MISSED THIS? Read Section
17.9 (Pages 765-769).
Consider a 0.14 M solution of a weak polyprotic
acid (H₂A) with the possible values of Kai and
K₁2 given here. Calculate the contributions to
[H3O+] from each ionization step. At what point
can the contribution of the second step be
neglected?
K₁1 = 1.0 × 10 and K₁2=1.0 x 10-6
Express your answers in molarity to two significant figures separated by a comma.
VE| ΑΣΦ 3 -C?
[H₂O₁. [H₂O] 3.7 10,6.03 101
Submit
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M
x Incorrect; Try Again; 4 attempts remaining
You have correctly determined the contribution to [H₂O] from the first ionization step. T
determine the contribution to [H₂O] from the second ionization step, prepare an ICE ta
(where ) represents initial, C represents change, and E represents equilibrium), showing
change in the H₂O concentration with the variable y and defining the equilibrium
concentrations in terms of the concentrations produced in the first step and the variable y
Substitute these expressions into the expression for Kazi

The standard reduction potentials is Ba metal will be oxidized to Ba2+ and the Li+ and Mg2+ ions will be reduced to Li and Mg metal.

When Ba metal is added to an aqueous solution containing dissolved LiCl and MgCl2, Ba metal will be oxidized to Ba2+ ions based on the standard reduction potentials. However, the species that will undergo reduction depends on their respective reduction potentials.

According to the standard reduction potentials, Li+ has a more positive reduction potential than Mg2+, which means Li+ has a greater tendency to undergo reduction compared to Mg2+. Therefore, Li+ ions will be reduced to Li metal while Mg2+ ions will remain in solution.

The overall reaction can be represented as follows:

Ba(s) + 2Li+(aq) → Ba2+(aq) + 2Li(s)

Therefore, the correct answer is Ba metal will be oxidized to Ba2+ and the Li+ and Mg2+ ions will be reduced to Li and Mg metal. Mg2+ ions will not be reduced to Mg metal is incorrect. The formation of H2 gas and -OH ions, which are not supported by the standard reduction potentials is incorrect.  -OH ions are not formed when Li+ ions undergo reduction is incorrect. A reaction does occur based on the standard reduction potentials is incorrect.

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The rate constant for the reaction can be found out using Arrhenius equation.

Arrhenius equation can be stated as:

[tex]ln\frac{k2}{k1}=\frac{Ea}{R}[\frac{1}{T1}-\frac{1}{T2}][/tex]

i.e [tex]log\frac{k2}{k1} = \frac{Ea}{2.303R} [\frac{1}{T1}-\frac{1}{T2} ][/tex]

i.e [tex]log(k2)-log(k1) = \frac{Ea}{2.303R} [\frac{T2-T1}{T1xT2}][/tex]

From the given data, k1 = 6.1 s⁻¹, T1 = 600K, T2 = 805K, Ea = 262 kJ/mol and R = 8.314 J/molK

Substituting in the Arrhenius equation, we get

[tex]log\frac{k2}{6.1x10^-^8}= \frac{262}{2.303 x 8.314} [\frac{805-600}{600 x 805}][/tex]

[tex]log (k2) = log (k1) + \frac{Ea}{2.303R} [\frac{T2-T1}{T1xT2}][/tex]

[tex]log (k2)= log(6.1x10^-^8) + \frac{262}{2.303x8.314} x \frac{805-600}{600x805}[/tex]

[tex]log(k2)= log (6.1x10^-^8) + 5.81 x 10^-^3\\log(k2) = -7.214 + 0.00581\\log(k2) = -7.21[/tex]

[tex]k2 = antilog (-7.21) = 6.17 x 10^-^8[/tex]

Thus, on solving for k2, we get k2 = 6.17 × 10⁻⁸ s⁻¹

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The area of a circle depends on the length of the radius, and the area of a sector depends on the ratio of the central angle to the entire circle, hence options A, B, D, E are correct.

A circle is the location of a point such that it is always a constant distance from a fixed point known as the center.

The statements true regarding the area of circles and sectors are:

The area of a circle depends on the length of the radius.

The area of a sector depends on the ratio of the central angle to the entire circle.

The area of the entire circle can be used to find the area of a sector.

The area of a sector can be used to find the area of a circle.

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(a) Electroless deposition is a type of electrochemical deposition that does not require electrons - X.

(b) Increasing metal ion concentration increases the deposition current density on the electrodeposited structures - X.

(c) The standard reaction Gibbs energy change for water electrolysis is negative, not positive, and generates 1.23V during electrolysis under standard conditions - X.

(d) When the system is under mass transport limitation, the electrodeposited structures are normally dense and uniform - O.

(e)  If you deposit metal A on metal B with a huge lattice misfit between them, the deposition process follows the island growth mechanism rather than the layer-by-layer growth mechanism - O.

(f) To make an AxBy alloy from metals A and B with standard reduction potentials of 1.0V and -1.0V, respectively, you need to apply a potential between -1.0V and 1.0V, depending on the desired stoichiometry - O.

(g) The length of metal nanowires made using AAO templated electrodeposition can be controlled by the anodization time and the thickness of the AAO template - O.

(h) The membrane electrolyte for PEMFC should only allow for ionic movement, not electronic movement  - O.

(i) The fuel cell electric vehicle generates less CO2 than traditional vehicles but still produces some CO2 during operation  - O.

(j) The organic leveler used in the electrodeposition process interacts with the substrate or growing deposits through chemical bonding rather than van der Waals interaction  - O.

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The molar solubility of AgbPO₄ in a 0.223 M Na₃PO₄ solution is  2.3 x 10⁻⁷ M.

Given:

The value of Ksp for Ag₃PO₄ = 1.3 x 10⁻²°

The balanced equation is:

Ag₃PO₄(s) ⇌ 3 Ag⁺(aq) + (PO₄)³⁻(aq)

The solubility product expression for this reaction is:

Ksp = [Ag+]³ [PO₄⁻³]

Initial: [Ag⁺] = 0 [PO₄⁻³]

= 0.223 M

Change: +3x +x

Equilibrium: [Ag₊] = 3x [PO₄⁻³]

= 0.223 + x

Substituting these values into the Ksp expression:

Ksp = (3x)³ (0.223 + x)

= 1.3 x 10⁻²⁰

Ksp = 27 x³ (0.223) ≈ 6.0 x 10⁻²⁰

Solving for x:

x ≈ 2.3 x 10⁻⁷ M

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Answer:So, about 0.093 g of the isotope will remain after 19.8 days.

Explanation:

The first step is to find the number of half-lives that have passed during 19.8 days:

Number of half-lives = time elapsed / half-life

Number of half-lives = 19.8 days / 3.8 days per half-life

Number of half-lives ≈ 5.21

This means that the initial amount of the isotope has been halved 5.21 times. The remaining fraction of the original amount can be calculated using the following formula:

Remaining fraction = (1/2)^(number of half-lives)

Substituting the values, we get:

Remaining fraction = (1/2)^5.21

Remaining fraction ≈ 0.031

Therefore, the amount of the isotope remaining after 19.8 days is:

Remaining amount = Remaining fraction x Initial amount

Remaining amount = 0.031 x 3.00 g

Remaining amount ≈ 0.093 g

So, about 0.093 g of the isotope will remain after 19.8 days.

The isotope in grams will remain after 19.8 days would be 0.081 grams.

The formula to calculate the left mass of a radioactive element can be deduced as -

[tex] \qquad\star\longrightarrow \underline{\boxed{\sf{m =m_{o} \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{t}{T½}} }}} \\[/tex]

Where-

According to the given specific parameters -

Now that we have all the required values, so we can plug them into the formula and solve for the left mass of a radioactive element-

[tex] \qquad \longrightarrow \sf \underline{m =m_{o} \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{t}{T½} }} \\[/tex]

[tex] \qquad\longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{19.8}{3.8} } \\[/tex]

[tex]\qquad \longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{\cancel{19.8}}{\cancel{3.8}} } \\[/tex]

[tex] \qquad\longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ 5.21052..... } \\[/tex]

[tex] \qquad\longrightarrow \sf m =3 \times 0.02700... \\[/tex]

[tex] \qquad\longrightarrow \sf m =0.081020....\;g \\[/tex]

[tex] \qquad\longrightarrow \sf \underline{m =\boxed{\sf{0.081\;g}}} \\[/tex]

Answer:

Paired electrons are the electrons in an atom that occur in an orbital as pairs.

⇒paired electrons always occur as a couple of electrons

unpaired electrons are the electrons in an atom that occur in an orbital alone.

⇒unpaired electrons occur as single electrons in the orbital.

Answer:

Paired electrons are the electrons in an atom that occur in an orbital as pairs whereas unpaired electrons are the electrons in an atom that occur in an orbital alone. Therefore, paired electrons always occur as a couple of electrons while unpaired electrons occur as single electrons in the orbital.

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The cell potential of a galvanic cell is determined by the difference in the reduction potentials of the two half-cells involved. The larger the difference, the higher the cell potential. The half-reaction with the highest reduction potential will give the largest cell potential when paired with the Ni2+/Ni electrode.

When looking at the reduction potentials, Al3+/Al has a standard reduction potential of -1.66 V, whereas Ni2+/Ni has a standard reduction potential of -0.25 V. Therefore, the reaction with the highest reduction potential difference (i.e., the largest cell potential) when paired with the Ni2+/Ni electrode would be the one that has a reduction potential greater than -0.25 V.
Out of the options given, Al3+/Al has the highest reduction potential and thus it would give the largest cell potential when paired with the Ni2+/Ni electrode. This is because the reduction potential difference between Al3+/Al and Ni2+/Ni is 1.41 V, which is the largest among the given options.
In conclusion, the half-reaction that would give the largest cell potential when paired with a Ni2+/Ni electrode is Al3+/Al.

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The required amount in moles are 3 moles of H₂O to produce 164 g of H₃PO₃.

To solve the problem, use the balanced chemical equation to relate the amount of H₃PO₃ formed to the amount of H₂O used. From the balanced chemical equation:

P2O₃ + 3H₂O → 2H₃PO₃

3 moles of H₂O are required to produce 2 moles of H₃PO₃. This can be written as:

2 moles H₃PO₃ / 3 moles H₂O

To find the number of moles of H₂O required to produce 164 g of H₃PO₃, use the molar mass of H₃PO₃:

1 mole H₃PO₃ = 82 g

So, 164 g of H₃PO₃   is equal to:

164 g H₃PO₃   / 82 g/mol = 2 moles H₃PO₃    

Using the ratio above, calculate the number of moles of H₂O required:

2 moles H₃PO₃ × (3 moles H2O / 2 moles H₃PO₃) = 3 moles H₂O

Therefore, 3 moles of H₂O are required to produce 164 g of H₃PO₃.

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The balanced half reaction is obtained as [tex]SO_{3} ^2- (aq) + H_{2} O (l) --- > SO_{4}^2- (aq) + 2H^+ (aq).[/tex]

We know that for the reaction to be seen as balanced we would haver to look at the masses and the charges and now we are going to have that, two protons are added to the product side to balance the charge. To balance the amount of hydrogen atoms on the reactant side, water  is also supplied.

Then when we look at the balanced reaction for an acid medium as have been required by the question then we are going to have;

[tex]SO_{3} ^2- (aq) + H_{2} O (l) --- > SO_{4}^2- (aq) + 2H^+ (aq).[/tex]

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The balanced chemical equation for the combustion of propane with oxygen is: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

To balance the equation, first balance the carbon atoms on both sides of the equation. There are three carbon atoms in the propane molecule and three in the carbon dioxide molecule, so balance the carbon atoms by putting a coefficient of 3 in front of the CO₂ molecule.

C3H8 + 5O2 → 3CO₂

Next, balance the hydrogen atoms. There are eight hydrogen atoms in the propane molecule and four in the water molecule, so balance the hydrogen atoms by putting a coefficient of 4 in front of the H₂O molecule.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Finally, balance the oxygen atoms. There are five oxygen atoms on the left side and 10 on the right side, so balance the oxygen atoms by putting a coefficient of 5 in front of the O₂ molecule.

Therefore, the correct whole number coefficient for propane, C3H8, is 1.

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According to a balanced chemical equation, solid lead (II) sulphide reacts with aqueous hydrochloric acid to produce solid lead (II) chloride and dihydrogen sulphide gas.

PbCl2 + H2S (s) PbS + 2HCl PbS stands for solid lead (II) sulphide, 2HCl for aqueous hydrochloric acid, PbCl2 for solid lead (II) chloride, and H2S (s) for dihydrogen sulphide gas in this equation. Parentheses represent the stages of the reactants and products. While the products are solid and gaseous, the reactants are in the solid and aqueous phases.

Since each element has an equal amount of atoms on both sides of the equation in the sulfide and in the other one as  hydrochloric well, the equation is balanced. There is one atom of lead on the left.

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The orbital's orientation in space is described by the magnetic quantum number (m). Any number between -l and +l may represent the value of m.

The electron's orbital form is determined by a quantum number called the azimuthal quantum number. Any integer between 0 and n-1 can be used to represent the value of l, and as it rises, the orbital's form becomes more complex.

The quantum numbers that are involved have been shown above.

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Answer:

Let me explain

Explanation:

The balanced equation for the given reaction is:

3Cu + 8NO3- + 4H+ → 3Cu2+ + 2NO + 4H2O

Here's how to balance the equation using the half-reaction method:

Half-reaction for oxidation: Cu → Cu2+

To balance this half-reaction, we need to add two electrons to the left side:

Cu → Cu2+ + 2e-

Half-reaction for reduction: NO3- → NO

To balance this half-reaction, we need to add three electrons to the right side:

NO3- + 3e- → NO

Now, we need to balance the number of electrons in both half-reactions. To do this, we need to multiply the oxidation half-reaction by three and the reduction half-reaction by two:

3Cu → 3Cu2+ + 6e-

2NO3- + 6e- → 2NO

Now, we can combine the two half-reactions by adding them together and canceling out the electrons:

3Cu + 8NO3- + 4H+ → 3Cu2+ + 2NO + 4H2O

This is the balanced equation for the given reaction.

The specific heat of aluminum is  0.92 J/g°C.

Use the principle of conservation of energy.

Q aluminum = -Qwater-calorimeter

(m aluminum)(c aluminum)(ΔT aluminum) = -(m water + m calorimeter)(c water)(ΔT water)

where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Calculate the heat lost by the aluminum.

Q aluminum = (m aluminum)(c aluminum)(ΔT aluminum)

where ΔT aluminum is the change in temperature of the aluminum when it was heated in boiling water:

ΔT aluminum = 98.7°C - 100°C

= -1.3°C

Q aluminum = (73.5 g)(c aluminum)(-1.3°C)

Q water-calorimeter = -(m water + m calorimeter)(c water)(ΔT water)

where ΔT water is the change in temperature of the water in the calorimeter:

ΔT_water = 28.2°C - 25.4°C

= 2.8°C

Q water-calorimeter = -(1500 g + m calorimeter)(4.18 J/g°C)(2.8°C)

Q water = -(1500 g)(4.18 J/g°C)(2.8°C)

Substitute the values and solve for c aluminum:

(73.5 g)(c aluminum)(-1.3°C) = -(1500 g)(4.18 J/g°C)(2.8°C)

c aluminum = -(1500 g)(4.18 J/g°C)(2.8°C) / (73.5 g)(-1.3°C)

c aluminum = 0.92 J/g°C

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Aerobic respiration is the term used to describe the process of energy release during the oxidation of food molecules in the presence of oxygen. In addition to energy (ATP), carbon dioxide gas and water molecules are also produced during respiration. The product is oxide. The correct option is C.

An element or compound gains oxygen atoms during an oxidation reaction. When a substance interacts with the element oxygen to form an oxide, an oxidation reaction takes place. An illustration of an oxidation reaction is combustion, or burning.

A molecule undergoes oxidation when it loses electrons or increases its oxidation status. A separate molecule that undergoes reduction in the process gains the electrons that are lost by the oxidising molecule.

Thus the correct option is C.

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Answer: 156.75 J/C

Explanation:

Q=CT (C is specific heat and T is change in temperature)

Disclaimer: I am assuming the initial temperature is 0 degrees Celsius because the question has been worded improperly. This assumption is not made because it is true, but because there is not enough information in the current question to solve for the specific heat. Also, the mass of the sample is given, which is unnecessary for solving the specific heat but necessary to solve for the specific heat capacity. Either the question has not been worded properly or the mass given is just to trick students.

Rearrange to isolate C: C = Q/T

Solve for C: C = 3135/(20-0) = 3135/20 = 156.75 J/C

Answer: 0.137

Explanation:

acellus

If the Sun's surface became much hotter (while the Sun's size remained the same), the Sun would emit more ultraviolet light but less visible light than it currently emits. The statement is True.

Solar radiation is radiant (electromagnetic) energy from the sun. It provides light and heat for the Earth and energy for photosynthesis. This radiant energy is necessary for the metabolism of the environment and its inhabitants 1. The three relevant bands, or ranges, along the solar radiation spectrum are ultraviolet, visible (PAR), and infrared.

Ultraviolet radiation makes up just over 8% of the total solar radiation.

When heat increases, so does the frequency and energy of the wavelengths. Because of this, some visible light would be converted to ultraviolet light.

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The appropriate response is: based on the facts provided.

A 2e-reducing agent, Na2S2O5, only produces SO3 when interacting with the ion Au3+.(option-4)

Sodium metabisulfite (Na2S2O5) is used, which implies that it is working as a reducing agent. In this case, it is reversibly reducing the Au3+ ion to gold (Au), an elemental form that precipitates as a brown solid.

The following is a representation of the reaction's balanced equation:

2Au (s) + 2SO3 2- (aq) + 2Na+ (aq) + 6H+ (aq) = 2Au3+ (aq) + Na2S2O5 (aq) + 3H2O (l)

The electrons required for reducing Au3+ to Au are here provided by Na2S2O5 acting as a 2e- reducing agent. H+ ions from the excess HCl are also present during the process.(option-4)

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Answer:

Over time, as generations of individuals with the trait continue to reproduce, the advantageous trait becomes increasingly common in a population, making the population different than an ancestral one.

Explanation:

have a nice day.

Answer:

Explanation:

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The single replacement reaction of iron and silver nitrate can be represented by the chemical equation Fe + [tex]2AgNO_{3}[/tex] → [tex]Fe(NO_{3} )_{2}[/tex] + 2Ag

In this reaction, a single element, iron (Fe), replaces another element, silver (Ag), in the compound silver nitrate [tex](AgNO_{3} )[/tex], resulting in the formation of iron (II) nitrate [tex](Fe(NO_{3} )_{2} )[/tex] and elemental silver (Ag). The balanced equation shows that one iron atom reacts with two silver nitrate molecules to produce one molecule of iron (II) nitrate and two silver atoms.

To balance the equation, we need to ensure that the same number of each type of atom is present on both sides of the equation. We begin by counting the number of atoms of each element present in the reactants and products. The left side of the equation has one Fe atom, two Ag atoms, two N atoms, and six O atoms. The right side has one Fe atom, two Ag atoms, two N atoms, and six O atoms. Therefore, the equation is already balanced.

Overall, this reaction is an example of a single replacement reaction in which a more reactive element (iron) replaces a less reactive element (silver) in a compound. The resulting products are an ionic compound (iron (II) nitrate) and a pure element (silver). This type of reaction can be used to extract metals from their compounds or to convert one metal into another.

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Cargnello and coworkers developed a novel catalyst that advances this objective by boosting the formation of long-chain hydrocarbons during chemical processes.

Thus, The same amounts of carbon dioxide, hydrogen, catalyst, pressure, heat, and time as the standard catalyst, it created 1,000 times more butane—the longest hydrocarbon it could produce at its maximum pressure—than the standard catalyst.

The novel catalyst is made of ruthenium, a platinum group rare transition metal that is coated in a thin coating of plastic. This idea accelerates chemical processes without being consumed in the process, much like any catalyst.

Another benefit of ruthenium is that it is less expensive than other platinum- and palladium-based high-quality catalysts.

Thus, Cargnello and coworkers developed a novel catalyst that advances this objective by boosting the formation of long-chain hydrocarbons during chemical processes.

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Here are 0.686 moles of hydrogen chloride in 25 grams of HCl.
Number of moles = Mass of substance (in grams) / Molar mass of substance

The molar mass of HCl can be calculated by adding the atomic masses of hydrogen (1.008 g/mol) and chlorine (35.45 g/mol), which gives a molar mass of 36.458 g/mol.
Now we can plug in the values:
Number of moles = 25 g / 36.458 g/mol
Number of moles = 0.686 moles
It's important to note that the molar mass of a substance is the mass in grams of one mole of that substance. This means that if we know the mass of a substance and its molar mass, we can find the number of moles present in that mass. This is a useful calculation in chemistry as it allows us to make accurate measurements and carry out calculations involving the reactions and properties of different substances.

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The water absorbed 1.998 kilocalories of heat, which is equivalent to the calorie content of the Flamin' Hot Cheeto.

Assuming that all the heat generated by the combustion of the Cheeto was absorbed by the water, we can calculate the amount of heat transferred to the water using the formula: q = m·C·∆T

where q is the amount of heat transferred, m is the mass of water, C is the specific heat of water, and ∆T is the change in temperature of the water.

First, we need to calculate the mass of water: m_water = 44.84 g

Next, we need to calculate the change in temperature of the water:

∆T = 66.5 °C - 22.0 °C = 44.5 °C

The specific heat of water is 1 calorie/(g·°C), so we can substitute these values into the formula to get:

q = (44.84 g) · (1 cal/(g·°C)) · (44.5 °C) = 1998.18 cal

Converting this to kilocalories, we get:

q = 1998.18 cal / 1000 cal/kcal = 1.998 kcal

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The molecule that represents butene is option C: H2C=CHCH2CH3.

Butene is an alkene with four carbon atoms and a double bond between the second and third carbon atoms. In the given structure, the carbon atoms are connected in a linear chain with hydrogen atoms attached to them. The double bond between the second and third carbon atoms is denoted by the "=" symbol.

To identify butene, we can count the number of carbon atoms in the molecule. Butene has four carbon atoms, and option C satisfies this requirement. Additionally, the presence of a double bond between the second and third carbon atoms is another characteristic feature of butene, which is represented by the "=" symbol in option C.

Option A, H3C-O-C-C-C-H3, represents an ether molecule, not butene. Option B, HC≡CH, represents acetylene, a different hydrocarbon. Option D, H3C-EC-C-C-H3, does not correctly represent a recognizable organic molecule. option C, H2C=CHCH2CH3, is the structure that represents butene accurately. Therefore, Option C is correct.

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Answer:

To solve these problems, we can use the formula:

q = mcΔT

where q is the heat transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the temperature change.

The mass of the sample of tin can be calculated as:

q = mcΔT

36298 J = m × 0.227 J/(g.°C) × (-160.56 °C)

m = 708.2 g

The temperature change of the sample of ammonia can be calculated as:

q = mcΔT

33834 J = 13.66 mol × 80.08 J/(mol.°C) × ΔT

ΔT = 31.7 °C

The mass of the sample of cobalt can be calculated as:

q = mcΔT

455500 J = m × 0.4187 J/(g.°C) × (-1132.52 °C)

m = 27.4 g

The specific heat capacity of indium can be calculated as:

q = mcΔT

12505 J = 372.4 g × c × 140.73 K

c = 0.238 J/(g.°C)

The temperature change of the sample of molybdenum can be calculated as:

q = mcΔT

35961 J = 4.721 mol × 24.06 J/(mol.°C) × ΔT

ΔT = 31.9 °C

The heat transferred by the sample of ethanol can be calculated as:

q = mcΔT

q = 56.2 g × 2.44 J/(g K) × (-110.56 K)

q = -15,585 J

The temperature change of the sample of chromium can be calculated as:

q = mcΔT

38674 J = 5.774 mol × 23.35 J/(mol.°C) × ΔT

ΔT = 27.4 °C

The heat transferred by the sample of magnesium can be calculated as:

q = mcΔT

q = 1.008 mol × 24.9 J/(mol K) × (-683.83 K)

q = -17,134 J

The heat transferred by the sample of tin can be calculated as:

q = mcΔT

q = 0.2687 mol × 27.112 J/(mol K) × 222.48 K

q = 1676.7 J

The specific heat capacity of neon can be calculated as:

q = mcΔT

14738 J = 1.008 mol × c × (-703.43 K)

c = 36.8 J/(mol.°C)

Explanation:

Answer:

B. Ozone

Explanation:

Answer: B

Explanation:

Ozone consists of three oxygen atoms covalent bonded together. Ozone is quite reactive.

The mass of the number of the reacting agent in the reaction producing Cr₂S₃ and H₂O are;

a. The mass of the H₂S is about 126.2 grams

b. Mass of Cr₂S₃ produced is about 242.242 grams

A reacting agent in a chemical reaction are the elements, compounds or molecules on the reaction side of a chemical reaction.

Whereby the chemical reaction is; Cr₂O₃ + 3·H₂S → Cr₂S₃ + 3·H₂O

We get;

5 a. The molar mass of H₂O is; 18.01528 g/mol

The number of moles of H₂O in 66.6 grams of H₂O is; 66.6/18.01528 ≈ 3.7 moles

The number of moles H₂S required to produce 3 moles of H₂O = 3 moles

Therefore, the number of moles H₂S required to produce 3.7 moles of H₂O = 3.7 moles

The molar mass of H₂S =- 34.1 g/mol

5 b. The molar mass of Cr₂S₃ = 200.2 g/mol

The molar mass H₂S = 34.1 g/mol

The mass of the H₂S = 123.7 grams

The number of moles of H₂S is; 123.7 grams/(34.1 g/mol) ≈ 3.63 moles

The stoichiometry of the reaction indicates that the mole ratio of the number of moles of Cr₂S₃ to the number of moles of the H₂S is; 1 : 3

Therefore, the number of moles of the Cr₂S₃ in the reaction is; 3.63 moles/3 ≈ 1.21 moles

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The amount of energy needed to heat 500 g of ice at  0⁰C to 60⁰C is 292,400 J. Option C.

In order to calculate the energy required to heat the ice, we need to consider two stages: first, we need to calculate the energy required to melt the ice, and second, we need to calculate the energy required to heat the resulting liquid water to 60°C.

To melt the ice, we need to supply energy equal to the heat of fusion of ice. The heat of fusion of ice is 334 J/g. Therefore, the energy required to melt 500 g of ice is:

Q1 = (334 J/g) x (500 g) = 167,000 J

Once the ice is melted, we need to heat the resulting liquid water to 60°C. The specific heat capacity of water is 4.184 J/(g°C). Therefore, the energy required to heat 500 g of water from 0°C to 60°C is:

Q2 = (4.184 J/(g°C)) x (500 g) x (60°C - 0°C) = 125,520 J

The total energy required to melt the ice and heat the resulting liquid water to 60°C is the sum of Q1 and Q2:

Q = Q1 + Q2 = 167,000 J + 125,520 J = 292,520 J

Thus, the amount of energy needed to heat 500 g of ice at  0⁰C to 60⁰C is 292,400 J.

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In between conductors, which are typically metals, and not-conductors or insulators, such as ceramics, exist materials known as semiconductors. Semiconductors can be pure elements like germanium or silicon or compounds like gallium arsenide.

In the given pictures, 'As' is a N-type semiconductor whereas 'Ga' is a P-type semiconductor.

When pentavalent impurities (P, As, Sb, and Bi) are added to a pure semiconductor (germanium or silicon), four of the five valence electrons form a bond with the four electrons of the pure semiconductor.

The dopant's fifth electron is liberated and used for conduction in the lattice are called N-type semiconductors.

When a trivalent impurity (B, Al, In, or Ga) is added into a pure semiconductor, three of the semiconductor's four valence electrons form a bond with the impurity's three valence electrons.

In the impurity, this results in an electron (hole) being missing called P -type semiconductors.

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