State whether the entropy of the system increases or decreases in each of the following processes:
(b) Gold is extracted and purified from its ore.

Answer:

Explanation:

Answer:

either expolshion or hide

Explanation:

so be ware :O

Answer:

61 neutrons.

Answer:

61 neutrons

Explanation:

the number of neutrons in silver is 61

The correct statement concerning the given balanced chemical equation is If we have more mass of C₂H₂, then O₂ must be limiting. The correct option is d.

To determine the limiting reactant in a chemical reaction, we compare the stoichiometry of the balanced equation to the amounts of reactants available. The limiting reactant is the one that is completely consumed, thereby limiting the amount of product formed.

In the given balanced chemical equation:

2C₂H₂(g) + 7O₂(g) → 6H₂O(l) + 4CO₂(g)

We can analyze the equation to determine the relationship between C₂H₂ and O₂.

From the coefficients of the balanced equation, we see that 2 moles of C₂H₂ react with 7 moles of O₂. This means that the stoichiometric ratio between C₂H₂ and O₂ is 2:7.

If we have more mass of C₂H₂, it means we have an excess amount of C₂H₂ compared to the stoichiometric ratio. In this case, C₂H₂ will be the excess reactant, and O₂ will be the limiting reactant. The excess C₂H₂ will not be completely consumed, and the reaction will be limited by the availability of O₂.

Therefore, the correct statement is d. If we have more mass of C₂H₂, then O₂ must be limiting. This is because the stoichiometry of the balanced equation indicates that a larger amount of C₂H₂ requires a corresponding larger amount of O₂ to be completely consumed. Option d. is the correct answer.

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Answer:

8.3

Explanation:

It would be the first one. I'm sorry if I'm wrong. I used to be awesome at this kind of stuff.

Answer:

Explanation:

Fourth option

ʕ◉ᴥ◉ʔ Gimme crown please

Answer: Coastlines.

Explanation: South America's and Africa's shoreline interlocks perfectly. Which implies they must have been joined together in history.

The pH at the half-equivalence point of this titration is 3.3.

The balanced chemical equation for the reaction between nitrous acid (HNO2) and lithium hydroxide (LiOH) is:

HNO2 + LiOH → LiNO2 + H2O

At the half-equivalence point, half of the nitrous acid has reacted with the lithium hydroxide. This means that 7.5 ml of the 0.150 M LiOH solution has been added to the 15.0 ml sample of 0.150 M nitrous acid.

To find the pH at the half-equivalence point, we need to calculate the concentrations of the nitrous acid and the nitrite ion (NO2-) at this point.

Before any LiOH is added, the concentration of nitrous acid is 0.150 M. At the half-equivalence point, half of the nitrous acid has reacted, so the concentration is now 0.075 M.

The balanced equation shows that one mole of nitrous acid reacts with one mole of LiOH to form one mole of nitrite ion. Therefore, at the half-equivalence point, the concentration of nitrite ion is also 0.075 M.

To find the pH, we need to calculate the pKa of nitrous acid. The pKa of nitrous acid is 3.3.

Using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where [A-] is the concentration of the conjugate base (nitrite ion) and [HA] is the concentration of the acid (nitrous acid).

At the half-equivalence point, [A-] = 0.075 M and [HA] = 0.075 M.

pH = 3.3 + log(0.075/0.075)

pH = 3.3 + log(1)

pH = 3.3

Therefore, the pH at the half-equivalence point of this titration is 3.3.

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Answer:

\(d.blood\)

Explanation:

Hope it helps!!!

Answer:

Explanation:

The Tacoma Narrows Bridge is the historical name given to the twin suspension bridge—originally built in 1940—that spanned the Tacoma Narrows strait. It collapsed just four months later due to aeroelastic flutter. Since then, this topic has become popular, with several case studies discussing the failure phenomenon of suspension cable bridges.

In the state of Washington, the construction of the Tacoma Narrows Bridge was completed and opened to the traffic on July 1st, 1940. It was the very first bridge to incorporate a series of plate girders as roadbed support, and the first bridge of its type (cable suspension). It was also the third largest suspension bridge of its time, with a 2800-foot central span and two side spans of 1100 feet each.

A west-side approach had a continuous steel girder of 450ft, while the east side had a long reinforced concrete frame of 210ft.  It had two cable anchorages of 26ft. along roadways, two 5ft. sidewalks and two 8ft. deep stiffening girders. Among several other structural details, the suspension cable anchorages to which the cables were connected were made of 20,000 cubic yards of concrete, 6 lakh pounds of structural steel, and 2.7 lakh pounds of reinforcing steel. Because of its extremely long length, it was considered a ‘narrow bridge’. The overall construction cost was estimated to be a whopping $6 Million in 1940. Considering inflation, this is equivalent to almost $1 Billion, and all of this for something that lasted just four months and seven days. Yet, this remains a great engineering feature for civil engineers to ponder over.

Answer:

Flammable liquid is any liquid having a flashpoint at or below 199.4 °F (93 °C).

Explanation:

:)

Flashpoints for flammable liquids are at 100°F. Lower flash points allow for easier ignition of liquids. A liquid's flash point is the lowest temperature at which a concentrated enough layer of vapor accumulates on top of its surface to allow for ignition.

A liquid that is flammable has a flash point* below 37.8 ° C (100 ° F). A flammable liquid has a flash point that ranges from 37.8 to 93.3° C (100 to 200° F), which is above the standard working temperature. Flammable liquids emit a vapor that, at standard working temperatures, is easily ignitable.

The substance ignites more readily the lower the flash point. For instance, petrol is more flammable than ethylene glycol and has a flash point of about -40 degrees C (-40 °F).

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Answer:

no i cant Your file couldn’t be accessed It may have been moved, edited, or deleted.

Explanation:

1) To react with 0.460 mole of SO₂, 1.15 moles of C are needed.

2) When 2.0 moles of C reacts, 1.6 moles of CO are produced.

3) To produce 0.35 mole of CS₂, 0.70 moles of SO₂ are required.

4) When 2.4 moles of C reacts, 0.48 moles of CS₂ are produced.

1.

From the balanced equation, the stoichiometric ratio between C and SO₂ is 5:2. Therefore, to calculate the moles of C required, we can set up a proportion:

(5 moles C / 2 moles SO₂) = (x moles C / 0.460 moles SO₂)

Solving for x, we find:

x = (5/2) × 0.460 = 1.15 moles C

2.

From the balanced equation, the stoichiometric ratio between C and CO is 5:4. Therefore, to calculate the moles of CO produced, we can set up a proportion:

(5 moles C / 4 moles CO) = (2.0 moles C / x moles CO)

Solving for x, we find:

x = (4/5) × 2.0 = 1.6 moles CO

3.

From the balanced equation, the stoichiometric ratio between SO₂ and CS₂ is 2:1. Therefore, to calculate the moles of SO₂ required, we can set up a proportion:

(2 moles SO₂ / 1 mole CS₂) = (x moles SO₂ / 0.35 moles CS₂)

Solving for x, we find:

x = (2/1) × 0.35 = 0.70 moles SO₂

4.

From the balanced equation, the stoichiometric ratio between C and CS₂ is 5:1. Therefore, to calculate the moles of CS₂ produced, we can set up a proportion:

(5 moles C / 1 mole CS₂) = (2.4 moles C / x moles CS₂)

Solving for x, we find:

x = (1/5) × 2.4 = 0.48 moles CS₂

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In this example, the limiting reagent is the alkyl οr aryl halide (RX), as it is present in a lοwer quantity and will be cοmpletely cοnsumed during the reactiοn.

In οrganic chemistry, an alkyl grοup is an alkane missing οne hydrοgen. The term alkyl is intentiοnally unspecific tο include many pοssible substitutiοns. An acyclic alkyl has the general fοrmula οf −CnH₂n+1.

A cyclοalkyl grοup is derived frοm a cyclοalkane by remοval οf a hydrοgen atοm frοm a ring and has the general fοrmula −CnH₂n+1. Typically an alkyl is a part οf a larger mοlecule. In structural fοrmulae, the symbοl R is used tο designate a generic (unspecified) alkyl grοup. The smallest alkyl grοup is methyl, with the fοrmula −CH₃

In the preparatiοn οf Grignard reagents, the reactiοn typically invοlves the reactiοn between an alkyl οr aryl halide and magnesium metal in an ether sοlvent. The general equatiοn fοr the fοrmatiοn οf a Grignard reagent can be represented as:

RX + Mg -> RMgX

Tο determine the limiting reagent in this reactiοn, we need tο cοmpare the mοles οf the alkyl οr aryl halide (RX) and the mοles οf magnesium (Mg) and identify which reactant is present in the lοwer quantity. The reactant that is cοnsumed cοmpletely and limits the amοunt οf prοduct fοrmed is the limiting reagent.

Let's assume we have a specific example where we are preparing a Grignard reagent by reacting 2 mοles οf alkyl οr aryl halide (RX) with 3 mοles οf magnesium (Mg).

Mοles οf RX = 2 mοles

Mοles οf Mg = 3 mοles

Since the stοichiοmetric ratiο between RX and Mg is 1:1 (1 mοle οf RX reacts with 1 mοle οf Mg), we can see that we have an excess οf Mg in this example.

The stοichiοmetry indicates that 2 mοles οf RX wοuld require 2 mοles οf Mg fοr cοmplete reactiοn. Hοwever, we have 3 mοles οf Mg, which is mοre than enοugh tο react with the 2 mοles οf RX.

Therefοre, in this example, the limiting reagent is the alkyl οr aryl halide (RX), as it is present in a lοwer quantity and will be cοmpletely cοnsumed during the reactiοn.

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Answer:

Concept of chemi-fluorescence

Explanation:

A glow stick usually consists of two chemicals in a larger plastic tube: , a base catalyst (mostly sodium salicylate), and a suitable dye (sensitizer, or fluorophor). This creates an exergonic reaction when mixed together.

When a glow stick is bent, the flurophor which is a chemical that easily re-emits light upon excitation in smaller capsules is released into the other substance, there by causing it to emit radiation/light in the uv-visible region. The brightness and longevity of the glow stick is determined by varying the concentration of these chemicals.

I hope this explanation clarifies things.

Answer:

Ionic bonds are stronger than covalent bonds

The main answer for matching each six-electron group designation to the correct molecular shape is as follows:
1. AX6: Octahedral
2. AX5E: Square pyramidal
3. AX4E2: Square planar

In these designations, "A" represents the central atom, "X" represents the bonding atoms, and "E" represents lone pairs of electrons.

The number following each letter indicates the count of each respective element in the molecule.

Based on these designations, we can match them with the correct molecular shapes:
1. AX6 has six bonding atoms, giving it an octahedral shape.
2. AX5E has five bonding atoms and one lone pair, forming a square pyramidal shape.
3. AX4E2 has four bonding atoms and two lone pairs, resulting in a square planar shape.

In summary, AX6 corresponds to an octahedral shape, AX5E to a square pyramidal shape, and AX4E2 to a square planar shape.

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Answer:

Answer:

na + B2 = t

Explanation:

Answer:

2NaN3→ 2Na + 3N2 is the balanced equation

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Answer A. The electronegativity of one atom is less than 1.7

Answer: its C

Explanation: i just got it wrong the right answer is c

The difference between the IR spectrum of the ketone product and the alcohol starting material can be seen in the absorption bands present in each spectrum.

In the IR spectrum of the alcohol starting material, there will be a strong, broad absorption band between 3200 and 3600 cm-1, which corresponds to the O-H stretch of the alcohol functional group.

In the IR spectrum of the ketone product, there will be a strong, sharp absorption band between 1670 and 1780 cm-1, which corresponds to the C=O stretch of the ketone functional group.

These spectra can help determine whether the reaction worked by comparing the absorption bands present in each spectrum. If the alcohol starting material has a strong O-H stretch and the ketone product has a strong C=O stretch, then it can be concluded that the reaction worked and the alcohol was successfully converted into a ketone.

However, if the ketone product does not have a strong C=O stretch or if the alcohol starting material still has a strong O-H stretch, then it can be concluded that the reaction did not work and the alcohol was not successfully converted into a ketone.

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Answer:

The number of neutrons in nucleus

The correlation between cognitive ability and job performance is represented by  D: strong, positive. Numerous studies have demonstrated that individuals with higher cognitive abilities tend to perform better in various job tasks and roles.

Cognitive ability refers to an individual's mental capacity to learn, solve problems, reason, and process information. It plays a crucial role in determining job performance, as it affects the ability to acquire job-related knowledge and skills.

This strong, positive relationship indicates that as cognitive ability increases, so does job performance. Employers often use cognitive ability tests during the recruitment process to predict an applicant's potential job performance. However, it is essential to note that cognitive ability is not the sole predictor of job success.

Other factors, such as personality traits, motivation, and job-specific skills, also contribute to overall job performance. In conclusion, there is a strong, positive correlation between cognitive ability and job performance, but other factors should also be considered when evaluating an individual's potential success in a job role.

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a. To determine the initial investment outlay for the spectrometer, we need to calculate the cash flows at Year 0. The base price of the spectrometer is $140,000, and the modification cost is $30,000.

Thus, the initial cost is the sum of these two amounts, which is

$170,000 ($140,000 + $30,000).

Additionally, we need to consider the increase in net operating working capital, which is $8,000. Therefore, the initial investment outlay for the spectrometer is

$178,000 ($170,000 + $8,000).

b. In Years 1, 2, and 3, we need to calculate the annual cash flows. Firstly, we consider the savings in before-tax labor costs, which is $50,000 per year. Next, we calculate the depreciation expense using the MACRS depreciation rates provided. In Year 1, the depreciation expense is

$140,000 * 0.33 = $46,200. In Year 2,

it is $140,000 * 0.45 = $63,000.

In Year 3, it is $140,000 * 0.15 = $21,000.

Finally, we subtract the depreciation expense from the before-tax labor cost savings to get the annual cash flows. In Year 1, it is

$50,000 - $46,200 = $3,800. In Year 2,

it is $50,000 - $63,000 = -$13,000. In Year 3,

it is $50,000 - $21,000 = $29,000.

c. To determine whether the spectrometer should be purchased, we need to calculate the net present value (NPV) of the project's cash flows. Using the WACC of 9%, we discount the cash flows in each year. The NPV is the sum of the discounted cash flows. If the NPV is positive, the project is considered favorable.

If it is negative, the project should not be pursued. In this case, we calculate the NPV by discounting the cash flows in Years 1, 2, and 3 at a rate of 9%. The NPV is calculated as: NPV = ($3,800 / (1 + 0.09)^1) + (-$13,000 / (1 + 0.09)^2) + ($29,000 / (1 + 0.09)^3). If the NPV is positive, the spectrometer should be purchased; if it is negative, it should not be purchased.

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The chemical that lies on alcohol group is alcohol with functional group'-OH'.

Functional group is defined as a substituent or group of toms or an atom which causes chemical reactions.Each functional group will react similarly regardless to the parent carbon chain to which it is attached.This helps in prediction of chemical reactions.

The reactivity of functional group can be enhanced by making modifications in the functional group .Atoms present in functional groups are linked to each other by means of covalent bonds.They are named along with organic compounds according to IUPAC nomenclature.

Functional group inter conversion is also possible by retro -synthesis.In some cases , functional groups can be charged molecules.

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The percent composition of carbon in C₆H₁₈ is 80 percent. It can be calculated by using the following formula:

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The IUPAC name for the given compound \((CH_3)_2C=CHCH_2CH_2OH\) is 3-hydroxy-2-methylbut-2-ene.

First, we identify the longest continuous carbon chain, which consists of five carbon atoms. This forms the parent chain, which is a butene.

Next, we locate the double bond, which is between the second and third carbon atoms. Since the double bond starts at the second carbon atom, we indicate it as "but-2-ene."

Moving on, we need to indicate the presence of substituents. In this case, we have a methyl group attached to the second carbon atom. Therefore, we include the prefix "2-methylbut-2-ene."

Lastly, there is a hydroxyl group (-OH) attached to the fourth carbon atom. We denote this as "hydroxy" and specify the position of the hydroxyl group, resulting in the full name "3-hydroxy-2-methylbut-2-ene."

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There are 20 grams of Sodium hydroxide in a molar solution that has a volume of 500 cm3.

Caustic soda and lye are two other names for sodium hydroxide. It is a typical component of soaps and cleansers. Sodium hydroxide seems to be a white, unscented solid at room temperature. Sodium hydroxide is a colorless, odorless liquid. Strong acids and water can cause explosive reactions with it.

Molarity is calculated by

moles (solute) / volume of solution in L

No. of moles = mass / molar mass

We have to find mass of sodium hydroxide (NaOH)

As we know, in 1 L (1000 cm³) of molar solution there is 40 gram of NaOH present.

From using this statement, 500 cm³ of a molar solution, means half liter of a molar solution contain -

= 40 / 2

= 20 gram of NaOH.

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