Which of these is an example of an agricultural use for radiation?

A. making heavy isotopes to find new elements.

B. irradiating wheat to kill fungus.

C. diagnostic procedures that image inside the body, such as a PET scan.

D. locating leaks in a water line in a building. ​​​

If 17 mol of sodium (Na) react in this reaction, 8.5 mol should be the end product i.e., sodium oxide (Na2O) according to the stoichiometry.

Stoichiometry is a technique used by scientists to quantify and control the quantity of reactants and products in chemical reactions on a big scale. Without it, reactions might not be complete, wasting costly ingredients and producing dangerous byproducts.

With so many applications in everyday life, stoichiometry is regarded as the core of chemistry. Stoichiometric calculations are used to determine the chemical makeup of every chemical product we use on a regular basis, including shampoos, cleansers, fragrances, soaps, and fertilizers. Stoichiometry is necessary for the chemical industry to function.

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Answer:

In this experiment, the amount of ice is the independent variable, and the temperature change is the dependent variable.

The pressure exerted by the nitrogen gas in the container of volume 10.0L  at a temperature of 298K is 12.14 atm.

Pressure in chemistry is defined as the force per unit area exerted by a gas on the walls of its container. It is the result of the constant, random motion of gas molecules colliding with the walls of the container.

To find the pressure exerted by the N₂ gas, we can use the ideal gas law:

PV = n*R*T

R is the universal gas constant, whose value is constant and is given by 0.0821 L·atm/mol·K

To solve for P:

P = n*R*T/V

Substituting the given values, we get:

P = (0.50 mol) *  (298 K) / (10.0 L)* (0.0821 L·atm/mol·K)

P = 12.14 atm

Therefore, the pressure exerted by the N₂ gas in the 10.0L container at 298K is 12.14 atm.

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Answer:

The gas that is at room temperature reacts violently with other elements without heating is ammonium chloride ammonia, mercury or sodium. Basically, the electronegative element will be adopting -1 oxidation state.

Explanation:

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Answer:

1 mole of Cl2 reacts with 2 moles of NaBr to produce 1 mole of Br2.

So, to calculate the number of moles of Br2 produced, we first need to find the number of moles of Cl2 required to react with 5.2 moles of NaBr. Since the stoichiometric ratio of Cl2 to NaBr is 1:2, we need half as many moles of Cl2 as moles of NaBr:

Number of moles of Cl2 = 5.2 moles NaBr / 2 = 2.6 moles Cl2

Now we can use the stoichiometric ratio between Cl2 and Br2 to calculate the number of moles of Br2 produced:

1 mole Cl2 produces 1 mole Br2

Therefore, 2.6 moles Cl2 will produce 2.6 moles Br2.

Explanation:

1.083 moles of Al are needed to react completely with 52.00 g O2.

To begin, we can use the molar mass of O2 to convert 52.00 g to moles.

52.00 g O2 / (32.00 g/mol) = 1.625 mol O2

The balanced equation shows that two moles of aluminum react with three moles of oxygen to form two moles of aluminum oxide. Therefore, we need to know how many moles of aluminum are required to react with 1.625 mol of oxygen.

Using the stoichiometric ratio from the balanced equation:

2 mol Al : 3 mol O2

We can set up a proportion to determine the number of moles of Al needed:

2 mol Al / 3 mol O2 = x mol Al / 1.625 mol O2

Solving for x:

x = (2 mol Al / 3 mol O2) x (1.625 mol O2) = 1.083 mol Al

Therefore, 1.083 moles of aluminum must react completely with 52.00 g of O2.

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We can start by using the given enthalpy change and the balanced equation to calculate the amount of moles of carbon dioxide produced:

1 mol of C2H5OH produces 2 mol of CO2

-1367 kJ/mol is released when 1 mol of C2H5OH is combusted

370 kJ of energy are used in the reaction

Now we can use a proportion to find the number of moles of C2H5OH that produce 370 kJ of energy:

-1367 kJ/mol / 1 mol of C2H5OH = -x kJ / (2 mol of CO2)

Solving for x gives:

x = (370 kJ) (1 mol of C2H5OH) / (-1367 kJ/mol) (2 mol of CO2)

x = 0.1355 mol of CO2

Finally, we can use the molar mass of CO2 (44.01 g/mol) to find the mass of CO2 produced:

mass of CO2 = number of moles of CO2 * molar mass of CO2

mass of CO2 = 0.1355 mol * 44.01 g/mol = 5.96 g of CO2

Therefore, when 370 kJ of energy are used in the reaction, 5.96 g of carbon dioxide are produced.

A. The mole fraction of sodium chloride (NaCI) is 0.042

B. The mole fraction of water (H₂O) is 0.958

We'll begin by obtaining the mole of each component in the solution. This is shown below:

For sodium chloride (NaCI)

Mole = mass / molar mass

Mole of NaCI = 3.57 / 58.5

Mole of NaCI = 0.061 mole

For water (H₂O)

Mole = mass / molar mass

Mole of H₂O = 25 / 18

Mole of H₂O = 1.389 mole

Finally, we shall determine the mole fraction of each component. Details below:

For sodium chloride (NaCI)

Mole fraction of NaCI = Mole of NaCI / total mole

Mole fraction of NaCI = 0.061 / 1.45

Mole fraction of NaCI = 0.042

For water (H₂O)

Mole fraction of H₂O = 1 - Mole fraction of NaCI

Mole fraction of H₂O = 1 - 0.042

Mole fraction of H₂O = 0.958

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Answer:

In a perfect world, where the calorimeter does not exchange any heat with its contents, the amount of heat lost by the hot water will be equal to the amount of heat gained by the cool metal pieces. This is due to the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. Therefore, the number of calories gained by the metal pieces will be equal to the number of calories lost by the hot water, which is 75 calories.

Explanation:

In summary, according to the principle of conservation of energy, in a perfect world where a calorimeter does not exchange any heat with its contents, the number of calories gained by the cool metal pieces will be equal to the number of calories lost by the hot water, which is 75 calories.

The reaction of 2,2-dichloro-3-methylbutane with water results in the formation of 3-methyl-2-butanol, which can react with OH- in a nucleophilic substitution reaction to form 3-methyl-2-butyl alcohol and water.

The reaction can be represented by the following equation:

2,2-dichloro-3-methylbutane + H2O → 3-methyl-2-butanol + H+ + Cl-

The reaction of 2,2-dichloro-3-methylbutane with water can result in the formation of an alcohol and a hydrogen ion. The product of this reaction is 3-methyl-2-butanol.

The reaction can be represented by the following equation:

2,2-dichloro-3-methylbutane + H2O → 3-methyl-2-butanol + H+ + Cl-

In this reaction, one of the chlorine atoms from 2,2-dichloro-3-methylbutane is replaced by a hydroxyl group (-OH) from water, resulting in the formation of an alcohol group (-OH) in the product, 3-methyl-2-butanol. The other chlorine atom remains as an ion, Cl-.

The product, 3-methyl-2-butanol, can react with OH- in a nucleophilic substitution reaction. In this reaction, the hydroxide ion (OH-) acts as a nucleophile and attacks the carbon atom that is attached to the leaving group (the -OH group) in the 3-methyl-2-butanol molecule. The leaving group then departs with its pair of electrons, forming a new bond with the nucleophile (OH-). The result is the formation of a new alcohol molecule.

The reaction can be represented by the following equation:

3-methyl-2-butanol + OH- → 3-methyl-2-butyl alcohol + H2O

In this reaction, the -OH group of 3-methyl-2-butanol is replaced by the hydroxide ion (OH-) to form 3-methyl-2-butyl alcohol, and a water molecule is formed as a byproduct.

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Answer:

a. The solute is the substance that is being dissolved, in this case, the red Gatorade mix.

b. The solvent is the substance in which the solute is being dissolved, in this case, water.

c. The solution is the resulting homogeneous mixture of the solute (red Gatorade mix) and the solvent (water).

d. To increase the molarity of the Gatorade in the water, you could add more Gatorade mix to the water while keeping the volume of the solution constant. Alternatively, you could decrease the volume of water while keeping the amount of Gatorade mix constant, which would increase the concentration of Gatorade in the solution.

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Answer:

To determine the lattice energy of CaCl₂, we can use the Born-Haber cycle, which relates the lattice energy to other thermodynamic quantities such as the enthalpy of sublimation, dissociation, and formation.

The Born-Haber cycle for CaCl₂ is as follows:

Ca(s) → Ca(g) ΔHsub

1/2 Cl₂(g) → Cl(g) ΔHdiss

Ca(g) + Cl(g) → CaCl(g) ΔHf

CaCl(g) → CaCl₂(s) ΔHlattice

We can use the following equation to calculate the lattice energy:

ΔHlattice = ΔHsub + ΔHdiss + ΔHf - IE1 - IE2 - EA

where IE1 and IE2 are the first and second ionization energies of Ca, and EA is the electron affinity of Cl.

Substituting the given values, we get:

ΔHlattice = (+178.2) + (+243.4) + (-795.8) - (+590) - (+1145) - (-348.7)

ΔHlattice = -57.1 kJ/mol

Therefore, the lattice energy of CaCl₂ is -57.1 kJ/mol.

Number of moles of S2 needed to produce 750 moles of SO2 is 375.

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We need to calculate the stoichiometric ratios of the reactants and choose the set that provides the required ratios with the least amount of excess or unused reactants.

To determine the most efficient set of reactants for a reaction, we should consider the stoichiometry of the reaction, which tells us the ratios of the reactants and products that are involved in the reaction. The most efficient set of reactants will be the one that produces the desired product with the least amount of excess or unused reactants, and hence the least amount of waste.

2A + 3B → 4C If we have 4 moles of A and 6 moles of B, we have the exact stoichiometric amounts required for the reaction to proceed, and all the reactants will be consumed completely to form 8 moles of product C. In this case, there will be no waste of materials, and the reaction will be the most efficient. On the other hand, if we have an excess of one of the reactants, for example, if we have 6 moles of A and 6 moles of B, then only 4 moles of A can react with 6 moles of B to produce 8 moles of C, and the remaining 2 moles of A will be unused and wasted. In this case, the reaction will be less efficient.

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Tο determine which catiοn will precipitate first, we need tο cοmpare the sοlubility prοducts (Ksp) οf their respective sulfates. The catiοn with the smaller Ksp value will precipitate first. The Ksp values fοr BaSO₄ and CaSO₄ are:

Ksp(BaSO₄) = 1.1 × 10⁻¹⁰

Ksp(CaSO₄) = 2.4 × 10⁻⁵

A catiοn is an iοn with a pοsitive charge that is fοrmed by the lοss οf οne οr mοre electrοns frοm a neutral atοm. Catiοns are fοrmed when atοms lοse electrοns tο achieve a mοre stable electrοn cοnfiguratiοn, typically by becοming isοelectrοnic with a nearby nοble gas.

Since the Ksp value fοr BaSO4 is smaller, it will precipitate first. Tο calculate the minimum cοncentratiοn οf Na₂SO₄ required tο precipitate all οf the Ba2+ iοns, we need tο use the sοlubility prοduct expressiοn fοr BaSO₄:

BaSO₄(s) ⇌ Ba2+(aq) + SO₄₂-(aq)

Ksp = [Ba2+][SO₄₂-]

Since all οf the Ba2+ iοns will be precipitated, the cοncentratiοn οf Ba2+ will be zerο οnce precipitatiοn is cοmplete. Thus, the Ksp expressiοn simplifies tο:

Ksp = [SO₄₂-]²

Sοlving fοr [SO₄₂-], we get:

[SO₄₂-] = sqrt(Ksp) = sqrt(1.1 × 10⁻¹⁰) = 1.05 × 10⁻⁵ M

This means that the cοncentratiοn οf Na₂SO₄ needs tο be at least 1.05 × 10⁻⁵M tο prοvide enοugh SO₄₂- iοns tο precipitate all οf the Ba₂+ iοns. We rοund this tο twο significant figures, giving a final answer οf 1.1 × 10⁻⁵ M Na₂SO₄.

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The amount of water is needed to dilute the stock solution to 0.25 M with a total volume of 547 mL is 548 mL

To solve this problem, we can use the formula:

M1V1 = M2V2

where M1 and V1 are the initial concentration and volume, and M2 and V2 are the final concentration and volume.

Given:

M1 = initial concentration = ?

V1 = initial volume = 250 mL

M2 = final concentration = 0.25 M

V2 = final volume = 547 mL

Mass of KCl used = 10.2 g

First, we need to calculate the initial concentration:

M1 = (moles of solute) / (volume of solution in liters)

The molar mass of KCl is 74.55 g/mol. Therefore, the number of moles of KCl used is:

moles of KCl = (mass of KCl) / (molar mass of KCl)

moles of KCl = 10.2 g / 74.55 g/mol

moles of KCl = 0.137 mol

The volume of the initial solution in liters is:

V1 = 250 mL = 0.25 L

Using the formula above, we can calculate the initial concentration:

M1 = (moles of KCl) / (volume of solution in liters)

M1 = 0.137 mol / 0.25 L

M1 = 0.548 M

Now, we can use the formula to calculate the volume of water needed to dilute the solution:

M1V1 = M2V2

(0.548 M)(250 mL) = (0.25 M)(547 mL + V_water)

137 mL = 0.25 M V_water

V_water = 548 mL

Therefore, 548 mL of water is needed to dilute the stock solution to a final volume of 547 mL with a concentration of 0.25 M using M1V1 = M2V2.

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Answer:

Explanation:Before dividing the initial solution:

Initial volume = 1.0 L

Initial concentration = 0.40 M

After dividing into two equal parts:

Each part has a volume of 0.5 L

The first part has a concentration of 0.40 M

The second part has a concentration of 0.20 M (diluted by 50% with 1.5 L of water)

When the two parts are mixed:

Total volume = 1.0 L + 0.5 L + 0.5 L + 1.5 L = 3.5 L

Total moles of sulfuric acid = (1.0 L x 0.40 M / 1000) + (0.5 L x 0.40 M / 1000) + (0.5 L x 0.20 M / 1000) + (1.5 L x 0 / 1000) = 0.5 mol

Final concentration before adding more sulfuric acid = 0.5 mol / 3.5 L = 0.14 M

When 2.0 L of 0.1 M sulfuric acid are added:

Total volume = 3.5 L + 2.0 L = 5.5 L

Total moles of sulfuric acid = 0.5 mol + (2.0 L x 0.1 M / 1000) = 0.7 mol

Final concentration = 0.7 mol / 5.5 L = 0.13 M

Therefore, the final molar concentration of the solution is 0.13 M.

Answer: C) wood cross out

Answer: 0.1334983576 g

Explanation:

The mass must be conserved on both sides of the equation. Mass put in must must = mass put out. So we must start by finding the mass of the reactants and then the mass of 2H2O. Then we must subtractio the mass of the reactants from the mass of 2H2O which will gives us the mass of CO2.

Mass of CH4:

(12.0107)+ 4(1.00794) = 16.48246 g/mol

we are given three mols of CH4 so divide 3 mol by 16.48246 g/mol

3 mol/ 16.48246 g/mol = 0.1820116657 g CH4

Mass of 2O2:

2(15.99994) = 31.99988 g/mol

we are given 2 moles of 2O2 so divide 2 mol by 31.99988 g/mol  

2 mol/ 31.99988 g/mol = 0.0625002344 g 2O2

Mass of 2H2O:

2(1.00794)+(15.99994) = 18.01582 g/mol

we are given 2 moles of 2H2O so divide 2 mol by 18.01582 g/mol

2 mol/ 18.01582 g/mol = 0.1110135425 g 2H20

Now we add up the grams on the reactatnt side and subtract that number from the mass of 2H2O:

0.1820116657 g CH4 + 0.0625002344 g 2O2 = 0.2445119001 g (total of g of reactants)

0.2445119001 g -  0.1110135425 g =  0.1334983576 g CO2

Systematic name : 5-chloro-2-pentanol (or 5-chloropentan-2-ol)

Systematic name : 1,2-difluoro-3-heptanol   (or 1,2-difluoroheptan-3-ol).

The active site of numerous reactions is the hydroxyl group (OH). Pentyl butyrate, which has an apricot-like aroma, is the ester that results from the reaction of 1-pentanol and butyric acid. Amyl acetate, also known as pentyl acetate, is the ester that is created when 1-pentanol and acetic acid are combined.

A research evaluating the efficacy of diesel fuel blends with different amounts of pentanol as an additive was done in 2014. Higher pentanol concentrations resulted in higher gaseous emissions at the expense of lower particulate emissions.

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If you start with 4.55 x 10²⁵ molecules of C₅H₁₂, you would produce 1.82 x 10²⁶ g of H₂.

The balanced equation shows that for every one molecule of C₅H₁₂ that reacts, 2 molecules of H₂ are produced.

C₅H₁₂ --> C₅H₈ + 2H₂

To determine the mass of H₂ produced, we need to convert the number of molecules of C₅H₁₂ to molecules of H₂ using the stoichiometry of the balanced equation:

1 molecule of C₅H₁₂ produces 2 molecules of H₂

4.55 x 10²⁵ molecules of C₅H₁₂ x (2 molecules of H₂ / 1 molecule of C₅H₁₂) = 9.10 x 10²⁵ molecules of H₂

Now, to find the mass of H₂ produced, we need to use the molar mass of H₂, which is approximately 2 g/mol:

9.10 x 10²⁵ molecules of H₂ x (2 g/mol) = 1.82 x 10²⁶ g of H₂

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Answer:

9.10 x 10²⁵ molecules of H₂ x (2 g/mol) = 1.82 x 10²⁶ g of H₂

Explanation:

Answer:

density = mass/volume

Rearranging, we get:

volume = mass/density

Substituting the given values, we get:

volume = 108g / 0.90 g/mL

volume = 120 mL

Therefore, the volume of the material is 120 mL. Answer: C

The density of a substance is its mass by volume. Then, the volume of the 108 g of material with a density of 0.90 g/ml is 120 mL.

Density is a physical property of matter that describes the amount of mass in a given volume of a substance. It is defined as the ratio of an object's mass to its volume.

Mathematically, density is expressed as:

Density = Mass / Volume

The unit of density depends on the units of mass and volume used. For example, in the SI system, the unit of mass is kilograms (kg) and the unit of volume is cubic meters (m³), so the unit of density is kg/m³.

Given that, density = 0.90 g/ml

mass = 108 g

then volume = mass/densiy

v = 108 g/0.900 g/ml = 120 mL.

Therefore, density of the material is 120 mL.

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Answer:

How much heat will be released when. 8.21 g of sulfur reacts with excess O1 according to the following equation? 25 +302 → 2SO3. AH° = -791.4 kJ.

Explanation:

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76.0 kJ, According to the provided reaction and stoichiometry, 76.0 kJ of heat will be emitted when 8.21 g of sulphur interacts with too much oxygen to create sulphur dioxide.

We must first determine how many moles of sulphur are present:

S mass divided by S's molar mass equals moles of S.

8.21 g/32.06 g/mol = 0.256 mol are the moles of sulphur.

We may infer that all of the sulphur will react because there is an excess of oxygen, hence the reaction will result in:

1 mol SO2 divided by 0.256 mol S results in 0.256 mol SO2.

q = nΔH

where n is the number of moles of the product (SO2), q is the quantity of heat released, and H is the reaction's enthalpy change.

q = (0.128 mol)(-296.8 kJ/mol) = -76.0 kJ

We can use the ideal gas law to solve for the volume of oxygen gas produced:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.08206 L atm/mol K), and T is the temperature in Kelvin.

First, we need to calculate the number of moles of oxygen gas produced from the given mass of potassium chlorate. We can use the molar mass of KClO3 (122.55 g/mol) and the stoichiometry of the balanced equation:

15.0 g KClO3 x (1 mol KClO3/122.55 g KClO3) x (3 mol O2/2 mol KClO3) = 0.184 mol O2

Now we can solve for the volume of oxygen gas at 756 torr (convert to atm) and 27°C (convert to Kelvin):

P = 756 torr = 0.996 atm

T = 27°C + 273.15 = 300.15 K

n = 0.184 mol

R = 0.08206 L atm/mol K

V = nRT/P = (0.184 mol)(0.08206 L atm/mol K)(300.15 K)/(0.996 atm) = 4.76 L

Therefore, the volume of O2 collected at 27°C and 756 torr pressure is 4.76 L.

Part B:

If the oxygen gas is collected over water, it will be saturated with water vapor. This means that the total pressure in the container will be the sum of the pressure of the oxygen gas and the pressure of the water vapor. The pressure of the water vapor can be calculated using the vapor pressure of water at the given temperature.

At 27°C, the vapor pressure of water is 26.7 torr. Therefore, the total pressure in the container will be:

P total = P oxygen gas + P water vapor = 756 torr + 26.7 torr = 782.7 torr

Since the problem provided the pressure in torr, we need to convert to atm before using the ideal gas law. Therefore, we need to adjust the pressure to:

P = 782.7 torr x (1 atm/760 torr) = 1.03 atm

We can then use the same equation as in part A to calculate the volume of oxygen gas:

V = nRT/P = (0.184 mol)(0.08206 L atm/mol K)(300.15 K)/(1.03 atm) = 4.32 L

Therefore, if the oxygen gas is collected over water, the volume of oxygen gas produced would be 4.32 L instead of 4.76 L (as in part A).

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The statement is false that the chemical compound ethanol, with a molecular mass of 46.07 g/mol, has the same molecular formula as Ethanol. The organic liquid, on the other hand, has molecules that weigh 32 g/mol.

Ethylene glycol is a substance that is frequently used as antifreeze. 38.7% of it is carbon, 9.75% hydrogen, and the remainder is oxygen. Ethylene glycol has a molecular weight of 62.07 g.

In the molecules of carbohydrates, the proportion of carbon to hydrogen to oxygen is 1:2:1. This class of organic compounds gets its name from the parts water (water, -hydrate) and carbon (carbon, carbo-).

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Capturing quality data and using analytics are critical for the fire and emergency services to improve situational awareness, enhance resource management, improve incident management, and develop better planning and prevention strategies.

What is fire?

Fire is a chemical reaction that occurs when a fuel (such as wood, paper, or gasoline) combines with oxygen in the air, producing heat and light. The reaction is exothermic, meaning it releases energy in the form of heat and light. Fire requires three elements to exist: fuel, oxygen, and heat. These elements are often referred to as the "fire triangle."

Capturing quality data and using analytics are critical for the fire and emergency services for several reasons:

Improved situational awareness: Fire and emergency services need accurate, up-to-date information to make informed decisions and respond effectively to emergencies.

Enhanced resource management: Fire and emergency services often operate under tight budget constraints and need to make the most of their resources.

Improved incident management: The ability to capture and analyze data in real-time can help fire and emergency services manage incidents more effectively.

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The formula for iron(II) bromide is FeBr2, while the formula for barium chloride is BaCl2. The products of the reaction are barium bromide (BaBr2) and iron(II) chloride (FeCl2).

What is Balanced Equation ?

A balanced chemical equation is a symbolic representation of a chemical reaction that shows the relative numbers of reactant and product molecules or ions involved. In a balanced equation, the number of atoms of each element must be the same on both the reactant and product sides, in order to obey the law of conservation of mass.

The balanced chemical equation for the reaction between iron(II) bromide and barium chloride is:

FeBr2 + BaCl2 → BaBr2 + FeCl2

In this reaction, the iron(II) ion (Fe2+) in iron(II) bromide (FeBr2) is replaced by the barium ion (Ba2+) from barium chloride (BaCl2), forming barium bromide (BaBr2) and iron(II) chloride (FeCl2). The balanced equation shows that two bromide ions (Br-) and two chloride ions (Cl-) are involved in the reaction, which ensures that the equation is balanced in terms of both mass and charge.

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According to the question Therefore, 81.00 mg/L CaCO3 is equivalent to 162.00 mg/L H2PO4-.

Calcium Carbonate, also known as CaCO3, is a naturally occurring mineral which is found in rocks, shells, and pearls. It is composed of the elements calcium, carbon, and oxygen, and is a major component of sedimentary rocks. Calcium Carbonate is an important building material, and is used in a range of industries, from paper and plastic to pharmaceuticals and cement. It is also used as an additive in many food products, such as baking powder and cheese.

To calculate this, we must first convert 81.00 mg/L CaCO3 to moles.

81.00 mg/L CaCO3 = 0.00045 moles CaCO3

Next, we need to convert from moles of CaCO3 to moles of H2PO4-. The molar ratio between CaCO3 and H2PO4- is 1:2, so for every mole of CaCO3, there are two moles of H2PO4-.

Therefore, 0.00045 moles CaCO3 = 0.00090 moles H2PO4-

Finally, we must convert the moles of H2PO4- back to mg/L:

0.00090 moles H2PO4- = 162.00 mg/L H2PO4-

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The mass of the unknown element cobalt is obtained as 40 g. This can be seen from the calculation that we have in the solution.

If you have a compound that contains an unknown element, you can use the molar mass of the compound to find the relative atomic mass of the unknown element

We can see that the question has already given us the mass of the sample as we have and the percentage of the cobalt that we have in the sample.

50 = x/80 × 100

x = 50/100 × 80

= 40 g

Thus we would have a total of about 40 g of cobalt in the sample.

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Missing parts;

A sample of the compound weighs 80 grams.  If the mass percent of cobalt is 50%, How many grams of cobalt are in the sample?