Select all of the equations which demonstrate the law of conservation of mass.
A Mg + S → MgS2

B C + O2 → C2O

C 4Cu + O2 → 2Cu2O

D 2H2 + O2 → 2H2O

E H2SO4 + Zn → 4ZnSO + H2

From the uncompleted equation, we have:

LiOH + HBr ->

Since we have two ionic substance reacting, we can conclude that the type of reaction is double displacement reaction as the reaction will involve exchange of ions between the reacting species.

The products of the reaction can be obtained by balancing the equation. This is shown below:

LiOH + HBr ->

By exchange of ion, we have

LiOH + HBr -> LiBr + H₂O

Now, observing the equation, we can see that the equation is balanced.

Thus, the products of the reaction are LiBr and H₂O

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Based on Table G, the mass of KCl that must be dissolved in 200 grams of H₂O at 10 °C to make a saturated solution is 60 g.

Based on Table G, the solubility of KCl at 10°C is given as 30 g/100 g water.

To calculate the mass of KCl that can be dissolved in 200 grams of water at 10°C, we can set up a proportion:

(30 g KCl / 100 g water) = (x g KCl / 200 g water)

Cross-multiplying and solving for x, we get:

x g KCl = (30 g KCl / 100 g water) * (200 g water)

x g KCl = 60 g KCl

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The limiting reactant will be the one that produces fewer moles of BaSO4. The excess reactant will be the one that has moles left over after the reaction.

To determine the grams of BaSO4 produced and the limiting reactant, we need to compare the stoichiometry of the balanced chemical equation for the reaction between Ba(NO3)2 and Na2SO4, which is:

Ba(NO3)2 + Na2SO4 → BaSO4 + 2NaNO3

First, calculate the number of moles for each reactant:

Moles of Ba(NO3)2 = 200.0 g / molar mass of Ba(NO3)2

Moles of Na2SO4 = 100.0 g / molar mass of Na2SO4

Then, calculate the moles of BaSO4 formed by comparing the stoichiometric coefficients:

Moles of BaSO4 formed = Moles of Ba(NO3)2 (according to the stoichiometry ratio)

Next, calculate the grams of BaSO4 formed:

Grams of BaSO4 formed = Moles of BaSO4 formed × molar mass of BaSO4

To identify the limiting reactant, compare the moles of BaSO4 formed from each reactant. The reactant that produces fewer moles of BaSO4 is the limiting reactant.

To determine the excess reactant remaining, calculate the moles of excess reactant and then convert it to grams.

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The density of the block of wood is approximately 38.40 g/mL.

To calculate the density of the block of wood, we need to use the formula:

Density = Mass / Volume

First, let's convert the mass of the block from grams (g) to milliliters (mL). Since the density is expressed in g/mL, the mass and volume need to have the same units.

Given:

Mass of the block = 450 g

Change in water level = 16.22 mL - 4.50 mL = 11.72 mL

Density = 450 g / 11.72 mL

Calculating the density:

Density ≈ 38.40 g/mL

Therefore, the density of the block of wood is approximately 38.40 g/mL.

The density of a substance represents its mass per unit volume. In this case, the mass of the block of wood is 450 g, and the volume is determined by the change in water level when the block is dropped into the graduated cylinder. By subtracting the initial water level (4.50 mL) from the final water level (16.22 mL), we find that the block occupies a volume of 11.72 mL. Dividing the mass by the volume gives us the density of the block, expressed in grams per milliliter.

It's important to note that the density of wood can vary depending on factors such as the type of wood and its moisture content. The value calculated here represents the density of the specific block used in the given scenario.

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Answer:

Thermal Conduction

Explanation:

The boiling point of a solution is influenced by the presence of solute particles, which can cause a change in the boiling point compared to the pure solvent. This phenomenon is known as boiling point elevation.

The magnitude of boiling point elevation depends on the concentration of the solute and the nature of the solute particles. In general, the greater the concentration of solute particles, the greater the boiling point elevation.

Comparing a 0.5m sodium chloride (NaCl) solution to a 0.3m aluminum sulfate ([tex]Al_2(SO_4)_3[/tex]) solution, we can determine the relative boiling point elevation.

Sodium chloride (NaCl) dissociates into two ions in solution (Na+ and Cl-), while aluminum sulfate ([tex]Al_2(SO_4)_3[/tex])dissociates into three ions (2[tex]Al_3[/tex]+ and 3[tex]SO_4[/tex]2-). This means that the aluminum sulfate solution will have a greater concentration of solute particles per mole than the sodium chloride solution.

Therefore, the boiling point of the 0.5m sodium chloride solution will be lower than the boiling point of the 0.3m aluminum sulfate solution.

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The following chemical equations must be balanced:

1. N2 + 3 H2 → 2 NH3

Type: Synthesis

2. 2 KClO3 → 2 KCl + 3 O2

Type: Single Replacement

3. 2 NaF + ZnCl2 → ZnF2 + 2 NaCl

Type- Decomposition

4. 2 AlBr3 + 3 Ca(OH)2 → Al2(OH)6 + 6 CaBr2

Type- Double Replacement

5. 2 H2 + O2 → 2 H2O

Type: Combustion

6. 2 AgNO3 + MgCl2 → 2 AgCl + Mg(NO3)2

Type: Synthesis

7. 2 Al + 6 HCl → 2 AlCl3 + 3 H2

Type: Decomposition

8. C3H8 + 5 O2 → 3 CO2 + 4 H2O

Type: Combustion

9. 2 FeCl3 + 6 NaOH → Fe2O3 + 6 NaCl + 3 H2O

Type: Double Replacement

10. 4 P + 5 O2 → 2 P2O5

Type: Synthesis

11. 2 Na + 2 H2O → 2 NaOH + H2

Type: Single Replacement

12. 2 Ag2O → 4 Ag + O2

Type: Decomposition

13. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

Type: Combustion

14. 2 KBr + MgCl2 → 2 KCl + MgBr2

Type: Double Replacement

15. 2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O

Type: Double Replacement

16. C5H12 + 8 O2 → 5 CO2 + 6 H2O

Type: Combustion

17. 4 Al + 3 O2 → 2 Al2O3

Type: Synthesis

18. Fe2O3 + 2 Al → 2 Fe + Al2O3

Type: Single Replacement

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Answer:

Explanation: the answer is in the picture

Answer:

[tex]\huge\boxed{\sf V_1=0.16 \ L}[/tex]

Explanation:

Initial Molarity =[tex]M_1[/tex] = 6.00 M

Final Volume = [tex]V_2[/tex] = 0.32 L

Final Molarity = [tex]M_2[/tex] = 3.0 M

Initial Volume = [tex]V_1[/tex] = ?

[tex]M_1V_1=M_2V_2[/tex]

Put the given data in the above formula,

Finding initial volume.

[tex]6 \times V_1=0.32 \times 3\\\\6 \times V_1 = 0.96\\\\Divide \ both \ sides \ by \ 6\\\\V_1=0.96/6\\\\V_1=0.16 \ L\\\\\rule[225]{225}{2}[/tex]

To produce 11.00 moles of MgCl2, you would need 858.00 grams of NaOH.

To determine the amount of NaOH needed to produce 11.00 moles of MgCl2, we need to use stoichiometry and the balanced chemical equation:

[tex]MgCl_2 + 2 NaOH[/tex] → [tex]2 NaCl + Mg(OH)_2[/tex]

From the balanced equation, we can see that the mole ratio between [tex]MgCl_2[/tex]and NaOH is 1:2.

Therefore, for every 1 mole of[tex]MgCl_2[/tex], we need 2 moles of NaOH.

Given: Moles of [tex]MgCl_2[/tex]= 11.00 moles

Using the mole ratio, we can calculate the moles of NaOH required:

moles of NaOH = 2 * moles of MgCl2

moles of NaOH = 2 * 11.00 moles

moles of NaOH = 22.00 moles

Now, we need to convert the moles of NaOH to grams using the molar mass of NaOH:

The molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol

grams of NaOH = moles of NaOH * molar mass of NaOH

grams of NaOH = 22.00 moles * 39.00 g/mol

grams of NaOH = 858.00 grams

Therefore, to produce 11.00 moles of [tex]MgCl_2[/tex], you would need 858.00 grams of NaOH.

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The solubility chart provides information about the solubility of various compounds in water. Here are the solubilities of the given compounds:

a. (NH₄)₂CO₃: According to the solubility chart, most carbonate (CO₃²⁻) salts are insoluble, except for those of Group 1 metals (alkali metals) and ammonium (NH₄⁺). Therefore, (NH₄)₂CO₃ is soluble.

b. Fe(OH)₂: Hydroxide (OH⁻) salts of transition metals, including iron (Fe), are generally insoluble, except for those of Group 1 metals and ammonium. Therefore, Fe(OH)₂ is insoluble.

c. Ca(OH)₂: Calcium hydroxide (Ca(OH)₂) is soluble. However, the given compound "CaOH" appears to be missing the subscript ₂, indicating two hydroxide ions. If it should be Ca(OH)₂, then it is soluble.

d. PbCl₂: According to the solubility chart, chloride (Cl⁻) salts, including lead chloride (PbCl₂), are generally soluble, except for those of silver (Ag⁺), lead (Pb²⁺), and mercury (Hg₂²⁺). Therefore, PbCl₂ is insoluble.

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Based on the given reaction, the acid-base pairs in this reaction are:

An acid-base pair refers to a set of two chemical species that are related through the transfer of a proton (H+ ion) during a chemical reaction.

One species acts as an acid by donating a proton, while the other acts as a base by accepting that proton.

In the given reaction:

HCO₃⁻ (aq) + NH₃ (aq) → NH₄⁺ + CO₃²⁻

An acid-base pair can be identified as follows:

HCO₃⁻ (bicarbonate ion) can act as an acid by donating a proton (H⁺), becoming CO₃⁻.

NH₃ (ammonia) can act as a base by accepting a proton (H⁺), becoming NH₄⁺ (ammonium ion).

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The name of LiBr is lithium bromate and the charge of the cation (K) is +.

A cation is a positively charged ion, i.e. one that would be attracted to the cathode in electrolysis. The opposite of a cation is an anion.

Cations and anions make up an ionic compound and determine the charge on the compound. For example, an ionic compound; Lithium bromate is given in this question.

Lithium bromate is made up of Lithium (Li+) as the cation and chlorine (Cl-) as the anion.

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Balance the following chemical equations:

1. N2 + 3 H2 → 2 NH3

Ex: Synthesis reaction

2. 2 KClO3 → 2 KCl + 3 O2

Single Replacement reaction

3. 2 NaF + ZnCl2 → ZnF2 + 2 NaCl

Decomposition reaction

4. 2 AlBr3 + 3 Ca(OH)2 → Al2(OH)6 + 6 CaBr2

Double Replacement reaction

5. 2 H2 + O2 → 2 H2O

Combustion reaction

6. 2 AgNO3 + MgCl2 → 2 AgCl + Mg(NO3)2

Synthesis reaction

7. 2 Al + 6 HCl → 2 AlCl3 + 3 H2

Decomposition reaction

8. C3H8 + 5 O2 → 3 CO2 + 4 H2O

Combustion reaction

9. 2 FeCl3 + 6 NaOH → Fe2O3 + 6 NaCl + 3 H2O

Double Replacement reaction

10. 4 P + 5 O2 → 2 P2O5

Synthesis reaction

11. 2 Na + 2 H2O → 2 NaOH + H2

Single Replacement reaction

12. 2 Ag2O → 4 Ag + O2

Decomposition reaction

13. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

Combustion reaction

14. 2 KBr + MgCl2 → 2 KCl + MgBr2

Double Replacement reaction

15. 2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O

Double Replacement reaction

16. C5H12 + 8 O2 → 5 CO2 + 6 H2O

Combustion reaction

17. 4 Al + 3 O2 → 2 Al2O3

Synthesis reaction

18. Fe2O3 + 2 Al → 2 Fe + Al2O3

Single Replacement reaction

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At the usual working conditions of 101.5 kPa and 21°C, the chemical oxygen generator system would generate roughly 220.84 litres of oxygen using 500 g of LiClO4.

We may use the ideal gas law and stoichiometry to calculate how many litres of oxygen are created by the chemical oxygen generator system employing 500 g of LiClO4.

We must first determine the moles of LiClO4. LiClO4 has a molar mass of approximately 106.39 g/mol. As a result, 4.704 mol of LiClO4 are produced from 500 g of LiClO4 using the formula: 500 g / 106.39 g/mol

We can see from the chemical equation that 1 mole of LiClO4 results in 2 moles of O2. 4.704 mol of LiClO4 will therefore result in:

2 mol O2 / 1 mol LiClO4 4.704 mol LiClO4 = 9.408 mol O2

The moles of O2 under the specified conditions must then be converted to volume. The ideal gas law, which goes as follows:

PV = nRT

Where:

P = pressure = 101.5 kPa

V = volume (in liters)

n = moles of gas = 9.408 mol

R = ideal gas constant = 8.314 J/(mol·K)

T = temperature = 21°C = 294 K (converted to Kelvin)

Rearranging the equation to solve for V:

V = (nRT) / P

V = (9.408 mol × 8.314 J/(mol·K) × 294 K) / (101.5 kPa × 1000 Pa/kPa)

Simplifying the units:

V = (9.408 × 8.314 × 294) / 101.5

V ≈ 220.84 liters

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Answer:

a. P2O7 - This is a covalent compound. P and O have similar electronegativities and they form a covalent bond between them, rather than an ionic bond.

b. SnBr2 - This is a covalent compound. Sn and Br have different electronegativities, but they still form a covalent bond due to their relatively small difference in electronegativity.

c. Fe(OH)2 - This is an ionic compound. Fe has a higher electronegativity than O and H, so it tends to donate its electrons and become positively charged. This results in the formation of ionic bonds between Fe and OH.

d. Cl3O8 - This is a covalent compound. Cl and O have similar electronegativities, so they form covalent bonds rather than ionic bonds.

Answer:

3rf or 5d cd Yu been successful of a new future

The name of the branched alkane shown below is 2- methylheptane that is in option D as alkane is a type of hydrocarbon, which is a compound consisting of hydrogen and carbon atoms only. 

Alkanes are characterized by having single bonds between carbon atoms and being saturated hydrocarbons, meaning they have the maximum number of hydrogen atoms bonded to each carbon atom. Alkanes are often referred to as "paraffins" and serve as the simplest and most basic form of hydrocarbons. They are relatively unreactive and are commonly found in petroleum and natural gas. The systematic names for alkanes are derived from the prefix "n-" or "normal-," followed by the Greek numerical prefix indicating the number of carbon atoms. For example, "n-pentane" refers to the straight-chain alkane with five carbon atoms.

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To calculate the molarity of the solution, we need to know the number of moles of BaI2 and the volume of the solution in liters.

First, let's calculate the number of moles of BaI2. We can use the formula:

Number of moles = Mass (in grams) / Molar mass

The molar mass of BaI2 can be calculated as follows:

Ba: atomic mass = 137.33 g/mol

I: atomic mass = 126.90 g/mol

2 x I = 2 x 126.90 g/mol = 253.80 g/mol

Total molar mass of BaI2 = 137.33 g/mol + 253.80 g/mol = 391.13 g/mol

Number of moles of BaI2 = 413 g / 391.13 g/mol ≈ 1.056 moles

Next, we need to convert the volume of the solution from milliliters to liters:

Volume of solution = 750 ml / 1000 = 0.75 L

Finally, we can calculate the molarity using the formula:

Molarity = Number of moles / Volume of solution

Molarity = 1.056 moles / 0.75 L ≈ 1.408 M

Therefore, the molarity of the BaI2 solution is approximately 1.408 M.

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Metamorphic rocks can be harder, less porous, and have crystals that can be lined, describing some of the ways in which metamorphic rocks differ from sedimentary rocks.

There are two different types of rocks: sedimentary rocks and metamorphic rocks. Igneous or sedimentary pre-existing rocks undergo changes under extreme heat and pressure to form metamorphic rocks. This process results in the recrystallization of minerals, leading to the formation of a new rock with distinct physical and chemical characteristics.

Therefore, the correct option is B.

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If a compound is found to contain 3.622 % carbon and 96.38 % bromine by weight. The molecular formula for the compound is CBr4.

First, get the empirical formula in order to calculate the molecular formula of the chemical. The empirical formula shows the atoms of a compound in their most straightforward whole number ratio.

Suppose 100 grams of the substance. To determine the mass of carbon and bromine in the compound using the provided percentages.

Mass of C = 3.622% of 100g

= 3.622g

Mass of Br = 96.38% of 100g

= 96.38g

The next step is to determine the atomic masses of carbon and bromine in order to determine the number of moles for each.

Atomic mass of carbon = 12.01 g/mol

Atomic mass of bromine = 79.90 g/mol

Moles of C = (mass of carbon) / (atomic mass of carbon)

= 3.622g / 12.01 g/mol

= 0.3016 mol

Moles of Br = (mass of bromine) / (atomic mass of bromine)

= 96.38g / 79.90 g/mol

= 1.205 mol

Divide the moles of each element by the fewest number of moles obtained, in this case the moles of carbon, to arrive at the empirical formula.

Empirical formula ratio:

C: (0.3016 mol) / (0.3016 mol)

= 1

Br: (1.205 mol) / (0.3016 mol)

= 4

The empirical formula for the compound is C₁Br4.

To determine the molecular formula, it is required to know the molecular weight of the compound. The molecular weight is  331.61 g/mol.

To find the number of empirical formula units in the molecular formula, divide the molecular weight by the empirical formula weight.

Empirical formula weight:

C = 12.01 g/mol × 1

= 12.01 g/mol

Br= 79.90 g/mol × 4

= 319.60 g/mol

Empirical formula weight = 12.01 + 319.60

= 331.61 g/mol

Now find the number of empirical formula units in the molecular formula:

Number of empirical formula units

= (molecular weight) ÷ (empirical formula weight)

Number of empirical formula units

= 331.61 g/mol / 331.61 g/mol

= 1

The number of empirical formula units is 1, the empirical formula C₁Br4 is would be  the molecular formula for this compound.

Thus, the molecular formula for the compound is CBr₄.

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The concentration of ethanol in units of volume/volume percent is 29.49%.

To calculate the concentration of ethanol in units of volume/volume percent, we need to determine the volume of ethanol relative to the total volume of the solution.

Total volume of the solution = volume of water + volume of ethanol

Total volume = 569 mL + 238 mL

Total volume = 807 mL

To express the concentration as volume/volume percent, we can calculate the ratio of the volume of ethanol to the total volume of the solution and multiply by 100 to obtain a percentage.

Concentration of ethanol = (volume of ethanol / total volume of solution) x 100

Concentration of ethanol = (238 mL / 807 mL) x 100

Concentration of ethanol = 0.2949 x 100

Concentration of ethanol = 29.49%

Therefore, the concentration of ethanol in the solution is approximately 29.49%.

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Answer:

The limiting reactant is BF3 because there is less of it than Li2SO4.

Explanation:

The following chemical equations must be balanced:

1. N2 + 3 H2 → 2 NH3

Type: Synthesis reaction

2. 2 KClO3 → 2 KCl + 3 O2

Type: Single Replacement reaction

3. 2 NaF + ZnCl2 → ZnF2 + 2 NaCl

Type- Decomposition reaction

4. 2 AlBr3 + 3 Ca(OH)2 → Al2(OH)6 + 6 CaBr2

Type- Double Replacement reaction

5. 2 H2 + O2 → 2 H2O

Type: Combustion reaction

6. 2 AgNO3 + MgCl2 → 2 AgCl + Mg(NO3)2

Type: Synthesis reaction

7. 2 Al + 6 HCl → 2 AlCl3 + 3 H2

Type: Decomposition reaction

8. C3H8 + 5 O2 → 3 CO2 + 4 H2O

Type: Combustion reaction

9. 2 FeCl3 + 6 NaOH → Fe2O3 + 6 NaCl + 3 H2O

Type: Double Replacement reaction

10. 4 P + 5 O2 → 2 P2O5

Type: Synthesis reaction

11. 2 Na + 2 H2O → 2 NaOH + H2

Type: Single Replacement reaction

12. 2 Ag2O → 4 Ag + O2

Type: Decomposition reaction

13. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

Type: Combustion reaction

14. 2 KBr + MgCl2 → 2 KCl + MgBr2

Type: Double Replacement reaction

15. 2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O

Type: Double Replacement reaction

16. C5H12 + 8 O2 → 5 CO2 + 6 H2O

Type: Combustion reaction

17. 4 Al + 3 O2 → 2 Al2O3

Type: Synthesis reaction

18. Fe2O3 + 2 Al → 2 Fe + Al2O3

Type: Single Replacement reaction

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The molarity of the sucrose in the iced tea mixture prepared by dissolving  0.750 moles of sucrose in the 2.00 L of iced tea is

The following data were obtained from the question:

The molarity of the solution can be obtained as illustrated below:

Molarity of solution = mole / volume

Molarity of solution = 0.750 mole / 2 liters

Molarity of solution = 0.375 M

Thus, we can conclude from the above calculation that the molarity of the solution is 2.5 M

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The dissociation of sodium chloride in water allows it to act as an electrolyte, conducting electricity through the movement of ions.

The bonding between sodium and chlorine is classified as ionic bonding. In this type of bonding, electrons are transferred from one atom to another, resulting in the formation of ions. Sodium (Na) readily donates one electron from its outermost shell to achieve a stable electron configuration, while chlorine (Cl) accepts this electron to fill its outermost shell. As a result, sodium forms a positively charged ion (Na+), known as a cation, while chlorine forms a negatively charged ion (Cl-), known as an anion. The electrostatic attraction between these oppositely charged ions creates a strong bond between sodium and chlorine, forming sodium chloride (NaCl) as the resulting compound.

When sodium chloride is dissolved in water, the compound dissociates into separate sodium cations and chloride anions. Water molecules, which have a polar nature, surround the individual ions due to their attraction to opposite charges. This process is called hydration or solvation. The water molecules effectively separate the sodium and chloride ions, leading to the compound's dissolution. This is because water molecules have a higher affinity for the charged ions compared to the ionic bond holding the compound together.

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The molar mass of the gas is approximately 20.72 g/mol.

To determine the molar mass of the gas, we can use the ideal gas law equation:

PV = nRT

where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = ideal gas constant (0.0821 L·atm/(mol·K))

T = temperature (in Kelvin)

First, let's convert the given values to the appropriate units:

Pressure = 432 torr = 432/760 atm

Volume = 333 mL = 333/1000 L

Temperature = 23°C = 23 + 273.15 K

Substituting the values into the ideal gas law equation:

(432/760) atm * (333/1000) L = n * 0.0821 L·atm/(mol·K) * (23 + 273.15) K

Simplifying the equation:

0.191 atm * 0.333 L = n * 0.0821 L·atm/(mol·K) * 296.15 K

Solving for the number of moles (n):

n = (0.191 atm * 0.333 L) / (0.0821 L·atm/(mol·K) * 296.15 K)

n ≈ 0.01012 moles

Finally, we can calculate the molar mass using the formula:

Molar mass = mass of the gas sample / moles of gas

Molar mass = 0.210 g / 0.01012 moles

Molar mass ≈ 20.72 g/mol

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