barium reacts with cobalt (iii) cyanide to produce

The pressure of a gas that occupies 50.5L at 69.0°C is 6.95 atm.

The pressure of an ideal gas can be calculated using Avogadro's equation as follows;

PV = nRT

Where;

According to this question, 12.5 mol of an ideal gas occupies 50.5 L at 69.00°C. The pressure can be calculated as follows:

P × 50.5 = 12.5 × 0.0821 × 342

50.5P = 350.9775

P = 6.95 atm

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Answer:

Explanation:

In the given reaction, carbon has a greater oxidation number in carbon dioxide (CO₂) than in carbon monoxide (CO). In CO₂, the oxidation number of carbon is +4, while in CO it is +2.

A silver bar with the mass of the 300 grams is heated from the 30 °C to 55 °C. The amount heat does the silver bar absorb in the joules is 1762.5 J.

The mass of the silver bar = 300 g

The initial temperature = 30 °C

The final temperature = 55 °C

The heat energy is expressed as :

Q = mc ΔT

Where,

The m is mass of the silver bar = 300 g

The c is the specific heat capacity = 0.235 J/g °C

The ΔT is the change in the temperature = final temperature - initial temperature

The ΔT is the change in the temperature = 55 °C - 30 °C

The ΔT is the change in the temperature = 25 °C

The heat energy, Q = 300 × 0.235 × 25

The heat energy, Q = 1762.5 J

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Answer:

0.0104 moles of gas in the flask.

Explanation:

To calculate the number of moles of gas in the flask, you can use the ideal gas law equation: PV = nRT. Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant and T is temperature.

First, you need to convert the pressure from mmHg to atm and the temperature from Celsius to Kelvin. The pressure in atm is 760.0 mmHg / 760 mmHg/atm = 1 atm. The temperature in Kelvin is 17.00°C + 273.15 = 290.15 K.

Next, you need to convert the volume from mL to L. The volume in L is 250.0 mL / 1000 mL/L = 0.2500 L.

Now you can plug all the values into the ideal gas law equation and solve for n: (1 atm)(0.2500 L) = n(0.08206 L·atm/mol·K)(290.15 K). Solving for n gives n = 0.0104 mol.

So there are approximately 0.0104 moles of gas in the flask.

The mass of the silver block, given that it was initially at 55.1 °C  and is submerged into 100.0 g of water at 25.0°C is 189.8 g

We'll begin our calculation by obtaining the heat absorbed by the water. Details below:

Q = MCΔT

Q = 100 × 4.184 × 2.9

Q = 1213.36 J

Finally, we shall determine the mass of the silver block. Details below:

Q = MCΔT

-1213.36 = M × 0.235 × -27.2

-1213.36 = M × -6.392

Divide both sides by -6.392

M = -1213.36 / -6.392

M = 189.8 g

Thus, we can conclude that the mass of the silver block is 189.8 g

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Complete question:

A silver block, initially at 55.1∘C, is submerged into 100.0 g of water at 25.0∘C in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 27.9∘C. The specific heat capacities for water and silver are Cs,water = 4.18J/(g⋅∘C) and Cs, silver = 0.235J/(g⋅∘C). What is the mass of the silver block?

A chemical interaction between an acid and a base is known as an acid-base reaction.

Thus, These are known as acid-base theories, such as the Brnsted-Lowry acid-base theory, and they offer alternative conceptions of the reaction mechanisms and their application in solving related problems.

When examining acid-base reactions for gaseous or liquid species, or when the acid or basic character may be less obvious, their significance becomes clear.

The relative potency of the conjugated acid-base pair in the salt controls the pH of its solutions when weak acids and bases react. The resulting salt or its solution can be basic, neutral, or acidic. A strong acid and a weak base can combine to generate an acid salt.

Thus, A chemical interaction between an acid and a base is known as an acid-base reaction.

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273.8 grams of zinc (Zn) are needed to produce 8.45 grams of hydrogen gas [tex](H_2).[/tex]

In this chemical reaction, zinc (Zn) reacts with sulfuric acid [tex](H_2SO_4)[/tex] to produce zinc sulfate  [tex](ZnSO_4)[/tex] and hydrogen gas [tex](H_2).[/tex]

From the balanced chemical equation, we can see that 1 mole of zinc (Zn) reacts with 1 mole of sulfuric acid [tex](H_2SO_4)[/tex] to produce 1 mole of hydrogen gas [tex](H_2).[/tex]

Hydrogen gas  has a molar mass of 2.016 g/mol.

Therefore, 8.45 grams of hydrogen gas is equal to 8.45 g / 2.016 g/mol = 4.19 moles of [tex]H_2[/tex].

Since 1 mole of Zn reacts with 1 mole of [tex]H_2[/tex] , we need 4.19 moles of Zn to produce 4.19 moles of [tex]H_2[/tex].

molar mass of Zinc = 65.38 g/mol.

Therefore, 4.19 moles of Zn has a mass of 4.19 moles x 65.38 g/mol = 273.8 grams.

Therefore, you would need 273.8 grams of zinc (Zn) to produce 8.45 grams of hydrogen gas [tex](H_2).[/tex]

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The solubility of NH₃ in water at 90°C is approximately 0.03 g per 100 g of water.

The solubility can be determined from a solubility table or by using the appropriate equilibrium constant.

According to a solubility table, the solubility of ammonia in water at 90°C is approximately 88 g per 100 g of water.

Alternatively, the equilibrium constant for the dissolution of ammonia in water at 90°C can be used to calculate the solubility.

The equilibrium constant (K) for the reaction:

NH3 (g) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)

is approximately 1.76 x 10⁻⁵ at 90°C.

Using the equilibrium constant expression:

K = [NH4+][OH-]/[NH3][H2O]

Assuming that the concentration of water remains constant at 100 g per 100 g of solution, and that the concentration of NH4+ and OH- are negligible compared to that of NH3, the solubility of NH3 can be calculated as:

[NH3] = K[H2O] = 1.76 x 10⁻⁵ x 100 = 1.76 x 10⁻³ mol/L

Converting to grams per 100 g of water:

1.76 x 10⁻³ mol/L x 17.03 g/mol = 0.03 g/100 g of water

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The complete question is below:

determine the solubility of NH₃ in water at 90° C​

The nucleus completing the following equation is option C: ₂₄⁵⁰Cr.

This reaction is a type of radioactive nuclei decay.

Radioactive decay is the process by which unstable atomic nuclei undergo spontaneous transformations in order to achieve a more stable state. This is accomplished by the emission of particles and/or electromagnetic radiation from the nucleus. The decay may occur by several mechanisms, including alpha decay, beta decay, gamma decay, and electron capture.

In alpha decay, the nucleus emits an alpha particle, which consists of two protons and two neutrons, resulting in a daughter nucleus that has two fewer protons and two fewer neutrons than the original nucleus.

In beta decay, a neutron in the nucleus is converted into a proton and an electron, and the electron is then emitted from the nucleus as a beta particle. This results in the daughter nucleus having one more proton and one fewer neutron than the original nucleus.

In gamma decay, the nucleus emits a gamma ray, which is a high-energy electromagnetic radiation, without changing the number of protons or neutrons in the nucleus.

In electron capture, an electron from the inner shell of the atom is captured by the nucleus, and a proton in the nucleus is converted into a neutron. This results in the daughter nucleus having one fewer proton and one more neutron than the original nucleus.

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[tex]FeCl_3[/tex] is added to ensure all reducing agent is oxidized, indicating endpoint in dichromate titration.

In a dichromate titration,   [tex]FeCl_3[/tex] is frequently added to the example arrangement as a sign of the endpoint. The expansion of [tex]FeCl_3[/tex] to the example arrangement assists with guaranteeing that the lessening specialist has been all oxidized by the potassium dichromate ([tex]K_2Cr_2O_7[/tex] ) arrangement.

[tex]FeCl_3[/tex] responds with any overabundance[tex]K_2Cr_2O_7[/tex]  in the answer for structure a red-earthy colored encourage of[tex]Fe(OH)_3[/tex] , demonstrating that the lessening specialist has been all oxidized. This response is known as a "back-titration" since overabundance[tex]K_2Cr_2O_7[/tex]   is added to the arrangement, trailed by the option of [tex]FeCl_3[/tex] to decide how much unreacted [tex]K_2Cr_2O_7[/tex] .

The response somewhere in the range of [tex]FeCl_3[/tex] and [tex]K_2Cr_2O_7[/tex] can be addressed as:

[tex]6FeCl_3 + K_2Cr_2O_7 + 7H_2SO_4 → 3Fe_2(SO_4)_3 + Cr_2(SO_4)_3 + K_2SO_4 + 7H_2O + 3Cl_2[/tex]

The [tex]FeCl_3[/tex] goes about as a pointer in this response since it responds with the overabundance [tex]K_2Cr_2O_7[/tex] until the decreasing specialist has been all oxidized. As of now, the red-earthy colored encourage of [tex]Fe(OH)_3[/tex]  structures, showing that the endpoint has been reached.

Without the expansion of [tex]FeCl_3[/tex], it would be challenging to precisely decide the endpoint of the titration. The expansion of [tex]FeCl_3[/tex] is important to guarantee that the lessening specialist has been all oxidized and that the endpoint has been reached, considering a more exact assurance of the grouping of the diminishing specialist in the example arrangement.

In outline, the expansion of [tex]FeCl_3[/tex] to the example arrangement in a dichromate titration is significant in light of the fact that it assists with guaranteeing that the decreasing specialist has been all oxidized, considering a more precise assurance of its focus.

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The complete question is -

What is the reason for a dichromate titration, and how does [tex]FeCl_3[/tex]support deciding the endpoint of the titration? Might you at any point give the compound condition to the response somewhere in the range of [tex]FeCl_3[/tex] and [tex]K_2Cr_2O_7[/tex] and make sense of why [tex]FeCl_3[/tex] goes about as a marker in this response?

We can use the equation:

q = m * c * ΔT

where q is the heat absorbed or released by the water, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature of the water.

Since we know that the calorimeter contains 100 mL (or 100 g, since 1 mL of water has a mass of 1 g) of water and that the temperature of the water increased by 9.3°C, we can plug in these values:

q = (100 g) * (4.184 J/g-°C) * (9.3°C)

q = 3896.68 J

However, this is not the total amount of heat produced by the reaction. We need to take into account the heat absorbed by the calorimeter itself, which has a heat capacity of 50.2 J/°C. If we assume that the temperature of the calorimeter did not change during the reaction (i.e., it remained constant), we can calculate the heat absorbed by the calorimeter:

q_calorimeter = (50.2 J/°C) * (9.3°C)

q_calorimeter = 466.86 J

The total heat produced by the reaction is then:

q_reaction = q_water + q_calorimeter

q_reaction = 3896.68 J + 466.86 J

q_reaction = 4363.54 J

Therefore, the heat produced by the reaction is 4363.54 J.

Answer:

true

Explanation:

The volume of N2 produced each day is approximately 24.56 L.

First, let's find the number of moles of NO3- that flow into the swamp each day:

NO3- molar mass = 14.01 (N) + 3 * 16.00 (O) = 62.01 g/mol

135 g / 62.01 g/mol ≈ 2.177 moles of NO3-

In the denitrification process, NO3- is reduced to N2 gas. The balanced equation for denitrification is:

2NO3- → N2 + 3O2

From the stoichiometry of the reaction, we can see that 2 moles of NO3- produce 1 mole of N2. Therefore, the moles of N2 produced each day can be calculated as follows:

moles of N2 = 2.177 moles of NO3- / 2 ≈ 1.0885 moles of N2

Now we can use the ideal gas law equation to find the volume of N2 produced:

PV = nRT

where:

P = Pressure (1.00 atm)

V = Volume (in Liters)

n = Moles of N2 (1.0885 moles)

R = Ideal gas constant (0.0821 Latm/molK)

T = Temperature (17.0°C or 290.15 K)

Solving for the volume of N2:

V = nRT / P

V = (1.0885 moles) * (0.0821 Latm/molK) * (290.15 K) / (1.00 atm)

V ≈ 24.56 L

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Final temperature: 677.4K. Work done: -7026J.

Heat exchanged: 0J. Change in internal energy: -7026J.

(i) For an adiabatic process, T1(V1)^γ-1 = T2(V2)^γ-1.

When we substitute the values (γ=1.4, T1=300K, V1=6dm³, V2=2dm³), we get T2 = 677.4K.

(ii) w = -(P1V1 - P2V2)/(γ-1) = -(nRT1 - nRT2)/(γ-1) = -5 * 8.314 * (677.4 - 300) / 0.4 = -7026J.

For adiabatic, q = 0. ΔE = q + w = -7026J (since q=0).

Final temperature: 677.4K. Work done: -7026J.

Heat exchanged: 0J. Change in internal energy: -7026J.

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The S-P difference (sec) is the time gap between the arrival of the S-wave and the arrival of the P-wave at a seismic station. The S-P discrepancy is depicted in the figure as 20 seconds.

The amplitude (mm) of a seismic wave is the largest displacement from its resting point. The amplitude of the waves is not depicted in the image and cannot be calculated based on the information provided.

Distance (km): Using the S-P time difference and the known velocity of seismic waves, the distance from the seismic station to the earthquake epicenter may be determined. Seismic wave velocity is determined by the type of wave and the features of the Earth's interior. The velocity of P-waves in the Earth's crust, for example, is around 6 km/s. We may compute the distance to the epicenter using this value and the S-P difference of 20 seconds as follows:

Distance = Speed x Time = 6 km/h x 20 seconds = 120 kilometres

As a result, the distance between the seismic station and the earthquake epicenter is about 120 km.

The magnitude of an earthquake (M) is a measurement of the energy generated by the earthquake based on the amplitude of the seismic waves and the distance to the epicenter. Magnitude is commonly measured on a logarithmic scale, with each whole number reflecting a factor of ten increase in energy release.

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According to the law of conservation of mass, the mass of the reactant must be equal to the mass of the product, hence the reaction is wrong

The law of conservation of mass states that mass within a closed system remains the same over time.

It states that the mass in an isolated system can neither be created nor be destroyed but can be transformed from one form to another.

Thus,  the mass of the reactants must be equal to the mass of the products for a low energy thermodynamic process.

From the information given, we have the reaction written as;

HCI + NaOH → NaCl

The mass of the reactant Hydrogen(H) is not found on the product

The mass of the reactant(Oxygen) is also not found

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We can use the combined gas law to determine the volume of the balloon at a higher altitude. The combined gas law relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the pressure, volume, and temperature of the gas at the initial state, and P2, V2, and T2 are the pressure, volume, and temperature of the gas at the final state.

We are given the initial pressure (P1 = 761 mmHg), volume (V1 = 56.0 L), and temperature (T1 = 23.1 °C = 296.25 K) of the gas, and the final pressure (P2 = 0.0772 atm), and temperature (T2 = -6.97 °C = 266.18 K) of the gas. We can solve for V2, the final volume of the gas:

(P1 x V1) / T1 = (P2 x V2) / T2

V2 = (P1 x V1 x T2) / (P2 x T1)

V2 = (761 mmHg x 56.0 L x 266.18 K) / (0.0772 atm x 296.25 K)

V2 = 2,040 L (rounded to three significant figures)

Therefore, the volume of the weather balloon at the higher altitude is approximately 2,040 L.

(a) To calculate the mass of sulfuric acid in the truck, we can multiply the volume of sulfuric acid by its density. Given that the truck carries 34,000 liters of sulfuric acid and the density of sulfuric acid is 1.84 kg/L.

we can use the formula:

Mass (m) = Volume (V) × Density (D)

Plugging in the given values:

Volume (V) = 34,000 L Density (D) = 1.84 kg/L

m = 34,000 L × 1.84 kg/L

m ≈ 62,560 kg (rounded to the nearest whole number)

Therefore, the mass of sulfuric acid in the truck is approximately 62,560 kg.

(b) The amount of sulfuric acid in the truck is already given in the question as 34,000 L (volume).

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(a) To find the mass of sulfuric acid in the truck, we need to use the formula:
mass = density x volume

The volume of sulfuric acid in the truck is given as 34,000 L. The density of sulfuric acid is 1.84 kg/L. Therefore, the mass of sulfuric acid in the truck is:

mass = 1.84 kg/L x 34,000 L = 62,560 kg
So there are 62,560 kg of sulfuric acid in the truck.
(b) To find the amount of sulfuric acid in the truck, we need to use the formula:
amount = mass / molar mass
The molar mass of sulfuric acid is 98.08 g/mol. To convert the mass from kg to g, we need to multiply by 1000. Therefore, the amount of sulfuric acid in the truck is:
amount = 62,560,000 g / 98.08 g/mol = 636,816.3 mol
So there are 636,816.3 moles of sulfuric acid in the truck.

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The oxidation number of the unknown chlorine in the compound is + 5

The oxidation number of an element in a compound is determined by a set of rules based on its position in the periodic table, as well as the charges of other atoms in the compound

We know that the oxidation number of the chlorine which we want to obtain would be designated as x and the total of the oxidation numbers of the elements in the compound is zero.

Thus we have that;

1 + x + 3(-2) = 0

1 + x - 6 = 0

-5 + x = 0

x = 5

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5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.

Molarity (M) is defined as the moles of solute per liter of the solution.

Balanced chemical equation is : 2Al + 6HCl → 2AlCl₃ + 3H₂

From the equation, we can see that 2 moles of Al react with 6 moles of HCl to produce 2 moles of AlCl₃ and 3 moles of H₂.

As moles of HCl = Molarity × Volume

moles of HCl = 1.65 mol/L × 0.352 L

moles of HCl = 0.58128 mol

and moles of Al = (2/6) × moles of HCl

moles of Al = (1/3) × 0.58128 mol

moles of Al = 0.19376 mol

mass of Al = moles of Al × molar mass of Al

mass of Al = 0.19376 mol × 26.98 g/mol

mass of Al = 5.221 g

So, 5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.

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A 6.50-g sample of copper metal at 25.0 °C is heated by the addition of 145 J of energy. The final temperature of the copper is 83.7 °C. The specific heat capacity of copper is 0.38 J/g-K.

The correct answer choice is "83.7"

To solve this problem, we can use the equation:

q = mcΔT

where q is the amount of energy absorbed by the copper, m is the mass of the copper, c is the specific heat capacity of copper, and ΔT is the change in temperature of the copper.

Rearranging this equation to solve for ΔT, we get:

ΔT = q / (mc)

Substituting the given values, we get:

ΔT = 145 J / (6.50 g x 0.38 J/g-K)

ΔT = 58.7 K

Therefore, the final temperature of the copper is:

25.0 °C + 58.7 °C = 83.7 °C

So the correct option is 83.7.

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Answer: D

Explanation:

When 12.50 mL of 1.05 M KOH is added to 50.0 mL of 0.225 M HBr, the resulting solution has a pH of 13.35.

Here’s how to calculate it:

First, we need to determine the number of moles of KOH and HBr in the solution:

moles of KOH = (12.50 mL) * (1.05 mol/L) * (1 L/1000 mL) = 0.013125 mol moles of HBr = (50.0 mL) * (0.225 mol/L) * (1 L/1000 mL) = 0.01125 mol

KOH is a strong base and HBr is a strong acid, so they will react completely to form water and a salt (KBr):

KOH + HBr -> KBr + H2O

The number of moles of KOH is greater than the number of moles of HBr, so there will be an excess of KOH in the solution after the reaction is complete:

moles of excess KOH = moles of KOH - moles of HBr = 0.013125 mol - 0.01125 mol = 0.001875 mol

The total volume of the solution is the sum of the volumes of KOH and HBr:

total volume = 12.50 mL + 50.0 mL = 62.5 mL

The concentration of excess OH- ions in the solution is:

[OH-] = moles of excess KOH / total volume = 0.001875 mol / (62.5 mL * (1 L/1000 mL)) = 0.03 M

The pOH of the solution can be calculated using the formula pOH = -log[OH-]:

pOH = -log(0.03) = 1.52

The pH can be calculated using the formula pH + pOH = 14:

pH = 14 - pOH = 14 - 1.52 = 13.35

So the correct answer is D. 13.35.

Answer:

Explanation:

[tex]\frac{918.7 g}{1} *\frac{1}{216.59m } = 4.241 mol[/tex] To start off the mol of HgO must be found.

After that the molar ratio between HgO and O must be found but in this case its 1:1

[tex]4.241 mol HgO*\frac{1 molO}{1molHgO} = 4.241 mol O[/tex] the mols of HgO is put on the bottom to cancel out  with the other one leaving just mols of oxygen. Finally to find g of oxygen it must be multiplied by its molar mass.

[tex]\frac{4.241 molO}{1} * \frac{15.999 g}{mol} = 67.85 g[/tex] Oxygen

Answer:

OB. formation of new elements.

Nuclear reactions involve changes in the nucleus of an atom, such as the splitting of a nucleus or the combining of two nuclei. These reactions can result in the formation of new elements, as the number of protons in the nucleus determines the element. In contrast, chemical reactions involve the rearrangement of electrons between atoms to form new compounds, but do not involve changes to the nucleus.

The mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.

we can use the principles of chemical equilibrium and the mole fraction formula.

First, we need to write the balanced chemical equation for the reaction involving A, B, C, and D. Let's assume that the reaction is:

A + 2B <=> C + D

where A, B, C, and D are the chemical species, and the coefficients indicate their stoichiometric ratios.

Next, we need to write the expression for the equilibrium constant, Kc, for this reaction:

Kc = [C][D] / [A][B]²

where [X] denotes the molar concentration of species X at equilibrium.

Since we know the initial moles of A, B, and D, we can calculate their total moles in the mixture:

Total moles = 1 mol A + 2 mol B + 1 mol D = 4 mol

We also know that the final mixture contains 0.9 mol of C. Therefore, the molar concentration of C at equilibrium is:

[C] = 0.9 mol / 4 L = 0.225 M

Since we have only one unknown, we can use the equilibrium constant expression to calculate the molar concentration of D:

Kc = [C][D] / [A][B]²

0.9 = (0.225)(D) / (1)(2²)

D = 1.8

Therefore, the molar concentration of D at equilibrium is 1.8 M.

Using the law of conservation of mass, we can also calculate the molar concentration of A and B at equilibrium:

[A] = 1 mol / 4 L = 0.25 M

[B] = 2 mol / 4 L = 0.5 M

Mole fraction of X = moles of X / total moles

Mole fraction of A = 1 mol / 4 mol = 0.25

Mole fraction of B = 2 mol / 4 mol = 0.5

Mole fraction of C = 0.9 mol / 4 mol = 0.225

Mole fraction of D = 1 mol / 4 mol = 0.25

Therefore, the mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.

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The equation that shows that the total mass during a chemical reaction stays the same is 2) H₂ + O₂ → H₂O.

This is an example of a balanced chemical equation where the number of atoms of each element on both the reactant and product side is equal. In other words, the total number of atoms of each element is conserved, and therefore the total mass is conserved. In the other reactions, either the number of atoms on the product side is different from the reactant side or there is no reaction at all.

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The mass (in grams) of aluminum, Al that must react with iodine to form AlI₃, given that -836.0 KJ of heat is relaesd is 149.0 g

The mass of aluminum required to react with iodine to produce AlI₃ can be obtain as shown below:

2Al(s) + 3I₂(s) → 2AlI₃(s) ∆H = -302.9 KJ

From the balanced equation above,

When -302.9 KJ of heat energy is released, 54 g of aluminum, Al reacted.

Therefore,

When -836.0 KJ of heat energy will be release = (-836.0KJ × 54 g) / -302.9 KJ = 149.0 g of aluminum, Al will react.

Thus, from the above calculation, we can conclude that the mass of aluminum, Al required is 149.0 g

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