If 11.5 ml of vinegar sample (d=1g/ml) is titrated with 18.5 ml of standardized Sodium hydroxide
solution. What is the percent of acetic acid (by mass) in the vinegar?

Answer:

pH is 8.52

Explanation:

According to the working in the photo

Answer: c) This process is an example of sublimation.

Explanation:

When liquid is further heated, the molecules gain more kinetic energy and molecules move farther and convert to gaseous state and the process is called vaporization.

Sublimation is a process of converting a substance from solid state to gaseous state without the formation of liquid at constant temperature.

Deposition is a process of converting a substance from gaseous state to solid state without the formation of liquid at constant temperature.

In liquid the particles are loosely packed as the inter molecular forces of attraction are not that strong. As they are cooled, the kinetic energy of the molecules decreases and thus the molecules start getting closer and convert to solid state. The process is called freezing.

Answer: Schedule (2) two drug

Explanation: Although it’s virtually difficult or impossible to design a set of defining drug classification standards because even experts have struggled on which drugs should be and not be on a particular schedule but drugs are generally categorised based on their abuse rate (misuse and physical dependency) and their medical use. Drugs with no medical use and higher abuse rate which has physical and mental implications are placed under schedule one, drugs with medical use but also higher abuse rate with physical and mental implications are placed under schedule two, drugs with lower or moderate abuse rate are placed under schedule three, drugs with low potential for dependency are place under schedule four and drugs which are mostly use for antidiarrheal, analgesic and antitussive are placed under schedule five.

Chang's drug has medical use because it was prescribed by his doctor, it also has a high abuse rate because he started overdosing on them which led to a severe physical and mental implications.

Therefore Cheng's drugs will mostly likely be categorised as a schedule (2) two drug.

Answer:

The correct equation to calculate the heat of this reaction is:

ΔH = m*s*∆T

Explanation:

During any chemical reaction, heat can either be absorbed from the environment or released to the environment through the reaction. The heat exchange between a chemical reaction and its environment is known as the reaction enthalpy, or H. However, H cannot be measured directly; the change in temperature of a reaction over time is used to find the enthalpy change over time (denoted as ΔH).

In general ΔH = m*s*∆T, where m is the mass of the reactants, s is the specific heat of the product, and ΔT is the change in the reaction temperature.

Answer:

1. Hot water

2. A pile of wood shavings

Explanation:

Sugar can dissolve more quickly in hot water than in cold water because there is more energy in hot water molecules. Because they are moving faster, they have more energy to break the bonds that hold sugar together. There is also more energy available to break the hydrogen bonds that hold water together.

Wood shavings have a greater contact surface than the solid hunk of wood, which is why they have a higher calorific value and then they will catch fire more easily.

Answer:

The remaining concentration of H2SO4 is 0.152 M

Explanation:

Step 1: Data given

Volume of H2SO4 = 0.950 L

Molarity H2SO4 = 0.410 M

Volume of KOH = 0.900 L

Molarity of KOH = 0.240 M

Step 2: The balanced equation

H2SO4 + 2KOH → K2SO4 + 2H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles H2SO4 = 0.410 M * 0.950 L

Moles H2SO4 = 0.3895 moles

Moles KOH = 0.240 M * 0.900L

Moles KOH = 0.216 moles

Step 4: Calculate the limiting reactant

For 1 mol H2SO4 we need 2 moles KOH to produce 1 mol K2SO4 and 2 moles H2O

The limiting reactant is KOH. It will completely be consumed (0.216 moles).

H2SO4 is in excess. There will react 0.216/2 = 0.108 moles. There will remain 0.3895 moles - 0.108 moles = 0.2815 moles

Step 5: Calculate the concentration of H2SO4 remaining

[H2SO4] = moles / volume

[H2SO4] = 0.2815 moles / 1.85 L

[H2SO4]= 0.152 M

The remaining concentration of H2SO4 is 0.152 M

Answer:

See explaination

Explanation:

methyl-3-nitrobenzoate = 3-NO2-C6H4-COOCH3

Singlet at 4 ppm = CH3 from ester (COOCH3)

Triplet at 7.6 ppm = Aromatic, H5 proton

Doublet at 8.2 ppm - Aromatic, H4 and H6 protons

Singlet at 8.8 ppm - Aromatic, H2 proton

See attached file for diagrammatic representation and further solution.

Methylbenzoate has 8 distinct carbon environments, therefore it would show 8 distinct 13C NMR signals. Methyl-3-nitrobenzoate, due to its symmetrical structure, has 7 distinct carbon environments, therefore it would show 7 distinct 13C NMR signals.

The product, methyl-3-nitrobenzoate, and the reactant, methylbenzoate, are both organic compounds, and their 13C NMR signals can be determined by the number of carbon environments they have.

Methylbenzoate has 8 distinct carbon environments: benzoate carbon framework (7) + methyl group (1). Hence, in the 13C NMR spectroscopy, methylbenzoate would show 8 distinct signals.

Methyl-3-nitrobenzoate on the other hand, due to the introduction of the nitro group, creates symmetry which changes the carbon environments, thus, it has 7 distinct carbon environments: nitrobenzoate carbon framework (6) + methyl group (1). So, methyl-3-nitrobenzoate would show 7 distinct signals in its 13C NMR spectroscopy.

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Answer:

5L

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

A 2.5 liter sample of gas is STP. When the temperature is raised to 546 K and the pressure remains constant the new volume of the gas will be 5 L.

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Pressure

V = Volume

n = number of moles  

R = Ideal gas constant

T = Temperature

STP is Standard Temperature and Pressure. At STP the temperature is 273 K or 0°C and the standard pressure is 1 atm.

Now, according to question the pressure is constant, then

At constant pressure

V ∝ T

[tex]\frac{V_1}{V_2} = \frac{T_1}{T_2}[/tex]

[tex]\frac{2.5\ L}{V_2} = \frac{273\ K}{546\ K}[/tex]

[tex]V_{2} = \frac{546 \times 2.5\ L}{273}[/tex]

V₂ = 5 L

Thus, from the above conclusion we can say that A 2.5 liter sample of gas is STP. When the temperature is raised to 546 K and the pressure remains constant the new volume of the gas will be 5 L.

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Answer:

Magnesium chloride and water  

Explanation:

Mg(OH)₂ + 2HCl ⟶                MgCl₂                +   2H₂O

                                    magnesium chloride        water

Answer: 4 moles of Na are needed to produce 4 moles of NaCl

Explanation:

The balanced chemical equation is :

[tex]2Na+Cl_2\rightarrow 2NaCl[/tex]

According to stoichiometry :

2 moles of [tex]NaCl[/tex] are formed from = 2 moles of [tex]Na[/tex]

Thus 4 moles of [tex]NaCl[/tex] will be formed from = [tex]\frac{2}{2}\times 4=4moles[/tex]  of [tex]Na[/tex]

Thus 4 moles of Na are needed to produce 4 moles of NaCl

Answer: There are 6.725 mols

Explanation:

Message me for extra information

parkguy786 snap

Answer:

[tex]n=0.286mol[/tex]

Explanation:

Hello,

In this case, we consider oxygen as an ideal gas, for that reason, we use yhe ideal gas equation to compute the moles based on:

[tex]PV=nRT\\\\n=\frac{PV}{RT}[/tex]

Hence, at 3.50 atm and 25 °C for a volume of 2.00 L we compute the moles considering absolute temperature in Kelvins:

[tex]n=\frac{3.50atm*2.00L}{0.082\frac{atm*L}{mol*K}(25+273)K} \\\\n=0.286mol[/tex]

Best regards.

Answer:

A. move materials through the body

Explanation:

The blood circulatory system (cardiovascular system) transports materials and delivers nutrients and oxygen to all cells in the body.

Answer:

see explaination

Explanation:

1. A metabolic pathway that OXIDIZES an energy-rich source to produce ATP from ADP.

Oxidation reactions are exergonic and can be coupled to produce ATP from ADP + Pi

2. Electrons are transferred from electron DONORS to compounds with a STRONGER reduction potential.

Electrons are moved from compounds with low reduction potential to compounds with high reduction potential.

3. As electrons move through the electron transport chain, ETC, the energy in the electron is used to pump PROTONS across a membrane.

In respiration, protons are pumped from the mitochondrial matrix to the perimitochondrial space.

4. The pumping of these molecules against their concentration gradient is a form of FACILITATED transport.

Since protons are charged particles, they require a carrier protein for their transport. The movement of these molecules back into the cell (down their concentration gradient) releases energy which the cell couples to the formation of ATP.

6. The energy in PHOTON/LIGHT is absorbed by an electron in a photocenter. This energy is converted from light energy into chemical energy into kinetic energy as the energized electron is used in back to back REDOX reaction in the electron transport chain, ETC.

7. The ETC creates A PROTON GRADIENT which is used by the cell to power the formation of ATP.

8. In CYCLIC ELECTRON TRANSPORT, the electron returns to the photocenter.

9. In NON-CYCLIC photosynthesis the electron reduces NADP+ to form NADPH.

Answer:

Check the explanation

Explanation:

1. A metabolic pathway that OXIDIZES an energy-rich source to produce ATP from ADP.

Oxidation reactions are exergonic and can be coupled to produce ATP from ADP + Pi

2. Electrons are transferred from electron DONORS to compounds with a STRONGER reduction potential.

Electrons are moved from compounds with low reduction potential to compounds with high reduction potential.

3. As electrons move through the electron transport chain, ETC, the energy in the electron is used to pump PROTONS across a membrane.

In respiration, protons are pumped from the mitochondrial matrix to the perimitochondrial space.

4. The pumping of these molecules against their concentration gradient is a form of ACTIVE transport.

6. The energy in PHOTON/LIGHT is absorbed by an electron in a photocenter. This energy is converted from light energy into chemical energy into kinetic energy as the energized electron is used in back to back REDOX reaction in the electron transport chain, ETC.

7. The ETC creates A PROTON GRADIENT which is used by the cell to power the formation of ATP.

8. In CYCLIC ELECTRON TRANSPORT, the electron returns to the photocenter.

9. In NON-CYCLIC photosynthesis the electron reduces NADP+ to form NADPH.

200 million year to 550 million years old

Final answer:

To calculate the length of each edge of the cube that would store your yearly CO2 emissions, convert the emissions from kg to metric tons and use the density of carbon dioxide at STP to find the volume. The length of each edge would be approximately 17.9 meters.

Explanation:

To calculate the length of each edge of the cube that would store your yearly CO2 emissions, we first need to convert the emissions from kg to metric tons. Since the average CO2 emissions per person per year is 10,000 kg, this is equal to 10 metric tons (1 metric ton = 1000 kg).

Next, we need to find the volume of the cube. The formula for the volume of a cube is V = s^3, where s represents the length of each edge.

Let's use the given data to solve for s:

CO2 emissions per person per year: 10 metric tons

Density of CO2 at STP: 1.98 kg/m³ (source: https://pubchem.ncbi.nlm.nih.gov/compound/carbon_dioxide)

Using the density, we can convert the metric tons of CO2 to the corresponding volume in cubic meters:

10 metric tons * 1000 kg/metric ton = 10,000 kg

10,000 kg / 1.98 kg/m³ ≈ 5,051.51 m³

Now, let's solve for s:

s^3 = 5,051.51 m³

s ≈ 17.9 meters (rounded to one decimal place)

Therefore, each edge of the cube that would store your yearly CO2 emissions at STP would be approximately 17.9 meters long.

Answer:

It is the sum of internal energies of two or more substances

Explanation:

Answer:

C. It is the portion of internal energy that can be transferred from one substance to another.

Explanation:

the other one was incorrect so this one was what it was

Answer:

0.72 M

Explanation:

Given data

Step 1: Calculate the final volume (V₂)

The final volume of the solution is equal to the sum of the initial volume of the solution and the volume of water.

[tex]V_2 = V_1 + VH_2O = 200 mL + 300mL = 500mL[/tex]

Step 2: Calculate the concentration of the diluted solution (C₂)

We will use the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{1.8M \times 200mL}{500mL}= 0.72 M[/tex]

Answer:

Oxidation number of Fe(iron) on right side of reaction arrow = (+2)

Explanation:

NADH+H+ + FMN   FMNH2 + NAD+

FMNH2 - Reduced FMN (1,5-Dihydroriboflavin 5'-(dihydrogen phosphate)

The reactant that is reduced is Flavin mononucleotide(FMN)

QH2 + 2cyt c(fe³+)  ----------------------->  Q + 2Cytc ( fe²+) + 2H+

Oxidation number of Fe(iron) on right side of reaction arrow = (+2)

The equation of the transfer of two electrons from NADH to FMN is given below:

NADH + H⁺ + FMN ⟶ NAD⁺ + FMNH₂

The oxidation number of iron on the right side of the reaction arrow is +2 (Fe²⁺)

In the electron transport chain, electrons are passed from electron carriers such as NADH through various carriers and eventually to oxygen. Water and energy in the form of ATP is produced.

The electron carriers are organized into complexes; Complex I, II, III, and IV

In complex I, also known as NADH Dehydrogenase, electrons are passed from NADH to ubiquinone through an FMN-containing flavoprotein and several iron-sulfur centers.

The equation below shows how two electrons are passed from NADH to FMN:

NADH + H⁺ + FMN ⟶ NAD⁺ + FMNH₂

In complex III, electrons are transferred from coenzyme Q to cytochrome c, a single electron-carrier which contains iron. The equation for the electron transfer is given below:

QH₂ + 2 cyt c ( Fe³⁺) ⟶ Q + 2 cyt c ( Fe²⁺ ) + 2H⁺

In the reaction above, the two electron-carrier ubiquinone transfers its two electrons to two molecules of the one electron-carrier cytochrome c containing iron in the oxidized iron (iii) state, Fe³⁺. The electrons accepted reduces the Fe³⁺ in cytochrome c to Fe²⁺.

Therefore, the oxidation number of iron on the right side of the reaction arrow (reduced cytochrome c) is +2.

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Answer:

Explanation:

Benzaldehyde can be prepared from toluene through Etard reaction .In this reaction benzaldehyde is treated with reagent chromyl chloride ( CrO₂Cl₂ ) which oxidises it to benzaldehyde.

to form benzaldehyde .

                      CrO₂Cl₂

C₆H₅CH₃ ----------------------->  C₆H₅CHO

Answer:

1.  Saturated hydrocarbons may be cyclic or acyclic molecules.

2.  An unsaturated hydrocarbon molecule contains at least one double bond.

Explanation:

Hello,

In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).

Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true

Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.

Therefore, first and second statements are correct.

Best regards.

Final answer:

Statements 1 and 2 are correct regarding saturated hydrocarbons being cyclic or acyclic and unsaturated hydrocarbons containing at least one double bond. Statement 3 is incorrect because ethylenediamine is not a hydrocarbon.

Explanation:

To address which statements concerning hydrocarbons are correct:

Saturated hydrocarbons may indeed be cyclic or acyclic molecules. When cyclic, they have the general formula CnH₂n, such as cycloalkanes, which are saturated compounds. Acyclic saturated hydrocarbons, also known as alkanes, have single bonds only and follow the general formula CHn₂+2.

An unsaturated hydrocarbon molecule contains at least one double or triple bond. Molecules with one or more double bonds are alkenes, with the simplest being ethene (or ethylene), C₂H₄.

Ethylenediamine, H₂NCH₂CH₂NH₂, is not an example of a saturated hydrocarbon. It contains amine groups (NH₂) and therefore is not a hydrocarbon.

Hence, statements 1 and 2 are correct, while statement 3 is incorrect.

Answer:

The pressure equilibrium constant  is  [tex]K_p = 323[/tex]

Explanation:

From the question we are told that

    The volume of the flask is  [tex]V = 50 mL = 50 *10^{-3} L[/tex]

    The pressure of sulfur dioxide is  [tex]P_s = 3.7 \ atm[/tex]  

     The pressure of  oxygen gas [tex]P_o = 2.3 \ atm[/tex]

     The pressure of sulfur trioxide at equilibrium is [tex]P_t = 2.2 \ atm[/tex]

The chemical equation for this reaction is

          [tex]2 SO_2_{(g)} + O_2_{(g)}[/tex]    ⇄   [tex]2SO_3_{(g)}[/tex]

The partial pressure of  oxygen at  equilibrium is mathematically evaluated as

         [tex]P_p__{O}} = P_o - P_t[/tex]

Substituting values

         [tex]P_p__{O}} = 2.3 -2.2[/tex]

         [tex]P_p__{O}} = 0.1 \ atm[/tex]

The partial pressure of  sulfur dioxide  at  equilibrium is mathematically evaluated as

         [tex]P_p__{s}} = P_s - P_t[/tex]

Substituting values

         [tex]P_p__{S}} = 3.7 -2.2[/tex]

         [tex]P_p__{O}} = 1.5 \ atm[/tex]

From the chemical equation  pressure constant is mathematically represented as

           [tex]K_p = \frac{[P_t]^2}{[P_p__{o}} ]^2 [P_p__{s}}]}[/tex]

Substituting values

          [tex]K_p = \frac{[2.2]^2}{[ 0.1 ]^2 [{ 1.5}]}[/tex]

          [tex]K_p = 323[/tex]

Answer:

18.018 seconds.

Explanation:

Given that the half life of Manganese, Mn = 3 seconds. The initial sample mass = 90.0 gram, the final sample mass = 1.40 gram.

The general idea to the question is to look for the time it will take to decay from the initial mass that is 90 gram to 1.40 gram.

Therefore, we will be making use of the formula below;

J(t) = J(o) × (1/2)^t/t(hL).

Where t(hL) is the half life, t is the time taken, J(t)= mass after time,t and J(o) is the initial mass. So, let us slot in the values into the equation above.

1.4 = 90 × (1/2)^ t/3.

1.4/90 = (1/2)^t/3.

t/3 = log(0.5) (1.4/90).

+Please note that the 0.5 of the log is at the subscript).

That is the base 0.5 logarithm of (1.4/90) 0.01556 is 6.0060141295.

t = 3 × 6.0060141295.

t = 18.018 seconds.

To find the time it takes for a 90 gram sample of Manganese-58 to decay to 1.4 grams, you need to calculate the number of half-lives by continuously halving 90 until you reach 1.4. Afterwards, multiplying the number of half-lives by the half-life duration (3 seconds) gives the total decay time.

This question refers to radioactive decay and the concept of a half-life. The half-life of a substance is the time it takes for half the substance to decay. In the case of Manganese-58, its half-life is approximately 3 seconds.

Looking to find the time it takes for a 90 gram sample to decay to approximately 1.4 grams, this would involve multiple half-lives. You would need to calculate how many half-lives it takes for 90.0 g to become 1.40g. With each half-life, the amount of the original substance decreases by 50%. The answer you will get by dividing 90 by 2 repeatedly until you reach 1.4 shows you the number of half-lives that have passed. Multiply the number of half-lives by the duration of a single half-life (in this case, 3 seconds) to get the total decay time.

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Answer:

The true statements include;

- The sample is initially a gas.

- The final state of the substance is a solid.

- One or more phase changes will occur.

The untrue/false statements include;

- The liquid initially present will vaporize.

- The final state of the substance is a liquid.

Explanation:

A couple pieces of informatton on Fluorine is imitially provided.

The substance fluorine has the following properties: normal melting point: 53.5 K normal boiling point: 85.0 K triple point: 1.6×10-4 atm, 53.4 K critical point: 55 atm, 144.1 K

So, a question is now attached about a sample of Fluorine. A sample of fluorine at a pressure of 1.00 atm and a temperature of 90.3 K is cooled at constant pressure to a temperature of 49.3 K.

We are then told to examine a group of options to find the ones that are correct/apply.

Taking the options one at a time

- The sample is initially a gas.

The initial state of the Fluorine sample has its temperature at 90.3 K, which is above the gas' boiling point. Hence, the sample can be concluded to initially be a gas.

- The liquid initially present will vaporize.

The sample doesn't initially contain liquid. And even of it did, the temperature is cooled, not heated , Hence, this statement is wrong.

- The final state of the substance is a solid.

The sample of Fluorine moves from a temperature higher than boiling point (85.0 K), with the sample in gaseous form, to one that is at a lower temperature (49.3 K) than the gas' normal melting point (53.5 K).

At temperatures lower than melting point, a substance exists in the solid form. Hence, this statement is true. The final state of the substance is solid.

- One or more phase changes will occur.

In moving from 90.3 K to 49.3 K for the sample and passing through the substance's boiling and melting points (85.0 K and 53.5 K respectively) along the way, it is logical to conclude that there would be one or more phase changes will occur. This statement is true.

- The final state of the substance is a liquid.

This is false as we already established that the final state of the substance is a solid. Hence, this statement is false.

Hope this Helps!!!

Final answer:

At 1.00 atm, fluorine starts as a gas at 90.3 K, then condenses to a liquid as it cools, and finally becomes a solid at 49.3 K, indicating that both condensation and freezing phase changes occur.

Explanation:

The substance fluorine has different states at various temperatures and pressures. To determine the state changes of fluorine when cooling from a temperature of 90.3 K to 49.3 K at constant pressure of 1.00 atm, we refer to the given melting and boiling points of fluorine. According to the information:

Normal melting point: 53.5 K

Normal boiling point: 85.0 K

At the starting temperature of 90.3 K and 1.00 atm, fluorine is above the boiling point, so the sample is initially a gas. As the temperature cools to below the boiling point but still above the melting point, any liquid that may be present will not vaporize; instead, the gas will condense to form a liquid. Since the final temperature of 49.3 K is below the melting point of 53.5 K, the final state of the substance is a solid. Throughout this process, one or more phase changes will occur; specifically, the gas will condense to a liquid and then freeze into a solid. Therefore, the final state of the substance will not remain a liquid; this statement is false.

Answer:

Scandium

Titanium

Vanadium

Chromium

Manganese

Iron

Cobalt

Nickel

Copper

Zinc

Yttrium

Zirconium

Niobium

Molybdenum

Technetium

Ruthenium

Rhodium

Palladium

Silver

Cadmium

Lanthanum

Hafnium

Tantalum

Tungsten

Rhenium

Osmium

Iridium

Platinum

Gold

Mercury

Actinium

Rutherfordium

Dubnium

Seaborgium

Bohrium

Hassium

Meitnerium

Darmstadtium

Roentgenium

Copernicium

Explanation:

all of those are transition metals lol

Transition metals, located in groups 3-11 of the periodic table, are elements with partially filled d orbitals and variable reactivity. Metals like scandium and iron are very active, while platinum is relatively inert. There are also inner transition metals, which occupy an f orbital and include the lanthanide and actinide series.

Transition metals are elements with partially filled d orbitals, located in the d-block of the periodic table. They are characterized by their various reactivity, with metals such as scandium and iron being very active, and metals like platinum being relatively inert. Transition metals are located in groups 3-11 of the periodic table, and include elements like copper, gold, and oxidized iron, among others.

There are also inner transition metals, which are metallic elements where the last electron added occupies an f orbital. They have two series: the lanthanide series, from lanthanum (La) to lutetium (Lu), and the actinide series, from actinium (Ac) to lawrencium (Lr).

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Answer:

I have the same question

Explanation:

The question is incomplete; the complete question is:

Write electron configurations for each of the following elements. Use the symbol of the previous noble gas in brackets to represent the core electrons. Express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. For example, (He)2s^22p^2. For: Te, Br, I, Cs

Answer:

Te- [Kr]4d^10 5s^2 5p^4

Br- [Ar] 3d^10 4s^2 4p^5

I - [Kr] 4d^10 5s^2 5p^5

Cs - [Xe]6s^1

Explanation:

We could write a short hand electron configuration for any element. All we need to do is to study the long hand electron configuration of the element and decode its noble gas core. Every element is composed of an inert gas core configuration showing the inner electrons followed by the outermost shell electrons.

The inert gas core could be shown by writing the symbol of the particular noble gas involved within square brackets followed by the element's outer electron configuration as shown in the answer above.

The electron configuration for Tellurium (Te) is [Kr]4d^105s^25p^4, determined by filling in the core electrons from the previous noble gas (Krypton) and by its position in the Periodic table.

The electron configuration for the element Te (Tellurium) can be determined by understanding its position in the periodic table. Tellurium belongs to Group 16 and Period 5. Therefore, filling the electrons until we reach Te, we start from the noble gas at the end of Period 4, which is Krypton [Kr]. From there, we proceed to fill the 5th orbital. In order, this looks like: [Kr]4d^105s^25p^4. Thus, the electron configuration of Te is [Kr]4d^105s^25p^4.

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Answer: 5535.1KJ/mol

Explanation:The equation for breaking of all the bonds in the benzene is written as follows:

C_6 H_6(g) --> 6C(g) +6H (g)

∆H_rxn= ∑moles of product X∆H_PRODUCT -∑moles of reactants x ∆H_REACTANT  

={6mols X ∆H_°f(C) + 6mols X ∆H_°f(H)} –{1 mol ∆H_°fC_6 H_6 (g)

={6 X 217.74+ 6X 718.4} – {82.9}KJ/mol

= 5618.04 – 82.9 KK/mol

=5535.1KJ/mol

b) The equation for the breaking of the carbon-carbon bonds in benzene is illustrated below as

C_6 H_6(g) --> 6C---H (g)

A) The standard enthalpy change for the reaction is ; 5535.1 KJ/mol

B) The chemical equation for the reaction that corresponds to breaking just the carbon-carbon bonds is  :  C₆H₆(g) --> 6C---H(g)

A) The standard enthalpy change for the reaction which represents breaking of all bonds

Reaction equation ; C₆H₆(g) --> 6C(g) +  6H (g)

∆H = ∑ ( moles of product * ∆Hpro ) - ( moles of reactant * ∆H reactant )

      = ( 6 * 217.94  +  6 *  718.4 ) –  ( 82.9)   KJ/mol

      = ( 5618.04 ) - ( 82.9 ) = 5535.1  KJ/mol

B) The chemical equation for the reaction that corresponds to breaking the carbon-carbon bonds

chemical equation  = C₆H₆(g) --> 6C---H(g)

Hence we can conclude that The answers to your questions are 5535.1 KJ/mol and C₆H₆(g) --> 6C---H(g)

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Answer:

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Explanation:

For a reversible reaction, the equilibrium constant for a backward reaction is reciprocal.

If the coefficients in a balanced equation are multiplied by a factor, n, the equilibrium expression is raised to the nth power.

K' = (K)^n

In the second reaction the value kf n is 2. The first reaction is multiplied by 2.

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Answer: -9  8  48  -6

Explanation: trust me