Calculate the values of pH and pOH, based on Kw and the concentration of all species present in a neutral solution of water.

Here are 0.686 moles of hydrogen chloride in 25 grams of HCl.
Number of moles = Mass of substance (in grams) / Molar mass of substance

The molar mass of HCl can be calculated by adding the atomic masses of hydrogen (1.008 g/mol) and chlorine (35.45 g/mol), which gives a molar mass of 36.458 g/mol.
Now we can plug in the values:
Number of moles = 25 g / 36.458 g/mol
Number of moles = 0.686 moles
It's important to note that the molar mass of a substance is the mass in grams of one mole of that substance. This means that if we know the mass of a substance and its molar mass, we can find the number of moles present in that mass. This is a useful calculation in chemistry as it allows us to make accurate measurements and carry out calculations involving the reactions and properties of different substances.

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Answer:

Explanation:

Therefore, the pH of the HCl solution is 5.

The molar solubility of AgbPO₄ in a 0.223 M Na₃PO₄ solution is  2.3 x 10⁻⁷ M.

Given:

The value of Ksp for Ag₃PO₄ = 1.3 x 10⁻²°

The balanced equation is:

Ag₃PO₄(s) ⇌ 3 Ag⁺(aq) + (PO₄)³⁻(aq)

The solubility product expression for this reaction is:

Ksp = [Ag+]³ [PO₄⁻³]

Initial: [Ag⁺] = 0 [PO₄⁻³]

= 0.223 M

Change: +3x +x

Equilibrium: [Ag₊] = 3x [PO₄⁻³]

= 0.223 + x

Substituting these values into the Ksp expression:

Ksp = (3x)³ (0.223 + x)

= 1.3 x 10⁻²⁰

Ksp = 27 x³ (0.223) ≈ 6.0 x 10⁻²⁰

Solving for x:

x ≈ 2.3 x 10⁻⁷ M

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The standard reduction potentials is Ba metal will be oxidized to Ba2+ and the Li+ and Mg2+ ions will be reduced to Li and Mg metal.

When Ba metal is added to an aqueous solution containing dissolved LiCl and MgCl2, Ba metal will be oxidized to Ba2+ ions based on the standard reduction potentials. However, the species that will undergo reduction depends on their respective reduction potentials.

According to the standard reduction potentials, Li+ has a more positive reduction potential than Mg2+, which means Li+ has a greater tendency to undergo reduction compared to Mg2+. Therefore, Li+ ions will be reduced to Li metal while Mg2+ ions will remain in solution.

The overall reaction can be represented as follows:

Ba(s) + 2Li+(aq) → Ba2+(aq) + 2Li(s)

Therefore, the correct answer is Ba metal will be oxidized to Ba2+ and the Li+ and Mg2+ ions will be reduced to Li and Mg metal. Mg2+ ions will not be reduced to Mg metal is incorrect. The formation of H2 gas and -OH ions, which are not supported by the standard reduction potentials is incorrect.  -OH ions are not formed when Li+ ions undergo reduction is incorrect. A reaction does occur based on the standard reduction potentials is incorrect.

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The specific heat of the object is approximately 0.930 J/g°C.To find the specific heat of an object, we can use the formula:Q = mcΔT,Where Q represents the heat energy transferred, m is the mass of the object, c is the specific heat, and ΔT is the change in temperature.

Given that the mass of the object is 75 g and the temperature change is from 50°C to 93°C, we can plug these values into the formula:

3000 J = (75 g) * c * (93°C - 50°C)

Simplifying the equation:

3000 J = (75 g) * c * 43°C

Now, we can solve for c by dividing both sides of the equation by (75 g * 43°C):

c = 3000 J / (75 g * 43°C)

Calculating this expression:

c ≈ 0.930 J/g°C

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The heat energy required to raise the temperature of the vapor q_total = q1 + q2 = 25.76 kJ + 84.39 J = 26.85 kJ.

The heat energy required to convert 66.3 g of liquid SO2 at 201.2 K to gaseous SO2 at 263.1 K is 26.85 kJ.

To solve this problem, we need to calculate the heat energy required to vaporize the given mass of liquid SO2 at its boiling point and then raise the temperature of the resulting vapor to the desired final temperature. The heat energy required for vaporization can be calculated using the molar heat of vaporization of SO2, which is given as 24.9 kJ/mol. The heat energy required to raise the temperature of the vapor can be calculated using the specific heat capacity of liquid SO2, which is given as 1.36 J/g K.

First, we need to calculate the number of moles of SO2 in 66.3 g of liquid SO2 using its molar mass. The molar mass of SO2 is 64.06 g/mol, so:

n = m/M = 66.3 g / 64.06 g/mol = 1.034 mol

The heat energy required for vaporization is then:

q1 = ΔHvap * n = 24.9 kJ/mol * 1.034 mol = 25.76 kJ

Next, we need to calculate the heat energy required to raise the temperature of the resulting vapor from 201.2 K to 263.1 K. The specific heat capacity of liquid SO2 is used because we are raising the temperature of the vapor from the boiling point of liquid SO2.

q2 = n * C * ΔT = 1.034 mol * 1.36 J/g K * (263.1 K - 201.2 K) = 84.39

The total heat energy required is the sum of the heat energy required for vaporization and the heat energy required to raise the temperature of the vapor:

q_total = q1 + q2 = 25.76 kJ + 84.39 J = 26.85 kJ

Therefore, the heat energy required to convert 66.3 g of liquid SO2 at 201.2 K to gaseous SO2 at 263.1 K is 26.85 kJ.

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Answer:So, about 0.093 g of the isotope will remain after 19.8 days.

Explanation:

The first step is to find the number of half-lives that have passed during 19.8 days:

Number of half-lives = time elapsed / half-life

Number of half-lives = 19.8 days / 3.8 days per half-life

Number of half-lives ≈ 5.21

This means that the initial amount of the isotope has been halved 5.21 times. The remaining fraction of the original amount can be calculated using the following formula:

Remaining fraction = (1/2)^(number of half-lives)

Substituting the values, we get:

Remaining fraction = (1/2)^5.21

Remaining fraction ≈ 0.031

Therefore, the amount of the isotope remaining after 19.8 days is:

Remaining amount = Remaining fraction x Initial amount

Remaining amount = 0.031 x 3.00 g

Remaining amount ≈ 0.093 g

So, about 0.093 g of the isotope will remain after 19.8 days.

The isotope in grams will remain after 19.8 days would be 0.081 grams.

The formula to calculate the left mass of a radioactive element can be deduced as -

[tex] \qquad\star\longrightarrow \underline{\boxed{\sf{m =m_{o} \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{t}{T½}} }}} \\[/tex]

Where-

According to the given specific parameters -

Now that we have all the required values, so we can plug them into the formula and solve for the left mass of a radioactive element-

[tex] \qquad \longrightarrow \sf \underline{m =m_{o} \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{t}{T½} }} \\[/tex]

[tex] \qquad\longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{19.8}{3.8} } \\[/tex]

[tex]\qquad \longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{\cancel{19.8}}{\cancel{3.8}} } \\[/tex]

[tex] \qquad\longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ 5.21052..... } \\[/tex]

[tex] \qquad\longrightarrow \sf m =3 \times 0.02700... \\[/tex]

[tex] \qquad\longrightarrow \sf m =0.081020....\;g \\[/tex]

[tex] \qquad\longrightarrow \sf \underline{m =\boxed{\sf{0.081\;g}}} \\[/tex]

Answer: 156.75 J/C

Explanation:

Q=CT (C is specific heat and T is change in temperature)

Disclaimer: I am assuming the initial temperature is 0 degrees Celsius because the question has been worded improperly. This assumption is not made because it is true, but because there is not enough information in the current question to solve for the specific heat. Also, the mass of the sample is given, which is unnecessary for solving the specific heat but necessary to solve for the specific heat capacity. Either the question has not been worded properly or the mass given is just to trick students.

Rearrange to isolate C: C = Q/T

Solve for C: C = 3135/(20-0) = 3135/20 = 156.75 J/C

Answer: 0.137

Explanation:

acellus

Answer:

To solve these problems, we can use the formula:

q = mcΔT

where q is the heat transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the temperature change.

The mass of the sample of tin can be calculated as:

q = mcΔT

36298 J = m × 0.227 J/(g.°C) × (-160.56 °C)

m = 708.2 g

The temperature change of the sample of ammonia can be calculated as:

q = mcΔT

33834 J = 13.66 mol × 80.08 J/(mol.°C) × ΔT

ΔT = 31.7 °C

The mass of the sample of cobalt can be calculated as:

q = mcΔT

455500 J = m × 0.4187 J/(g.°C) × (-1132.52 °C)

m = 27.4 g

The specific heat capacity of indium can be calculated as:

q = mcΔT

12505 J = 372.4 g × c × 140.73 K

c = 0.238 J/(g.°C)

The temperature change of the sample of molybdenum can be calculated as:

q = mcΔT

35961 J = 4.721 mol × 24.06 J/(mol.°C) × ΔT

ΔT = 31.9 °C

The heat transferred by the sample of ethanol can be calculated as:

q = mcΔT

q = 56.2 g × 2.44 J/(g K) × (-110.56 K)

q = -15,585 J

The temperature change of the sample of chromium can be calculated as:

q = mcΔT

38674 J = 5.774 mol × 23.35 J/(mol.°C) × ΔT

ΔT = 27.4 °C

The heat transferred by the sample of magnesium can be calculated as:

q = mcΔT

q = 1.008 mol × 24.9 J/(mol K) × (-683.83 K)

q = -17,134 J

The heat transferred by the sample of tin can be calculated as:

q = mcΔT

q = 0.2687 mol × 27.112 J/(mol K) × 222.48 K

q = 1676.7 J

The specific heat capacity of neon can be calculated as:

q = mcΔT

14738 J = 1.008 mol × c × (-703.43 K)

c = 36.8 J/(mol.°C)

Explanation:

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The pressure of the helium gas (He) in the wet gas mixture is 833.8 mmHg.

In order to find the pressure of the helium gas in the wet gas mixture, we need to use the concept of partial pressures. According to Dalton's Law of Partial Pressures, the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas present in the mixture. In this case, the wet gas mixture contains helium (He) and water vapor (H2O). The pressure of water vapor (H2O) is given as 21.2 mmHg, which means that the partial pressure of water vapor (H2O) in the mixture is 21.2 mmHg. We can now use the total pressure and the partial pressure of water vapor (H2O) to find the partial pressure of helium (He) in the mixture. To do this, we can subtract the partial pressure of water vapor (H2O) from the total pressure:

Partial pressure of helium (He) = Total pressure - Partial pressure of water vapor (H2O)

Partial pressure of helium (He) = 855 mmHg - 21.2 mmHg

Partial pressure of helium (He) = 833.8 mmHg

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The mass of the number of the reacting agent in the reaction producing Cr₂S₃ and H₂O are;

a. The mass of the H₂S is about 126.2 grams

b. Mass of Cr₂S₃ produced is about 242.242 grams

A reacting agent in a chemical reaction are the elements, compounds or molecules on the reaction side of a chemical reaction.

Whereby the chemical reaction is; Cr₂O₃ + 3·H₂S → Cr₂S₃ + 3·H₂O

We get;

5 a. The molar mass of H₂O is; 18.01528 g/mol

The number of moles of H₂O in 66.6 grams of H₂O is; 66.6/18.01528 ≈ 3.7 moles

The number of moles H₂S required to produce 3 moles of H₂O = 3 moles

Therefore, the number of moles H₂S required to produce 3.7 moles of H₂O = 3.7 moles

The molar mass of H₂S =- 34.1 g/mol

5 b. The molar mass of Cr₂S₃ = 200.2 g/mol

The molar mass H₂S = 34.1 g/mol

The mass of the H₂S = 123.7 grams

The number of moles of H₂S is; 123.7 grams/(34.1 g/mol) ≈ 3.63 moles

The stoichiometry of the reaction indicates that the mole ratio of the number of moles of Cr₂S₃ to the number of moles of the H₂S is; 1 : 3

Therefore, the number of moles of the Cr₂S₃ in the reaction is; 3.63 moles/3 ≈ 1.21 moles

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The answer is A: Oil.

Answer:

Explanation:

A. Oil

hope it helps!

The cell potential of a galvanic cell is determined by the difference in the reduction potentials of the two half-cells involved. The larger the difference, the higher the cell potential. The half-reaction with the highest reduction potential will give the largest cell potential when paired with the Ni2+/Ni electrode.

When looking at the reduction potentials, Al3+/Al has a standard reduction potential of -1.66 V, whereas Ni2+/Ni has a standard reduction potential of -0.25 V. Therefore, the reaction with the highest reduction potential difference (i.e., the largest cell potential) when paired with the Ni2+/Ni electrode would be the one that has a reduction potential greater than -0.25 V.
Out of the options given, Al3+/Al has the highest reduction potential and thus it would give the largest cell potential when paired with the Ni2+/Ni electrode. This is because the reduction potential difference between Al3+/Al and Ni2+/Ni is 1.41 V, which is the largest among the given options.
In conclusion, the half-reaction that would give the largest cell potential when paired with a Ni2+/Ni electrode is Al3+/Al.

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The required amount in moles are 3 moles of H₂O to produce 164 g of H₃PO₃.

To solve the problem, use the balanced chemical equation to relate the amount of H₃PO₃ formed to the amount of H₂O used. From the balanced chemical equation:

P2O₃ + 3H₂O → 2H₃PO₃

3 moles of H₂O are required to produce 2 moles of H₃PO₃. This can be written as:

2 moles H₃PO₃ / 3 moles H₂O

To find the number of moles of H₂O required to produce 164 g of H₃PO₃, use the molar mass of H₃PO₃:

1 mole H₃PO₃ = 82 g

So, 164 g of H₃PO₃   is equal to:

164 g H₃PO₃   / 82 g/mol = 2 moles H₃PO₃    

Using the ratio above, calculate the number of moles of H₂O required:

2 moles H₃PO₃ × (3 moles H2O / 2 moles H₃PO₃) = 3 moles H₂O

Therefore, 3 moles of H₂O are required to produce 164 g of H₃PO₃.

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The peak at m/z = 58 corresponds to the molecular ion, [M]

The peak at m/z = 43 corresponds to the fragment ion [M-15]

The peak at m/z = 29 corresponds to the fragment ion [M-29]

The peak at m/z = 15 corresponds to the fragment ion [M-43].

The two possible isomers for this molecule is:

iii)  The molecule is likely methylpropane.

The peaks in the mass spectrum are as follows:

Peak at m/z = 58: This corresponds to the molecular ion, [M], which has a mass of 58. This means that the entire molecule has been ionized and has a charge of +1.

Peak at m/z = 43: This represents the fragment ion [M-15], which has lost a methyl group (CH₃) from the molecular ion.

Peak at m/z = 29: This represents the fragment ion [M-29], which has lost a propyl group (C₃H₇) from the molecular ion.

Peak at m/z = 15: This represents the fragment ion [M-43], which has lost both a methyl group and a propyl group from the molecular ion.

There are two possible isomers for a hydrocarbon with the molecular formula C₄H₁₀:

Butane: CH₃CH₂CH₂CH₃

Methylpropane: CH₃CH(CH₃)CH₃

iii) Based on the relative intensities of the peaks at m/z = 43 and m/z = 29, the molecule is likely methylpropane.

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Freezing, condensation, and deposition are the phase changes that involve the release of energy.

The phase changes that involve the release of energy are exothermic processes, where energy is released from the system into the surroundings. The energy released can be in the form of heat, light, or sound, depending on the nature of the phase change.

Among the common phase changes, the ones that involve the release of energy are:

Freezing: The process of liquid water freezing into ice is an exothermic process, where energy is released in the form of heat.

Condensation: When water vapor condenses into liquid water, energy is released in the form of heat.

Deposition: When water vapor directly changes into solid ice without passing through the liquid phase, energy is released in the form of heat.

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The balanced chemical equation for the combustion of propane with oxygen is: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

To balance the equation, first balance the carbon atoms on both sides of the equation. There are three carbon atoms in the propane molecule and three in the carbon dioxide molecule, so balance the carbon atoms by putting a coefficient of 3 in front of the CO₂ molecule.

C3H8 + 5O2 → 3CO₂

Next, balance the hydrogen atoms. There are eight hydrogen atoms in the propane molecule and four in the water molecule, so balance the hydrogen atoms by putting a coefficient of 4 in front of the H₂O molecule.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Finally, balance the oxygen atoms. There are five oxygen atoms on the left side and 10 on the right side, so balance the oxygen atoms by putting a coefficient of 5 in front of the O₂ molecule.

Therefore, the correct whole number coefficient for propane, C3H8, is 1.

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Cargnello and coworkers developed a novel catalyst that advances this objective by boosting the formation of long-chain hydrocarbons during chemical processes.

Thus, The same amounts of carbon dioxide, hydrogen, catalyst, pressure, heat, and time as the standard catalyst, it created 1,000 times more butane—the longest hydrocarbon it could produce at its maximum pressure—than the standard catalyst.

The novel catalyst is made of ruthenium, a platinum group rare transition metal that is coated in a thin coating of plastic. This idea accelerates chemical processes without being consumed in the process, much like any catalyst.

Another benefit of ruthenium is that it is less expensive than other platinum- and palladium-based high-quality catalysts.

Thus, Cargnello and coworkers developed a novel catalyst that advances this objective by boosting the formation of long-chain hydrocarbons during chemical processes.

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The appropriate response is: based on the facts provided.

A 2e-reducing agent, Na2S2O5, only produces SO3 when interacting with the ion Au3+.(option-4)

Sodium metabisulfite (Na2S2O5) is used, which implies that it is working as a reducing agent. In this case, it is reversibly reducing the Au3+ ion to gold (Au), an elemental form that precipitates as a brown solid.

The following is a representation of the reaction's balanced equation:

2Au (s) + 2SO3 2- (aq) + 2Na+ (aq) + 6H+ (aq) = 2Au3+ (aq) + Na2S2O5 (aq) + 3H2O (l)

The electrons required for reducing Au3+ to Au are here provided by Na2S2O5 acting as a 2e- reducing agent. H+ ions from the excess HCl are also present during the process.(option-4)

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The orbital's orientation in space is described by the magnetic quantum number (m). Any number between -l and +l may represent the value of m.

The electron's orbital form is determined by a quantum number called the azimuthal quantum number. Any integer between 0 and n-1 can be used to represent the value of l, and as it rises, the orbital's form becomes more complex.

The quantum numbers that are involved have been shown above.

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The area of a circle depends on the length of the radius, and the area of a sector depends on the ratio of the central angle to the entire circle, hence options A, B, D, E are correct.

A circle is the location of a point such that it is always a constant distance from a fixed point known as the center.

The statements true regarding the area of circles and sectors are:

The area of a circle depends on the length of the radius.

The area of a sector depends on the ratio of the central angle to the entire circle.

The area of the entire circle can be used to find the area of a sector.

The area of a sector can be used to find the area of a circle.

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Among the given compounds, I would expect the compound F (CH₂CH₂CHCH₃) to rearrange. This expectation is based on the concept of hyperconjugation and the stability of carbocations.

In organic chemistry, carbocations are positively charged carbon species with only six electrons in their valence shell. The rearrangement of a compound occurs when a more stable carbocation can be formed through the shifting of atoms or groups. This rearrangement is driven by the desire to achieve greater stability.

In compound F, the carbon atom adjacent to the positively charged carbon (CH₃⁺) is a primary carbon (attached to only one other carbon atom), making it less stable. By undergoing a rearrangement, this carbon can shift its attachment to the adjacent carbon, creating a secondary carbocation (with two carbon attachments) and a more stable molecule.

The rearrangement would result in the formation of a more stable carbocation, which is energetically favorable. This rearrangement process is known as a 1,2-shift or alkyl shift.

Overall, compound F would be expected to rearrange due to the increased stability offered by the formation of a secondary carbocation, resulting in a more energetically favorable molecular structure.

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The amount of energy needed to heat 500 g of ice at  0⁰C to 60⁰C is 292,400 J. Option C.

In order to calculate the energy required to heat the ice, we need to consider two stages: first, we need to calculate the energy required to melt the ice, and second, we need to calculate the energy required to heat the resulting liquid water to 60°C.

To melt the ice, we need to supply energy equal to the heat of fusion of ice. The heat of fusion of ice is 334 J/g. Therefore, the energy required to melt 500 g of ice is:

Q1 = (334 J/g) x (500 g) = 167,000 J

Once the ice is melted, we need to heat the resulting liquid water to 60°C. The specific heat capacity of water is 4.184 J/(g°C). Therefore, the energy required to heat 500 g of water from 0°C to 60°C is:

Q2 = (4.184 J/(g°C)) x (500 g) x (60°C - 0°C) = 125,520 J

The total energy required to melt the ice and heat the resulting liquid water to 60°C is the sum of Q1 and Q2:

Q = Q1 + Q2 = 167,000 J + 125,520 J = 292,520 J

Thus, the amount of energy needed to heat 500 g of ice at  0⁰C to 60⁰C is 292,400 J.

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The two statements that are true about the reaction are:

A catalyst is described as  a substance that speeds up a chemical reaction, or lowers the temperature or pressure needed to start one, without itself being consumed during the reaction.

If we happen to increase the temperature, it provides more kinetic energy to the reactant molecules and this makes it  more likely to overcome the activation energy barrier and engage in the reaction.

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The rate constant for the reaction can be found out using Arrhenius equation.

Arrhenius equation can be stated as:

[tex]ln\frac{k2}{k1}=\frac{Ea}{R}[\frac{1}{T1}-\frac{1}{T2}][/tex]

i.e [tex]log\frac{k2}{k1} = \frac{Ea}{2.303R} [\frac{1}{T1}-\frac{1}{T2} ][/tex]

i.e [tex]log(k2)-log(k1) = \frac{Ea}{2.303R} [\frac{T2-T1}{T1xT2}][/tex]

From the given data, k1 = 6.1 s⁻¹, T1 = 600K, T2 = 805K, Ea = 262 kJ/mol and R = 8.314 J/molK

Substituting in the Arrhenius equation, we get

[tex]log\frac{k2}{6.1x10^-^8}= \frac{262}{2.303 x 8.314} [\frac{805-600}{600 x 805}][/tex]

[tex]log (k2) = log (k1) + \frac{Ea}{2.303R} [\frac{T2-T1}{T1xT2}][/tex]

[tex]log (k2)= log(6.1x10^-^8) + \frac{262}{2.303x8.314} x \frac{805-600}{600x805}[/tex]

[tex]log(k2)= log (6.1x10^-^8) + 5.81 x 10^-^3\\log(k2) = -7.214 + 0.00581\\log(k2) = -7.21[/tex]

[tex]k2 = antilog (-7.21) = 6.17 x 10^-^8[/tex]

Thus, on solving for k2, we get k2 = 6.17 × 10⁻⁸ s⁻¹

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The balanced equation of the half-reaction for the reduction of BrO3- to Br- in an acidic solution is:

The balanced equation of the half-reaction in an acidic solution is determined as follows:

Unbalanced equation of the  half-reaction: BrO₃⁻ (aq) → Br-

The oxidation number of Br changed from +5 to - 1

Hence, it gained six electrons and 6 electrons are added to the reactant side.

The oxygen atoms are balanced by adding water molecules to the right-hand side of the reaction while hydrogen ions are added to the left-hand since the reaction took place in acidic conditions.

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The specific heat of aluminum is  0.92 J/g°C.

Use the principle of conservation of energy.

Q aluminum = -Qwater-calorimeter

(m aluminum)(c aluminum)(ΔT aluminum) = -(m water + m calorimeter)(c water)(ΔT water)

where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Calculate the heat lost by the aluminum.

Q aluminum = (m aluminum)(c aluminum)(ΔT aluminum)

where ΔT aluminum is the change in temperature of the aluminum when it was heated in boiling water:

ΔT aluminum = 98.7°C - 100°C

= -1.3°C

Q aluminum = (73.5 g)(c aluminum)(-1.3°C)

Q water-calorimeter = -(m water + m calorimeter)(c water)(ΔT water)

where ΔT water is the change in temperature of the water in the calorimeter:

ΔT_water = 28.2°C - 25.4°C

= 2.8°C

Q water-calorimeter = -(1500 g + m calorimeter)(4.18 J/g°C)(2.8°C)

Q water = -(1500 g)(4.18 J/g°C)(2.8°C)

Substitute the values and solve for c aluminum:

(73.5 g)(c aluminum)(-1.3°C) = -(1500 g)(4.18 J/g°C)(2.8°C)

c aluminum = -(1500 g)(4.18 J/g°C)(2.8°C) / (73.5 g)(-1.3°C)

c aluminum = 0.92 J/g°C

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When the products and reactants do not alter over time, we say that a chemical is in equilibrium concentration.  2 mol/ L is the of concentration for each reactant.

When the products and reactants do not alter over time, we say that a chemical is in equilibrium concentration. In other words, a chemical reaction enters a state of equilibrium or equilibrium concentration when the rate of forward reaction equals the rate of backward reaction. CAO=CBO=2m0l/dm³ and KC=10dm³/mol. substituting all the given values we get 2 mol/ L of concentration for each reactant.

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