why is atomic emission more sensitive to flame instability than atomic absorption ?

The ideal gas law, also called the general gas equation, is the equation of state of a hypothetical ideal gas. It is a good approximation of the behavior of many gases under many conditions,then the answer is that 0.0025 moles of oxygen gas were created by your process.

When pressure and temperature are the same, the amount of oxygen gas created by your reaction will be 0.0025 moles.

In accordance with the equation for a balanced chemical reaction, hydrogen peroxide, or H₂O₂, breaks down to produce water and oxygen gas.

2H2O2(aq)→2H2O(l)+O2(g)

You have all the data necessary to solve for the amount of moles of oxygen gas created using the ideal gas law equation because you have collected 0.061 L of oxygen gas at 295.15 K and 1 atm.

PV=nRT n=PVRT nO₂=1atm * 0.061L / (0.082 (L * atm / mol * K)) =0.0025 moles

Hence, if this was your initial inquiry, then the answer is that 0.0025 moles of oxygen gas were created by your process.

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One mole of carbon dioxide would require 84.01 grams of NaHCO₃.

It is discovered that the ratio of moles of CO₂ generated to moles of NaHCO₃ reacted is 1:2.

We can observe from this equation that 1 mole of NaHCO₃ results in 1 mole of CO₂. As a result, NaHCO₃ and CO₂ have a molar ratio of 1:1.

Na2CO₃(s) + H₂O(g) + CO₂ = 2 NaHCO₃(s)(g)

CO₂ has a molar mass of about 44.01 g/mol. As a result, we must determine how much NaHCO₃ weighs in relation to one mole of CO₂. Using the molar mass of NaHCO₃, the following can be calculated:

Molar mass of NaHCO₃ is 84.01 g/mol.

The mass of NaHCO₃ needed to create one mole of CO₂ is as follows:

(84.01 g NaHCO₃/1 mole NaHCO₃) = 84.01 g CO₂/mol for 1 mole of NaHCO₃.

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Answer:

To solve the 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 are given P1 = 750 mmHg, V1 = 1.27 L, T1 = 298 K, V2 = 0.75 L, T2 = 448 K, and the amount of gas does not change.

First, we can solve for P2 by rearranging the equation as:

P2 = (P1V1T2) / (V2T1)

Substituting the values we get:

P2 = (750 mmHg x 1.27 L x 448 K) / (0.75 L x 298 K)

P2 = 1504 mmHg

Therefore, the final pressure at a volume of 0.75 L and a temperature of 448 K is 1504 mmHg.

The ratio of aluminum to hydrogen in the reaction is 2:3. For every 2 moles of aluminum that reacts, 3 moles of hydrogen are produced.

The balanced chemical equation for the reaction 2Al + 3H₂SO₄ = 3H₂ + Al₂(SO₄)₃ shows that 2 moles of aluminum (2Al) reacts with 3 moles of sulfuric acid (3H₂SO₄) to produce 3 moles of hydrogen gas (3H₂) and 1 mole of aluminum sulfate (Al₂(SO₄)₃).

o express the ratio in terms of the number of atoms of each element involved in the reaction, we need to consider the coefficients of the balanced chemical equation. The coefficient in front of each element or compound indicates the number of moles of that substance involved in the reaction.

In the given equation, the coefficient for aluminum (Al) is 2 and the coefficient for hydrogen (H) is 6 (3 on each side). Therefore, the ratio of aluminum atoms to hydrogen atoms in the reaction is 2:6, which simplifies to 1:3.

So, for every one atom of aluminum that reacts, three atoms of hydrogen are produced. This can also be expressed as the molar ratio of aluminum to hydrogen, which is 2:3.

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Answer:

9mol N

Explanation:

The balanced chemical equation is:

N₂ + 3H₂ → 2NH₃

According to the stoichiometry of this reaction, 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.

So, to produce 2 moles of NH₃, we need 1 mole of N₂.

Therefore, to produce 18 moles of NH₃, we would need:

18 mol NH₃ × (1 mol N₂ / 2 mol NH₃) = 9 mol N₂

Thus, 9 moles of nitrogen are required to produce 18 mol of NH₃.

Answer:

the pressure of the gas is 1.67 atm when the volume increased to 4.5 L.

Explanation:

Assuming the temperature and the amount of gas remain constant (i.e., the process is isobaric):

Using Boyle's law, we can relate the initial pressure (P1) and volume (V1) to the final pressure (P2) and volume (V2):

P1V1 = P2V2

Plugging in the given values:

P1 = 1.50 atm

V1 = 5.00 L

V2 = 4.5 L

Solving for P2:

P2 = (P1V1)/V2 = (1.50 atm x 5.00 L)/4.5 L = 1.67 atm

This reaction is endothermic because energy is absorbed during the reaction. This can be seen by looking at the change in the enthalpy of the reactants compared to the products.

Given the reaction is as follows: [tex][Co(H_2O)_6]_2(aq) + 4Cl^{-}(aq) < -- > [CoCl_4]_2^{-}(aq) + 6H_2O(l)[/tex]

The products have a higher enthalpy than the reactants, meaning that the reaction absorbs energy and is endothermic. This can also be determined by looking at the oxidation states of the elements involved. The oxidation state of the cobalt atom in[tex][Co(H_2O)_6]^{2+}[/tex] is +2, while the oxidation state of the cobalt atom in [tex][CoCl_4]^{2-[/tex] is +3. This indicates that the cobalt atom has been oxidized, which requires energy and makes the reaction endothermic.

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Answer:

5

Explanation:

Answer:

(a) Electrode reactions:

Pt/H+/H2: 2H+(aq) + 2e- -> H2(g) (reduction)

Ag/AgCl/Cl-: AgCl(s) + e- -> Ag(s) + Cl-(aq) (reduction)

Overall cell equation: 2AgCl(s) + H2(g) -> 2Ag(s) + 2HCl(aq)

(b) Direction of motion of ions and electrons:

In the Pt/H+/H2 half-cell, hydrogen ions (H+) move towards the platinum electrode and accept electrons to form hydrogen gas (H2). In the Ag/AgCl/Cl- half-cell, silver ions (Ag+) move towards the silver chloride (AgCl) electrode and accept electrons to form silver (Ag) metal while chloride ions (Cl-) move away from the electrode. Electrons move from the hydrogen electrode to the silver electrode through the external circuit.

(c) Electrode labeling:

The Pt/H+/H2 electrode is the cathode (-) and the Ag/AgCl/Cl- electrode is the anode (+).

(d) Standard cell potential (Eo):

The standard cell potential can be calculated using the standard reduction potentials for each half-cell:

Eo(cell) = Eo(reduction, Ag/AgCl/Cl-) - Eo(reduction, Pt/H+/H2)

Eo(reduction, Ag/AgCl/Cl-) = +0.222 V (from standard reduction potential tables)

Eo(reduction, Pt/H+/H2) = 0 V (by definition)

Eo(cell) = +0.222 V - 0 V = +0.222 V

(e) Actual cell potential (E):

E(cell) = Eo(cell) - (0.0592 V / n) * log[H3O+]

where n is the number of electrons transferred in the balanced equation (2 in this case)

E(cell) = +0.222 V - (0.0592 V / 2) * log(1.0 x 10^-4 M)

E(cell) = +0.222 V - (0.0296 V) = +0.1924 V

(f) Values of constants A and B:

E(cell) = A + B.pH

At pH 7 (neutral), E(cell) = Eo(cell) = +0.222 V

Therefore, A = +0.222 V and B = -0.0592 V/pH

Answer:

the rate constant of the reaction is approximately 0.0021 s^-1.

Explanation:

To calculate the rate constant of a first-order reaction, we can use the following equation:

ln([A]t/[A]0) = -kt

Where [A]t is the concentration of reactant at time t, [A]0 is the initial concentration of reactant, k is the rate constant, and t is time.

Given that the reaction is 45% complete after 400 s, we can assume that [A]t/[A]0 = 0.55 (since 100% - 45% = 55%). Plugging this value into the equation above, we get:

ln(0.55) = -k(400)

Solving for k, we get:

k = -ln(0.55)/400

k ≈ 0.0021 s^-1

A first-order reaction is 45% complete after 400 s. Therefore, 0.0021 s⁻¹ is the rate constant of the reaction.

The chemical kinetics rate law, which connects the molecular concentration of reacting substances to reaction rate, uses the rate constant as a proportionality factor. The letter k in an equation designates it, which is also referred to as either the resultant rate constant and reaction rate coefficient.

The link among the molecular concentration of reactants with the rate for a chemical reaction is shown by the proportionality constant k. Utilising the molecular weights of each of the reactants with the sequence of the reaction, the rate constant can be calculated experimentally. As an alternative, the Arrhenius equation can be used to compute it.

ln([A]t/[A]0) = -kt

[A]t/[A]0 = 0.55

ln(0.55) = -k(400)

k = -ln(0.55)/400

k ≈ 0.0021 s⁻¹

Therefore, 0.0021 s⁻¹ is the rate constant.

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In both fusion and fission, stability decreases as a result of the reaction. In fusion, two smaller nuclei combine to form a larger nucleus, releasing energy in the process.

The resulting nucleus may be unstable and undergo radioactive decay, which can further release energy. In fission, a larger nucleus is split into smaller nuclei, also releasing energy. The resulting nuclei may also be unstable and undergo radioactive decay. In both cases, the process of splitting or combining nuclei releases energy, but it also reduces the overall stability of the resulting nuclei.

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CH2, N2O3 are covalent

NH4ClO, Fe3(PO4)2 and CrBr2 are ionic

Ionic compounds are formed when a metal atom donates one or more electrons to a nonmetal atom. This transfer of electrons creates ions, which are charged particles that attract each other due to their opposite charges.

Covalent compounds, on the other hand, are formed when two or more nonmetal atoms share electrons. In this type of bonding, each atom contributes one or more electrons to a shared pair, creating a covalent bond.

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Based on public policy, the most recent scientific findings, and efficient management, we give leadership on matters relating to food, agriculture, energy wealth, rural development, nutrition, & related topics.

A few examples of how biotechnology is influencing medicine are synthetic growth hormone and insulin as well as diagnostic procedures to find various ailments. The improvement of industrial procedures, environmental remediation, and agricultural production have all benefited from the use of biotechnology.

The demand for biotechnology is growing in industries including pharmaceutical, animal husbandry, agricultural, healthcare, medicine, genetic engineering, etc. In addition, biotechnologists are well compensated for their work. Thus, it is undoubtedly a smart professional choice.

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Answer:

2) 110 g P2O3

Explanation:

If the green box is in the denominator of a fraction and represents the molar mass of P2O3, then the correct answer choice would be 2) 110 g P2O3. Here’s how you can use this value to solve the problem:

The balanced chemical equation for the reaction between P2O3 and H2O to produce H3PO3 is: P2O3 + 3H2O → 2H3PO3.

From this equation, we can see that 1 mole of P2O3 reacts with 3 moles of H2O to produce 2 moles of H3PO3. So the number of moles of H3PO3 produced is twice the number of moles of P2O3 that reacted.

The molar mass of P2O3 is approximately 110 g/mol. So 93.2 g of P2O3 is equivalent to (93.2 g) / (110 g/mol) = 0.848 mol of P2O3.

Since the number of moles of H3PO3 produced is twice the number of moles of P2O3 that reacted, the number of moles of H3PO3 produced is 0.848 mol × 2 = 1.696 mol.

The molar mass of H3PO3 is (3 × 1.01 g/mol) + (1 × 30.97 g/mol) + (3 × 16.00 g/mol) = 81.99 g/mol. So 1.696 mol of H3PO3 is equivalent to (1.696 mol) × (81.99 g/mol) = 139 g of H3PO3.

So the complete reaction of 93.2 g P2O3 would produce approximately 139 g of H3PO3.

Assuming the acid used for titration is a strong acid (such as HCl), the balanced chemical equation for the reaction would be:

HCl + NaOH → NaCl + H2O

From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the number of moles of HCl used for titration can be calculated as:

moles of HCl = moles of NaOH = M x V x n

where M is the molarity of NaOH, V is the volume of NaOH used (100 mL or 0.1 L), and n is the number of moles of NaOH per liter of solution (1 mole/L).

moles of HCl = 0.15 M x 0.1 L x 1 mol/L = 0.015 mol

Since the volume of acid used for titration is 150 mL or 0.15 L, we can calculate the molarity of the acid as:

Molarity of acid = moles of acid / volume of acid used

Molarity of acid = 0.015 mol / 0.15 L = 0.1 M

Therefore, the molarity of the acid solution is 0.1 M.

Explanation:

a)Take percentages and divide by mole wt ( from periodic table) of the corresponding element

          C     14.25 / 12            = 1.1875

          O     56.93 / 15.99     = 3.56

          Mg   28.83/24.3        = 1.186        Divide by the smallest number

                        C 1.1875/1.186      = 1

                         O 3.56 / 1.186     = 3

                         Mg  1.186 / 1.186 =  1

C O3  Mg         commonly written as Mg CO3   ( magnesium carbonate)

There are approximately [tex]1.2*10^17[/tex] molecules of food dye in the last solution.

Assuming that one drop of 100% solution is approximately 0.1 mL, then 10 drops of the 100% solution would be equivalent to 1 mL. If this 1 mL of the 100% solution was diluted to a final volume of 50 mL, then the dilution factor is 50.

Using the given approximation of 1 gram of food dye having approximately [tex]1.2*10^22[/tex] molecules, we can calculate the number of molecules in the original 1 mL of the 100% solution to be:

1 g * ([tex]1.210^22[/tex]molecules/g) = [tex]1.210^22[/tex]molecules

For the first dilution, the number of molecules would be 1/10th of this value, or:

(1/10) * [tex]1.210^221.210^22[/tex] = [tex]1.210^21[/tex]molecule

For the second dilution, the number of molecules would be 1/10th of this value, or:

[tex](1/10) * 1.210^21 = 1.210^20[/tex] molecules

If this process was repeated for a total of 5 dilutions, the number of molecules in the final solution would be:

[tex](1/10)^5 * 1.210^22 = 1.210^17[/tex]molecules

Therefore, there are approximately [tex]1.2*10^17[/tex] molecules of food dye in the last solution.

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Answer:

Explanation:

Gather materials: clear glue, water, borax, food coloring, measuring cups and spoons, mixing bowl, and stirring utensil.

Create two batches of slime, keeping all variables constant except for the amount of borax used. In one batch, use 1 tablespoon of borax, and in the other batch, use 2 tablespoons of borax.

Mix the ingredients together in separate bowls until they reach the desired consistency.

Compare the consistency of the two slimes.

Data:

Qualitative data: Observations about the texture, color, and smell of the two batches of slime.

Quantitative data: Measurements of the amount of borax used in each batch and any other measurements deemed important for analyzing the consistency of the slime.

Independent variable: The amount of borax used in the slime recipe.

Dependent variable: The consistency of the slime.

Problem Question: What is the effect of varying the amount of borax solution on the consistency of slime?

Hypothesis: If the amount of borax solution in the slime mixture is increased, then the consistency of the slime will become firmer.

Experiment Steps:

In this experiment, the independent variable is the amount of borax solution, as it is being varied to test its effect on the slime's consistency. The responding variable is the consistency of the slime, which is being observed and recorded as the dependent variable.

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The molecular formula of the compound is [tex]C_6H_{12}O_6[/tex] whose molar mass is 180.16g/mol with mass percent composition: C 40.00%, H 6.72%, O 53.28%.

Given the mass percent compositions as:

the mass percent composition of carbon (C) = 40.00%

the mass percent composition of hydrogen (H) = 6.72%

the mass percent composition of oxygen (O) = 53.28%

The molar mass of the compound is  = 180.16 g/mol.

Let us assume we have 100g of compound then,

we have 40g of carbon, 6.72g of hydrogen and 53.28g of oxygen.

The atomic masses of carbon, hydrogen, and oxygen, respectively, are 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol.

We need to calculate the number of moles of each element present in the compound.

Number of moles of carbon = 40.00/12.01 g/mol = 3.33 moles

Number of moles of hydrogen = 6.72/1.01 g/mol = 6.65 moles

Number of moles of oxygen = 53.28/16.00 g/mol = 3.33 moles

Total number of moles = 3.33 + 6.65 + 3.33 = 13.31 moles

Mole ratio of carbon, hydrogen and carbon = 3.33 : 6.65 : 3.33

Mole ratio of carbon, hydrogen and carbon = 1 : 2 : 1

The empirical formula is = [tex]CH_2O[/tex]

The empirical formula mass of [tex]CH_2O[/tex] is = 30.03

n = 180.18/30.03 = 6

The molecular formula is = [tex]C_6H_{12}O_6[/tex]

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Answer:

Explanation:

need points :>

The heat for the reaction per gram of ZnS used is -4.51 kJ/g.

To calculate the heat for the reaction per gram of ZnS used, we need to first calculate the amount of heat released per mole of ZnS used and then convert that to per gram.

The given balanced chemical equation shows that 2 moles of ZnS react with 3 moles of O2 to produce 2 moles of ZnO and 2 moles of SO2, and the amount of heat released during the reaction is -879 kJ/mol.

So, the amount of heat released per mole of ZnS used is:

(-879 kJ/mol) / 2 = -439.5 kJ/mol

Now, to calculate the amount of heat released per gram of ZnS used, we need to divide the amount of heat released per mole by the molar mass of ZnS:

-439.5 kJ/mol / 97.46 g/mol = -4.51 kJ/g.

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The correct answer is The percent yield of a chemical reaction is a measure of the efficiency of the reaction. It is calculated by comparing the actual yield.

Which is the amount of product obtained in a reaction, to the theoretical yield, which is the amount of product that would be obtained if the reaction proceeded perfectly and no losses occurred. To calculate the percent yield in this case, we need to first determine the theoretical yield based on the amount of starting material used. The theoretical yield can be calculated using the balanced chemical equation and the stoichiometry of the reaction. Once the theoretical yield is determined, we can then use the formula for percent yield: Percent yield = (actual yield / theoretical yield) x 100% In this case, the chemist calculated a theoretical yield of 216.4 g based on the amounts of starting materials used. However, after isolating the purified product, she found that she only obtained 199.6 g of product. To calculate the percent yield, we can plug these values into the formula: Percent yield =[tex](199.6 g / 216.4 g) x 100% = 92.2%\\[/tex]Therefore, the percent yield for this reaction is 92.2%, which means that the reaction was quite efficient, but there were some losses or inefficiencies during the reaction or purification process. By calculating percent yield, chemists can evaluate the efficiency of a reaction and make adjustments to improve the process in future experiments.

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The percentage of iron in ferric oxide is 70%.

Molecular formula of ferric oxide is [tex]Fe_{2} O_{3}[/tex].

First, we will calculate the total mass of ferric oxide.

Molecular weight of oxygen = 16

Molecular weight of iron = 56

Therefore, total mass of oxygen=3 × 16 =48 g

and total mass of iron=2 × 56 =112 g

Percentage of metal in the chemical = total mass of the metal/ total mass of the compound× 100

Now, total mass of ferric oxide = 48+ 112 =160 g

Percentage of iron in the chemical = total mass of the iron/ total mass of ferric oxide × 100= 112160 × 100=70 %

Therefore, the percentage of iron in ferric oxide is 70%.

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The amount of potassium permanganate (KMnO4) in the volumetric pipette and the 100 ml solution is the same.

When a volumetric pipette is used to take 10 ml of the stock solution, it is designed to deliver an accurate volume of liquid, ensuring that the amount of solute (KMnO4) is accurately transferred to the flask. By adding water to the volumetric flask to make a more dilute solution, the total amount of solute (KMnO4) remains the same, but it is now distributed throughout the larger volume of the flask.

Therefore, there is no more or less KMnO4 in the volumetric pipette or the 100 ml solution. Both contain the same amount of KMnO4, which was accurately transferred using the volumetric pipette. It is important to note that accuracy in transferring the correct volume is critical in ensuring that the concentration of the diluted solution is correct.

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Approximately 5.43 x [tex]10^{13}[/tex] moles of NaCl are extracted each year from the largest salt mine in the world.

What is NaCl?

NaCl is the chemical formula for table salt, which is a compound made up of sodium and chlorine ions. It is a white, crystalline solid that is commonly used as a seasoning and preservative in food, as well as in many industrial processes. NaCl is highly soluble in water and is an important electrolyte in the human body, helping to regulate many physiological processes.

First, we need to convert 7.00 million tons to kilograms:

7.00 million tons x 2,000 lb/ton x 1 kg/2.205 lb = 3.17 x [tex]10^{9}[/tex]kg

Next, we need to calculate the number of moles of NaCl present in this amount of halite:

Molar mass of NaCl = 58.44 g/mol

3.17 x [tex]10^{9}[/tex] kg x 1000 g/kg / 58.44 g/mol = 5.43 x [tex]10^{13}[/tex] moles of NaCl

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Explanation:

We can use the ideal gas law to calculate the mass of the N2 gas sample:

PV = nRT

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

First, we need to convert the temperature to Kelvin:

T = 50 °C + 273.15 = 323.15 K

Now we can rearrange the ideal gas law to solve for the number of moles:

n = PV/RT

n = (3 atm)(0.6 L)/(0.08206 L·atm/mol·K)(323.15 K)

n = 0.0705 mol

The molar mass of N2 is 28.02 g/mol, so we can calculate the mass of the N2 gas sample:

mass = n × molar mass

mass = 0.0705 mol × 28.02 g/mol

mass = 1.98 g

Therefore, the mass of the N2 gas sample is approximately 1.98 g.

The probability that two heterozygous individuals will have a child together with the short earlobes phenotype is 15%.

The likelihood of the genotype and phenotype of the progeny can be calculated using a Punnett square. Let's refer to the long earlobe gene as "L" and the short earlobe allele as "l."

The probability of each genotype is:

LL = 1/4

Ll = 1/2

ll = 1/4

Now we need to take into account the 60% penetrance of the short earlobes genotype. This means that even if an individual has the genotype for short earlobes (ll), they may not actually exhibit the phenotype.

So the probability of having the short earlobes phenotype is:

P(short earlobes phenotype) = P(ll) x 0.6

P(ll) = 1/4

Substituting in:

P(short earlobes phenotype) = (1/4) x 0.6

P(short earlobes phenotype) = 0.15 or 15%

The probability that two heterozygous individuals will have a child together with the short earlobes phenotype is 15%.

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Answer: 0.10

I'm bad at explaining but trust, I had the same question.

There is no reason to run these solutes at the 20 MWCO because they are both much smaller than the MWCO of the membrane.

The MWCO (molecular weight cut off) is the molecular weight of a solute at which it will be retained by a membrane during a process such as ultrafiltration or dialysis. If a solute has a molecular weight higher than the MWCO of a membrane, it will be retained and not pass through the membrane. If the molecular weight of a solute is lower than the MWCO, it will pass through the membrane.

In this case, glucose has a molecular weight of 180 g/mol (6 carbons x 12 g/mol per carbon + 6 oxygens x 16 g/mol per oxygen) and albumin has a molecular weight of approximately 81,942 g/mol (607 amino acids x 135 g/mol per amino acid). Both of these solutes have molecular weights that are much lower than 20,000 g/mol, which is a typical MWCO for ultrafiltration or dialysis membranes.

They would both easily pass through the membrane and be lost during the process. Instead, a membrane with a much lower MWCO would be needed if we wanted to retain these solutes during a process such as ultrafiltration or dialysis.

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The enthalpy change for the reaction Mn(s) + O₂(g) → MnO₂(s) is +1563 kJ/mol

What is Enthalpy?

Enthalpy is a measure of the total heat energy in a thermodynamic system. It is represented by the symbol H and is typically measured in units of joules or calories. Enthalpy can be used to describe the amount of heat that is absorbed or released during a chemical reaction or a phase change in a substance. It is a useful concept in thermodynamics and is commonly used in chemical and physical processes.

To determine AH for Equation b, we can use Hess's Law which states that if a reaction is carried out in a series of steps, the sum of the enthalpy changes for the individual steps will be equal to the enthalpy change for the overall reaction.

First, we need to manipulate Equation a to obtain the same number of moles of MnO₂ as in Equation b.

4Al(s) + 3MnO₂(s) → 2Al₂O3(s) + 3Mn(s) (multiply by 2/3)

8/3 Al(s) + 2MnO₂(s) → 4/3 Al₂O3(s) + 2Mn(s)

Next, we can write the overall reaction as:

8/3 Al(s) + 2MnO₂(s) + 3/2 O₂(g) → 4/3 Al₂O3(s) + 2Mn(s) + O₂(g)

The enthalpy change for this reaction can be calculated by adding the enthalpy change of Equation a and the opposite of the enthalpy change of Equation b (because Equation b is the reverse of the reaction in Equation a):

AH = (-3352 kJ/mol) + (-(-1) * (-1789 kJ/mol))

AH = -3352 kJ/mol + 1789 kJ/mol

AH = -1563 kJ/mol

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