Explain how the determination of the rate law equation significantly differs from the determination of the equilibrium constant keq expression.

In the given example, it led Sam’s theory to produce a hypothesis. It is because the educated guess that he has made has been linked to the theory where in the theory is about happiness and production that led him to think that having happy employees will make more products to be produced.

The molar concentration of the tartrazine solution is calculated by dividing the number of moles of tartrazine by the volume of the solution in liters. After converting the given mass of tartrazine to moles and the volume to liters, the molar concentration comes out to be approximately 0.000131 M.

To solve this problem, you must first identify the molar mass of the tartrazine, which is

534.37 g/mol

. The molar concentration, also known as molarity, is calculated by dividing the number of moles of solute by the volume of the solution in liters. First, convert the mass of tartrazine to moles by dividing the given mass (0.035g) by its molar mass,

0.035 g ÷ 534.37 g/mol ≈ 0.0000655 mol

. The volume is given as 500 mL, which is equal to 0.5 liters, To find molarity, divide the number of moles by the volume in liters,

0.0000655 mol ÷ 0.5 L = 0.000131 M

. Therefore, the molar concentration of the solution is approximately 0.000131 M.

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Answer:

pH=3.68

Explanation:

Hello,

At first, by knowing the 9.60-pH of the 0.070M solution of NaB, we can compute the Kb as the B contained into the NaB behaves as a base:

[tex]B^-+H_2O(l)<-->HB+OH^-[/tex]

Now, one can compute concentration of the OH ions because it is the same concentration of the HB based on the aforementioned chemical reaction:

[tex]pH=-log([H^+])\\\\ H^+=10^{-pH}=10^{-9.60}=2.51x10^{-10}\\OH^-=\frac{Kw}{[H^+]} =\frac{1x10^{-14}}{2.51x10^{-10}} =3.98x10^{-5}[/tex]

[tex][OH^-]=[HB]=3.98x10^{-5}M[/tex]

The Kb is then:

[tex]Kb=\frac{[HB][OH^-]}{[B^-]}=\frac{(3.98x10^{-5})^2}{0.070-3.98x10^{-5}} =2.264x10^{-8}[/tex]

After doing that, the Ka for the acid, taking into account its dissociation is:

[tex]HB<-->H^++B^-\\Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{2.264x10^{-8}} =4.42x10^{-7}[/tex]

Based on the ICE conditions and table, one states the change during the dissociation of HB:

[tex]Ka=\frac{[H^+][B^-]}{[HB]}\\4.42x10^{-6}=\frac{x^2}{0.010-x} \\x=2.08x10^-4M[/tex]

Finally, the found value for x equals to the H+ concentration, so we compute the pH:

[tex][H^+]=2.08x10^{-4}\\pH=-log(2.08x10^{-4})\\pH=3.68[/tex]

Best regards.

Th pH of a 0.010 M solution of HB is 3.68 and value of Kb for HB is 2.26×10⁻⁸.

pH of any solution is define as the negative logarithm of the concentration of H⁺ ion in the solution.

Given that, pH = 9.6

9.6 = -log[H⁺]

[H⁺] = [tex]10^{-9.6}[/tex] = 2.51×10⁻¹⁰

Also we know that,

[H⁺][OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.51×10⁻¹⁰ = 3.98×10⁻⁵

Now we calculate the value of Kb for HB following the below equation:
                                     B⁻  +  H₂O(l)  →  HB  +  OH⁻

Initial:                        0.070                      0         0

Equilibrium:        0.070-3.98×10⁻⁵   3.98×10⁻⁵ 3.98×10⁻⁵

Kb = [3.98×10⁻⁵]² / (0.070-3.98×10⁻⁵)

Kb = 2.26 × 10⁻⁸

Also we know the relation Ka × Kb = 10⁻¹⁴

Ka = 10⁻¹⁴/2.26 × 10⁻⁸ = 4.42×10⁻⁷

For the below equation, value of Ka will be:

                              HB  ↔  H⁺  +  B⁻

Initial:                   0.010      0       0

Equilibrium:        0.010-x    x        x

Ka = x² / (0.010-x)

We can neglect the value of x as it is very small as compare to 0.010.

4.42×10⁻⁷ = x² / 0.010

x = 2.08×10⁻⁴ M

So, the pH of the HB solution is:

pH = -log(2.08×10⁻⁴)

pH = 3.68

Hence required pH is 3.68.

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The IUPAC name for the given organic compound: CH₃CH₂CH(CH₃)CH₂CH₂OH is 3-methylhexan-2-ol.

IUPAC nomenclature refers to the chemical CH₃CH₂CH(CH₃)CH₂CH₂OH as 3-methylhexan-2-ol.

This term accurately describes its make-up: a six-carbon chain (hexane) with alcohol connected to the second carbon and a hydroxyl group (3-methyl) at the third carbon.

In order to precisely characterise organic compounds and enable unambiguous communication and identification of their structures, the IUPAC nomenclature adopts a systematic methodology.

Chemistry relies on this methodical naming strategy to describe and understand the precise atom arrangement within complicated compounds.

Thus, the answer is 3-methylhexan-2-ol.

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Your question seems incomplete, the probable complete question is:

"Identify the IUPAC name for the following organic compound: CH₃CH₂CH(CH₃)CH₂CH₂OH. Be sure to use correct punctuation and formatting."

Answer:

The most correct would be option A.

Explanation:

More correct would be to say that DNA contains, among other things, the information needed to make proteins, which have various functions within cells and organelles.

An example of a Physical Change is:

C. sugar dissolving in water

Physical Change:

To determine which of the options represents a physical change, it's important to understand the difference between physical and chemical changes:

1) Physical Change: A change that affects one or more physical properties of a substance without altering its chemical composition. Examples include changes in state (like melting or dissolving), shape, or size.

2) Chemical Change: A change that results in the formation of new chemical substances. This often involves a reaction that alters the chemical structure of the original substance.

Let's analyze each option:

A. Wood Burning

Type: Chemical Change

Reason: Burning wood transforms it into ash, smoke, and gases, changing its chemical composition.

B. A Piece of Metal Rusting

Type: Chemical Change

Reason: Rusting involves the reaction of metal with oxygen, resulting in the formation of iron oxide, which is a different substance.

C. Sugar Dissolving in Water

Type: Physical Change

Reason: When sugar dissolves, it forms a solution but retains its chemical composition. No new substances are formed.

D. A Mineral Weathering to Form Another Mineral

Type: Chemical Change

Reason: Weathering often involves chemical reactions that change the mineral's composition.

To make 900 L of a 0.140 M aqueous solution of nitric acid, 2142 g or 2.142 kg of ammonia (NH3) is required, assuming a 100% yield in the chemical reaction.

The production of aqueous nitric acid involves a sequence of chemical reactions, the first of which is the combustion of ammonia: 4NH3(g) + 502(g) => 4NO(g) + 6H₂O(g). This means that 4 moles of ammonia (NH3) react to form 4 moles of nitric oxide (NO) which then further reacts to form nitric acid (HNO3).

Given that the molarity (M) of the solution is defined as the number of moles of solute per liter of solution, a 0.140 M solution of nitric acid means there are 0.140 moles of nitric acid per liter of solution. So, if we require 900.00 L of this solution, we will need 900*0.140 = 126 moles of nitric acid.

From the combustion reaction, we know that 1 mole of NH3 reacts to produce 1 mole of NO, and therefore 1 mole of HNO3. As a result, to achieve 126 moles of HNO3, we will need the same amount of NH3. Ammonia has a molar mass of approximately 17 g/mol, hence we require 126 moles * 17 g/mol = 2142 g or 2.142 kg of ammonia, assuming 100% yield in the reaction.

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You would need approximately 2142 grams of ammonia to produce 900.00 l of a 0.140 M solution of nitric acid, assuming 100% yield in the chemical reactions.

To determine how many grams of ammonia you would need to form 900.00 l of a 0.140 M solution of nitric acid, you must first understand that the main chemical reaction involved in the production of nitric acid is the combustion of ammonia (4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)).

This balanced equation tells us that for every 4 moles of ammonia (NH3), 4 moles of nitric oxide (NO) are produced. Using the molarity of the nitric acid solution which is M = mol/L, we can calculate the moles of nitric acid to be 0.140 mol/l * 900.00 l = 126 mol. Assuming a 100% yield, we would need the same amount of moles of ammonia. The molar mass of NH3 is approximately 17 g/mol, so the mass of ammonia needed would be 126 mol * 17 g/mol = 2142 grams.

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The density of the unknown metal, calculated using the formula density = mass / volume with the given mass of 35.4 g and volume of 3.11 cm³, is 11.382 g/cm³.

To calculate the density of the unknown metal, you can use the density formula
which is density = mass / volume. Given that the mass of the metal is 35.4 g and the volume is 3.11 cm³, the calculation would be as follows:

Density = mass / volume
Density = 35.4 g / 3.11 cm³
Density = 11.382 g/cm³

Therefore, the density of the sample of the unknown metal is 11.382 g/cm³.

thx but it's actually 143

The last step is to calculate the percent by mass of each element in ammonium nitrate (NH4NO3). The masses of the elements in one mole of the compound are: mass N = 28.0 g mass H = 4.0 g mass O = 48.0 g The molar mass of the compound is 80.0 g/mol. What is the mass of one mole of the compound? 80.0g

Answer: 80 g

Explanation:Molar mass of the compound is the mass of 1 mole of compound which is the sum of masses of each element.

Mass of 1 mole of compound=mass of nitrogen + mass of hydrogen+ mass of oxygen= 28+4+48 = 80 g.

Percentage by mass of nitrogen=[tex]\frac{\text {mass of nitrogen}}{\text {Total mass}}\times 100\ %[/tex]

Percentage by mass of nitrogen=[tex]\frac{28}{80}\times 100\%=35\%[/tex]

Percentage by mass of hydrogen=[tex]\frac{\text {mass of hydrogen}}{\text {Total mass}}\times 100\%[/tex]

Percentage by mass of hydrogen=[tex]\frac{4}{80}\times 100\%=5\%[/tex]

Percentage by mass of oxygen=[tex]\frac{\text {mass of oxygen}}{\text {Total mass}}\times 100\%[/tex]

Percentage by mass of oxygen=[tex]\frac{48}{80}\times 100\%=60\%[/tex]

Answer: 0.84 atmosphere

Explanation:

Pressure of the gas is defined as the force exerted by the particles on the walls of the container. It is expressed in various terms like 'mmHg', 'atm', 'kiloPascals' etc..

All these units of pressure are inter convertible.

We are given:

Pressure of the gas = 640 mmHg

Converting this unit of pressure into 'atm' by using conversion factor:

[tex]760mmHg=1atm[/tex]

[tex]640mmHg=\frac{1}{760}\times 640=0.84atm[/tex]

Thus there will be 0.84 atm.

In the conformer I, both the substituents are in axial positions. Methyl group has two 1, 3 diaxial (H----CH_3) interactions and isopropyl group has two 1, 3 diaxial (H----pr) interactions. Therefore, the energy of these interactions is given as (2x0.9) + (2x1.1) = 4.0kCal/mol.

In conformer II both the substituents are equatorial positions. So, axial interactions are absent and one gauche interaction between isopropyl and methyl groups exists therefore, the energy of this interaction is given as 1.1kCal/mol. The difference in the energy of two conformations is calculated as

E = (1.0 – 1.1) kCal/mol = 2.9 kCal/mol

The answer to the question is 2.9 kCal/mol

The autoionization constant Kw can be calculated at different temperatures, given a pH value, using the relationships between pH, pOH, the ion concentrations and Kw itself. For a neutral solution with a pH of 7.78, Kw can vary from standard values due to temperature changes.

The subject of your question is related to the pH level, the autoionization constant Kw, and the temperature of a solution. These topics fall under

Chemistry

at the high school level. To calculate the Kw for a solution with a pH of 7.78, we first need to understand the relationship between the pH and pOH values. The sum of the pH and pOH is always equal to 14, a fact known as the ion-product constant for water. Using this relationship, we can solve for the pOH value, which is 14 - pH.

In this case, the pOH value would be 14 - 7.78 = 6.22. Then, we obtain the concentrations of OH- and H3O+ ions from the expressions [H3O+] = 10^-pH and [OH-] = 10^-pOH. However, given this is a neutral solution, they should be equal. Substituting these values in Kw = [H3O+][OH-], we can calculate the value of Kw which should be near to 1.0 × 10^-14, the standard Kw value at 25 °C, but may vary due to the different temperature causing the change in pH.

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The value of Kw at this temperature is approximately 2.8 × 10⁻¹⁶ .

To find the value of Kw at a given temperature where the pH of a neutral solution is 7.78, calculate pKw (which is the sum of pH and pOH), and then use it to determine Kw.

We can use the formula:

Since pKw = pH + pOH:

Thus,

Hence, the value of Kw at this temperature is approximately 2.8 × 10⁻¹⁶

Approximately 103.14 mL of concentrated HCl is necessary to make a 0.250 M HCl solution in a 5 L volume.

To determine the volume of concentrated HCl required to make a 0.250 M HCl solution in a 5 L volume, we can use the dilution formula:

[tex]\[ C_1V_1 = C_2V_2 \][/tex]

First, we need to ensure that all volumes are in the same unit, so we convert 5 L to milliliters:

[tex]\[ 5 \text{ L} \times 1000 \text{ mL/L} = 5000 \text{ mL} \][/tex]

Now we can plug the values into the dilution formula:

[tex]\[ (12.1 \text{ M})(V_1) = (0.250 \text{ M})(5000 \text{ mL}) \][/tex]

Solving for [tex]\( V_1 \):[/tex]

[tex]\[ V_1 = \frac{(0.250 \text{ M})(5000 \text{ mL})}{12.1 \text{ M}} \][/tex]

[tex]\[ V_1 = \frac{1250}{12.1} \text{ mL} \][/tex]

[tex]\[ V_1 \approx 103.14 \text{ mL} \][/tex]

Answer :

Electron-dot structure : It is also known as Lewis-dot structure. It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.  The valence electrons are represented by 'dot'.

The given molecule is, [tex]CHClO[/tex]

As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron, chlorine has '7' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in, [tex]CHClO[/tex] = 4 + 1 + 7 + 6 = 18

According to electron-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.

The electron-dot structure of [tex]CHClO[/tex] is shown below.

Lewis-dot structure is another name for the electron-dot structure. It displays the bonds that exist between a molecule's atoms, as well as the unpaired electrons that are present there. 'Dot' stands in for the valence electrons.

The electrons that make up an atom's valence shell are shown by an electron dot structure, commonly referred to as a Lewis dot formula. Knowing the compound's chemical formula makes drawing electron dot diagrams possible.

The diagram of the electron-dot structure for CHClO is attached in the image below.

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Beryllium and calcium have similar chemical properties because they both have two valence electrons as alkaline earth metals. Beryllium forms more covalent bonds due to its small ionic size and high charge density.

Beryllium and calcium have similar chemical properties because they are both alkaline earth metals and share the same number of valence electrons, which is two. This similarity is due to their position in the same group (Group 2) of the periodic table. Elements within the same group have the same number of valence electrons, which significantly influences their chemical behavior. It's the loss, gain, or sharing of these electrons that determines how elements react with one another.

However, due to the small size of the beryllium ion, which results in a high charge density, beryllium tends to form more covalent compounds, behaving somewhat differently than calcium. Beryllium is somewhat unique among its group in that it forms covalent bonds more readily than its group members, a property it shares with aluminum, which lies diagonally to it on the periodic table. Both beryllium and aluminum can form polar highly covalent compounds, for example, polymeric hydrides, chlorides, and alkyls.

Given the heat of reaction for C3H8 (-2219.9 kJ/mole), approximately 48.63 kg of C3H8 would need to react to release 2.45 x10^6 kJ of heat.

The balanced chemical reaction is C3H8 + 5O2 -> 3CO2 + 4H2O and its known enthalpy is -2219.9 kJ. If this amount of heat is released for the combustion of one mole of C3H8, to calculate the mass of propane (C3H8) needed to release 2.45 x10^6 kJ of heat, we need to consider the proportional relationship between heat and mass.

To release -2219.9 kJ of heat, 1 mole (or about 44.09 g) of C3H8 is needed. Therefore, to release -2.45 x10^6 kJ, you just multiply 2.45 x10^6 kJ by 44.09 g and then divide by 2219.9 kJ, which gives you approximately 48628.75 g or 48.63 kg. In summary, around 48.63 kg of C3H8 would need to react to release 2.45 x10^6 kJ of heat.

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To release 2.45×10⁶ kJ of heat, 2206.6 moles of C₃H₈ are required. Using the molar mass, this corresponds to 97,311.06 grams of C₃H₈. Therefore, the total mass needed is 97,311.06 grams.

To solve this problem, follow these steps:

2 C₃H₈(g) + 7 O₂(g) → 6 CO₂(g) + 8 H₂O(g)

moles of C₃H₈ = (2.45×10⁶ kJ) / (2219.9 kJ/2 moles) = 2206.6 moles of C₃H₈

mass (g) = 2206.6 moles × 44.1 g/mol = 97,311.06 g

Therefore, the mass of C₃H₈ required to release 2.45×10⁶ kJ of heat is 97,311.06 grams.

The electrode potential values of [tex]{\text{L}}{{\text{i}}^ + }[/tex], [tex]{\text{N}}{{\text{a}}^ + }[/tex] and [tex]{{\text{K}}^ + }[/tex] are [tex]\boxed{ - 3.045{\text{ V}}}[/tex], [tex]\boxed{ - {\text{2}}{\text{.714 V}}}[/tex] and [tex]\boxed{ - {\text{2}}{\text{.925 V}}}[/tex] respectively.

Further explanation:

Electrochemical cells are the devices that are used to generate an electric current from the energy released by the spontaneous redox reaction. Redox reactions involve the transfer of electrons between the chemical species by oxidation and reduction process. The oxidation process represents the loss of electrons and reduction process represents the gain of electrons.

In order to determine the standard reduction potential, the electrode is coupled with standard hydrogen electrode as illustrated in the attached image. The SHE acts as anode and the metal electrode acts as the cathode. The value of reduction potential for hydrogen has been assigned as 0.00 V.

The formula to calculate cell potential [tex]\left( {{E_{{\text{cell}}}}} \right)[/tex] of the reaction when the reduction occurs as cathode and oxidation occur at anode is as follows:

[tex]{E_{{\text{cell}}}} = {E_{{\text{cathode}}}} - {E_{{\text{anode}}}}[/tex]

The expression of Nernst equation that relates the reduction potential of a half or full electrochemical cell reaction to the standard electrode potential is given as,

[tex]{E_{{\text{cell}}}} = E_{{\text{cell}}}^\circ - \dfrac{{2.303RT}}{{nF}}\log \dfrac{{\left[ P \right]}}{{\left[ R \right]}}[/tex]               ……. (1)

Here,

[tex]{E_{{\text{cell}}}}[/tex] is cell potential.

[tex]E_{{\text{cell}}}^\circ[/tex] is standard cell potential.

[tex]R[/tex] is gas constant and is equal to [tex]8.314{\text{ J/K}} \cdot {\text{mol}}[/tex].

[tex]T[/tex] is temperature.

[tex]n[/tex] denotes the number of moles of electrons transferred in the reaction.

[tex]F[/tex] is Faraday constant and is equal to [tex]95484.56{\text{ C/mol}}[/tex].

[tex]\left[ P \right][/tex] is the concentration of product.

[tex]\left[ R \right][/tex] is the concentration of reactant.

At standard temperature [tex]25\;^\circ {\text{C}}\left( {298{\text{ K}}} \right)[/tex] the value of [tex]\dfrac{{2.303RT}}{F}[/tex] is equal to [tex]0.0592{\text{ V}}[/tex]. Thus the equation(1) is written as,

[tex]{E_{{\text{cell}}}} = E_{{\text{cell}}}^\circ - \dfrac{{0.0591}}{n}\log \dfrac{{\left[ P \right]}}{{\left[ R \right]}}[/tex]         …… (2)

Knowing the concentration of the electrolyte solutions taken in the cathodic compartment and substituting the values in the Nernst equation gives the reduction potential for the half cell.

The electrode potential values of [tex]{\mathbf{L}}{{\mathbf{i}}^{\mathbf{ + }}}[/tex], [tex]{\mathbf{N}}{{\mathbf{a}}^{\mathbf{ + }}}[/tex] and [tex]{{\mathbf{K}}^{\mathbf{ + }}}[/tex] are [tex]{\mathbf{ - 3}}{\mathbf{.045 V}}[/tex], [tex]{\mathbf{ - 2}}{\mathbf{.714 V}}[/tex] and [tex]{\mathbf{ - 2}}{\mathbf{.925 V}}[/tex] respectively as determined with respect to standard hydrogen electrode at [tex]{\mathbf{298 K}}[/tex].

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Answer details:  

Grade: Senior School

Subject: Chemistry

Chapter: Electrochemistry

Keywords: Electrochemical cells, Redox reactions, Oxidation, reduction, cell potential, half-cell, Nernst equation, Faraday constant.

In 2 moles of Na₃PO₄ the number moles of oxygen atom is c. 8.

To determine the number of moles of oxygen atoms in 2 moles of Na₃PO₄, we first need to understand the chemical formula of sodium phosphate (Na₃PO₄).

The chemical formula of sodium phosphate is Na₃PO₄.

Given that we have 2 moles of Na₃PO₄, we can calculate the total number of oxygen atoms present in these 2 moles by using the following steps:

1. Calculate the molar mass of Na₃PO₄:

2. Calculate the number of moles of oxygen atoms in 2 moles of Na3PO4:

Given that 1 mole of Na₃PO₄ contains 4 moles of oxygen atoms, we can set up a proportion:

1 mole of Na₃PO₄ / 4 moles of O = 2 moles of Na₃PO₄ / x moles of O

By cross-multiplying, we get:

Therefore, there are 8 moles of oxygen atoms in 2 moles of Na₃PO₄.

Correct question is: How many moles of oxygen atoms are in 2 moles of Na₃PO₄?

a. 3

b. 3

c. 8

d. 6

Answer: The correct answer is Option A.

Explanation:

There are 3 subatomic particles present in an atom. They are: protons, electrons and neutrons.

Protons carry positive charge, electrons carry negative charge and neutrons does not carry any charge.

Any neutral atom has an equal number of protons and electrons.  An ion is formed when atom looses or gain electron.

Thus, an ionized atom will always have unequal number of protons and electrons.

Hence, the correct answer is Option A.

Among the given statements, the true statement regarding ion is option A. An ionized atom has a number of protons that is unequal to the number of electrons.

An ion is an atom or molecule that has gained or lost one or more electrons, which changes the overall charge of the atom or molecule.

Therefore, an ionized atom has a number of protons that is unequal to the number of electrons. Option A is the correct answer.

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The number of Hg atoms present in 1.0 ×10⁻¹⁰ mol of mercury is 6.022 × 10¹³.

Avogadro’s number can represent the number of particles in one mole of any substance. Generally, these particles can be molecules, atoms, ions, electrons, or protons, which depends upon the type of chemical reaction or their reactant and product.

The value of Avogadro’s number is approximately equal to 6.022 × 10²³ per mol.

Given, the number of moles of mercury (Hg) =  1.0 ×10⁻¹⁰ mol

The number of Hg atoms of mercury in one mole = 6.022 × 10²³ atoms

Then 1.0 ×10⁻¹⁰ mol of mercury will have atoms = 6.022 × 10²³ × 1.0 ×10⁻¹⁰

The number of atoms of mercury =  6.022 × 10¹³ Hg atoms

Therefore, the number of Hg atoms present in 1.0 ×10⁻¹⁰ mol of mercury is equal to  6.022 × 10¹³.

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