Now let's try one without any help from the simulation. Write a balanced chemical equation for the combustion reaction of ethane gas (C2H6) and oxygen gas (O2), then answer the following questions:(o) How many moles of carbon dioxide are produced from the combustion of 6.20 moles of ethane gas? (You may assume you have an excess of oxygen gas)(p) How many moles of carbon dioxide are produced from the combustion of 3.92 moles of oxygen gas? (You may assume you have an excess of ethane gas)(q) How many moles of carbon dioxide are produced from the combustion of 6.20 moles of ethane gas with 3.92 moles of oxygen gas?(r) How much excess reactant remains after the reaction described in (q)?(s) How much excess reactant remains after the combustion of 6.10 moles of ethane gas with 5.69 moles of oxygen gas?

Answer: D: a volcano forming a new island in a chain

Explanation:

Answer:

\(5.450 \times {10}^{ - 4 } \times 3.550 \times {10}^{ - 7} \\ 19.3475 \times {10 }^{ - 11} \\ 1.935 \times {10}^{?10} \)

Answer:

d.

Explanation:

Answer:

D!

Explanation:

A is not true! When big meteors fall into the ocean, they cause mega-tsunamis I think, which does a lot of damage.

B, no, Uranus does too.

C, I think this is untrue, red wavelength and reflects the blue the blue!

D, is so true! A lot of reliable sources on this one!

Sorry if I did any wrong, I haven't learn chemistry in school yet.

The answer is B) Decrease. The equilibrium constant (Keq) for the reaction will decrease when the temperature is increased by 20 K while the volume is kept constant.

When the temperature of a chemical reaction at equilibrium is increased, the equilibrium constant (Keq) can change. In this case, the reaction is exothermic (∆H°rxn < 0), which means it releases heat.

According to Le Chatelier's principle, when the temperature is increased, the equilibrium will shift in the direction that absorbs heat. Since the reaction is exothermic, it will favor the reactant side in order to consume the excess heat.

In this reaction, the forward reaction (2NO2 ⇔ N2O4) is the exothermic direction. Therefore, when the temperature is increased, the equilibrium will shift to the left, favoring the formation of more reactants (NO2).

As a result, the concentration of NO2 will increase, while the concentration of N2O4 will decrease. This change in concentrations will lead to a decrease in the value of Keq.

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1: Mechanism: Diels-Alder, extrusion: carbon monoxide.

2: Overcome steric hindrance, increase reactivity.

1: Mechanism of the Diels-Alder Reaction and Subsequent Extrusion of Carbon Monoxide:

The Diels-Alder reaction involves the cycloaddition of a conjugated diene and a dienophile to form a cyclic compound called a cycloadduct. The reaction proceeds through a concerted mechanism, meaning that all bond formations and bond breaking occur simultaneously.

Let's consider the reaction between tetraphenylcyclopentadienone (diene) and a dienophile, such as maleic anhydride. The mechanism involves the following steps:

Step 1:  Formation of the π-complex

The diene and dienophile approach each other, and the electron-rich diene forms a weak bond with the electron-deficient dienophile. This interaction is called the π-complex. In the case of tetraphenylcyclopentadienone, the carbonyl group of the dienophile interacts with the π-system of the diene.

Step 2: Formation of the transition state

The π-complex undergoes a concerted reaction, leading to the formation of a transition state. In this transition state, the π-bonds between the diene and dienophile begin to break, and new σ-bonds start to form.

Step 3: Formation of the cycloadduct

The transition state collapses, resulting in the formation of the cycloadduct. The π-bonds are fully broken, and new σ-bonds are formed, resulting in the closure of a new ring. In the case of tetraphenylcyclopentadienone and maleic anhydride, the cycloadduct formed is 9,10-dihydroanthracene-9,10-α,β-succinic anhydride.

Subsequent Extrusion of Carbon Monoxide:

In some cases, after the formation of the cycloadduct, further rearrangements may occur. In this specific example, the cycloadduct formed contains a strained cyclic anhydride moiety. Upon heating, this cyclic anhydride undergoes thermal decomposition, resulting in the extrusion of carbon monoxide (CO). The extrusion of CO occurs due to the release of strain in the cyclic system, and it helps stabilize the final product. The resulting compound is the desired product of the reaction.

2: The Need for Strong Heating in Initiating the Diels-Alder Reaction:

Heating the mixture of tetraphenylcyclopentadienone and a dienophile strongly is necessary to initiate the Diels-Alder reaction due to the following reasons:

a) Activation of Reactants: The elevated temperature provides the kinetic energy required to overcome the energy barrier of the reaction. The diene and dienophile need sufficient thermal energy to break their existing π-bonds and form new σ-bonds.

b) Overcoming Steric Hindrance: In the case of tetraphenylcyclopentadienone, the presence of bulky phenyl groups creates steric hindrance. This steric hindrance makes it more difficult for the diene and dienophile to approach each other and react. By heating the reaction mixture, the thermal energy helps to overcome the steric hindrance and allows the reactants to come into close proximity for the reaction to occur.

c) Increasing Reaction Rate: Heating the reaction mixture increases the rate of the Diels-Alder reaction. The rate of most chemical reactions, including the Diels-Alder reaction, typically increases with temperature due to the greater number of collisions between reactant molecules and the higher proportion of molecules possessing sufficient energy to undergo the reaction.

Overall, the strong heating of the mixture of tetraphenylcyclopent

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The pH before any base is added to the butanoic acid solution is 10. The concentration of H₂SO₄ is 0.234 M.

a) To find the pH before any base is added to the butanoic acid solution, we need to calculate the concentration of H⁺ ions using the dissociation of butanoic acid:

CH₃CH₂CH₂COOH ⇌ H⁺ + CH₃CH₂CH₂COO⁻

The initial concentration of butanoic acid is 0.150 M, and the Ka of butanoic acid is \(1.5 \times 10^{-5}\).

Using the equation for Ka:

\(Ka = \frac {[H^+] [CH_{3}CH_{2}CH_{2}COO^-]}{[CH_{3}CH_{2}CH_{2}COOH]}\)

Since the initial concentration of butanoic acid is equal to the concentration of CH₃CH₂CH₂COOH, we can assume that the concentration of H⁺ at equilibrium will be negligible compared to the initial concentration of butanoic acid. Therefore, we can approximate the initial concentration of H⁺ as 0.

Using the equation for Ka:

\(Ka = \frac {[H^+] [CH_{3}CH_{2}CH_{2}COO^-]}{[CH_{3}CH_{2}CH_{2}COOH]}\)

Since [H⁺] = 0, we can rearrange the equation to solve for [CH₃CH₂CH₂COO⁻]:

\([CH_{3}CH_{2}CH_{2}COO^-] = \frac {Ka}{[CH_3CH_2CH_2COOH]}\)

             \(= \frac {(1.5 \times 10^{-5})}{(0.150)}\)

            \(= 1 \times 10^{-4}\)

Taking the negative logarithm (pOH) of [CH₃CH₂CH₂COO⁻]:

\(pOH = -log_{10}([CH_{3}CH_{2}CH_{2}COO^-])\)

   \(= -log_{10}(1 \times 10^{-4})\)

   = 4

Since pH + pOH = 14, we can find the pH:

pH = 14 - pOH

  = 14 - 4

  = 10

Therefore, the pH before any base is added to the butanoic acid solution is 10.

b) To determine the concentration of H2SO4, we can use the stoichiometry of the reaction between H2SO4 and NaOH:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

The balanced equation shows that 1 mole of H₂SO₄ reacts with 2 moles of NaOH. From the volume of NaOH required to reach the equivalence point (62.5 mL), we can calculate the number of moles of NaOH used:

moles of NaOH = 0.375 M NaOH (0.0625 L NaOH)

                          = 0.0234 mol NaOH

Since the stoichiometry is 1:2 for H2SO4 to NaOH, the number of moles of H₂SO₄ present in the solution is half the number of moles of NaOH used:

moles of H₂SO₄ = 0.0234 mol NaOH / 2

                          = 0.0117 mol H2SO4

To calculate the concentration of H2SO4, we divide the moles of H₂SO₄ by the volume of the solution:

concentration of H₂SO₄ = 0.0117 mol H2SO4 / 0.0500 L solution

                                        = 0.234 M

Therefore, the concentration of H₂SO₄ is 0.234 M.

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The pH of the solution containing 2.80 M acetic acid is 2.34.

Given, The Ka value for acetic acid, CH3COOH(aq), is 1.8x10^-5.Molar concentration of acetic acid, CH3COOH(aq), is 2.80 M.

Step 1 The equation for the ionization of acetic acid is as follows.CH3COOH(aq) + H2O(l) ⇆ H3O+(aq) + CH3COO-(aq)

Step 2Expression for Ka isKa = [H3O+][CH3COO-]/[CH3COOH(aq)]1.8 x 10-5 = [H3O+][CH3COO-]/2.80[H3O+] = √(Ka [CH3COOH(aq)]) = √(1.8 x 10-5 x 2.80) = 0.00462 M

Step 3pH = -log[H3O+] = -log(0.00462) = 2.34

So, the pH of the solution containing 2.80 M acetic acid is 2.34.

Acetic acid (CH3COOH) is a weak acid with a Ka value of 1.8x10⁻.

By utilizing this Ka value and the molar concentration of acetic acid, the pH of a 2.80 M acetic acid solution can be calculated.

Using the equation Ka = [H3O+][CH3COO-]/[CH3COOH(aq)], and after simplifying,

it can be determined that [H3O+] = √(Ka [CH3COOH(aq)]).

After substituting the values for Ka and [CH3COOH(aq)], [H3O+] is found to be 0.00462 M.

Finally, pH can be calculated by the expression pH = -log[H3O+], and we obtain the answer of pH=2.34.

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The trend which depicts the reactivity of most nonmetals show in a periodic table is that it  increases from left to right across the periodic table and is denoted as option C.

This is referred to as the tabular arrangement of elements into groups and periods which is according to their similar features such as reactivity, number of electron shells etc.

When considering the periodic table, the reactivity of most nonmetals such as oxygen decreases top down within groups and increases from left to right across period which is therefore the reason why it was chosen as the most correct choice.

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Answer:

B

Explanation:

c isnt right

The byprοducts in this reactiοn are acetic acid (οptiοn A) and ethanοl (οptiοn C).

In acid catalysis and base catalysis, a chemical reactiοn is catalyzed by an acid οr a base. By Brønsted–Lοwry acid–base theοry, the acid is the prοtοn (hydrοgen iοn, H+) dοnοr and the base is the prοtοn acceptοr. Typical reactiοns catalyzed by prοtοn transfer are esterificatiοns and aldοl reactiοns.

The acid-catalysed hydrοlysis οf diethyl acetamidοbenzylmalοnate can lead tο variοus by prοducts depending οn the reactiοn cοnditiοns and specific chemical pathways. Hοwever, withοut mοre detailed infοrmatiοn οr a specific reactiοn mechanism, it is difficult tο prοvide a cοmprehensive list οf the by prοducts that may fοrm.

Based οn the infοrmatiοn prοvided, the acid-catalysed hydrοlysis οf diethyl acetamidοbenzylmalοnate delivers the desired (+)-phenylalanine hydrοchlοride prοduct and the fοllοwing byprοduct(s):

E. Bοth A and C: Acetic acid and ethanοl.

The hydrοlysis οf diethyl acetamidοbenzylmalοnate invοlves the cleavage οf ester bοnds, resulting in the fοrmatiοn οf acetic acid as a byprοduct. Additiοnally, since diethyl acetamidοbenzylmalοnate is an ester, hydrοlysis οf the ester bοnds can alsο prοduce ethanοl as anοther byprοduct.

Therefοre, the byprοducts in this reactiοn are acetic acid (οptiοn A) and ethanοl (οptiοn C).

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Complete Question :

Answer: Chromium is both an element and a compound

Explanation:

Answer:

Three Oxygen atoms gains two electrons each ( total of six) from two Al atoms, each Al atom loses three electrons (total of six) to three oxygen atom , in this process each Al atom becomes Al3+ ion and Oxygen atom after gaining two electrons becomes O2− ion

Explanation:

Three Oxygen atoms gains two electrons each ( total of six) from two Al atoms, each Al atom loses three electrons (total of six) to three oxygen atom , in this process each Al atom becomes Al3+ ion and Oxygen atom after gaining two electrons becomes O2− ion

Answer:

Approximately 7.65

Explanation:

math.

skillz.

Answer:

0.20 mol

Explanation:

Let's consider the reduction of iron from an aqueous solution of iron (II).

Fe²⁺ + 2 e⁻ ⇒ Fe

The molar mass of Fe is 55.85 g/mol. The moles corresponding to 5.6 g of Fe are:

5.6 g × 1 mol/55.85 g = 0.10 mol

2 moles of electrons are required to deposit 1 mole of Fe. The moles of electrons required to deposit 0.10 moles of Fe are

0.10 mol Fe × 2 mol e⁻/1 mol Fe = 0.20 mol e⁻

0.20 mol of electrons is required to deposit 5.6g of iron from a solution of iron (2) tetraoxosulphate(6)

 The reduction of iron from an aqueous solution of iron (II).

\(Fe^{+2} +2e^{-} \rightarrow Fe\)

The formula for number of moles is as follows:-

\(Number \ of \ moles=\frac{Mass}{Molar\ mass}\)

The molar mass of Fe is 55.85 g/mol. The moles corresponding to 5.6 g of Fe are:

\(5.6 g \times\frac{1\ mol}{55.85\ g} = 0.10 \ mol\)

2 moles of electrons are required to deposit 1 mole of Fe. The moles of electrons required to deposit 0.10 moles of Fe are:-

\(0.10 mol Fe\times\frac{2\ mol\ e^{-} }{1\ mol\ e^{-}} = 0.20 \ mol e^{-}\)

Hence, 0.20 mol of electrons is required to deposit 5.6g of iron.

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Answer:

Evaporation

Explanation:

because the ammonia is left unclosed

The smell of ammonia has been spread from a corner to entire room by the process of diffusion. Thus, option B is correct.

Ammonia has been the liquid compound with the formula \(\rm NH_4\). It has a pungent smell. Ammonia has been used in several synthesis reactions, and in various industrial processes.

The diffusion has been the process of transfer of molecules from higher concentration to lower concentration. The smell of ammonia has been spread in the entire laboratory from a corner.

The ammonia molecules in the bottle are at higher concentration while there has been lower concentration of ammonia in the environment. With the process of diffusion there has been the transfer of ammonia molecules in the environment.

Thus, diffusion has been responsible for spreading the smell of ammonia in the laboratory from a corner. Thus, option B is correct.

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The axon involved in the transmission of pain that is thinly myelinated and conducts the first feeling of pain that is often felt as coming on as a sharp, rapid feeling is called the A-delta fiber.

A-delta fibers are one of the two main types of sensory neurons involved in the transmission of pain signals to the brain. The other type is called C fibers, which are unmyelinated and transmit a slower, more diffuse and longer-lasting feeling of pain. A-delta fibers are responsible for the initial, sharp sensation of pain that is felt immediately after an injury.

In contrast, C fibers are responsible for the more persistent, dull, and aching sensation of pain that is felt after the initial sharp sensation has subsided. Both A-delta fibers and C fibers play important roles in the experience of pain, and both types of fibers are involved in the transmission of pain signals to the brain.

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Answer:

bases are sour in taste. second one is that when it react with acid ot produces salt and water.

Answer:

The answer to your question is molecules.

Answer:

acid +base gives salt and water

acid and carbon gives salt and water and carbon dioxide

acid and metal gives salt and hydrogen

The interspecific chemical messenger that benefits the producer is called an allelochemical.

Allelochemicals are chemical compounds produced by one species that affect the growth, survival, or reproduction of other species. These chemicals can have positive or negative effects on the recipient species, but they always benefit the producer.

For example, some plants produce allelochemicals that inhibit the growth of competing plants, giving them an advantage in the struggle for resources. Other allelochemicals attract predators or parasites of herbivores, providing protection for the producer. Allelochemicals can also be used in communication between species, such as pheromones used by insects to attract mates.

Overall, allelochemicals play an important role in shaping interactions between species in ecological communities.

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Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

\(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\)  where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation:  \(T_C+273=T_K\)

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, \(T_1 = 273[K]\), and \(T_2 = (49.4+273)[K]=322.4[K]\).

\(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\)

\(\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}\)

\(\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})\)

\(1810.04571428[mL]=V_2\)

Adjusting for significant figures, this gives \(V_2=1810[mL]\)

Answer:

This organelle has two major functions: it stores the cell's hereditary material, or DNA, and it coordinates the cell's activities, which include growth, intermediary metabolism, protein synthesis, and reproduction (cell division).

You would need to add 250 mL of water.

Answer:

Endothermic

Explanation:

Because Endothermic is cold and exothermic is hot and if you are using an ice pack it would be Endothermic and if you were using something that was hot it would be exothermic

Answer:

Explanation:

Particles in both liquids and gases (collectively called fluids) move randomly. This is called Brownian motion. They do this because they are bombarded by the other moving particles in the fluid. Larger particles can be moved by light, fast-moving molecules.

The molecular formula of dye:

A major textile dye manufacturer developed a new yellow dye with a molecular formula \(C_{15}N_{3}H_{15}\).

Given:

Percentage composition:

C = 75.95%

N = 17.72%

H = 6.33%

Molar mass = 240 g

To find: molecular formula of dye

Calculation:

Let's assume the mass of the compound is 100 gm.

So,  Mass of C = 75.95 gm

Mass of N = 17.72 gm

Mass of H = 6.33 gm

The number of moles = given mass/mass

Therefore, the number of moles of C = 75.95/12 = 6.33

The number of moles of N = 17.72/14 = 1.27

Number of moles of H =6.33/1 = 6.33

Simplest ratio:

C = 6.33/1.27 = 4.98 =5

N = 1.27/1.27 = 1

H = 6.33/1.27 = 4.98 =5

Therefore, the empirical formula is \(C_{5}NH_{5}\)

The mass of empirical formula = 12 x 5 + 14 + 1 x 5 = 79 gm

Number of moles= molar mass/ empirical formula

n = 240/79 = 3.04 =3

Therefore, molecular formula = empirical formula x n

Hence, molecular formula = \(C_{15}N_{3}H_{15}\)

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Considering  the following elementary gas-phase reversible reaction to be carried out isothermally with no pressure drop,

              1)  0% = 0

             2)  50%= 0.5

             3)   0.99Xe= 99%

When a chemical reaction is reversible, the reactants and products both react to create the final product at the same time.

Such a reaction involves constant forward and backward reactions between the reactants and products.

Simply said, when molecules collide, reactant bonds are disrupted, and the energy released from this broken connection is used to create new product molecules.

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Answer:

City B has the larger annual temperature range

Explanation:

This is correct option because generally the northern side of the equator is high in temperature than the southern hemisphere part.

Since the southern side of the equator or Southern Hemisphere, where city A resides will generally have higher altitude or rise, so this creates higher average temperature.

Ionic and Covalent Bonds.

Ionic bonds are between metals and non-metals, where electrons are transferred from the metal to the non-metal.

Covalent bonds are between non-metals, where electrons are shared between atoms and are not lost or gained.

In Anions, electrons are gained

Anions are negatively charged ions. They gain electrons to achieve a stable electron configuration.

b) Electrons are negatively charged subatomic particles that orbit the nucleus of an atom. They are responsible for chemical reactions and the formation of chemical bonds between atoms.

The most common examples of anions are halide ions such as chloride (Cl-), bromide (Br-), and iodide (I-), which are formed by the addition of an extra electron to the outermost shell of a halogen atom.

Other examples of anions include hydroxide (OH-), sulfate (SO42-), and nitrate (NO3-), which are important in many chemical reactions and biological processes. Anions are usually attracted to positively charged ions or molecules, called cations, to form ionic compounds.

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