If a propane heater burns 38.95 g C3H8, it consumes 38.95 mol C3H8. 0.8830 mol C3H8. 1 mol C3H8. 44.10 mol C3H8.

Explanation:

It is known that compound that contain carbon atom and hydrogen atom which are combined together are known as organic compounds. Carbon atom is the main element present in an organic compound.

For example, [tex]C_{6}H_{6}[/tex] is an organic compound.

Whereas compounds that does not contain carbon atom are known as inorganic compounds.

Thus, we can conclude that given compounds are classified as follows.

Organic compounds : [tex]C_{6}H_{6}[/tex], [tex]C_{2}H_{6}[/tex], [tex]CH_{2}O[/tex], and [tex]MgCO_{3}[/tex].

Inorganic compounds : [tex]Ca_{3}(PO_{4})_{2}[/tex].

The solubility of O2 in water is directly proportional to the partial pressure of the gas. To find the solubility of O2 at a pressure of 2.40 atm, we can use Henry's law and the given solubility at 0.370 atm.

Solubility of O2 in water at different pressures

The solubility of oxygen (O2) in water is directly proportional to the partial pressure of oxygen gas above the water. This relationship is described by Henry's law. According to Henry's law, the solubility of a gas is equal to the product of the Henry's law constant (kH) and the partial pressure of the gas. Therefore, to find the solubility of O2 at a pressure of 2.40 atm and 25 °C, we can use the following equation:

Solubility = kH * Pressure

Given that the solubility of O2 at 0.370 atm and 25 °C is 15.3 g/100 g H2O, we can rearrange the equation to solve for the Henry's law constant:

kH = Solubility / Pressure

Substituting the values into the equation, we get:

kH = (15.3 g/100 g H2O) / 0.370 atm

Solving this equation will give us the value of kH. Once we have the value of kH, we can use it to calculate the solubility of O2 at a pressure of 2.40 atm.

Answer:

Kc = 1.28

Explanation:

Let's consider the following reversible reaction.

H₂O(g) ⇄ H₂(g) + 0.5 O₂(g)

The equilibrium constant (Kc) is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

[tex]Kc=\frac{[H_{2}][O_{2}]^{0.5}}{[H_{2}O]} =\frac{(0.370).(0.750)^{0.5}}{0.250} =1.28[/tex]

Answer: The correct answer is Option d.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{eq}[/tex]

For a general chemical reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The expression for [tex]K_{eq}[/tex] is written as:

[tex]K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

There are 3 conditions:

From the above expression, the equilibrium constant is directly dependent on product concentration. Thus, more is the concentration of product, more will be the equilibrium constant.

The highest values of [tex]K_{eq}[/tex] will favor the product more.

Hence, the correct answer is Option d.

A solution containing fluoride ions will be weakly basic because fluoride reacts with water to produce hydroxide ions, raising the pH slightly above 7.

Unlike chloride ions, which make a neutral solution, fluoride ions make the solution weakly basic.

When determining the pH of a solution containing fluoride ions (F¯), it is important to understand how fluoride interacts with water. Sodium fluoride (NaF) dissolves in water, producing fluoride ions, which then weakly react with water:

F¯ (aq) + H₂O (l) ⇒ HF (aq) + OH¯ (aq)

This reaction creates hydroxide ions (OH¯), making the solution weakly basic. Thus, while a chloride ion solution is neutral, a fluoride ion solution will have a pH slightly above 7 due to the weak base formed.

The hydrogen ion concentration [H⁺] is 0.1 mol/L, making the pH of the solution 1. This pH level indicates that the solution is acidic.

Let's work through the problem step by step:

A. Calculate the hydrogen ion concentration [H⁺]

We know that the product of hydrogen ion concentration [H⁺] and hydroxide ion concentration [OH⁻] in water is equal to 1 x 10⁻¹⁴ at 25°C. This is represented by the equation:

[H⁺] * [OH⁻] = 1 x 10⁻¹⁴

Given that [OH⁻] = 1 x 10⁻¹³ mol/L, we can rearrange the equation to solve for [H⁺]:

[H⁺] = 1 x 10⁻¹⁴ / [OH⁻]

Substituting the given value:

[H⁺] = 1 x 10⁻¹⁴ / 1 x 10⁻¹³ = 1 x 10⁻¹ = 0.1 mol/L

B. Calculate the pH

The pH of a solution is calculated using the formula:

pH = -log[H⁺]

Substituting the value for [H⁺]:

pH = -log(0.1) = 1

C. Identify if the solution is an acid, base, or neutral

A solution with a pH less than 7 is acidic, a pH greater than 7 is basic, and a pH of 7 is neutral.

Since the pH of this solution is 1, it is an acidic solution.

Explanation:

A pH scale is a scale which helps in identifying whether a given substance is basic, acidic or neutral in nature.

So, when a substance has pH value less then 7 then it means the substance is acidic in nature. Whereas when a substance has pH value more then 7 then it means the substance is basic in nature.

On the other hand, if pH value of a substance is equal to 7 then it means the substance is neutral in nature.

Hence, we can conclude that the pH scale runs from 0 to 14, and is an indication of how acidic or basic a solution is. A pH close to zero indicates an acidic solution, and a pH near 14 indicates a basic solution.

The atomic number of tin is 50, and the mass numbers of the tin isotopes are 115, 117, and 126. Each isotope has 50 protons, and the number of neutrons varies for each isotope. All isotopes of tin have 50 electrons.

The atomic number of an element represents the number of protons in the nucleus. So, for tin, which has an atomic number of 50, it means there are 50 protons in its nucleus.

The mass number of an isotope is the sum of the number of protons and neutrons in the nucleus. For the isotopes given, 115Sn has a mass number of 115, 117Sn has a mass number of 117, and 126Sn has a mass number of 126.

Since the number of protons in an atom is equal to the atomic number, each isotope of tin will have 50 protons. The number of neutrons is equal to the mass number minus the atomic number. So, 115Sn has 65 neutrons, 117Sn has 67 neutrons, and 126Sn has 76 neutrons. The number of electrons in a neutral atom is equal to the number of protons, so each isotope will also have 50 electrons.

Answer : The density of a piece of metal is [tex]2.7008g/cm^3[/tex]

Explanation :

To calculate the volume of cube, we use the formula:

[tex]V=a^3[/tex]

where,

a = edge length of cube

Given :

Edge length of cube = 2.50 cm

Volume of cube = [tex](2.50cm)^3=15.625cm^3[/tex]

Given :

Mass of metal = 42.20 g

To calculate density of a substance, we use the equation:

[tex]Density=\frac{Mass}{Volume}[/tex]

Putting values in above equation, we get:

[tex]\text{Density of metal}=\frac{42.20g}{15.625cm^3}=2.7008g/cm^3[/tex]

Hence, the density of a piece of metal is [tex]2.7008g/cm^3[/tex]

Final answer:

Equilibrium constant values are crucial in determining the extent of a reaction. Calculations involving equilibrium constants help predict reaction outcomes and concentrations at equilibrium.

Explanation:

Equilibrium Constant (K) is a value that indicates the extent to which a reaction will proceed at a certain temperature. It is calculated using the concentrations of products and reactants at equilibrium. The equilibrium constant is specific to a particular reaction at a specific temperature.

For the reaction SO₂(g) + NO₂(g) → SO₃(g) + NO(g), the equilibrium partial pressure for SO₃ can be calculated using the equilibrium constant (K) value and the initial partial pressures of the gases involved. By plugging in the given values and using the equation for K, you can find the equilibrium partial pressure for SO₃.

Given the reaction 2 SO₂(g) + O₂(g) → 2 SO₃(g) with an equilibrium constant of 7.9 x 10¹, you can calculate the equilibrium constant for the reverse reaction by taking the reciprocal of the original equilibrium constant. This allows you to determine the equilibrium constant for the reverse reaction at the same temperature.

Explanation:

Since the given formula is [tex]H_{3}PO_{4}[/tex]. It consists of three hydrogen atoms, one phosphorous atom and four oxygen atoms.

[tex]H_{3}PO_{4}[/tex] is known as phosphoric acid. It is a weak acid and it dissociates to give hydrogen ions when dissolved in a solvent.

Phosphoric acid ([tex]H_{3}PO_{4}[/tex]) is also known as orthophosphoric acid.

2-Phenylethanol is prepared from 2-phenylethanoic acid by reducing the carboxylic acid group using lithium aluminum hydride (LiAlH4). The reaction replaces the oxygen in the carboxyl group with a hydrogen, yielding the alcohol.

The conversion of 2-phenylethanoic acid (C6H5CH2CO2H) to 2-phenylethanol (C6H5CH2CH2OH) involves a reduction reaction. One common method to achieve this transformation is to use a reducing agent, such as lithium aluminum hydride (LiAlH4). Here's the reaction:

[tex]\[ \text{C}_6\text{H}_5\text{CH}_2\text{CO}_2\text{H} + \text{LiAlH}_4 \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{OH} + \text{LiAl(OH)}_4 \][/tex]

In this equation:

- C6H5CH2CO2H represents 2-phenylethanoic acid.

- LiAlH4 is lithium aluminum hydride, a powerful reducing agent.

- C6H5CH2CH2OH represents 2-phenylethanol.

- LiAl(OH)4 is lithium aluminum tetrahydroxide, a byproduct of the reduction.

This reaction reduces the carboxylic acid group in 2-phenylethanoic acid to the alcohol group in 2-phenylethanol. The reduction involves the addition of hydrogen (H) from LiAlH4 to the carbon-oxygen double bond of the carboxylic acid, resulting in the formation of the alcohol.

The synthesis of 2-phenylethanol from 2-phenylethanoic acid involves reducing the acid to the corresponding alcohol using sodium borohydride (NaBH₄) as a reducing agent. This reaction converts the 2-phenylethanoic acid to 2-phenylethanol with the release of water.

The preparation of 2-phenylethanol (C₆H₅CH₂CH₂OH) from 2-phenylethanoic acid (C₆H₅CH₂CO₂H) can be carried out through the following steps:

2-phenylethanoic acid is reduced to 2-phenylethanol using sodium borohydride:

C₆H₅CH₂CO₂H + 4[H] → C₆H₅CH₂CH₂OH + H₂O

In this reaction, sodium borohydride (NaBH₄) serves as the source of hydrogen (H).

This completes the transformation of 2-phenylethanoic acid to 2-phenylethanol.

Final answer:

Bubbles forming during a reaction involving calcium and water suggests a chemical reaction is occurring, likely the formation of solid calcium hydroxide or the release of carbon dioxide gas if an acid is present.

Explanation:

The student has observed the formation of bubbles while titrating a solution of calcium and water. This observation indicates that a chemical reaction is taking place. When solid calcium oxide reacts with liquid water, as per the information provided, the result is the formation of solid calcium hydroxide. This exothermic reaction may release gas and cause bubbling. Another possibility could be the formation of calcium carbonate and the release of carbon dioxide gas when reacting with an acid, which also leads to bubbling.

Final answer:

At 393 K, which is 119.85°C, acetic acid exists in a liquid state, as it is above its melting point and below its boiling point.

Explanation:

Acetic acid, also known by its systematic name ethanoic acid, exists in different physical states depending on the temperature. Pure acetic acid solidifies at 16.6°C and is referred to as glacial acetic acid because it often solidified in cold laboratories lending the appearance of ice. At 393 K (which is equivalent to 119.85°C), acetic acid is well above its melting point and well below its boiling point of about 118°C. Hence, at 393 K, acetic acid would exist in a liquid state. It is important to note that the compound forms dimers due to strong intermolecular hydrogen bonding, which affects its physical properties, including melting and boiling points.

Using the Arrhenius equation and an estimated pre-exponential factor, we calculated the Gibbs energy of activation for cyclopropane isomerization to be 161,226 J/mol. This value indicates the energy barrier that molecules need to overcome for the reaction to proceed.

Calculation of Gibbs Energy of Activation for Cyclopropane Isomerization

The thermal isomerization of cyclopropane to propene requires overcoming an energy barrier, represented by the Gibbs energy of activation, ΔG⁰‡. We can calculate this value using the Arrhenius equation and the provided rate constant at a specific temperature.

Steps:

Convert temperature to Kelvin: T = 500°C + 273.15 K = 773.15 K

Plug values into the Arrhenius equation:

k = A * exp(-ΔG⁰‡ / RT)

where:

k = rate constant (5.95 × 10^-4 s^-1)

A = pre-exponential factor (unknown)

R = gas constant (8.314 J/mol·K)

T = temperature (773.15 K)

Rearrange the equation to solve for ΔG⁰‡:

ΔG⁰‡ = -RT * ln(k / A)

Estimate the pre-exponential factor: A for this reaction is typically in the range of 10^8 - 10^13 s^-1. For our calculation, assume A = 10^10 s^-1.

Calculate ΔG⁰‡:

ΔG⁰‡ = -8.314 J/mol·K * 773.15 K * ln(5.95 × 10^-4 s^-1 / 10^10 s^-1) = 161,226 J/mol

Therefore, the Gibbs energy of activation for the thermal isomerization of cyclopropane to propene is approximately 161,226 J/mol.

Answer : The final concentration was 0.199 M

Explanation :

The expression used for second order kinetics is:

[tex]kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}[/tex]

where,

k = rate constant = [tex]0.0265M^{-1}min^{-1}[/tex]

t = time = 180.0 min

[tex][A_t][/tex] = final concentration = ?

[tex][A_o][/tex] = initial concentration = 4.25 M

Now put all the given values in the above expression, we get:

[tex]0.0265\times 180.0=\frac{1}{[A_t]}-\frac{1}{4.25}[/tex]

[tex][A_t]=0.199M[/tex]

Therefore, the final concentration was 0.199 M

Answer:

7.5 M/s

Explanation:

The equation of Reaction is given as;

BrO^3- + 5Br^- + 6H^+ --------------> 3Br2 + 3H2O. ----------------------------(1)

Rate of Reaction is the speed in which reaction take place for the transformation of reactants to products.

For the reaction (1) above;

===> - (∆[BrO^3-]/ ∆t) = -(∆Br^-])/∆t = -∆[H^+] /∆t = ∆[Br2]/∆t.

Notice from above that there is negative signs at the Change in the reacti species. This is so, because the Reactants is been used up in the reaction to produce the products

NB : I did not raise the number of the moles to the power of each species to avoid clumsiness.

We are given that; - (∆[BrO^3-]/ ∆t) = 1.5 × 10^-2 m/s; -(∆Br^-])/∆t = x m/s.

1 mole of BrO3^- = 5 mole of Br^-

So, 5 × 1.5 ×10^-2 .

= 7.5 × 10 ^-2 M/s