rank each set of compounds from most acidic (i) to least acidic (iii): a) 2,4-dichlorobutyric acid i.) most b) 2,3-dichloro butyric acid ii.) c.) 3,3-dimethylbutryic acid iii.) least 3b. explain why you chose this order:

Answers

Answer 1

Answer:

Explanation:

i) Most acidic: 2,4-dichlorobutyric acid

ii) Intermediate acidity: 2,3-dichlorobutyric acid

iii) Least acidic: 3,3-dimethylbutyric acid

The acidity of a compound is determined by the stability of its conjugate base. A stronger acid will have a more stable conjugate base. In this case, the presence of electron-withdrawing groups like chlorine atoms in the carboxylic acid group increases the acidity of the compound by stabilizing the negative charge on the conjugate base.

Comparing the three compounds, 2,4-dichlorobutyric acid has two chlorine atoms which are more electronegative than the methyl groups present in the other compounds. The presence of these electron-withdrawing groups increases the acidity of the compound, making it the most acidic of the three.

2,3-dichlorobutyric acid has only one chlorine atom in the carboxylic acid group, making it less acidic than 2,4-dichlorobutyric acid but more acidic than 3,3-dimethylbutyric acid.

3,3-dimethylbutyric acid does not have any electron-withdrawing groups in the carboxylic acid group, making it the least acidic of the three compounds.


Related Questions

determine the standard enthalpy change for the decomposition of hydrogen peroxide per mole of hydrogen peroxide.

Answers

The standard enthalpy change for the decomposition of hydrogen peroxide per mole of hydrogen peroxide is -98.2 kJ/mol.

when 1 mole of hydrogen peroxide (H2O2) ( H 2 O 2 ) undergoes decomposition, the heat evolved (ΔH) is −98.2kJ. − 98.2 k J . The molar mass of H2O2 H 2 O 2 is 34.015 g/mol. This means that the mass of 1 mole of H2O2 H 2 O 2 is 34.015 g.

This value is obtained from the standard enthalpy of formation of the products (H2 and O2) and the standard enthalpy of formation of the reactant (H2O2). Enthalpy of formation is the energy change that occurs when a compound is formed from its elements, in their standard states.

The difference between the enthalpies of formation of the products and the reactant is the enthalpy change for the reaction. In this case, the enthalpy change for the decomposition of hydrogen peroxide is -98.2 kJ/mol. This indicates that the decomposition of hydrogen peroxide is an exothermic reaction and it releases 98.2 kJ/mole of energy.

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an 80 proof bottle of vodka is equal to ___ bv.

Answers

An 80-proof bottle of vodka is equal to 40% alcohol by volume (ABV).

Proof, which is twice the percentage of alcohol by volume (ABV), is a unit of measurement for the amount of alcohol in a liquid. As a result, 40% of the content of an 80-proof bottle of vodka is alcohol. Accordingly, only 40% of the liquid in the bottle is actual alcohol, while the other 60% is made up of water and other chemicals.

The ABV of a bottle of alcohol is crucial to understand since it establishes the potency and potential consequences of the beverage. Drinks with a higher ABV are stronger and may affect the body more strongly.

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aldehydes and ketones prefer to fragment by ___ which produces a resonance stabilized acylium ion

Answers

Aldehydes and ketones prefer to fragment by cleavage of the C-C bond adjacent to the carbonyl group, which produces a resonance-stabilized acylium ion.

Aldehydes and ketones have a carbonyl gathering (C=O) in their sub-atomic design, which is energized because of the distinction in electronegativity among carbon and oxygen particles. The carbonyl gathering can go through different compound responses, for example, nucleophilic expansion, decrease, and fracture. Discontinuity of aldehydes and ketones includes the cleavage of the C bond neighboring the carbonyl gathering, which prompts the development of a reverberation settled acylium particle.

This response is leaned toward on the grounds that the subsequent acylium particle is settled by reverberation structures, which disperse the positive charge among various iotas in the particle. This adjustment makes the response exceptionally exothermic and expands its rate.

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Aldehydes and ketones prefer to fragment by cleavage of the C-C bond adjacent to the carbonyl group, which produces a resonance-stabilized acylium ion.

Aldehydes and ketones have a carbonyl gathering (C=O) in their sub-atomic design, which is energized because of the distinction in electronegativity among carbon and oxygen particles. The carbonyl gathering can go through different compound responses, for example, nucleophilic expansion, decrease, and fracture. Discontinuity of aldehydes and ketones includes the cleavage of the C bond neighboring the carbonyl gathering, which prompts the development of a reverberation settled acylium particle.

This response is leaned toward on the grounds that the subsequent acylium particle is settled by reverberation structures, which disperse the positive charge among various iotas in the particle. This adjustment makes the response exceptionally exothermic and expands its rate.

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How many moles are in 670 g of gold (|||) chloride

Answers

There are 2.208 moles of gold (III) chloride in 670 g.

To determine the number of moles in 670 g of gold (III) chloride, we need to first calculate the molar mass of gold (III) chloride, which is AuCl3.

The atomic mass of gold is 196.97 g/mol and the atomic mass of chlorine is 35.45 g/mol. Since there are three chlorine atoms in each molecule of gold (III) chloride, we multiply the atomic mass of chlorine by 3:

35.45 g/mol x 3 = 106.35 g/mol

Adding the atomic masses of gold and chlorine together gives us the molar mass of gold (III) chloride:

196.97 g/mol + 106.35 g/mol = 303.32 g/mol

Now, we can use this molar mass to convert 670 g of gold (III) chloride into moles:

670 g / 303.32 g/mol = 2.208 moles

Therefore, there are 2.208 moles of gold (III) chloride in 670 g.


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How many 1H NMR signals does CH3OCH2CH(CH3)2 show? How many^1H NMR signals does CH_3OCH_2CH(CH_3)_2 show? Enter your answer in the provided box.
.......................

Answers

The number of the NMR signals compound CH3OCH2CH(CH3)2 shows are:

3 H with singlet.6 H with doublet.1 H with muliplet.2 H with doublet.

A spectroscopic method for observing the local magnetic fields around atomic nuclei is nuclear magnetic resonance spectroscopy, sometimes referred to as magnetic resonance spectroscopy (MRS) or NMR spectroscopy.

This spectroscopy's foundation is the measurement of electromagnetic radiations' absorption in the radio frequency range between 4 and 900 MHz. Nuclear Magnetic Resonance Spectroscopy is the name given to the form of spectroscopy that is used to measure the absorption of radio waves in the presence of a magnetic field.

The sample is put in a magnetic field, and the nuclear magnetic resonance (NMR) signal is generated by radio waves excitation of the sample's nuclei, which is detected by sensitive radio receivers.

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The number of the NMR signals compound CH3OCH2CH(CH3)2 shows are:

3 H with singlet.
6 H with doublet.
1 H with muliplet.
2 H with doublet.

A spectroscopic method for observing the local magnetic fields around atomic nuclei is nuclear magnetic resonance spectroscopy, sometimes referred to as magnetic resonance spectroscopy (MRS) or NMR spectroscopy.

This spectroscopy's foundation is the measurement of electromagnetic radiations' absorption in the radio frequency range between 4 and 900 MHz. Nuclear Magnetic Resonance Spectroscopy is the name given to the form of spectroscopy that is used to measure the absorption of radio waves in the presence of a magnetic field.

The sample is put in a magnetic field, and the nuclear magnetic resonance (NMR) signal is generated by radio waves excitation of the sample's nuclei, which is detected by sensitive radio receivers.

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what is the ph of a solution prepared by mizing 100ml of 0.020m ba(oh)2 with 50ml of 0.400m of koh? assume that the volumes are addative

Answers

The pH of the solution is approximately 12.73.

First, we need to find the moles of each solution:

moles of Ba(OH)2 = 0.020 mol/L x 0.100 L = 0.002 mol

moles of KOH = 0.400 mol/L x 0.050 L = 0.020 mol

Next, we need to find the total volume of the solution:

Vtotal = 100 mL + 50 mL = 150 mL = 0.150 L

Now, we can find the total concentration of OH- ions:

[OH-] = moles of Ba(OH)2 + moles of KOH / Vtotal

[OH-] = (0.002 mol + 0.020 mol) / 0.150 L = 0.187 mol/L

Finally, we can find the pH of the solution using the following formula:

pH = 14 - log([OH-])

pH = 14 - log(0.187) = 12.73

Therefore, the pH of the solution is approximately 12.73.

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a carving in metal that is soaked with acid, inked, and stamped on paper

Answers

The process you are referring to is called etching. Etching is a technique in which a design is carved into a metal plate using tools such as needles or acid. Once the design is carved, the plate is soaked in an acid solution, which eats away at the exposed metal to create grooves.

After the acid bath, the plate is cleaned and dried, and ink is applied to the surface. The ink is worked into the grooves created by the acid, and any excess ink is wiped away from the surface. The plate is then placed on a press, and a sheet of paper is carefully placed on top of it. Pressure is applied to the paper and the plate, which transfers the ink from the grooves onto the paper, creating a print.

Etching allows for great flexibility in creating fine art prints, as the artist can use a variety of techniques to create different line qualities, textures, and tonal effects. Additionally, multiple copies of the same image can be made from a single plate, making etching a popular printmaking technique among artists.

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The term for a carving in metal that is soaked with acid, inked, and stamped on paper is called etching.

What is the process of Etching?

Etchings are a type of printmaking where the artist creates a design by using acid to etch lines into a metal plate. Once the plate is inked, the ink is pushed into the etched lines, and the plate is stamped onto paper, transferring the ink and creating a print. Etchings can be highly detailed and precise and are often used in fine art prints. The acid bites into the exposed metal areas, creating recessed lines and textures on the plate. The plate is then inked and wiped, leaving ink only in the etched lines and textures. Finally, the plate is pressed onto paper to transfer the ink, creating a print. Etching is a versatile printmaking technique that allows for detailed and intricate designs to be transferred onto paper, and it has been used by artists for centuries to create a wide range of artistic prints.

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which observation best describes the physical appearance of a compound when the end of its melting point range is reached? the compound begins to convert to a liquid. the compound completely converts to a liquid. the compound begins to evaporate.

Answers

A compound turns completely into a liquid this observation best describes the physical appearance of a compound when it reaches the end of its melting point range. Here option B is the correct answer.

When a solid compound is heated, it undergoes a process called melting in which it transforms into a liquid state. The melting point of a compound is the temperature at which it changes from a solid to a liquid state. The melting process is characterized by a range of temperatures over which the compound is observed to be partially or fully melted.

The observation that best describes the physical appearance of a compound when the end of its melting point range is reached is B - the compound completely converts to a liquid. At the end of the melting point range, the compound has absorbed enough heat energy to fully overcome the intermolecular forces that hold its constituent particles together in a solid state, resulting in the complete transformation of the compound into a liquid.

This state is characterized by the loss of a crystalline structure, where the particles are free to move about and slide past each other, leading to an increased fluidity and mobility of the compound. At this stage, the compound is fully melted and can be poured or transferred into a new container in its liquid form.

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Complete question:

Which observation best describes the physical appearance of a compound when the end of its melting point range is reached?

A - the compound begins to convert to a liquid.

B - the compound completely converts to a liquid.

C - the compound begins to evaporate.

when you boil water, bubbles begin to form before the water boils. this happens because . question 12 options: the vapor pressure is increasing the water has salt dissolved in it it is simmering the dissolved air is coming out of the water

Answers

The dissolved air is coming out of the water, causing bubbles to form before the water boils. Option 4 is correct.

As the water is heated, the solubility of gases, such as air, decreases, causing the dissolved gases to be released as bubbles. This process is called nucleation and occurs at sites of imperfections in the container or impurities in the water, which provide a surface for the bubbles to form.

Once the water reaches its boiling point, the vapor pressure of the liquid equals atmospheric pressure, causing bubbles to form throughout the liquid, not just at the nucleation sites. Hence Option 4 is correct.

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Base your answer on the information and illustrations below and on your knowledge of biology. The illustrations represent cross sections of two different plant stems.
A student compared two stem cross sections. Stem cross section A is from a plant that can be used to produce products with valuable medicinal properties. Stem cross section B is from a plant growing in the same area of the forest and its usefulness for producing medicines is unknown. The student concluded that the stem cross sections had many structural similarities and that the plant that produced cross section B would produce the same valuable medicinal products.

Is the student's conclusion valid?

A) Yes, because the structural similarities indicate a close relationship between the organisms.
B) Yes, because these plants grow in the same regions of the forest ecosystem and look similar.
C) No, because he did not evaluate soil conditions, such as pH, with chemical indicators.
D) No, because this structural evidence alone is insufficient and molecular evidence should be obtained.

Answers

Option D is the correct answer. This is because the production of medicinal compounds is determined by the plant's genetics and biochemistry, which may not be reflected in the plant's structural features alone.

What is the students conclusion?

The student's conclusion is not valid. While the two stem cross sections may have many structural similarities, this is not sufficient evidence to conclude that the plant that produced cross section B will produce the same valuable medicinal products as the plant that produced cross section A.

Option A and B are incorrect because structural similarities do not necessarily indicate a close relationship between organisms or their biochemical properties. Option C is also incorrect because while soil conditions may affect plant growth, they do not necessarily determine a plant's ability to produce specific medicinal compounds.

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you have 400 grams (g) of a substance with a half life of 10 years. how much is left after 100 years?

Answers

After 100 years, there will be 6.25 grams of the substance remaining.

What is half life?

Half-life is the time it takes for half of the radioactive atoms in a sample to decay or for the concentration of a substance to decrease by half.

Amount remaining = initial amount x (1/2)^(number of half-lives)

In this case,  half-life of the substance is 10 years, which means that after 10 years, half of the substance will have decayed. After another 10 years (20 years total), half of remaining substance will decay, leaving 1/4 of the original amount. After another 10 years (30 years total), half of that remaining amount will decay, leaving 1/8 of the original amount. This process continues every 10 years.

To find the amount of substance remaining after 100 years, we need to know how many half-lives have occurred in that time: 100 years / 10 years per half-life = 10 half-lives

Amount remaining = 400 g x (1/2)¹⁰= 6.25 g

Therefore, after 100 years, there will be 6.25 grams of the substance remaining.

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In this list of elements, which one would have the least lone pairs in its Lewis structure?A) Ge B) Si C) Pb D) In.

Answers

Indium (In), option D, would have the fewest lone pairs in its Lewis structure of the elements listed.

An element is represented in a Lewis structure by its symbol, and valence electrons are shown as dots or lines. Valence electron pairs known as lone pairs don't participate in chemical bonding.

Subtracting the total number of electrons involved in bonding from the total number of valence electrons for that element yields the amount of lone pairs in a Lewis structure.

Indium (In) is the element with the lowest atomic number and the fewest valence electrons in the list of elements. As a result, of the above structures, its Lewis structure would have the fewest lone pairs.

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The element that would have the least lone pairs in its Lewis structure is D) In (indium).

In this list of elements (Ge, Si, Pb, In), the one with the least lone pairs in its Lewis structure would be Si (Silicon). To understand why, let's briefly discuss the concept of lone pairs and Lewis structures. Lone pairs are pairs of valence electrons that do not participate in bonding, while Lewis structures represent the arrangement of atoms, bonding electrons, and lone pairs in a molecule or ion. Now, let's consider the elements in your list: A) Ge (Germanium) has 4 valence electrons and typically forms 4 covalent bonds with no lone pairs. B) Si (Silicon) has 4 valence electrons and generally forms 4 covalent bonds with no lone pairs. C) Pb (Lead) has 4 valence electrons but can form 2 or 4 covalent bonds, which could leave 1 or 0 lone pairs. D) In (Indium) has 3 valence electrons and generally forms 3 covalent bonds, leaving 1 lone pair. Comparing the elements, both Si and Ge have no lone pairs in their typical Lewis structures. However, Si is the better answer due to its smaller atomic size and higher electronegativity, which make it less likely to form structures with lone pairs compared to Ge. Pb and In typically have lone pairs in their Lewis structures, making them less suitable choices for this question

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What volume of chlorine gas at 46.0◦C and
1.60 atm is needed to react completely with
5.20 g of sodium to form NaCl?

Answers

The volume of chlorine gas at 46.0°C and 1.60 atm that is needed to react completely with 5.20 g of sodium to form NaCl is 1.85 L

How do i determine the volume of chlorine gas needed?

We'll begin by obtaining the mole of 5.20 g of sodium. Details below:

Mass of Na = 5.20 gMolar mass of Na = 23 g/mol Mole of Na =?

Mole = mass / molar mass

Mole of Na = 5.20 / 23

Mole of Na = 0.226 mole

Next, we shall determine the mole of chlorine gas needed. Details below:

2Na + Cl₂ -> 2NaCl

From the balanced equation above,

2 moles of Na reacted with 1 mole of Cl₂

Therefore,

0.226 mole of Na will react with = (0.226 × 1) / 2 = 0.113 mole of Cl₂

Finally, we shall determine the volume of chlorine gas, Cl₂ needed. This is shown below:

Temperature (T) = = 46 °C = 46 + 273 = 319 KPressure (P) = 1.60 atmGas constant (R) = 0.0821 atm.L/molKNumber of mole (n) = 0.113 moleVolume of chlorine gas, Cl₂ (V) =?

PV = nRT

1.6 × V = 0.113 × 0.0821 × 319

Divide both sides by 1.6

V = (0.113 × 0.0821 × 319) / 1.6

V = 1.85 L

Thus, the volume of chlorine gas, Cl₂ needed is 1.85 L

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a sample of nobr was placed on a 1.00l flask containing no no or br 2 at equilibrium the flask contained

Answers

At equilibrium, the concentrations of NO, Br2, and NOBr in the flask will remain constant. However, without specific values for the initial concentration of NOBr or the equilibrium constant (Kc), it's not possible to determine.

.Based on the provided information, it seems that a sample of NOBr was placed in a 1.00 L flask at equilibrium, which means that the NOBr has decomposed into NO and Br2.

At equilibrium, the concentrations of NO, Br2, and NOBr in the flask will remain constant. However, without specific values for the initial concentration of NOBr or the equilibrium constant (Kc), it's not possible to determine the exact concentrations of these substances in the flask.

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A sample of NOBr being placed in a 1.00 L flask containing no NO or Br2 at equilibrium, I'll first provide the balanced chemical equation for the reaction:

[tex]2 NOBr (g) ⇌ 2 NO (g) + Br2 (g)[/tex]

At equilibrium, the concentrations of the reactants and products remain constant. To determine the concentrations of NOBr, NO, and Br2 at equilibrium, we need to follow these steps:

1. Write the expression for the equilibrium constant (Kc) based on the balanced chemical equation:
[tex]Kc = [NO]^2 [Br2] / [NOBr]^2[/tex]

2. Set up an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations of the species involved in the reaction. The initial concentrations of NO and Br2 are 0 since they are not initially present in the flask.

      NOBr      NO      Br2
I      C0        0        0
C     -2x        +2x      +x
E     C0-2x     2x       x

3. Substitute the equilibrium concentrations from the ICE table into the Kc expression:
[tex]Kc = (2x)^2 * x / (C0-2x)^2[/tex]


4. To solve for x, you need the value of Kc for the reaction. Look up the Kc value for this reaction in a reference or use provided information. Once you have Kc, substitute it into the equation and solve for x.

5. Calculate the equilibrium concentrations of NOBr, NO, and Br2 by substituting the value of x back into the ICE table:

[NOBr] = C0-2x
[NO] = 2x
[Br2] = x

By following these steps, you can determine the concentrations of NOBr, NO, and Br2 in the 1.00 L flask at equilibrium.

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k of 0.02911(m hr). if the initial concentration is 3.13 m, what is the concentration after 3.00 hours? your answer should have three significant figures (round your answer to two decimal places).

Answers

The concentration after 3.00 hours is 2.88 m.

To solve this problem, we will use the formula for the rate of a first-order reaction:

rate = k[A]

where k is the rate constant and [A] is the concentration of the reactant. We are given k = 0.02911(m/hr) and [A] = 3.13 m. We want to find the concentration after 3.00 hours, which we'll call [A'].

We can use the integrated rate law for a first-order reaction:

ln[A'] = -kt + ln[A]

where ln is the natural logarithm. Plugging in the given values, we get:

ln[A'] = -0.02911(m/hr) * 3.00 hr + ln[3.13 m]

Simplifying, we get:

ln[A'] = -0.08733 + 1.147

ln[A'] = 1.059

To solve for [A'], we'll take the inverse natural logarithm of both sides:

[A'] = e^(1.059)

[A'] = 2.884

Rounding to three significant figures, we get:

[A'] = 2.88 m

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one of the techniques used in this experiment was that of crystallization. when cooling a solution in the process of crystallization, why would an ice bath be preferable over cold water or ice alone? none of the answers shown are correct. ice is too cold and will freeze any solution. cold water would dilute the solution making it impossible for crystals to form. a mixture of ice and water will keep the temperature above freezing and will contact the entire portion of the container immersed in the ice/water mixture.

Answers

When conducting a crystallization process, it is important to cool the solution at a slow and controlled rate to encourage crystal formation.

An ice bath is preferable over cold water or ice alone because it can maintain a consistent low temperature without causing the solution to freeze solid. Ice alone is too cold and can cause the solution to freeze rapidly, preventing the formation of crystals. Cold water, on the other hand, is not able to maintain a consistent low temperature as the heat from the solution will quickly dissipate into the surrounding water, resulting in a slower cooling rate.

An ice bath, which is a mixture of ice and water, provides a more stable and uniform cooling environment for the solution, allowing for the crystals to form at a slower rate. Additionally, an ice bath can contact the entire portion of the container immersed in the mixture, ensuring that the solution is evenly cooled. Overall, an ice bath is the preferred method for cooling a solution during the process of crystallization.

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complete question is:-

one of the techniques used in this experiment was that of crystallization. when cooling a solution in the process of crystallization, why would an ice bath be preferable over cold water or ice alone? none of the answers shown are correct. ice is too cold and will freeze any solution. cold water would dilute the solution making it impossible for crystals to form. a mixture of ice and water will keep the temperature above freezing and will contact the entire portion of the container immersed in the ice/water mixture.  EXPLAIN.

for the dyes synthesized from a naphthol starting material, did the position of the hydroxyl group an effect on the wavelength of light that was absorbed by the dyes? explain g

Answers

Yes, the position of the hydroxyl group does have an effect on the wavelength of light absorbed by the dyes synthesized from a naphthol starting material.

This is because the position of the hydroxyl group determines the electronic properties of the molecule, which in turn affects the energy levels and transitions that occur when the molecule absorbs light. In general, molecules with hydroxyl groups attached to positions closer to the aromatic ring will absorb light at shorter wavelengths (higher energy), while those with hydroxyl groups attached to positions farther from the ring will absorb light at longer wavelengths (lower energy).

This phenomenon is known as the bathochromic or hypsochromic effect, depending on whether the shift is toward longer or shorter wavelengths, respectively.

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When 1 g gaseous I2 is heated to 1000 K in a
1.00 L sealed container, the resulting equilibrium mixture contains 0.83 g of I2. Calculate
Kc for the dissociation equilibrium
I2(g) ⇀↽ 2 I(g).

Answers

The equilibrium constant, Kc, for the dissociation of I2(g) to 2I(g) at 1000 K is approximately 0.000567 (rounded to three significant figures).

What is Equilibrium?

In chemistry, equilibrium refers to a state of balance or stability in a chemical system where the rates of forward and reverse reactions are equal, and the concentrations of reactants and products remain constant over time. It is a dynamic process, as reactions continue to occur, but the overall concentrations of species in the system do not change.

To calculate the equilibrium constant, Kc, for the dissociation of I2(g) to 2I(g), we can use the concentrations of the species at equilibrium.

Given:

Initial moles of I2(g) = 1 g / molar mass of I2 = 1 g / 253.8 g/mol = 0.00395 mol

Final moles of I2(g) = 0.83 g / molar mass of I2 = 0.83 g / 253.8 g/mol = 0.00327 mol

Since 1 mole of I2 dissociates to form 2 moles of I(g), the change in moles of I(g) is 2 times the change in moles of I2:

Change in moles of I(g) = 2 * (Initial moles of I2 - Final moles of I2)

= 2 * (0.00395 mol - 0.00327 mol)

= 0.00136 mol

Now, we can calculate the equilibrium concentration of I2, [I2], and the equilibrium concentration of I(g), [I], in mol/L.

[I2] = Final moles of I2 / Volume of container

= 0.00327 mol / 1.00 L

= 0.00327 mol/L

[I] = Change in moles of I(g) / Volume of container

= 0.00136 mol / 1.00 L

= 0.00136 mol/L

Finally, we can use the concentrations of I2 and I at equilibrium to calculate the equilibrium constant, Kc, using the following expression:

Kc = [tex]l^{2}[/tex] / [I2]

= [tex](0.00136 mol/L)^{2}[/tex]^2 / 0.00327 mol/L

= 0.000567

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154.42g of oxygen gas (O2) react with an excess of ethane (C2H6) produces how many moles of water vapor (H2O)?

Answers

2.77 moles of water vapour (H2O) are created when 154.42 g of oxygen gas (O2) reacts with an excess of ethane (C2H6).

Calculation-

In order to create water vapour [tex](H_2O)[/tex], ethane [tex](C_2H_6)[/tex]and oxygen gas (O2) must be burned. The chemical equation for this reaction is:

[tex]C_2H_6 + 7O_2 -- > 4H_2O + 6CO_2[/tex]

We may deduce from the equation that when 1 mole of ethane (C2H6) interacts with 7 moles of oxygen gas (O2), 4 moles of water vapour (H2O) are created.

We must utilise its molar mass to translate the 154.42 g of oxygen gas (O2) consumed into moles. 32 g/mol (16 g/mol for each oxygen atom multiplied by two for O2) is the molar mass of oxygen gas.

Moles of oxygen gas (O2) = Mass of oxygen gas (O2) / Molar mass of oxygen gas (O2)

Moles of oxygen gas (O2) = 154.42 g / 32 g/mol

Moles of oxygen gas (O2) = 4.83 mol (rounded to two decimal places)

The balanced equation's stoichiometry predicts that 7 moles of oxygen gas [tex](O_2)[/tex]and 4 moles of water vapour [tex](H_2O)[/tex] will react. We can thus calculate the moles of water vapour [tex](H_2O)[/tex] created using the stoichiometric principle.

Moles of water vapor [tex](H_2O)[/tex] = Moles of oxygen gas [tex](O_2)[/tex] × (4 moles of [tex]H_2O[/tex] / 7 moles of O2)

Moles of water vapor [tex](H_2O)[/tex] = 4.83 mol × (4/7)

Moles of water vapour[tex](H_2O)[/tex] = 2.77 mol (rounded to two decimal places)

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What is the density of hydrogen sulfide (H2S) at 0.7 atm and 322 K?

Answers

Answer:

0.9g/L.

Explanation:

To calculate the density of hydrogen sulfide (H2S) at 0.7 atm and 322 K, we can use the ideal gas law:

PV = nRT

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

We can rearrange this equation to solve for the number of moles of gas:

n = PV / RT

Next, we can use the molar mass of H2S (34.08 g/mol) to convert the number of moles to mass:

mass = n × molar mass

Finally, we can divide the mass by the volume to obtain the density:

density = mass/volume

Let's assume a volume of 1 L (since the volume is not given in the question). Then we have:

P = 0.7 atm

T = 322 K

R = 0.08206 L·atm/(mol·K)

molar mass of H2S = 34.08 g/mol

First, we calculate the number of moles of H2S using the ideal gas law:

n = PV / RT

n = (0.7 atm) (1 L) / (0.08206 L·atm/(mol·K) × 322 K)

n = 0.0265 mol

Next, we calculate the mass of H2S using the number of moles and the molar mass:

mass = n × molar mass

mass = 0.0265 mol × 34.08 g/mol

mass = 0.9 g

Finally, we calculate the density of H2S:

density = mass/volume

density = 0.9g/1 L

density = 0.9 g/L

Therefore, the density of hydrogen sulfide (H2S) at 0.7 atm and 322 K is approximately 0.9g/L.

A 5. 0 L sample of gas is collected at 400. MmHg at 727 C. What is the volume if the temperature were cooled to 77 C and the pressure increased to 700. MmHg?

Answers

The volume would be approximately 0.71 L if the temperature were cooled to 77 °C and the pressure increased to 700 mmHg.

We will use the combined gas law to solve this problem;

P₁V₁/T₁ = P₂V₂/T₂

where P₁, V₁, as well as T₁ are the initial pressure, volume, and the temperature, respectively, and P₂, V₂, and T₂ will be the final pressure, volume, as well as temperature, respectively.

Plugging in the given values, we get;

(400 mmHg)(5.0 L)/(1000 K) = (700 mmHg)(V₂)/(350 K)

Simplifying and solving for V₂, we get;

V₂ = (400 mmHg)(5.0 L)(350 K)/(700 mmHg)(1000 K)

V₂ ≈ 0.71 L

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which acid in table 14.2 is most appropriate for preparation of a buffer solution with a ph of 3.7? explain your choice.

Answers

We can create a buffer solution with a pH of 3.7 by using formic acid as the buffer system's acid component.

What pH does a buffer solution have?

To keep fundamental conditions in place, these buffer solutions are used. A weak base and its salt are combined with a strong acid to create a basic buffer, which has a basic pH. Aqueous solutions of ammonium hydroxide and ammonium chloride at equal concentrations have a pH of 9.25. These solutions have a pH greater than seven.

Why may the pH of a buffered solution resist changing?

When little amounts of acid or base are supplied, buffers can resist pH changes, because they have an acidic component (HA) to neutralise OH- ions and a basic component (A-) to neutralise H+ ions, they are able to accomplish this.

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A balloon is rubbed against a wall. The picture on the left shows the balloon and the wall before rubbing. The picture on the right shows the balloon and the wall after rubbing.

What happened when the balloon was rubbed against the wall? (5.b)


2. A balloon is rubbed against a wall. The picture on the left shows the balloon and the wall before rubbing. The picture on the right shows the balloon and the wall after rubbing.




What happened when the balloon was rubbed against the wall?


A. Electrons were transferred from the wall to the balloon.

B. Protons were transferred from the wall to the balloon.

C. Electrons were transferred from the balloon to the wall.

D. Protons were transferred from the balloon to the wall.

Answers

Answer: The answer should be A

Explanation:

What mass (grams) of nitrogen dioxide gas, NO2, is there in 67.2 liters at stop conditions

Answers

At STP (Standard Temperature and Pressure) conditions, 1 mole of gas occupies 22.4 L of volume.

What mass of nitrogen dioxide gas is present in STP conditions?

We can use the following conversion factor to find the number of moles of NO₂ gas:

1 mole NO₂ = 22.4 L at STP

To find the mass of NO₂ gas, we need to use the molar mass of NO₂, which is 46.0055 g/mol.

Putting all this together, we get:

(67.2 L) / (22.4 L/mol) = 3 moles of NO₂ gas

3 moles of NO₂ gas x 46.0055 g/mol = 138.02 g of NO₂ gas

Therefore, there are 138.02 grams of nitrogen dioxide gas in 67.2 liters of gas at STP conditions.

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What is the total number of oxygen atoms on the right-hand side of this chemical equation? 3 ΗNO, (α )- ΗNO, (α g) +H,0()+2NO (g)

Answers

The total number of oxygen atoms on the right-hand side of the balanced equation is 8.

The compound condition gave isn't adjusted, so it should be adjusted first prior to deciding the absolute number of oxygen iotas on the right-hand side. Here is the fair condition:

3 HNO2 (α) + H2O (l) → 2 NO (g) + 2 HNO3 (aq)

Presently, we can count the absolute number of oxygen particles on the right-hand side of the situation. There are two NO particles, every one of which contains one oxygen iota, for a sum of 2 oxygen molecules.

There are likewise two HNO3 particles, every one of which contains three oxygen iotas, for a sum of 6 oxygen molecules. So the complete number of oxygen iotas on the right-hand side of the situation is:

2 + 6 = 8

Thusly, there are a sum of 8 oxygen particles on the right-hand side of the reasonable substance condition.

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what is the voltage of a galvanic cell that does 788 j of work when 255 coulomb of charge is transferred?

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The voltage of the  galvanic cell is 3.09 volts when the work done to  transfer the charge of 255 colombs is 788 joules.

The voltage of a galvanic cell can be calculated using the formula:
[tex]Voltage (V) = Work (J) / Charge (C)[/tex]
Given that the galvanic cell does 788 J of work and transfers 255 coulombs of charge, we can plug  these values into the formula:

[tex]Voltage (V) = Work (J) / Charge (C)[/tex]
[tex]Voltage (V) = 788 J / 255 C = 3.09 V[/tex]
So, the voltage of the galvanic cell is approximately 3.09 volts.

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consider a reaction between two gaseous reactants (4 mol of a and 4 mol of b) in the closed flasks shown below. assume that the two reactions are both at room temperature. which reaction will occur faster?

Answers

Answer:

....................................................

Factors such as pressure, volume, and the presence of catalysts can affect the rate of the reaction.

Figure out the reaction between two gaseous reactants?

The two gaseous reactants (4 mol of A and 4 mol of B) in the closed flasks shown below will occur faster, I would need more information about the specific conditions in each flask. Factors such as pressure, volume, and the presence of catalysts can affect the rate of the reaction.

If you could provide more details about the flasks and the conditions, I would be happy to help you determine which reaction will occur faster.

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the molar solubility of pbi 2 is 1.5 × 10 −3 m. calculate the value of ksp for pbi 2 .4.5 x 10 -6

Answers

The value of Ksp for PbI2 is 4.05 × 10^-8 if the molar solubility of PBI 2 is 1.5 × 10 −3 m.

The molar solubility of PBI 2 = 1.5 × 10 −3 m

The solubility product constant  = 2 .4.5 x 10 -6

The solubility product constant (Ksp) for PbI2 can be estimated using the molar solubility of PbI2, the stoichiometry of the equilibrium equation is:

[tex]PbI2(s) = Pb2+(aq) + 2I-(aq)[/tex]

The equation for Ksp is:

Ksp = [tex][Pb2+][I-]^2[/tex]

[Pb2+] = S = 1.5 × 10−3 M,

[I-] = 2S = 3 × 10−3 M

The stoichiometric coefficient of I- is 2. Substituting these values into the Ksp equation we get:

Ksp =[tex](1.5 × 10^-3) × (3 × 10^-3)^2[/tex]

Ksp = 4.05 × 10^-8

Therefore, we can conclude that the value of Ksp for PbI2 is 4.05 × 10^-8.

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The value of Ksp for PbI2 is 3.375 × 10^-9 or 4.5 x 10 -6. The expression for the solubility product constant (Ksp) of a sparingly soluble salt such as PbI2 is: Ksp = [Pb2+][I-]^2

where [Pb2+] and [I-] are the molar concentrations of the lead ion and iodide ion, respectively, in a saturated solution of PbI2.

Given that the molar solubility of PbI2 is 1.5 × 10^-3 M, we can assume that [Pb2+] and [I-] in the saturated solution are also equal to 1.5 × 10^-3 M. Therefore, we can substitute these values into the Ksp expression and solve for Ksp:

Ksp = (1.5 × 10^-3 M)(1.5 × 10^-3 M)^2
Ksp = 3.375 × 10^-9

So the value of Ksp for PbI2 is 3.375 × 10^-9 or 4.5 x 10 -6 (if that was a typo in the question).

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if you theoretically performed the bromination of phenol with only one equivalent of br2 which product do you think would predominate

Answers

The product that would predominate in the bromination of phenol with only one equivalent of Br2 is the para-bromophenol.

If the bromination of phenol was performed with only one equivalent of Br2, it is more likely that the para product would predominate due to steric hindrance effects that make it difficult for the ortho product to form. The reaction of phenol with Br2 is an electrophilic aromatic substitution where Br+ attacks the electron-rich aromatic ring.

The ortho position is sterically hindered by the presence of the bulky -OH group, making it difficult for the incoming Br+ ion to attack this position. On the other hand, the para position is less hindered, and the incoming Br+ ion can easily attack this position, leading to the predominance of the para product.

Although some ortho product may still form due to the statistical probability of the reaction, it would not be as significant as the para product.

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The complete question is:

Had you performed the bromination of phenol with only one equivalent of Br2, which product (ortho or para) do you think would predominate? Hint: think about probability and statistics.

F-actin is a polymer of G-actin monomers and exhibits symmetry. (T/F)

Answers

F-actin is a polymer of G-actin monomers and exhibits symmetry is a False statement.

A class of globular, multifunctional proteins called actin creates the thin filaments in muscle fibrils as well as the microfilaments in the cytoskeleton. Its mass is around 42 kDa, and its diameter ranges from 4 to 7 nm; it is present in almost all eukaryotic cells, where it may be detected in concentrations of over 100 M.

The monomeric subunit of two different types of filaments in cells—thin filaments, a component of the contractile apparatus in muscle cells, and microfilaments, one of the three main elements of the cytoskeleton—is an actin protein. Both G-actin and F-actin, which are present either as a free monomer termed G-actin (globular) or as a component of a linear polymer microfilament known as F-actin (filamentous), are necessary for such crucial cellular processes.

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F-actin is a polymer of G-actin monomers and exhibits symmetry is a False statement.

A class of globular, multifunctional proteins called actin creates the thin filaments in muscle fibrils as well as the microfilaments in the cytoskeleton. Its mass is around 42 kDa, and its diameter ranges from 4 to 7 nm; it is present in almost all eukaryotic cells, where it may be detected in concentrations of over 100 M.

The monomeric subunit of two different types of filaments in cells—thin filaments, a component of the contractile apparatus in muscle cells, and microfilaments, one of the three main elements of the cytoskeleton—is an actin protein. Both G-actin and F-actin, which are present either as a free monomer termed G-actin (globular) or as a component of a linear polymer microfilament known as F-actin (filamentous), are necessary for such crucial cellular processes.

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