order the elements s, cl, and f in terms of increasing atomic radii.

Answer: B

Explanation:

(s) means the compound is in its solid form, and (aq) means that the compound is dissolved in a solvent (the solvent is often water).

40.5 g of N2 occupies 32.7 L at STP. The correct answer is (C) 32.4 L.

We can use the ideal gas law to solve this problem:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

At STP, the pressure is 1 atm and the temperature is 273.15 K. The ideal gas constant is 0.08206 L·atm/(mol·K). We can calculate the number of moles of N2 using its molar mass, which is 28.02 g/mol:

n(N2) = 40.5 g / 28.02 g/mol = 1.446 mol

Substituting these values into the ideal gas law equation:

V = nRT / P = (1.446 mol)(0.08206 L·atm/(mol·K))(273.15 K) / 1 atm = 32.7 L

Therefore, 40.5 g of N2 occupies 32.7 L at STP.

The correct answer is (C) 32.4 L.

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The process in which a nucleus spontaneously breaks down by emitting radiation is known as radioactive decay.

During this process, the unstable nucleus emits radiation in the form of alpha particles, beta particles, or gamma rays until it becomes stable. Radioactive decay is a random process and occurs at a specific rate, known as the half-life, which varies for each radioactive substance.

This process is used in many applications, including carbon dating and medical imaging. It is important to note that fusion and transformation are not related to radioactive decay, as they refer to the process of combining two nuclei to form a heavier nucleus and the process of changing one element into another, respectively. Answering more than 100 words, we can say that radioactive decay is a fundamental process that helps us understand the behavior of atoms and how they change over time.

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Solving equations with variables on both sides requires a strategic approach to ensure that the correct steps are taken to isolate the variable on one side of the equation.

One key strategy is to simplify the equation by combining like terms on each side. This can be done by adding or subtracting terms as necessary, while ensuring that the equation remains balanced.

Another strategy is to move all the variable terms to one side of the equation and all the constant terms to the other side. This can be done by adding or subtracting terms to both sides as necessary.

It's important to remember that when adding or subtracting terms, the operation must be applied to both sides of the equation to keep it balanced.

Once the variable terms are on one side of the equation and the constant terms are on the other, the equation can be solved by isolating the variable and determining its value.

It's important to check the solution by substituting the value back into the original equation and verifying that both sides are equal.

By following a strategic approach, equations with variables on both sides can be successfully solved.

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The best evidence that a chemical reaction occurred during the ignition of fireworks is the release of light, heat, and new colored substances.

When fireworks are ignited, several chemical reactions take place that result in the release of energy in the form of light, heat, and the formation of new substances. The different colored powders mixed together in the tube contain metal salts, which produce specific colors when heated.

As the heat energy causes these metal salts to react, they release energy in the form of light and heat, producing the bright and colorful display we associate with fireworks. Additionally, the formation of new substances, such as gases and solid particles, is a key indicator that a chemical reaction has taken place during the ignition of the fireworks.

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Fe3O4 contains 72.4% iron by mass.The formula of the oxide of iron is Fe3O4, which means it contains 3 atoms of iron and 4 atoms of oxygen.

To calculate the mass percent of iron in Fe3O4, we need to first determine the molar mass of Fe3O4:

Molar mass of Fe3O4 = (3 x molar mass of Fe) + (4 x molar mass of O)

= (3 x 55.845 g/mol) + (4 x 15.9994 g/mol)

= 231.5332 g/mol

Next, we need to determine the mass of iron in one mole of Fe3O4:

Mass of iron in one mole of Fe3O4 = 3 x molar mass of Fe

= 3 x 55.845 g/mol

= 167.535 g/mol

Finally, we can calculate the mass percent of iron in Fe3O4:

Mass percent of iron = (mass of iron ÷ total mass of Fe3O4) x 100%

= (167.535 g/mol ÷ 231.5332 g/mol) x 100%

= 72.4%

Therefore, Fe3O4 contains 72.4% iron by mass.

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Answer:

Out of the following properties, the least useful for identifying a sample of calcite would be its color. Calcite is a mineral that can be found in a variety of colors, including white, gray, black, brown, red, orange, yellow, green, and blue. This means that its color is not a reliable indicator of its identity.

Other properties that can be used to identify calcite include its hardness (3 on the Mohs scale), specific gravity (2.71), and cleavage (rhombohedral). Calcite is also a birefringent mineral, which means that it splits light into two rays of different polarization when viewed through a polarizing filter.

Explanation:

To convert -196°C to Fahrenheit, you can use the formula: F = (C x 1.8) + 32. Plugging in -196 for C, you get:

F = (-196 x 1.8) + 32
F = -320.8

Therefore, nitrogen boils at -320.8°F. It's important to note that Fahrenheit and Celsius scales have different zero points and different degrees of size. This means that -196°C is a much lower temperature than -320.8°F, as the Fahrenheit scale has a smaller degree of size. This conversion is useful for comparing temperatures across different measurement systems, and for understanding the temperature ranges of different substances and materials.

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The balanced nuclear equation for the alpha decay of polonium-218 to give lead-214 is,

[tex]^218Po[/tex] -> [tex]^214Pb[/tex] + [tex]^4He[/tex]

In the given nuclear equation, the nuclide polonium-218 (symbolized as [tex]^218Po[/tex]) undergoes alpha emission, which means it releases an alpha particle. An alpha particle consists of two protons and two neutrons, which is symbolized as [tex]^4He[/tex].

After the alpha emission, the resulting nuclide is lead-214 (symbolized as [tex]^214Pb[/tex]). Lead-214 is formed by subtracting the alpha particle ([tex]^4He[/tex]) from the original polonium-218 nucleus.

Overall, the nuclear equation represents the radioactive decay process where polonium-218 emits an alpha particle, resulting in the formation of lead-214.

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When, a compound having an enthalpy of vaporization of 28.4 kj/mol. at 274 k it has a vapor pressure of 122 mm hg. Then, the vapor pressure of the compound at its normal boiling point is 1.037 x 10⁻³ mm Hg.

To solve this problem, we can use the Clausius-Clapeyron equation;

ln(P₂/P₁) = (ΔHvap/R) x (1/T₁ - 1/T₂)

where;

P₁ = vapor pressure at temperature T₁

P₂ = vapor pressure at temperature T₂

ΔHvap = enthalpy of vaporization

R = gas constant (8.314 J/(mol·K))

ln = natural logarithm

We are given P₁ = 122 mm Hg at T₁ = 274 K, and we want to find P₂ at the normal boiling point, which we can assume is the boiling point at 1 atm of pressure. At this pressure, the boiling point is equal to the normal boiling point.

We will convert the pressure from mm Hg to atm by dividing by 760;

P₁ = 122/760 = 0.1605 atm

We can assume that the enthalpy of vaporization is constant over the small temperature range between T₁ and the normal boiling point, so we can use the given value of ΔHvap.

We can also assume that the boiling point of the liquid increases linearly with pressure, so we can use the boiling point at 1 atm as an approximation for the boiling point at P₂. We can find the boiling point at 1 atm from a table or calculator;

Boiling point of compound = 337 K

Now we put all the values into the Clausius-Clapeyron equation and solve for P;

ln(P₂/0.1605) = (28.4 x 10³ J/mol / (8.314 J/(mol·K))) x (1/274 K - 1/337 K)

ln(P₂/0.1605) = -9.36

P₂/0.1605 = [tex]e^{(-9.36)}[/tex]

P₂ = 1.37 x 10⁻⁵ atm

Finally, we can convert the pressure back to mm Hg;

P₂ = 1.037 x 10⁻³ mm Hg

Therefore, the vapor pressure of the compound at its normal boiling point is approximately 1.037 x 10⁻³ mm Hg.

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The covalent bond between an amino acid and its tRNA is a specific type of covalent bond called an ester linkage.

The covalent bond between an amino acid and its tRNA is formed through a specific type of covalent bond called an ester linkage. This linkage is formed between the carboxyl group of the amino acid and the 3' hydroxyl group of the tRNA.

The process of forming this bond is called aminoacylation, and it is catalyzed by an enzyme called aminoacyl-tRNA synthetase.

Each amino acid has a specific aminoacyl-tRNA synthetase enzyme that catalyzes the formation of the ester bond between the amino acid and the tRNA molecule with the corresponding anticodon.

Once the amino acid is attached to the tRNA, the resulting aminoacyl-tRNA can then be used in protein synthesis, where the tRNA delivers the amino acid to the ribosome and the amino acid is incorporated into the growing polypeptide chain.

Therefore, the correct answer is an ester linkage, which is formed between the carboxyl group of the amino acid and the 3' hydroxyl group of the tRNA during the process of aminoacylation.

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The concentration of [tex]2H_3PO_4[/tex] remaining after 5.20 seconds is approximately 0.254 M, the correct option is A.

To determine how much [tex]2H_3PO_4[/tex] is left after 5.20 seconds, we can use the integrated rate equation for a second-order reaction:

1/[A]t - 1/[A]0 = kt,

where;

[A]t = concentration of [tex]2H_3PO_4[/tex] at time t

[tex][A]_0[/tex] = initial concentration

k = rate constant

t = time elapsed.

Substituting the given values:

1/[A]t - 1/1.02 = (0.566 [tex]M^{-1}s^{-1}[/tex]) × 5.20 s,

Simplifying the equation:

1/[A]t = 1/1.02 + (0.566 [tex]M^{-1}s^{-1}[/tex]) × 5.20 s,

Calculating the right side:

1/[A]t = 0.9804 [tex]M^{-1}[/tex] + 2.9452 [tex]M^{-1}[/tex],

1/[A]t = 3.9256 [tex]M^{-1}[/tex].

Taking the reciprocal of both sides:

[A]t = 1 / (3.9256 [tex]M^{-1[/tex]),

[A]t = 0.254 M.

Thus, the correct option is A.

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The complete question is:

The following reaction follows second-order kinetics with a rate constant of 0.566 [tex]M^{-1}s^{-1}[/tex]. Suppose a vessel initially contains [tex]2H_3PO_4[/tex] at a concentration of 1.02 m. How much is left 5.20 seconds later?

[tex]2H_3PO_4[/tex] → [tex]P_2O_5+ 3H_2O[/tex]

(group of answer choices)

A. 0.25 M

B. 0.51 M

C. 0.56 M

D. 0.91 M

The pH of the acetate buffer solution, which is a mixture of equal volumes of 0.33 M acetic acid and 0.15 M sodium acetate, is approximately 4.42.

To calculate the pH of an acetate buffer solution, we can use the Henderson-Hasselbalch equation;

pH = pKa + log([A⁻]/[HA])

where; pH is the pH of the buffer solution

pKa will be the acid dissociation constant of the weak acid (acetic acid in this case)

[A⁻] will be the concentration of the conjugate base (acetate ion)

[HA] will be the concentration of the weak acid (acetic acid)

Given; Concentration of acetic acid (HA) = 0.33 M

Concentration of sodium acetate (A⁻) = 0.15 M

We need to determine the pKa of acetic acid to proceed. The pKa value of acetic acid is 4.76.

Now, let's substitute the given values into the Henderson-Hasselbalch equation;

pH = 4.76 + log(0.15/0.33)

pH = 4.76 + log(0.4545)

To calculate the pH, we can evaluate the logarithm;

pH = 4.76 + (-0.343)

pH ≈ 4.42

Therefore, the pH of the acetate buffer solution will be 4.42.

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The pH of a 0.30 M solution of a weak base is 10.66. We can use this information to calculate the pOH of the solution, which is the negative logarithm of the hydroxide ion concentration. The pOH can then be used to calculate the pKb, which is the negative logarithm of the base dissociation constant, Kb. By taking the antilog of the pKb value, we can determine the value of Kb for the weak base.

To calculate the pOH of the solution, we use the equation: pOH = 14 - pH. Therefore, pOH = 14 - 10.66 = 3.34. Using the relationship between pOH and Kb, we have pKb = 14 - pOH = 10.66. Taking the antilog of pKb, we have Kb = 2.5 x 10^-4.

Therefore, the Kb of the weak base in the 0.30 M solution is 2.5 x 10^-4. This value indicates the strength of the base as a proton acceptor in solution. A higher value of Kb indicates a stronger base, which means it is more likely to accept a proton from water and generate hydroxide ions. The calculation of Kb is an essential step in understanding the properties of weak bases and their behavior in solution.

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Explanation:

Plugging into the following equations will give you the answer (the answer is the attached image):

[tex]pH+pOH=14[/tex]

[tex]pH=-log_{10}([H^+])[/tex]

[tex]pOH=-log_{10}([OH^-])[/tex]

[tex][H^+][OH^-]=10^{-14}[/tex]

[tex][H^+]=10^{-pH}[/tex]

[tex][OH^-]=10^{-pOH}[/tex]

Answer:

answer is a

Explanation:

I just did that question

The volume of a gas originally at standard temperature and pressure was recorded as 488.8 mL, the gas would occupy a volume of 5.97 mL at a pressure of 100.0 atm and temperature of -245.0 °C.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

(P₁V₁)/T₁ = (P₂V₂)/T₂

where P₁, V₁, and T₁ are the initial pressure, volume, and temperature, respectively, and P₂, V₂, and T₂ are the final pressure, volume, and temperature, respectively.

We can first convert the initial temperature to Kelvin:

T₁ = 273.15 K (since it is at standard temperature)

Next, we can convert the final temperature to Kelvin:

T₂ = (-245.0 °C + 273.15) K = 28.15 K

We can then plug in the values and solve for V₂:

(1 atm x 488.8 mL) / 273.15 K = (100.0 atm x V₂) / 28.15 K

V₂ = (1 atm x 488.8 mL x 28.15 K) / (100.0 atm x 273.15 K) = 5.97 mL

Therefore, the gas would occupy a volume of 5.97 mL at a pressure of 100.0 atm and temperature of -245.0 °C.

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When heated, potassium permanganate (KMⁿO4) decomposes into:

Potassium permanganate is a strong oxidizing agent that decomposes when heated and releases oxygen gas. The reaction is observed by placing small amount of potassium permanganate in a test tube and gently heating it over a Bunsen burner.

As the temperature increases, the purple color of the potassium permanganate fades and brown-black manganese dioxide is formed. Oxygen gas can be seen bubbling out of the test tube and can be confirmed with a glowing splint test.

After the reaction is complete, the residue left in the test tube is potassium manganate which can be identified by its green color.

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Acid deposition forms from sulfur dioxide when it combines with oxygen and water in the atmosphere, producing sulfuric acid.

Sulfur dioxide is a gas that is emitted from the burning of fossil fuels, particularly coal and oil. When it is released into the atmosphere, it can react with oxygen and water to form sulfuric acid. This chemical reaction occurs naturally in the atmosphere, but it can be accelerated by human activities such as industrial processes and transportation.

Once formed, the sulfuric acid can be carried by wind and deposited on the ground as acid rain or snow. Acid deposition can have significant negative impacts on the environment, including harming plants, animals, and aquatic life. It can also contribute to the deterioration of buildings and monuments made of stone or metal. Therefore, it is important to reduce sulfur dioxide emissions through policies and technologies that promote cleaner energy sources and reduce pollution.

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The molecules that can form hydrogen bonds with another identical molecule are HOOH (hydrogen peroxide), CH₃CH₂NH₂(ethylamine), and CH₃CH₂F (ethyl fluoride).                                                                  

Hydrogen bonding occurs between a hydrogen atom bonded to an electronegative atom (such as oxygen, nitrogen, or fluorine) and another electronegative atom in a different molecule. Based on this criterion, the molecules that can form hydrogen bonds with another identical molecule are:

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A mixed solvent of EtOH and water is used in recrystallization to improve solubility and selectivity, resulting in better product purity and yield.

A mixed solvent system of EtOH and water is preferred over a single solvent system of water or ethanol in recrystallization because it can improve solubility and selectivity, leading to better product purity and yield. EtOH is a good solvent for organic compounds, while water is a good solvent for polar compounds.

When these two solvents are mixed, they can dissolve a wider range of compounds than either solvent alone. Additionally, the ratio of EtOH to water can be adjusted to fine-tune the solubility and selectivity of the mixed solvent system. This allows for better control over the recrystallization process, resulting in higher-quality products.

In summary, the use of a mixed solvent system in recrystallization can enhance the efficiency and effectiveness of the process, ultimately leading to improved product quality.

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we need 56.11 grams of KOH to make 1.00 liter of a 1 molar solution of potassium hydroxide.

To make a 1 molar solution of potassium hydroxide (KOH), we need to dissolve enough KOH in water to make a solution where the concentration of KOH is 1 mole per liter of solution. The molarity (M) is defined as the number of moles of solute (KOH in this case) per liter of solution, so we can use this equation:

M = moles of solute / liters of solution

We can rearrange this equation to solve for the moles of solute:

moles of solute = M x liters of solution

Since we want to make 1.00 liter of a 1 molar solution of KOH, we can substitute those values into the equation:

moles of solute = 1.00 mol/L x 1.00 L = 1.00 mol

So we need 1.00 mole of KOH to make 1.00 liter of a 1 molar solution. To find the mass of KOH needed, we need to use its molar mass:

KOH molar mass = 39.10 g/mol (for K) + 16.00 g/mol (for O) + 1.01 g/mol (for H) = 56.11 g/mol

So, the mass of KOH needed to make 1.00 liter of a 1 molar solution is:

mass of KOH = moles of KOH x molar mass of KOH

= 1.00 mol x 56.11 g/mol

= 56.11 g

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The false statement about disulfide bond formation is (d) disulfide bonds form spontaneously within the ER because the lumen of the ER is oxidizing. Disulfide bonds do form within the ER, but not spontaneously.

Instead, they are formed by the action of enzymes called protein disulfide isomerases (PDIs). PDIs catalyze the oxidation of cysteine residues to form disulfide bonds. Disulfide bonding (b) stabilizes the structure of proteins, and (a) disulfide bonds do not form under reducing environments. Additionally, disulfide bonds do not (c) grow from both ends, the growth rate is faster at the plus ends. Instead, they are formed between two cysteine residues on the same polypeptide chain or between different polypeptide chains.

The false statement about disulfide bond formation among the given choices is (a) disulfide bonds do not form under reducing environments grow from both ends, the growth rate is faster at the plus ends. This statement is unrelated and incorrect. In reality, disulfide bonds (b) form by oxidation of cysteine pairs, (c) stabilize protein structures, and (d) form spontaneously within the ER due to its oxidizing environment. Disulfide bonds play a vital role in maintaining the proper folding and stability of proteins, especially those secreted or located in extracellular environments.

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The reaction mixture needs to be filtered while warm after reflux because the product may solidify and become difficult to filter if left to cool.


During reflux, the reaction mixture is heated to boiling and maintained at that temperature for a period of time. This process can cause the product to dissolve in the solvent and react with other components in the mixture. After reflux, the reaction mixture needs to be filtered to separate the solid product from the solvent and other components.  

Filtering the mixture while it is still warm prevents the product from solidifying and becoming difficult to filter. If left to cool, the product may form crystals that clog the filter and slow down or even stop the filtration process. By filtering the mixture while it is still warm, the product remains in a liquid state and is easier to separate from the rest of the mixture. This ensures that the product is obtained in the desired form and purity.

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There are 6 moles of O-atoms in 126 amu of HNO3.

To find the number of moles of O-atoms in 126 amu of HNO3, we need to first calculate the number of moles of HNO3 in 126 amu and then multiply it by the number of O-atoms per molecule of HNO3.

The molecular weight of HNO3 is 63 g/mol (1+14+48=63), which means that 1 mole of HNO3 has a mass of 63 g. To find the number of moles of HNO3 in 126 amu, we divide 126 by the molar mass of HNO3:

126 amu / 63 g/mol = 2 moles

So, there are 2 moles of HNO3 in 126 amu.

Each molecule of HNO3 contains 3 O-atoms, so the total number of O-atoms in 2 moles of HNO3 is:

2 moles x 3 O-atoms/mole = 6 moles of O-atoms

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The solution that contains the 0.10 M sodium hydroxide and the 0.10 M sodium cyanide. The formula of the substance which will precipitates first is the Zn(OH)₂.

The chemical equation for the reaction of the zinc acetate with the solutions and with the solubility products is as :

Zn(C₂H₃O₂)₂   + 2KOH   --->  Zn(OH)₂  +  2KC₂H₃O₂

The ksp of the  Zn(OH)₂  = 1.2 × 10⁻¹⁷

Zn(C₂H₃O₂)₂   +  2NaCN  -->  Zn(CN)₂   +  2C₂H₃O₂Na

The ksp of the  Zn(OH)₂  = 2.6 × 10⁻¹³

The higher the value of the Ksp of the solute, then the more soluble the solute  in the solvent.

The Ksp value of the Zn(OH)₂ is less as compared to the Ksp of the Zn(CN)₂,  Therefore, the Zn(OH)₂ precipitates first.

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The average atomic mass of this element is 12.0107 amu.Therefore, the average atomic mass of this element is 12.0107 amu.

To find the average atomic mass, we need to take into account the abundance and mass of each isotope. We can use the following formula:

Average atomic mass = (abundance of isotope 1 x mass of isotope 1) + (abundance of isotope 2 x mass of isotope 2)

Plugging in the values given in the question, we get:

Average atomic mass = (0.0107 x 13) + (0.9893 x 12)
Average atomic mass = 0.1391 + 11.8716
Average atomic mass = 12.0107 amu

Therefore, the average atomic mass of this element is 12.0107 amu.

To calculate the average atomic mass of an element with two stable isotopes, you need to multiply the mass of each isotope by its abundance (in decimal form) and then add the results together. Here's the calculation:

Isotope 1: 13 amu * 0.0107 = 0.1391
Isotope 2: 12 amu * 0.9893 = 11.8716

Average atomic mass = 0.1391 + 11.8716 = 12.0107 amu

So, the average atomic mass of this element is 12.0107 amu.

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The net charge on the 247 g piece of sulfur would be -1.84 x 10^13 elementary charges.

To solve this problem, we first need to calculate the number of sulfur atoms in the 247 g piece of sulfur:

number of sulfur atoms = (mass of sulfur / atomic mass of sulfur) x Avogadro's number

number of sulfur atoms = (247 g / 32.1 g/mol) x 6.022 x 10^23 atoms/mol

number of sulfur atoms = 1.84 x 10^25 atoms

Next, we need to determine how many atoms have an extra electron:

number of atoms with an extra electron = (1 / 1 x 10^12) x number of sulfur atoms

number of atoms with an extra electron = (1 x 10^12)^-1 x 1.84 x 10^25

number of atoms with an extra electron = 1.84 x 10^13 atoms

Each of these atoms with an extra electron has a net charge of -1, so the total net charge on the sulfur piece would be:

total net charge = -1 x number of atoms with an extra electron

total net charge = -1 x 1.84 x 10^13

total net charge = -1.84 x 10^13

Therefore, the net charge on the 247 g piece of sulfur would be -1.84 x 10^13 elementary charges.

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The majority of valence electrons are negatively charged particles, and they are all grouped in various orbitals or shells. Additionally, these electrons are in charge of how atoms interact with one another and create chemical bonds.

An element is defined as the pure substance which consists of only one type of atom which all have the same numbers of protons in their nuclei. Elements are the simplest chemical forms which cannot be broken down through chemical reactions.

Here the element potassium has the atomic number 19 and its electronic configuration is 2, 8, 8, 1. It contains the four energy levels, they are 'K', 'L', 'M' and 'N'. The number of valence electrons in potassium is 1.

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Your question is incomplete, most probably your full question was:

Which element has 1 valence electron and 4 energy levels?