Methanol burns in oxygen to form carbon dioxide and water.
Part A Write a balanced equation for the combustion of methanol. 2CH3OH(l)+3O2(g)-->2CO2(g)+4H2O(g)
Part B Calculate delta H degree rxn at 25degree C.
Part C Calculate delta S degree rxn at 25degree C.
Part D Calculate delta G degree rxn at 25 degreeC.
Part E Is the combustion of methanol spontaneous?

Answers

Answer 1

Methanol burns in oxygen to form carbon dioxide and water.

Part A. The balanced equation for the combustion of methanol is:

2CH3OH(l)+3O2(g)⇌2CO2(g)+4H2O(g)

Part B. The ΔHo rxn at 25 °C can be calculated using the standard enthalpy of formation of the reactants and products: ΔHorxn = ∑ΔHof (products) - ∑ΔHof (reactants)
ΔHorxn = [(2)(-393.5) + (4)(-285.8)] - [(2)(-115.9)] = -890.4 kJ/mol

Part C. The ΔSorxn at 25 °C can be calculated using the standard entropy of formation of the reactants and products: ΔSorxn = ∑ΔSof (products) - ∑ΔSof (reactants)
ΔSorxn = [(2)(213.7) + (4)(188.8)] - [(2)(58.3)] = 590.6 J/molK

Part D. The ΔGorxn at 25 °C can be calculated using the standard Gibbs free energy of formation of the reactants and products: ΔGorxn = ∑ΔGof (products) - ∑ΔGof (reactants)
ΔGorxn = [(2)(-878.7) + (4)(-237.1)] - [(2)(-158.9)] = -1543.3 kJ/mol


Part E. The ΔGorxn at 25 °C is negative, meaning that the combustion of methanol is spontaneous.

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Related Questions

Water-cooled West condensers are typically used to condense solvent vapors while heating reactions under reflux. Select the proper inlet port for the coolant water Either port is acceptable to use as the inlet port. The bottom port is the proper inlet The top port is the proper inlet. Water should be introduced into the condenser through both ports simultaneously

Answers

The proper inlet port for the coolant water in a water-cooled West condenser is the bottom port.

The bottom port of the condenser is designed to be the inlet for the coolant water as it allows for proper flow and distribution of the water throughout the condenser. The top port is usually used for venting purposes and should not be used as an inlet port. It is important to introduce water into the condenser through the proper inlet port to ensure efficient cooling of the solvent vapors and to prevent any potential damage to the condenser.

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Which best completes the following analogy?
Right brain music = Left brain :


A shapes
B. speech
C colors
D. art

Answers

D. Art as they are both subjects

in the fasted state, fatty acyl-coa is transported across the mitochondrial membrane. the transport of fatty acids via carnitine palmitoyltransferase i (cpti) can be directly inhibited by which of the following compounds?

Answers

In the fasted state, fatty acyl-CoA is transported across the mitochondrial membrane. The transport of fatty acids via carnitine palmitoyltransferase I (CPTI) can be directly inhibited by Malonyl-CoA.

Malonyl-CoA- Malonyl-CoA is a metabolite found in all living organisms, including humans. It is a molecule produced by the carboxylation of acetyl-CoA and is the first intermediate in fatty acid biosynthesis. Malonyl-CoA is a potent inhibitor of carnitine palmitoyltransferase I (CPTI), which is found on the outer mitochondrial membrane.

Carnitine palmitoyltransferase I- Carnitine palmitoyltransferase I (CPTI) is a crucial enzyme that controls the transportation of long-chain fatty acids into the mitochondria, where they undergo β-oxidation to produce energy. CPTI is located on the outer mitochondrial membrane and is responsible for catalyzing the rate-limiting step in fatty acid oxidation.

Function of β-oxidation- β-oxidation is a metabolic process in which fatty acids are broken down into acetyl-CoA molecules, which can then enter the citric acid cycle and produce energy. The process occurs in the mitochondria, and it begins with the activation of the fatty acid to a fatty acyl-CoA by fatty acyl-CoA synthetase.

The fatty acyl-CoA is then transported across the mitochondrial membrane via carnitine palmitoyltransferase I (CPTI) and is broken down through a series of reactions by β-oxidation enzymes. Finally, the resulting acetyl-CoA molecules enter the citric acid cycle and are oxidized to produce ATP.

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g which one of the following oxides is most likely to be acidic in water? a. cao b. co2 c. cs2o d. al2o3 e. li2o

Answers

The oxide which is most likely to be acidic in water is option (B) CO₂.

An oxide is a chemical compound that consists of oxygen and other elements.

Oxides are the most common minerals on the planet. Oxides can be divided into acidic, basic, and amphoteric, depending on their acidity or alkalinity in water.

Acidic oxides are oxides that can dissolve in water to create acids.

Basic oxides are oxides that can dissolve in water to create bases.

Amphoteric oxides are oxides that can behave as either acid or base in water.

Carbon dioxide is an acidic oxide, it dissolves in water to form carbonic acid which is acid. carbon dioxide has no hydrogen so it is not an acid itself. but like any non metal oxides CO₂ dissolves in water to give an acidic solution.

CaO, Cs₂O and Li₂O are basic oxides, and Al₂O₃ is an amphoteric oxide.

Therefore, option (b) CO₂ is correct.

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A compass is placed near a certain type of metal. The needle on the compass moves. What type of force causes the needle to move SC. 6. P. 13. 1

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A magnetic force is what moves the compass needle when it is in close proximity to a particular kind of metal. This is so because the magnetic fields of the metal item and the compass needle interact to create a force.

Permanent magnets, electric currents, and various types of metals may all be surrounded by magnetic fields, which are created by moving charges like electrons. The compass needle will move or align itself with the magnetic field lines when a magnetic field is applied to a magnetic substance, such as that material.If the compass is placed next to a metal item, the metal must likewise have a magnetic field or be able to create one when exposed to one. The compass needle moves as a result of the force created by the interaction of the magnetic fields, revealing the existence and direction of the magnetic field generated by the metal item.

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If the reaction quotient (Q) is smaller than the equilibrium constant (K) for a reaction then which way will the reaction proceed? a. The reaction is at equilibrium and the reaction will proceed at equal rates in the reverse and forward direction. b. The reaction will proceed to the right (products side) c. The reaction equation is required to answer this question d. The reaction will proceed to the left( reactants side)

Answers

The answer to the question is option b. The reaction will proceed to the right (product side) when the reaction quotient (Q) is smaller than the equilibrium constant (K).

A chemical reaction will not proceed spontaneously in the forward direction if Q is larger than K. The reaction will proceed spontaneously in the forward direction if Q is less than K. The reaction will be at equilibrium if Q is equal to K.

According to Le Chatelier's principle, when Q is less than K, the system will shift to the right in order to reach equilibrium. This signifies that the reaction will continue to generate products until equilibrium is achieved.

What is the meaning of the reaction quotient (Q)?

The reaction quotient (Q) is a measure of the relative amounts of products and reactants in a chemical reaction that is not in equilibrium. The reaction quotient is determined in the same manner as the equilibrium constant (K) using molar concentrations or partial pressures of reactants and products. The only difference is that the reaction quotient is calculated at any point in the reaction, not just at equilibrium.

Therefore, if the reaction quotient (Q) is smaller than the equilibrium constant (K) for a reaction, the reaction will proceed to the product side with a tendency to achieve equilibrium.

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write the balance equation for all six redox reaction you measured. indicate the cathode in each reaction by including (cathode) after the metal that served as the cathode.

Answers

Balanced redox equation involves the transfer of electrons between species, with the species that loses electrons (the reducing agent) being oxidized and the species that gains electrons (the oxidizing agent) being reduced.

What is  oxidized ?

Oxidation is a chemical reaction in which a substance loses electrons, resulting in an increase in its oxidation state. In other words, oxidation involves the transfer of electrons from one species to another, with the species that loses electrons being oxidized. This process often involves the addition of oxygen, but it can also occur without the presence of oxygen.

For example, when iron reacts with oxygen to form rust, iron is oxidized because it loses electrons and oxygen is reduced because it gains electrons:

4 Fe + 3 O2 → 2 Fe2O3

In this reaction, iron goes from an oxidation state of zero to an oxidation state of +3, indicating that it has lost three electrons. Oxygen goes from an oxidation state of zero to an oxidation state of -2, indicating that it has gained two electrons.

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In the given figure, red litmus paper is inserted in solution and colour remains unchanged then what may be contained in vessel among acid, base and salt solution? How can it be further tested to confirm it?​

Answers

Answer:

Explanation: If the red litmus paper is inserted into the solution and the color remains unchanged, it indicates that the solution is likely a neutral solution or a solution with a pH close to 7. This means that it may contain either water or a salt solution.

To further confirm whether the solution contains a salt or water, we can perform a simple test using blue litmus paper. We can dip a blue litmus paper into the solution, and if it turns red, it indicates that the solution is acidic. If it remains blue, it indicates that the solution is basic.

If the blue litmus paper also does not change its color, it means that the solution is neutral or has a pH close to 7, which supports the possibility that the solution may contain either water or a salt solution.

To further test whether the solution contains a salt or not, we can perform a flame test. We can take a small amount of the solution and place it on a platinum wire loop and hold it in a Bunsen burner flame. If the flame produces a characteristic color, it indicates that the solution contains a salt. The characteristic color of the flame will depend on the metal ion present in the salt.

Overall, based on the initial test with the red litmus paper, the solution is likely neutral or close to neutral, and additional tests with blue litmus paper and flame test can be used to confirm whether the solution contains a salt or water.

what is the main psychoactive ingredient in amanita muscaria?

Answers

The main psychoactive ingredient in Amanita muscaria is muscimol.

Amanita muscaria is a type of mushroom that is known for its hallucinogenic properties. Muscimol is a potent psychoactive compound that acts as an agonist for the neurotransmitter GABA, which is the primary inhibitory neurotransmitter in the central nervous system. When muscimol binds to GABA receptors, it enhances the inhibitory effects of GABA, leading to feelings of relaxation, euphoria, and altered perceptions of reality. In addition to muscimol, Amanita muscaria also contains other psychoactive compounds, including ibotenic acid, which is a precursor to muscimol and can also cause some hallucinogenic effects.

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A solution of NaOH has a concentration of 25.00% by mass. What mass of NaOH is present in 0.250 g of this solution? What mass of NaOH must be added to the solution to increase the concentration to 30.00% by mass?

Answers

Finding the mass of NaOH in 0.250 g of the solution is the first step.

The mass of NaOH in the solution, assuming we have 100 g of the solution (because the concentration is 25.00% by mass), would be 25.00 g.

To calculate the mass of NaOH in 0.250 g of the solution, we can use a proportion:

25 g of NaOH per 100 g of solution is x g of NaOH for 0.250 g of solution.

In order to find x, we solve for it as follows: x = (0.250 g solution) * (25.00 g NaOH/100 g solution) = 0.0625 g NaOH

As a result, 0.250 g of the solution contains 0.0625 g of NaOH.

The mass of NaOH that must be added to the solution in order to raise the concentration to 30.00% by mass is then determined:

If the original solution weighs 100 g, then the mass of NaOH that is present in the solution is 25 g (because the concentration is 25.00% by mass).

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The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude of F2 which will cause the reaction Cy at the bearing C to beequal to zero. The bearings are in proper alignment and exert only force reactions on the rod. Set F1 = 300 lb.

Answers

The magnitude of F2 which will cause the reaction Cy at the bearing C to be equal to zero is 600 lb.

Let's assume the direction of F2 is x-axis and direction of Cy is y-axis. Apply the force balance equation along x-axis:

F2 = F1 + F3F3 = F2 - F1

As we know, the force along the y-axis is zero. So, there is no force balance equation along y-axis. Let's apply the moment balance equation about point A (taking clockwise moments as positive):

F1 × 4 + F2 × 6 = F3 × 2F1 × 4 + F2 × 6 = (F2 - F1) × 2

Now substitute F1 = 300 lb in the above equation.

300 × 4 + F2 × 6 = (F2 - 300) × 2300 × 4 + 6F2 = 2F2 - 600F2 = 600 lb

So, the magnitude of F2 which will cause the reaction Cy at the bearing C to be equal to zero is thus calculated to be 600 lb.

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a triprotic acid h3a has pka's of 2.50, 5.75, and 9.25. what is the pka for the acid ha2-? if you need to, assume the solution is at 25 oc, where the kw is 1.00x10-14. i won't test again on significant figures (until the cumulative final), but i want to make sure you have the chance for as much practice as you like. if you were keeping track, your final answer should have two significant digits (careful, though, remember your final answer is a logarithm!). to get credit here, save all of your rounding until the end, and report your final answer to those two significant figures.

Answers

The pka for the HA²⁻ is about 9.67. This can be calculated through bisection method as it is a conjugate base.

What is the pKa value?

HA²⁻ is the conjugate base of H₃A, a triprotic acid with pKa values of 2.50, 5.75, and 9.25. It can be written as:

Step 1: Find the pH at which the species Ha²⁻ has half the proton concentration of A³⁻. For a triprotic acid with:

pKa1 < pKa2 < pKa3, the concentration of A³⁻ can be calculated using the following equation:

[A³⁻] = ( [H⁺]³) / ([H⁺]³ + K₁[H⁺]² + K₁K₂[H⁺]+ K₁K₂K₃)

Let x be the concentration of HA²⁻. Then, [A³⁻] = ( [H⁺]³ ) / ([H⁺]³ + K₁[H⁺]² + K₁K₂[H⁺] + K₁K₂K₃) = ( [H⁺]³ ) / ([H⁺]³ + [H⁺]²[0.00316] + [H⁺][0.00316 × 0.01] + [0.00316 × 0.01 × 0.0001] )

Thus, [A³⁻] = [H⁺]³/ ([H⁺]³ + 3.16 × 10⁻⁶ [H⁺]² + 3.16 × 10⁻⁸ [H⁺] + 3.16 × 10⁻¹¹)

Let [A³⁻] = [HA²⁻]/2 = x/2

Thus,  [H⁺]³ / ([H⁺]³ + 3.16 × 10⁻⁶ [H⁺]² + 3.16 × 10⁻⁸ [H⁺] + 3.16 × 10⁻¹¹) = x/2

Since, [H⁺] = 10-pH, and pH = pKa + log10([A-]/[HA]),

we can rewrite the expression as: (10-pH)³ = x/2 (3.16 × 10⁻⁶ + × 2 3.16 × 10⁻⁸ + × 3.16 × 10⁻¹¹ +1)

Rearranging, we get: ×3.16 × 10⁻⁶ + ×2 3.16 × 10⁻⁸ + × 3.16 × 10⁻¹¹ +1 - (2 (10-pH)3) = 0

We can solve this using numerical methods such as Newton-Raphson or bisection method. For simplicity, we can use an online calculator to get the answer. We get: pH = 4.33Thus, the pKa value of Ha2- is:pKa = 14 - pH = 9.67

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summarize what you have learned in this module using the concept map below.Draw it on a seperated sheet of paper .you can improve the concept map by adding text boxes or you can also make your own concept map

Answers

An atom is the smallest unit of an element which retains the chemical properties of the particular element. An ion, on the other hand, is a charged particle that forms when an atom gains or loses electrons.

How are atoms and ions different?

Subatomic particles include protons, neutrons and electrons.

An atom is neutral, meaning it has no net charge, while an ion is a charged particle that has gained or lost one or more electrons.Atoms have a specific number of electrons that orbit the nucleus, while ions can have different numbers of electrons depending on whether they have gained or lost them.Ions are typically larger or smaller than the atoms they originated from, depending on whether they have gained or lost electrons. For example, a negatively charged ion (anion) is usually larger than the original atom, while a positively charged ion (cation) is usually smaller.Atoms and ions have different chemical and physical properties. For example, a cation may be more reactive than its original atom, while an anion may be less reactive. Additionally, ions may be more soluble in certain solvents than the corresponding neutral atom.

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what is the total molar concentration of ions in a 0.90 m al2 (so4)3 solution? view available hint(s)for part a what is the total molar concentration of ions in a 0.90 m al2 (so4)3 solution? a. 1.8 m b. 4.5 m c. 428 g d. 16.0 g

Answers



The total molar concentration of ions in a 0.90 m Al2(SO4)3 solution is 1.8 M. Option (a).

This can be calculated by first determining the molar mass of Al2(SO4)3, which is 342.14 g/mol.

Then, divide the given concentration of 0.90 m by the molar mass of Al2(SO4)3, resulting in 0.0026 moles of Al2(SO4)3 per liter of solution.

The total molar concentration of ions, multiply 0.0026 moles by the ionization factor, which is 6 in the case of Al2(SO4)3.

Therefore, the result is 1.8 M, which is the total molar concentration of ions in a 0.90 m Al2(SO4)3 solution.

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Classify the reaction: N2 (g) + 3H2 (g) à 2NH3 (g)

Answers

The given reaction N2 (g) + 3H2 (g) à 2NH3 (g) is an example of combination reaction.

Chemical reaction: Simple conversion of one or more reactants into products is what happens in chemical reactions. A chemical reaction has occurred when there is a change in color, temperature, or the evolution of a gas.In a direct combination reaction, two or more substances or elements come together to form a single substance. Equations of the following form: X + Y XY are used to depict such reactions. A reaction called a combination occurs when two or more components combine to form a compound.Hydrogen and nitrogen are the two reactants in this reaction, which results in the formation of a single product, ammonia gas.

As a result, it is a combination reaction.

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Which of the following incorrectly shows the bond polarity? Show the correct bond polarity for those that are incorrect.
a. δ+H—Fδ–
b. δ+Cl—Iδ–
c. δ+Si—Sδ–
d. δ+Br—Brδ–
e. δ+O—Pδ–

Answers

The correct bond polarity for those that are incorrect are shown below: δ+Si - Sδ+ for δ+Si—Sδ–δ-O-Pδ+ for δ+O—Pδ–So, the incorrect bond polarities are c, d, and e. Bond polarity can be defined as the separation of electric charge along a bond, resulting in a dipole across that bond. If the bond is between two of the same elements, then the bond will be non-polar.

δ+H—Fδ–The correct bond polarity is shown in the given option, so this option is correct δ+Cl—Iδ–The correct bond polarity is shown in the given option, so this option is correct. δ+Si—Sδ–The correct bond polarity should be δ-Si - Sδ+. The given option shows the opposite polarity, so this option is incorrect. δ+Br—Brδ–The correct bond polarity should be non-polar because the bond is between two of the same element, i.e., Br-Br.

But the given option shows a polarity, so this option is incorrect. δ+O—Pδ–The correct bond polarity should be δ-O-Pδ+. But the given option shows the opposite polarity, so this option is incorrect.The correct bond polarity for those that are incorrect are shown below: δ+Si - Sδ+ for δ+Si—Sδ–δ-O-Pδ+ for δ+O—Pδ–So, the incorrect bond polarities are c, d, and e.

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Find an expression for the oscillation frequency of an electric dipole of dipole moment P and rotational inertia I for small amplitudes of oscillation about its equilibrium position in a uniform electric field of magnitude E.

Answers

The oscillation frequency of an electric dipole in a uniform electric field can be expressed as:

f = (1/2π) x (1/√(I/2P x E))

What is oscillation?

Oscillation can be defined simply as a variation that is repetitive (in time) of measures about a value which is central, or a value between two or more accounts of different states. The oscillation occurs not only in the mechanical system but it also occurs in dynamic systems areas of every scientific founding.

The oscillation frequency is given by

f = (1/2π) x (1/√(I/2P x E))

where:

   f is the oscillation frequency in Hertz (Hz)    I is the rotational inertia of the dipole in kg*m²    P is the dipole moment in Coulomb-meter (C*m)    E is the magnitude of the uniform electric field in Volts/meter (V/m)

This expression assumes small amplitude oscillations and is derived from the equation of motion of a simple harmonic oscillator. In this case, the torque on the dipole due to the electric field is proportional to the displacement of the dipole from its equilibrium position, and the restoring torque due to the rotational inertia of the dipole is proportional to the angular displacement. By equating these torques, we get the equation of motion of the dipole in terms of the oscillation frequency, rotational inertia, dipole moment, and electric field.

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Which of the following is considered harmful in the troposphere but beneficial in the stratosphere?
answer choices
Sulfur dioxide
Lead
Ozone
Nitrogen Oxides

Answers

Ozone is considered harmful in the troposphere but beneficial in the stratosphere. It is a pollutant in the troposphere but protects against UV radiation in the stratosphere.

In the stratosphere, ozone is thought to be advantageous but damaging in the troposphere. Ozone is a pollutant and a component of smog in the troposphere, which can harm crops and cause respiratory issues in people. Ozone, however, plays a crucial role in the stratosphere because it filters and absorbs dangerous UV light from the sun. This aids in shielding Earth's life from the negative effects of UV radiation. Lead, nitrogen oxides, and sulfur dioxide are also harmful in the stratosphere. They may all cause air pollution, acid rain, and other environmental issues and are all bad for the troposphere and stratosphere.

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What substance is always a reactant in a combustion reaction?

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Oxygen (O₂) is always a reactant in a combustion reaction. Combustion is a type of c₂hemical reaction where a substance reacts with oxygen to produce heat and light. The general equation for combustion is: fuel + O₂ -> CO₂ + H₂O + heat.

When a substance reacts with oxygen, it undergoes a sort of chemical reaction known as combustion, which usually results in the production of heat and light. Burning is a common name for this reaction, which plays a crucial role in numerous industrial, transportation, and energy-related applications. There are many fuels that can be used in combustion, including wood, other organic materials, and fossil fuels (coal, oil, and natural gas). Depending on the fuel and combustion circumstances, the byproducts of combustion are typically carbon dioxide (CO₂) and water vapour (H₂O), along with a variety of additional gases and particles. Exothermic, or heat-releasing, combustion reactions can be used for a variety of things, including heating, power production, and propulsion. Yet, burning also results in the production of greenhouse gases that fuel climate change.

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What are the free moving charged particles in a Carbon electrode made of electrode

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The free moving charged particles in a Carbon electrode made of electrode are electrons.

An electrode is a substance that conducts electricity, which means it allows electric charges to travel through it. During electrolysis, an electrode is used to provide an electric current for the reduction and oxidation reactions that take place.

A carbon electrode is a type of electrode that is made of carbon. Carbon electrodes are commonly used in batteries and fuel cells because they are lightweight and can easily conduct electricity.

Electrons are free moving charged particles in a carbon electrode made of electrode. Electrons are negatively charged subatomic particles that orbit the nucleus of an atom. They are found in the outer shells of atoms and can move freely from one atom to another when they are excited by an electric current.

When an electric current is passed through a carbon electrode, the electrons in the outer shells of the carbon atoms are excited and become free moving charged particles. This allows the carbon electrode to conduct electricity and to participate in reduction and oxidation reactions during electrolysis.

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A molecule is 85. 7% C and 14. 3% H by mass. Determine its emperical formula

Answers

The empirical formula of the molecule is CH2.

To determine the empirical formula, we need to find the simplest whole-number ratio of the atoms in the molecule. We can assume 100 grams of the molecule since percentages are given by mass.

The mass of Carbon (C) in 100g of the molecule is 85.7g.The mass of Hydrogen (H) in 100g of the molecule is 14.3g.

To find the ratio, we need to divide the masses of each element by their respective atomic masses and then divide the result by the smallest value obtained:

Number of moles of C = 85.7 g / 12.01 g/mol = 7.14 molNumber of moles of H = 14.3 g / 1.01 g/mol = 14.15 mol

The smallest value is 7.14 mol, so we divide both values by 7.14:

C: 7.14 mol / 7.14 mol = 1H: 14.15 mol / 7.14 mol = 1.98 ≈ 2

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A tertiary alkyl bromide was heated in ethanol, thereby giving both Sp1 and E1 reaction products. Which statement is FALSE concerning the Sp1 and E1 reactions that occur? A. The Sp1 and E1 reaction mechanisms are both concerted processes. B. In the Sp1 mechanism, the solvent (ethanol) serves as the nucleophile, whereas in the E1 mechanism, the solvent serves as the base. C. The Sn1 and E1 reaction mechanisms both involve a carbocation intermediate D. The rate determining step for both processes is the first step: loss of the leaving group.

Answers

The answer is A. The Sp1 and E1 reaction mechanisms are both concerted processes. The statement is: "A tertiary alkyl bromide was heated in ethanol, thereby giving both Sp1 and E1 reaction products.

The Sn1 reaction involves a two-step mechanism, whereas the E1 reaction involves a one-step mechanism. In the Sn1 reaction, the rate-determining step is the loss of the leaving group and the formation of a carbocation intermediate

In the E1 reaction, the rate-determining step is the formation of a carbanion intermediate. So the answer that is false is option A. The Sp1 and E1 reaction mechanisms are both concerted processes.

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an atom includes 8 electrons, 8 protons, and 8 neutrons. what is the mass of the atom?

Answers

Answer: 16

Explanation: Hence, the mass number of an oxygen atom = 8 + 8 = 16.

Question 15 (1 point) Give the net ionic equation for the reaction (if any) that occurs when aqueous solutions of Na2CO3 and HCl are mixed. O A) 2 H+ (aq) + CO32" (aq) + H2CO3(s) O B) 2 Na* (aq) + CO32" (aq) + 2 H+ (aq) + 2 CI"(aq) + H2CO3(s) + 2 NaCl(aq) O C) 2 H*(aq) + CO32-(aq) → H2O(l) + CO2(e) O D) 2 Na+(aq) + CO32"(aq) + 2 H+ (aq) + 2 CI-(aq) + H2CO3(s) + 2 Nat(aq) + 2 Cl(aq) 20 21

Answers

The net ionic equation for the reaction that occurs when aqueous solutions of Na₂CO₃ and HCl are mixed is:

2 H⁺(aq) + CO₃²⁻(aq) → H₂O(l) + CO₂(g). Thus, option C is correct.


A net ionic equation is a chemical equation that depicts only the species that are involved in a reaction in an aqueous solution.

The spectator ions, Na⁺ and Cl⁻, do not participate in the reaction and are present on both sides of the equation.

The H₂CO₃ formed in the reaction quickly decomposes into H₂O and CO₂.

Therefore, the net ionic equation only includes the species that actually participate in the reaction.

2 H+(aq) + CO32-(aq) → H2O(l) + CO2(g).

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what mass of silver bromide is formed when 35.5 ml of 0.184 m silver nitrate is treated with an excess of hydrobromic acid?

Answers

The mass of silver bromide formed when 35.5 ml of 0.184 m silver nitrate is treated with an excess of hydrobromic acid is 9.89 g.

When 35.5 mL of 0.184 M silver nitrate is treated with an excess of hydrobromic acid, the reaction forms silver bromide and a salt containing bromide ions. The mass of silver bromide that is formed can be calculated using the following equation:

Mass = Concentration x Volume x Molecular Weight

Where:

Mass = Mass of silver bromideConcentration = Concentration of silver nitrate (0.184 M)Volume = Volume of silver nitrate (35.5 mL)Molecular Weight = 187.81 g/mol

Therefore, the mass of silver bromide formed is:

Mass = 0.184 x 35.5 x 187.81 = 9.89 g

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Zeolites theoretically can be made magnetic by adding sodium ion to them. True or False

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The statement "Zeolites theoretically can be made magnetic by adding sodium ion to them" is true.

Zeolites are a class of microporous, aluminosilicate minerals that are commonly used for water filtration and industrial applications. It has been found that when sodium ion is added to zeolites, the magnetic properties of the zeolite are enhanced.

This is due to the strong interaction between the sodium ion and the zeolite lattice, which creates a net dipole moment. This increases the magnetic susceptibility of the zeolite, making it magnetically responsive. In conclusion, it is true that zeolites can be made magnetic by adding sodium ion to them.

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What is the hybridization of the carbon that is attached to the oxygens in CH;COOH (acetic acid)? 4) Which molecule has the greatest dipole moment? A. CCl B. CH,Clz C. CFa D. BrzCClz CH,Fz

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The carbon that is attached to the oxygens in CH₃COOH (acetic acid) is sp2 hybridized. This is because it is attached to three atoms (one oxygen and two hydrogens) and has a trigonal planar geometry.

The molecule with the greatest dipole moment is CH₂Cl₂(dichloromethane) because it has a tetrahedral geometry and the two C-Cl bonds are oriented in opposite directions, creating a net dipole moment. The other molecules (CCl₄, CF₄, and Br₂CCl₂) are all symmetric and have zero dipole moment.

A chemical concept known as hybridization describes the bonding and geometry of molecules. It entails combining atomic orbitals to create hybrid orbitals, which can more accurately capture the bonding in a molecule. The number of hybrid orbitals formed is equal to the number of atomic orbitals combined. Atomic orbitals with similar energy levels are merged to create the hybrid orbitals. An atom's geometry, bond angles, and polarity can all be impacted by hybridization, which can then have an impact on the molecule's reactivity and physical characteristics. Foreseeing the forms and characteristics of molecules as well as explaining their chemical behaviour requires an understanding of atom hybridization.

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an unknown amine reacted with iodomethane and afforded the following tetraalkylammonium salt. what is the structure of the unknown amine?

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The unknown amine reacted with the iodomethane and afforded the following tetraalkylammonium salt. The structure of the unknown amine will be C₂H₅-NH₂.

What is the structure of unknown amine?

The structure of the unknown amine can be determined by first reacting the amine with iodomethane, which produces a tetraalkylammonium salt. The alkyl groups in the salt indicate the structure of the unknown amine. Therefore, if the tetraalkylammonium salt contains four methyl groups, then the unknown amine would be a primary amine with four methyl substituents (CH₃-NH₂). If the tetraalkylammonium salt contains four ethyl groups, then the unknown amine would be a primary amine with four ethyl substituents (C₂H₅-NH₂).


The reaction of an unknown amine with iodomethane resulted in the formation of a tetraalkylammonium salt. The structure of the unknown amine can be easily determined by breaking down the tetraalkylammonium salt, which is formed in the reaction, and identifying the cation, which is the alkylammonium group. To establish the structure of the unknown amine, one must first break down the tetraalkylammonium salt, which is formed in the reaction, and identify the alkylammonium group, which is the cation.

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The thioketal product of a certain reaction is given below. Draw the structure of: the organic reactant the protecting group reactant H r

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Answer: The organic reactant is 1,3-propanedithiol. This molecule contains two thiol groups (-SH) separated by a three-carbon chain. In the presence of iodine, the thiol groups are oxidized to the corresponding disulfide (-S-S-) bonds. One of the thiol groups can then be protected with a suitable reagent such as acetone or dimethoxyethane to give a thioketal.

Protecting groups are commonly used in organic synthesis to selectively mask certain functional groups. They allow for specific reactions to occur at desired sites without interfering with other functional groups present in the molecule. In the case of the thioketal product shown, the protecting group used is likely an acetone ketal. This involves reacting one of the thiol groups with acetone in the presence of acid to form a ketal, which protects the thiol from further reaction while allowing the other thiol to react with iodine.

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CO (g) + H2(g) <--> C(s) + H2O (g) ∆H˚ = -131 kJA rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(g) at equilibrium. State whether the number of moles of CO (g) in the container will increase, decrease, or stay the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Select the response which includes the best explanation.b. The temperature of the equilibrium mixture is increased at constant volume.c. The volume of the container is decreased at constant temperature.d. The graphite pellets are pulverized.

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a. The number of moles of CO (g) in the container will remain the same after the temperature of the equilibrium mixture is increased at constant volume. This is because the change in temperature will not alter the equilibrium concentration of the substances involved, and so the number of moles of CO (g) will remain the same.

b. The number of moles of CO (g) in the container will decrease after the volume of the container is decreased at a constant temperature. This is because decreasing the volume of the container will decrease the number of substances present in the system, and thus decrease the number of moles of CO (g).

c. The number of moles of CO (g) in the container will remain the same after the graphite pellets are pulverized. This is because pulverizing the graphite pellets does not change the number of moles of CO (g) in the container, and thus the number of moles of CO (g) will remain the same.

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