based on the six given solvents, which of the following answers lists all of the solvents that are not suitable for liquid-liquid extraction of an aqueous layer? 1. n-pentane 2. dimethyl sulfoxide 3. hexanes 4. dichloromethane 5. toluene 6. isopropanol

The correct statement is that A, B, and C all have different freezing points. This is because the freezing point of a substance depends on the number of solute particles per unit volume of solution. Since A, B, and C each contain different amounts of sucrose, ethanol, and sodium hydroxide, they each have different amounts of solute particles, and therefore different freezing points.

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The amount of benzene in the sample that burned was [tex]4.14*10^{26}[/tex] molecules.

The molecular weight of benzene is 78.11 g/mol. This means that the number of moles of benzene in the sample can be calculated by dividing the mass of the sample (522 liters of CO2 is approximately 522 kg at STP) by 78.11 g/mol.

The number of moles of benzene in the sample can then be used to calculate the number of benzene molecules. This is given by the equation:

Number of benzene molecules = Number of moles x Avogadro's Constant

The number of benzene molecules in the sample that underwent combustion is:

Number of benzene molecules

[tex]=\frac{ 522 kg }{ 78.11 g/mol} * 6.02 * 10^{23} molecules/mol\\\\= 4.14 * 10^{26 }molecules[/tex]

Therefore,[tex]4.14*10^{26}[/tex] molecules of benzene molecules were in the sample that underwent combustion.

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Everyone is aware that the first billion years of Earth's existence were marked by intense volcanic activity. Because of the circumstances, there were so many gases in our atmosphere that life could not exist there (Very similar like the atmosphere of Mars & Venus of today).

Carbon dioxide was most prevalent, whereas oxygen was scarce. Next, the modifications that affected the ratio of gases were:

1) To increase oxygen levels. "Photosynthesis by Plants" was the sole and most significant source of increasing oxygen.

2) In order to reduce carbon dioxide. - I Absorbable by plants for the photosynthesis process.

(ii) The majority of the CO2 is dissolved in the oceans, and a large portion of it is trapped in sedimentary rocks like limestone and fossil fuels.

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1.8 moles of nitrogen react with 3 moles of hydrogen to form 2 moles of ammonia.

The chemical equation for the reaction is:

N2 + 3H2 → 2NH3

Therefore, 2 moles of ammonia are formed.

Ammonia is an odourless gas that has a potent scent.. It is the main component of many industrial and household cleaning products, and is also used widely in agriculture as a fertilizer. Ammonia can be hazardous to human health if inhaled, ingested, or if it comes in contact with the skin or eyes. Inhalation of high concentrations of ammonia can cause irritation of the eyes, nose, throat, and lungs. Ingestion of ammonia can cause severe damage to the digestive system.

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B) 52.000 amu is the right response because it comes nearest to the calculated average atomic mass.

To calculate the average atomic mass of chromium, we need to take into account the mass and abundance of each of its isotopes. We can use the following formula to calculate the average atomic mass:

Average atomic mass = (mass of isotope 1 x abundance of isotope 1) + (mass of isotope 2 x abundance of isotope 2) + (mass of isotope 3 x abundance of isotope 3) + (mass of isotope 4 x abundance of isotope 4)

Substituting the values given in the problem, we get:

Average atomic mass = (49.946 amu x 0.0435) + (51.941 amu x 0.8379) + (52.941 amu x 0.0950) + (53.939 amu x 0.0236)

Average atomic mass = 2.173027 + 43.592439 + 5.02999 + 1.272404

Average atomic mass = 52.06786 amu

Therefore, the correct answer is B) 52.000 amu, which is the closest value to the calculated average atomic mass.

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The graphical solution to the above system of inequalities is attached accordingly.

A system of inequalities is a set of two or more inequalities that are considered together. The solution to a system of inequalities is the set of all values that satisfy all the inequalities simultaneously.

To graph the system of inequalities y < 6x + 4 and y > 5x + 4, we first graph the boundary lines y = 6x + 4 and y = 5x + 4.

To graph the line y = 6x + 4:

Plot the y-intercept (0, 4).

Use the slope, rise over run, to find additional points. For example, if we move up 6 units and right 1 unit from the y-intercept, we get the point (1, 10).

Draw a straight line through the points.

To graph the line y = 5x + 4:

Plot the y-intercept (0, 4).

Use the slope, rise over run, to find additional points. For example, if we move up 5 units and right 1 unit from the y-intercept, we get the point (1, 9).

Draw a straight line through the points.

Now we need to shade the area that satisfies the system of inequalities.

To do this, we can choose a test point, such as (0, 0), and plug it into both inequalities:

y < 6x + 4:

0 < 6(0) + 4

0 < 4 (true)

y > 5x + 4:

0 > 5(0) + 4

0 > 4 (false)

Since (0, 0) satisfies the first inequality but not the second, we shade the area below the line y = 6x + 4 but above the line y = 5x + 4.

The shaded area is the region between the two lines, below y = 6x + 4, and above y = 5x + 4.

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The most likely route to forming a cyclopentyl reagent from an amide would involve deprotonation of the amide with a strong base, such as lithium diisopropylamide (LDA), followed by treatment with an alkyl halide. The amide would be deprotonated, resulting in a carbanion that could be reacted with the alkyl halide, forming a new alkyl carbocation. A subsequent nucleophilic attack from a Lewis base, such as the conjugate base of the amide, would lead to formation of the cyclopentyl reagent.

This reaction can also be accomplished by a single-step procedure involving a Grignard reagent, such as ethylmagnesium bromide, and the amide. The Grignard reagent is deprotonated by the amide, and the resulting carbanion is reacted with the alkyl halide, resulting in a new alkyl carbocation. A subsequent nucleophilic attack from a Lewis base, such as the conjugate base of the amide, would lead to formation of the cyclopentyl reagent.

Therefore, the most likely route to forming the cyclopentyl reagent from the starting amide would involve deprotonation of the amide with a strong base followed by treatment with an alkyl halide. Alternatively, a single-step procedure using a Grignard reagent and the amide could also be used.

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Answer: Volume will become one third .

Explanation:

For volume of gas the equation is  given as

             P V  =  n R T

                V   =   [tex]\frac{n R T}{P}[/tex]

now increasing P by three times we have new volume V' as

           →    V'  = [tex]\frac{n R T}{3P}[/tex]

         →     V'   =   [tex]\frac{1}{3} \frac{n R T}{P}[/tex]

          →     V'  =  [tex]\frac{1}{3} V[/tex]

Therefore the volume will become one third of the previous volume.

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Answer:

To answer this question, we need to first write the balanced chemical equation for the combustion of C2H2:

C2H2 + 2.5 O2 → 2 CO2 + H2O

From the equation, we can see that for every mole of C2H2 burned, we produce one mole of water (H2O).

To find the number of moles of C2H2 in 25.0 grams, we need to divide the mass by the molar mass:

molar mass of C2H2 = 2(12.01 g/mol) + 2(1.01 g/mol) = 26.04 g/mol

moles of C2H2 = 25.0 g / 26.04 g/mol = 0.961 mol

Therefore, the number of moles of H2O produced is also 0.961 mol.

So the total number of moles present in the 52.0-gram sample of Na(s) is 0.8 moles.

How to calculate mole  ?

Divide the sample's mass by its molar mass to determine the total number of moles in a 52.0-gram sample of sodium (Na).

As Na has an atomic mass of 23.0 atomic mass units (amu) and each element has an average of 6.02 x 1023 atoms per mole, Na has a molar mass of 23.0 grams/mole.

The gram-formula mass of Na

which is reported as 65.0 grams/mole, must be used since the inquiry pertains to a sample of Na in its solid state (Na(s)).

moles = mass molar mass moles

= 52.0 g 65.0 g/mol moles

= 0.8 mol

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After calculating the mass by using the molar mass of calcium (Ca) and Avogadro's number (Nₐ), Rounding to the nearest whole number, the mass of the sample is approximately 365 g.

How to determine the mass of the sample?

To determine the mass of the sample, we need to use the molar mass of calcium (Ca) and Avogadro's number (Nₐ) to convert the number of atoms to mass.

The molar mass of calcium is approximately 40.08 g/mol, which means that one mole of calcium contains 6.022 x 10²³ atoms.

So, first we need to find out how many moles of calcium are in the sample:

moles of calcium = number of atoms / Avogadro's number

moles of calcium = 5.5 x 10²⁴ / 6.022 x 10²³

moles of calcium = 9.13

Now we can use the molar mass of calcium to convert moles to grams:

mass of sample = moles of calcium x molar mass of calcium

mass of sample = 9.13 x 40.08

mass of sample = 365.4 g

Rounding to the nearest whole number, the mass of the sample is approximately 365 g.

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The four mixtures listed above can be separated by filtration. Filtration is a technique commonly used to separate a suspended solid from a liquid, or from another solid, in order to purify or clarify the product or solution. In this respect, mayonnaise, muddy water, shaving cream and gelatin can be separated by filtration.

First, mayonnaise can be separated by filtration. Mayonnaise is a combination of oil and egg yolks, and it is a mixture that can be separated using a cheesecloth or a paper filter. This process works because the large particles of egg yolks and oil can be caught in the filter, while the smaller liquid components can be left behind.

Muddy water can also be separated by filtration. Muddy water typically contains suspended solids such as dirt and clay, which can be removed by passing the water through a filter. These solids are collected on the filter while the remaining water is clarified and clean.

Shaving cream can be separated by filtration as well. By using a filter paper, the oils contained in the shaving cream will stick to the filter, while the rest of the ingredients in the cream will be trapped behind the filter.

Finally, gelatin mixtures can also be separated by filtration. Gelatin is a mixture that contains proteins, fats, sugars and minerals. Separating these components can be achieved by passing the mixture through a filter. The proteins and fats will be collected on the filter, while the minerals and sugars will remain behind.

In conclusion, filtration is a process used to separate different components from a mixture. This process can be used to separate mayonnaise, muddy water, shaving cream and gelatin mixtures. By using filtration, the suspended solids in these mixtures can be removed, while the remaining components can be considered purified or clarified.

To prepare a 0.40M solution using 90.0g of Kool-Aid, we would need to dissolve the Kool-Aid in 657.5 mL of water.

How to determine the volume of water needed to prepare a 0.40M solution using 90.0g of Kool-Aid?

To determine the volume of water needed to prepare a 0.40M solution using 90.0g of Kool-Aid, we need to know the molar mass of Kool-Aid and the desired volume of the final solution.

Since Kool-Aid is a mixture of various compounds, we do not have a single molar mass value for it. However, we can estimate its average molar mass by assuming that the main component of Kool-Aid is sugar (sucrose), which has a molar mass of 342.3 g/mol.

Assuming that the Kool-Aid powder is 100% sucrose, we can calculate the number of moles of Kool-Aid present in 90.0 g as follows:

number of moles = mass

/ molar mass

number of moles = 90.0 g / 342.3 g/mol

number of moles = 0.263 mol

To prepare a 0.40M solution using this amount of Kool-Aid, we need to dissolve it in a volume of water that will result in a final solution with a total volume of 1 L (1000 mL) and a concentration of 0.40 mol/L. We can calculate the required volume of water as follows:

moles of Kool-Aid = concentration x volume

0.263 mol = 0.40 mol/L x volume

volume = 0.263 mol / 0.40 mol/L

volume = 0.6575 L = 657.5 mL

Therefore, to prepare a 0.40M solution using 90.0g of Kool-Aid, we would need to dissolve the Kool-Aid in 657.5 mL of water.

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Answer:

 5 moles H2O

Explanation:

We need a balanced equation for the reaction.

Ca(OH)2 + 2HCl =  CaCl2 + 2H20

Everything should be balanced.  This tells us we should expect 2 moles of H2O from 2 moles of HCl.  That is a 1:1 molar ratio, or

(1 mole H2O/1 mole HCl)

If we start with 5 moles of HCl, we can use the molar ratio:

(5 moles HCl)*((1 mole H2O/1 mole HCl) = 5 moles H2O  (the unit "moles HCl" cancels out, leaving just moles H2O)

The combustion reaction of sucrose (C12H22O11) with oxygen (O2) can be represented by the following equation:

C12H22O11 + 12 O2 → 12 CO2 + 11 H2O

From the balanced equation, we can see that 1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of carbon dioxide.

The molar mass of sucrose is approximately 342.3 g/mol (12 x 12.01 g/mol + 22 x 1.01 g/mol + 11 x 16.00 g/mol), which means that 84 g of sucrose is equivalent to:

84 g / 342.3 g/mol = 0.2456 moles of sucrose

According to the balanced equation, the combustion of 1 mole of sucrose produces 12 moles of carbon dioxide. Therefore, the combustion of 0.2456 moles of sucrose will produce:

0.2456 moles x 12 moles CO2 / 1 mole sucrose = 2.9472 moles of carbon dioxide

Finally, we can convert moles of carbon dioxide to mass using the molar mass of carbon dioxide, which is approximately 44.01 g/mol. Therefore, the mass of carbon dioxide produced from burning 84 g of sucrose in Skittles would be:

2.9472 moles x 44.01 g/mol = 129.5 g of CO2

So, if you ate an entire bag of Skittles (84 g of sucrose), you would breathe out approximately 129.5 g of carbon dioxide.

The correct electron configuration for Cl in noble-gas notation is option D, [Ne]₃s²³p⁵.

This notation indicates that the electron configuration of Cl is the same as that of the noble gas neon ([He]₂s²²p⁶), except for the three additional electrons in the 3rd shell, occupying the 3s and 3p orbitals.

The electron configuration for Y in noble-gas notation can be determined by finding the noble gas that precedes Y in the periodic table, which is Kr. The electron configuration for Kr is [Ar]₃d¹⁰⁴s²⁴p⁶

Yttrium (Y) has an atomic number of 39, which means it has 39 electrons. The electron configuration for Y in noble-gas notation is therefore:

[Kr]₅s²⁴d¹

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Answer:

0.0415mol

Explanation:

C = n/v

n= 4.15ml x 10

n = 41.5mml

An ester is ethyl butyrate (C6H12O2), usually referred to as ethyl butanoate or butyric ether. In addition to function as a solvent, ethanol butyrate is often employed as an extractor, flavouring, or fragrance.

It is a flammable, colourless liquid with a smell of banana, orange, or pineapple. Alcoholic beverages contain the volatile ethyl ester ethyl butanoate, which saccharomyces produces during fermentation. When ethanol and a fatty acid interact, ethanol esters are created.

The chemical name for ethyl butyrate, popularly known as ethyl butanoate or butyric ether, is CH3CH2CH2COOCH2CH3. Kerosene, paraffin oil, and propylene glycol are all soluble in it. It is an important factor used as a taste enhancer in processed oranges and provides a delicious aroma comparable to banana.

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Answer:

B. 5 to 3

Explanation:

There's 5 O2 and 3 CO2

5:3 or 5 to 3

(It isn't A because of the order it was asked in. it always matters what you put first, so generally speaking,

"A to B" or "B to A" ratio wouldn't be the same.)

Answer: But

Explanation:
1 carbon atom - meth
2 carbon atoms - eth
3 carbon atoms - prop
4 carbon atoms - but

this needs to be memorised

The heat energy needed to raise the temperature of 2.11 kg of the cooking oil from 23°C to 191°C is 609960 J.

To calculate the heat energy needed to raise the temperature of the cooking oil, we can use the formula:

Q = m × c × ΔT

where:

Q = heat energy (in joules)

m = mass of oil (in grams)

c = specific heat of the oil (in J/(g°C))

ΔT = change in temperature (in °C)

First, we need to convert the mass of the oil from 2.11 kg to grams:

2.11 kg = 2.11 x 1000 g/kg = 2110 g

Substituting the given values into the formula, we have:

Q = 2110 g × 1.75 J/(g°C) × (191°C - 23°C)

Q = 2110 g × 1.75 J/(g°C) × 168°C

Q = 609960 J

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This reddish-brown irritating gas that gives photochemical smog its brownish color and is also a major component of acid deposition is Nitrogen dioxide (NO2).

Nitrogen dioxide (NO2) is a reddish-brown gas that is a photochemical smog component. It's also a component of acid deposition when it's converted to an acid in the atmosphere. Nitrogen dioxide (NO2) is one of the major air pollutants that contribute to the formation of smog. It's generated from the burning of fossil fuels, particularly in high-temperature combustion processes that produce smog.

Nitrogen dioxide is formed when nitrogen in the atmosphere reacts with oxygen in the atmosphere.Nitrogen dioxide is one of the primary pollutants produced by motor vehicle exhausts. When nitrogen oxide (NO) and volatile organic compounds (VOCs) are released into the atmosphere by burning fossil fuels, they react with sunlight to create photochemical smog. Nitrogen dioxide, a secondary air pollutant, is produced as a result of this reaction. Nitrogen dioxide (NO2) is a reddish-brown gas that absorbs light at wavelengths below 400 nanometers, giving photochemical smog its brownish color.

Nitrogen dioxide (NO2) is a major component of acid deposition when it is converted in the atmosphere into an acid. Acid deposition is a type of precipitation that contains acidic substances, such as sulfuric acid and nitric acid. When acid deposition falls to the ground, it can cause damage to plant and animal life, as well as buildings and other structures.

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Answer: 3 hydrogen Atoms

Answer:3 atoms

Explanation:

Therefore, 0.2062 mol of NO can be produced from 3.3 g of O2. In order to determine the number of moles of NO that can be produced from 3.3 g of O2.

2NO + O2 → 2NO2 From the balanced equation, we can see that the stoichiometric ratio between O2 and NO is 1:2. This means that for every one mole of O2 that reacts, two moles of NO are produced.

To calculate the number of moles of O2 present in 3.3 g, we need to use the molar mass of O2, which is 32.00 g/mol:

moles of O2 = mass of O2 / molar mass of O2

moles of O2 = 3.3 g / 32.00 g/mol

moles of O2 = 0.1031 mol

Now, we can use the stoichiometric ratio to determine the number of moles of NO produced:

moles of NO = 2 x moles of O2

moles of NO = 2 x 0.1031 mol

moles of NO = 0.2062 mol

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The average speed of SO2 molecules at a certain temperature and pressure will be the same as the average speed of chlorine molecules, which is: 0.0380 m/s

At a certain temperature and pressure, chlorine molecules have an average speed of .0380 m/s. Under the same conditions, the average speed of SO2 molecules will be the same. This is because temperature and pressure are constant parameters and do not change the speed of molecules. Therefore, the average speed of SO2 molecules at the same temperature and pressure is .0380 m/s.

To further explain, the temperature is related to the kinetic energy of molecules and the average speed of a molecule is proportional to the square root of its absolute temperature. Therefore, when the temperature is constant, the average speed of the molecules is constant too.

Similarly, pressure is related to the number of collisions between molecules, but it does not have an effect on the average speed of the molecules. Therefore, when the temperature and pressure are constant, the average speed of the molecules is also constant.

In conclusion, the average speed of SO2 molecules at a certain temperature and pressure will be the same as the average speed of chlorine molecules, which is .0380 m/s.

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The total molar mass of hydrogen in 8 moles of water is 16 g/mole.

1 molecule of water contains 2 atoms of H and 1 atom of O.

What is molar mass?

The ratio between the mass and the substance content (measured in moles) of any sample of a chemical compound is known as the molar mass (M). The molar mass of a substance is a bulk attribute rather than a molecular one. The compound's molar mass is an average over numerous samples, which frequently have different masses because of isotopes.

The molar mass is a crucial characteristic of the material that is independent of sample size. The coherent unit of molar mass in the International System of Units (SI) is kg/mol.

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Answer:

The balanced chemical equation for the combustion of ethane (C2H6) is:

C2H6 + O2 → CO2 + H2O

According to the stoichiometry of the balanced equation, 1 mole of C2H6 reacts with 3 moles of O2 to produce 2 moles of CO2. Therefore, the ratio of moles of O2 consumed to moles of CO2 produced is:

3 mol O2 / 2 mol CO2

To determine how many moles of O2 are consumed per every 50 moles of CO2 produced, we can set up a proportion:

3 mol O2 / 2 mol CO2 = x mol O2 / 50 mol CO2

Cross-multiplying and solving for x, we get:

x = (3 mol O2 / 2 mol CO2) x 50 mol CO2 = 75 mol O2

Therefore, 75 moles of O2 are consumed per every 50 moles of CO2 produced during the combustion of ethane.

Explanation:

The concentration of the final solution, rounded to one significant figure, is 0.5 M.

To find the concentration of the final solution, we need to use the formula:

C1V1 = C2V2

where;

In this problem, we can plug in the given values and solve for C2:

C1 = 12 M

V1 = 20 mL = 0.02 L

V2 = 500 mL = 0.5 L

C1V1 = C2V2

(12 M)(0.02 L) = C2(0.5 L)

0.24 = 0.5C2

C2 = 0.48 M

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The amount of energy produced when 93.5 g of oxygen reacts with 13.2 g of hydrogen is -31,640.8 kJ.T.

When 93.5 g of oxygen reacts with 13.2 g of hydrogen in the reaction 2H2 + O2 → 2H2O, the amount of energy produced is -572 kJ. To calculate this, we can use the following equation:

Energy (in kJ) = -572 x (moles of oxygen) x (moles of hydrogen)

First, we need to find the moles of oxygen and hydrogen. To do this, we need to use the molar mass of each element, which can be found in the periodic table. The molar mass of oxygen is 16 g/mol, and the molar mass of hydrogen is 2 g/mol. Therefore, we can calculate the moles of oxygen as follows:

93.5 g of oxygen / 16 g/mol = 5.84 mol of oxygen

And the moles of hydrogen can be calculated as follows:

13.2 g of hydrogen / 2 g/mol = 6.6 mol of hydrogen

Now, we can plug these values into the equation:

Energy (in kJ) = -572 x (5.84 mol of oxygen) x (6.6 mol of hydrogen) = -31,640.8 kJ

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