form an ionic compound. It shows two atoms of sodium, each having one valence electron, and one atom of oxygen, having six valence electrons.

The formation of an ionic compound between sodium and oxygen involves the transfer of electrons from sodium to oxygen, resulting in the formation of oppositely charged ions. In the initial state, sodium (Na) has one valence electron while oxygen (O) has six valence electrons. Sodium will lose one electron to become a positively charged ion (Na+), and oxygen will gain two electrons to become a negatively charged ion (O2-). Option D depicts the initial atoms when sodium and oxygen form an ionic compound. It shows two atoms of sodium, each having one valence electron, and one atom of oxygen, having six valence electrons. This arrangement represents the transfer of electrons from sodium to oxygen, resulting in the formation of Na+ and O2- ions. Options A, B, and C do not depict the correct arrangement of atoms in the initial state before the formation of the ionic compound between sodium and oxygen.

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