2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)
If 538 mol
of octane combusts, what volume of carbon dioxide is produced at 31.0 ∘C
and 0.995 atm?

The Gibbs free energy change for the given reaction at 25°C and the given concentrations is -25.5 kJ/mol

The Gibbs free energy change (∆G) of a reaction can be calculated using the equation:

∆G = -RT ln(K)

Where R is the gas constant (8.314 J/molK), T is the temperature in Kelvin, and K is the equilibrium constant.

The equilibrium constant (K) can be calculated from the acid dissociation constant (Ka) as:

K = [C] ÷ ([A] × [B])

Substituting the given values, we get:

K = (5.00 x 10⁻⁵) ÷ (1.50 x 1.00) = 3.33 x 10⁻⁵

Therefore,

∆G = - (8.314 J/molK) × (298 K) × ln(3.33 x 10⁻⁵)

= 25.5 kJ/mol

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1. The star that has a temperature of approximately 9000 K and luminosity about 20 times greater than the Sun’s luminosity is Vega.

2. The type of star that is considered part of the main sequence is red dwarfs.

3. The star that is cooler than the Sun is Barnard’s Star.

4. The Sun is classified as a main sequence star.

5. The force most responsible for the formation of stars is gravity.

6. The star that has a higher luminosity and a lower temperature than the Sun is Aldebaran.

7. Compared to the temperature and luminosity of the star Polaris, the star Sirius is hotter and more luminous.

The theoretical yield of sodium sulfate, Na₂SO₄, formed from the reaction of 4.9 g of sulfuric acid, H₂SO₄ and 5.0 g of sodium hydroxide, NaOH is 7.1 g

First, we shall determine the limiting reactant. This is shown below:

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

From the balanced equation above,

98 g of H₂SO₄ reacted with 80 g of NaOH

Therefore,

4.9 g of H₂SO₄ will react with = (4.9 × 80) / 98 = 4 g of NaOH

From the above calculation, we can see that only 4 g of NaOH out of 5 g is needed to react with 4.9 g H₂SO₄.

Thus, the limiting reactant is H₂SO₄

Finally, we shall determine theoretical yield of sodium sulfate, Na₂SO₄ formed. Details below:

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

From the balanced equation above,

98 g of H₂SO₄ reacted to produce 142 g of Na₂SO₄

Therefore,

4.9 g of H₂SO₄ will react to produce = (4.9 × 142) / 98 = 7.1 g of Na₂SO₄

Thus, the theoretical yield of sodium sulfate, Na₂SO₄ formed is 7.1 g

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Charles's law of gases states that the density of an ideal gas is inversely proportional to its temperature at constant pressure.

The equation is as follows;

Va/Ta = Vb/Tb

Where;

0.67/362 = 1.12/Tb

0.00185Tb = 1.12

Tb = 605.41K

This temperature in °C is 605.41 - 273 = 332°C

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We would need about 16 hydrogen atoms so that we can convert the compound to a saturated fat.

In animal products like meat and dairy, saturated fat is a form of dietary fat that is normally solid at room temperature. It is known as being "saturated" because each molecule of fat has the most hydrogen atoms possible, giving it a stable structure.

We can see this by counting the number of double bonds in the fat and there are eight of them so sixteen hydrogen atoms are needed for saturation.

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The equation that represents a balanced reaction involving 14 atoms of oxygen is:

2C2H6 + 7O2 --> 4CO2 + 6H2O

a. The relationship between variables in this equation is that 2 moles of C2H6 (ethane) react with 7 moles of O2 (oxygen) to produce 4 moles of CO2 (carbon dioxide) and 6 moles of H2O (water). This equation follows the law of conservation of mass, where the total number of atoms on both sides of the equation is the same, indicating a balanced reaction.

b. The graph is linear, as the coefficients of the reactants and products in the equation are whole numbers and form a consistent ratio. The coefficients of 2, 7, 4, and 6 represent the stoichiometry of the reaction, indicating a fixed relationship between the reactants and products.

c. An example of a situation where this balanced equation could be applicable is the combustion of ethane (C2H6) in the presence of excess oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), which is a common reaction in the combustion of hydrocarbon fuels. The equation represents the balanced stoichiometry of this reaction, where 2 moles of ethane react with 7 moles of oxygen to produce 4 moles of carbon dioxide and 6 moles of water, involving a total of 14 atoms of oxygen in the reaction.

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Plants rely on their leaves to produce food through a process called photosynthesis. This involves converting light energy into organic compounds, like sugars, using chloroplasts that contain the pigment chlorophyll. By combining carbon dioxide and water with light energy, plants create glucose, which serves as an energy source and building material. Along with stomata, which help regulate gas exchange with the environment, the leaf acts as the plant's primary kitchen for food production.

The value of K at equilibrium, for the reaction is 0.0049

First, we shall list out the given parameters from the question. This is shown below:

Equilibrium constant is defined as:

Equilibrium constant = [Product]ᵐ / [Reactant]ⁿ

Where

With the above formula, we can obtain the equilibrium constant, K as follow:

Equilibrium constant, K = [B₂][AC] / [AB₂C]

K = (0.007 × 0.0118) / 0.0168

K = 0.0049

Thus, the equilibrium constant, K for the reaction is 0.0049

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The rate constant, k = 5.27 E-2 s⁻¹, determines the rate law for the reaction P → E + Z.

The rate of the reaction P → E + Z can be expressed as:

Rate = - d[P]/dt = d[E]/dt = d[Z]/dt

where d[P], d[E], and d[Z] = changes in the concentrations of P, E, and Z, respectively, over a small time interval dt.

Use the experimental data to determine the rate constant and the order of the reaction.

Calculate the initial rate of the reaction in each trial by dividing the change in concentration of P by the time interval:

rate1 = (d[P]/dt)1 = (0.30 M - 0 M)/(20 s) = 0.015 M/s

rate2 = (d[P]/dt)2 = (0.60 M - 0.30 M)/(20 s) = 0.015 M/s

rate3 = (d[P]/dt)3 = (0.90 M - 0.60 M)/(20 s) = 0.015 M/s

The initial rates are the same in all three trials, which suggests that the reaction is first-order with respect to P.

Now using any of the three trials to determine the value of the rate constant k, trial 1:

Rate1 = k[P]1

k = Rate1/[P]1 = (1.58 E-2 M/s)/(0.30 M) = 5.27 E-2 s⁻¹

Therefore, the rate law for the reaction P → E + Z is:

Rate = k[P]

where k = 5.27 E-2 s⁻¹ is the rate constant.

Use the rate law to calculate the expected rates of the reaction at different concentrations of P. For example:

Rate2 = k[P]2 = (5.27 E-2 s⁻¹)(0.60 M) = 3.16 E-2 M/s

Rate3 = k[P]3 = (5.27 E-2 s⁻¹)(0.90 M) = 4.74 E-2 M/s

These expected rates are close to the experimental rates, which suggests that the rate law is a good approximation for the reaction under these conditions.

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Answer:

solution resists changes in pH

Explanation:

the inherent property of buffers is to resist change to ph even when acids and bases are added. when the base is added, it is quickly neutralized by the conjugate acid, so the ph won't change.

Answer:

B: The solutions resists changes in pH.

Explanation:

Buffer reactions maintain stable pH of solutions.

The reaction is second order with a rate constant of 0.043 mol/l min.

For CA = 0.1 mol/l, -rA = 0.1 mol/l min

For CA = 0.2 mol/l, -rA = 0.3 mol/l min

For CA = 0.3 mol/l, -rA = 0.5 mol/l min

For CA = 0.4 mol/l, -rA = 0.6 mol/l min

The slope of this line is equal to the order of the reaction (n), and the y-intercept is ln(k).

Slope = (0.6931 - (-2.3026)) / (0.3010 - (-0.9163)) = 1.929

ln(k) = -2.3026 + 1.929 * (-0.3010)

ln(k) = -3.1504

k = e^(-3.1504) = 0.043 mol/l min

The reaction is second order with a rate constant of 0.043 mol/l min.

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The molar volume of a gas can be calculated using the ideal gas law:

PV = nRT

where P is the pressure of the gas in atmospheres (atm), V is the volume of the gas in liters (L), 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 of the gas in Kelvin (K).

To solve for the molar volume of CO2 at 39°C (312 K) and 652 torr (0.859 atm), we can rearrange the ideal gas law as follows:

V = (nRT) / P

First, we need to calculate the number of moles of CO2. We can use the following equation, which relates the pressure, volume, number of moles, and temperature of a gas:

PV = nRT

Solving for n, we get:

n = (PV) / (RT)

Substituting the given values, we get:

n = (0.859 atm * V) / (0.08206 L·atm/mol·K * 312 K)

Now we can substitute this expression for n into the equation for the molar volume:

V = (nRT) / P

V = [(0.859 atm * V) / (0.08206 L·atm/mol·K * 312 K)] * (0.08206 L·atm/mol·K * 312 K) / (0.859 atm)

Simplifying, we get:

V = 24.45 L/mol

Therefore, the molar volume of CO2 at 39°C and 652 torr is 24.45 L/mol.

Answer:

The answer is 966.67 mm of Hg.

Explanation:

To solve this problem, we can use Boyle's Law which states that the pressure of a gas is inversely proportional to its volume when the temperature and the amount of gas are kept constant. The formula for Boyle's Law is:

P1V1 = P2V2

where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.

Using the given values:

P1 = 580 mmHg

V1 = 500 cm³

V2 = 300 cm³

We can solve for P2:

P1V1 = P2V2

580 mmHg x 500 cm³ = P2 x 300 cm³

290,000 mmHg·cm³ = P2 x 300 cm³

P2 = 290,000 mmHg·cm³ / 300 cm³

P2 = 966.67 mmHg (rounded to the nearest hundredth)

Therefore, the pressure when the volume is reduced to 300 cm³ is approximately 966.67 mmHg.

The chemical molecule CrO2 is also known as chromium(IV) oxide or chromic acid. It has the molecular formula CrO2 and is an inorganic substance.

In the solid state, CrO2 exists as a solid with a layered structure, and it is considered a cationic compound. The cation present in CrO2 is chromium(IV) ion, denoted as Cr4+.

On the other hand, the anion present in CrO2 is oxide ion, denoted as O2-. The oxidation state of oxygen in this compound is -2.

So, the cations present in CrO2 are Cr 4+ ions, and the anions present are O2 -2 ions.

In CrO2, the cation present is Chromium (Cr) with a charge of +4, and the anion present is Oxygen (O) with a charge of -2.

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Answer:

zn + Hcl cual es su rreaccion

When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.

Thus, The adjustments required to reach equilibrium might not be as obvious if we have a mixture of reactants and products that have not yet reached equilibrium.

In this situation, we can compare the Q and K values for the system to forecast changes.

By adding or withdrawing one or more of the reactants or products, an equilibrium chemical system can be momentarily moved out of equilibrium. After that, additional adjustments are made to the reactant and product concentrations in order to bring the system back to equilibrium.

Thus, When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.

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The equilibrium constant is written as;

Keq = [tex][D]^2 [C]^5/[A] [B]^3[/tex]

The equilibrium constant's value is influenced by the reaction's chemical make-up and temperature.

The product of the product concentrations, each raised to the power of their stoichiometric coefficient, divided by the product of the reactant concentrations, each raised to the power of their stoichiometric coefficient, is known as the equilibrium constant.

The equilibrium constant is Keq = [tex][D]^2 [C]^5/[A] [B]^3.[/tex]

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If a gaseous sample has a density of 0.978 g/L at 30 °C and 615 torr,  the molar mass of the compound is 24.8 g/mol.

To calculate the molar mass of the compound, we first need to calculate the number of moles present in the gaseous sample using the ideal gas law:

PV = nRT

Where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.

Converting the given pressure of 615 torr to atm:
615 torr = 0.811 atm
Converting the given temperature of 30°C to Kelvin:
30°C + 273.15 = 303.15 K

Rounding off to two significant figures, we get:
P = 0.81 atm
T = 303 K

Now, rearranging the ideal gas law equation to solve for n:
n = PV/RT

Substituting the given values:

n = (0.978 g/L) x (1 L) / (0.081 atm x 0.0821 L atm/mol K x 303 K)

n = 0.0394 mol

Next, we can calculate the molar mass of the compound using the formula:

molar mass = mass / mole
molar mass = (0.978 g/L) x (1 L) / 0.0394 mol
molar mass = 24.8 g/mol
Therefore, 24.8 g/mol is the molar mass of the compound.

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15.66% of the first cation is still in solution as the second cation is just beginning to precipitate.

One of the three main nutrients that are most frequently used in fertilisers is phosphorous, which is obtained from processing phosphate rock (the other two are nitrogen and potassium).

You can also make phosphoric acid into phosphoric acids, which are utilised in everything from food and skincare to animal feed and electronics. Over the course of millions of years, organic matter accumulates to form the sedimentary rock known as phosphate.

When [tex]Na_{3}Po_{4}[/tex] added to the solution of Ca and Mg, [tex]Ca_{3}(Po_{4})_{2}[/tex] and [tex]Ag_{3}Po_{4}[/tex] are formed.

Ksp of [tex]Ca_{3}(Po_{4})_{2} = 2.07*10^{-33}[/tex]

Ksp of [tex]Ag_{3}Po_{4} = 0.09*10^{-17}[/tex]

Concentration of [tex][Ca^{2+}][/tex] = 0.040 M

Concentration of [tex][Ag^{+}][/tex] = 0.0990 M

[tex]Ag_{3} Po_{4} - > 3Ag^{+} + Po_{4}^{3-}[/tex]

Ksp = [tex][Ag^{+}]^{3} [Po4^{3-}][/tex]

[tex]0.09*10^{-17} = (0.099)^{3} [Po_{3-}][/tex]

[tex][Po_{3-}] = 9.16*10^{-14}M[/tex]

[tex]Ksp = [Ca^{2+}]^{3} [Po_{3-}][/tex]

[tex]2.07*10^{-33} = (0.040)^{3} [Po_{4}^{3-}]^{2}[/tex]

[tex][Po_{4}^{3-}] = 5.68*10^{-15} M[/tex]

[tex][Po_{4}^{3-}][/tex] is smaller in [tex]Ca_{3}(Po_{4})_{2}[/tex]

[tex]Ca_{3}(Po_{4})_{2}[/tex] will start precipitating first

[tex]Ksp = [Ca^{2+}]^{3} [Po_{4}^{3-}]^{2}[/tex]

[tex]2.07*10^{-33} = [Ca^{2+}]^{3} (9.16*10^{-14})^{2}[/tex]

[tex][Ca^{2+}] = 6.27*10^{-3} M[/tex]

[tex]\%\ of\ Ca^{2+}[/tex] remaining [tex]= 6.27*10^{-3}/0.040 * 100[/tex]

= 15.66 %

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The information given can be used to construct the empirical formula for a tin oxide. We must first determine the mass of tin in the sample. This may be achieved by deducting the mass of the crucible, cover, and sample (21.76 g) from the mass of the empty crucible and cover (19.66 g).

This gives us a mass of 2.10 g of tin in the sample. The mass of oxygen in the sample must then be determined. To achieve this, we must deduct the mass of the crucible, cover, and sample (21.76 g) from the mass of the same components (22.29 g) prior to protracted heating. This provides us with an oxygen mass of 0.53 g.

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Answer:

Explanation:

The change in entropy of a system can be determined by comparing the entropy of the reactants to the entropy of the products. The reaction that leads to an increase in the number of moles of gas or particles will generally have a positive change in entropy.

I. 2N2(g) + O2(g) → 2N2O(g)

The reactants have 3 moles of gas, while the product also has 3 moles of gas. Therefore, there is no change in the number of moles of gas, and the change in entropy is likely to be small.

II. CaCO3(s) → CaO(s) + CO2(g)

The reactant is a solid, while the products are a solid and a gas. The formation of a gas from a solid leads to an increase in the number of moles of particles, and therefore an increase in entropy.

III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

The reactants consist of a solid and a liquid, while the products consist of an aqueous solution and a gas. The formation of a gas leads to an increase in the number of moles of particles, and therefore an increase in entropy.

Therefore, reactions II and III have a positive change in entropyentropy

The average mass of chromium is 52.1. Isotopic mass is defined as the average mass of all the isotopes of a specific element.

The average atomic mass of an element is referred to as the sum of the masses of its isotopes, each multiplied by its natural abundance which can be also explained as the decimal associated with the percent of atoms of that element that are of a given isotope. Average atomic mass is equal to f1M1 + f2M2 and so on. Hydrogen, chromium, lithium, cobalt, oxygen, boron, plutonium, and carbon are some examples.

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The product that is formed first is product B and the product that will be the major product if given enough time would also be product B.

In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products.

The activation energy of the reaction can be shown on a diagram as the energy between the reactants that the transition state.

The product with lesser activation energy is the product that is formed first and the major product is decided by the stability of the product which depends on the energy. Lesser is the energy of the product, greater is its stability.

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If in a reaction between vinegar and antacid tablets, the antacid is the limiting reagent. cm of gas. At constant pressure and temperature, three tablets produce 600 cm³ . The  volume that four tablets will produce is: C. 800 cm³.

Since the antacid is the limiting reagent, the amount of gas produced will be directly proportional to the number of tablets used.

We know that three tablets produced 600 cm³ of gas. Therefore, we can set up a proportion:

3 tablets produce 600 cm³ of gas

4 tablets produce x cm³ of gas

To solve for x, we can use cross-multiplication:

3 tablets × x cm³ of gas = 4 tablets × 600 cm³ of gas

3x = 2400

x = 800 cm³

Therefore the answer is C. 800 cm³.

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Answer:

Explanation:

Here are the labels for each situation:

1.Usually conducts electricity & heat well - Metal (1)

2.At room temperature these are gases or liquids - Nonmetal (2)

3.Will lose valance electrons to form compounds - Metal (1)

4.Can be used as semiconductors - Metalloid (3)

5.Will gain valance electrons to form compounds - Nonmetal (2)

Answer:

The main difference between practical work inside and outside a laboratory is that the practical work inside the lab includes good equipment and chemicals which are very advanced and the practical outside a laboratory is more about the safety of life.

Explanation:

Practicals are set up at stations with lab equipment and chemicals, where students can learn, and researchers can experiment and find different new things.

Thus, the practical work inside the lab includes lab equipment and chemicals, and the practical outside a laboratory is more about conserving nature.  

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