what is the purpose or function of hexane in the synthesis of aspartame experiment? it serves as a solvent for unwanted nonpolar organic compounds. it is used to stabilize the dienophile intermediate. it is one of the major products of the reaction. it serves as an oxidizing agent. it is a drying agent.

Buffering capacity is the ability of a solution to resist changes in pH when an acidic or basic compound is added to it.

The greater the buffering capacity of a solution, the more resistant it is to changes in pH.

A solution prepared from a weak acid and the salt of its conjugate base is known as a buffer solution. In this case, the weak acid can donate a proton to the added base, while the conjugate base can accept a proton from the added acid, thus preventing the pH of the solution from changing significantly.

The buffering capacity of a buffer solution depends on the concentrations of the weak acid and its conjugate base. The optimal buffering capacity occurs when the concentrations of the weak acid and its conjugate base are approximately equal.

Therefore, the solution that would have the greatest buffering capacity is the one with the highest concentration of the weak acid and its conjugate base, as this would result in the highest total concentration of buffer components.

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When 3 ml of copper sulfate and 3 ml of sodium sulfide are mixed together, a chemical reaction occurs which results in the formation of a new compound. This reaction is known as a precipitation reaction.

The copper sulfate is a blue aqueous solution while the sodium sulfide is a yellow aqueous solution. When they are mixed together, a black precipitate of copper sulfide is formed.

The chemical equation for this reaction is:

CuSO4 + Na2S → CuS + Na2SO4

In this equation, CuSO4 represents copper sulfate, Na2S represents sodium sulfide, CuS represents copper sulfide, and Na2SO4 represents sodium sulfate.

The black precipitate of copper sulfide that forms in the reaction is insoluble in water, which means that it will settle at the bottom of the container in which the reaction is taking place. The colour of the solution also changes from blue to dark brown or black due to the formation of copper sulfide. The reaction is exothermic, which means that heat is released during the reaction.

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The 2-Methyl-1-propene is the most stable compound This is true because it has more substituents (alkyl groups) attached to the double bond, which increases its stability due to the electron-donating effect of the alkyl groups. This makes sp2 hybridized carbon atoms more electronegative.

The 1-Butene is the least stable compound This is true because it has fewer substituents attached to the double bond compared to 2-Methyl-1-propene, making it less stable. The more carbon atoms attached to the double bond, the more stable the alkene This statement is true as well. Alkenes with more carbon atoms attached to the double bond have increased stability due to the electron-donating effect of the alkyl groups. A trans isomer is less stable than a cis isomer due to more steric hindrance: This statement is false. In general, trans isomers are more stable than cis isomers because they have fewer steric hindrances and lower energy conformations. sp2 hybridized carbon atoms are more electronegative than sp3 hybridized atoms: This statement is true. sp2 hybridized carbon atoms have a greater proportion of "s" character (33% s-character and 67% p-character) than sp3 hybridized carbon atoms (25% s-character and 75% p-character). This makes sp2 hybridized carbon atoms more electronegative.

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Hot mix asphalt (HMA) is a famous paving material for roads and highways. Binder content is crucial for quality and performance.

The reason that is important to have optimum binder content in HMA is  so as  to keep the aggregates binded together.

The thing that happen if less than the optimum binder content is used in HMA is  that the aggregates will not bind well and this will cause failure.

Optimum binder content in HMA is important for achieving desired properties. Not enough binder weakens pavement, too much makes HMA soft and prone to rutting and bleeding.

So, it is  crucial to use the right amount of binder for durable HMA pavement. Using less-than-optimum binder can make the pavement brittle and prone to cracking. Excess binder in HMA causes costly pavement failure and deformity, like rutting and bleeding. This can make HMA pavement unsafe for traffic.

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Molarity is a measure of solute concentration in a solution, defined as the number of moles of solute per liter of solution.

To calculate the molarity of a 2.34 m aqueous Na2SO4 solution with a density of 1.11 g/ml, we must first calculate the volume of the solution. This can be done by dividing the mass of the solution (2.34 m) by its density (1.11 g/ml). The volume of the solution is then 2.11 L.

Next, we must calculate the number of moles of Na2SO4 in the solution. The molecular weight of Na2SO4 is 142 g/mol, so the number of moles in 2.34 m of Na2SO4 is 2.34 m divided by 142 g/mol, or 0.0164 moles.

Finally, we can calculate the molarity of the solution by dividing the number of moles (0.0164 moles) by the volume of the solution (2.11 L). The molarity of the 2.34 m aqueous Na2SO4 solution with a density of 1.11 g/ml is 0.0077 moles/L.

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In this equation, ^(15)O represents the isotope oxygen-15, ^(15)N represents the resulting nitrogen-15 isotope, and e^+ represents the positron emitted during the decay process.

Oxygen-15, an artificially produced radioactive isotope, undergoes decay by emitting a single positron, which is a positively charged electron. The nuclear equation for this decay can be represented as ^15O → ^15N + e⁺, where ^15O denotes oxygen-15, ^15N represents the resulting nitrogen-15 isotope, and e⁺ denotes the emitted positron. This radioactive decay process occurs as the oxygen-15 nucleus transforms into a nitrogen-15 nucleus, simultaneously releasing a positron. Such decay pathways are commonly observed in isotopes that have an excess of protons in their nuclei, aiming to achieve a more stable configuration through the emission of particles.

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The explanation of citric acid cycle as an aerobic process overall, implying that the citric acid cycle cannot take place without oxygen which creates Carbon dioxide, ATPs, and reductants like NADH and FADH₂.

Interesting hints, such as volcanic gases, massive iron ore deposits, and bubbles of old air trapped in amber, point to major shifts in the earth's atmosphere throughout its history. Two key conclusions may be drawn from combining these hints with the fossil record: first, that early life originated without oxygen, and second, that oxygen first formed between 2 and 3 billion years ago (see figure below) as a result of photosynthesis by the blue-green cyanobacteria. This history is reflected in the chemistry of cellular respiration. Glycolysis, its initial step, occurs everywhere and does not require oxygen.

We find it difficult to conceive that the oxygen gas's emergence must have been devastating for the anaerobic species that developed in its absence since they are completely dependent on it. However, because oxygen is very reactive, its initial impact on evolution was so detrimental that some have dubbed this time the "oxygen catastrophe." Life began to recover, though, when oxygen eventually created a shielding ozone layer.

The variety of aerobic creatures multiplied once the first species had the ability to utilise oxygen to their benefit. The Theory of Endosymbiosis states that the evolution of multicellularity, or the development of multicellular eukaryotic creatures, followed the engulfment of some of these aerobic bacteria. The majority of life today follows glycolysis with the 21% oxygen environment that exists today.

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The solubility of a gas in a liquid is affected by the temperature of the solution. As the temperature of the solution increases, the solubility of a gas in the liquid also increases.

This is because an increase in temperature causes the molecules of the gas to move faster, which increases the rate of diffusion and dissolution into the liquid.

Therefore, we would expect the solubility of N₂(g) in water to increase as the temperature increases from 20°C to 80°C. The exact increase in solubility may depend on the specific conditions of the solution, such as the pressure and the concentration of the gas in the liquid.

It's worth noting that there is a limit to the solubility of N₂(g) in water at high temperatures. At very high temperatures, the solubility of N₂(g) in water may decrease due to the formation of a supercritical fluid. However, at the temperatures mentioned in the question (20°C to 80°C), the solubility of N₂(g) in water is likely to increase.  

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The "digit of uncertainty" when using the electronic balance in this lab would be the hundredths place. This is because the electronic balance typically displays measurements to two decimal places, with the last digit representing the uncertainty in the measurement. Therefore, the hundredths place would be the digit of uncertainty.

The "digit of uncertainty" when using the electronic balance in your lab refers to the smallest unit of measurement that the balance can accurately report. In this context, you would need to look at the decimals to determine if the digit of uncertainty is in the tenths or the hundredths place. If the balance provides measurements up to one decimal place, then the digit of uncertainty is in the tenths place. If it provides measurements up to two decimal places, then the digit of uncertainty is in the hundredths place.

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Human-pushed expanded tiers of carbon dioxide withinside the atmosphere, there may be greater CO₂ dissolving into the sea.

The ocean's common pH is now round 8.1 , that's basic (or alkaline), however as the sea keeps to soak up greater CO₂, the pH decreases and the sea will become greater acidic. Carbon dioxide impacts the pH of blood with the aid of using reacting with water to shape carbonic acid (H₂CO₃), that could dissociate to shape a hydrogen ion (H+) and a hydrogen carbonate ion (HCO₃⁻). Increasing the awareness of carbon dioxide withinside the blood consequently outcomes in greater H+ ions and a decrease pH.

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The kind of compounds that we have in the question are;

[tex]CH_{4}[/tex] - Covalent

[tex]PBr_{2}[/tex] - Polar covalent

[tex]F_{2}[/tex] - Covalent

[tex]H_{2} O[/tex]- Polar covalent

[tex]C_{3} H_{8}[/tex] - covalent

[tex]Se_{2}[/tex] - Covalent

NaCl - ionic

[tex]AlF_{3}[/tex] - Ionic

MgO - ionic

[tex]Al_{2} O_{3}[/tex] - ionic

Gilbert N. Lewis first suggested this kind of structure in 1916, and it is now frequently used in chemistry to show how bonds and molecule structure interact.

Atoms are represented by symbols, while the bonds between them are shown by lines. Each atom's valence electrons are shown as dots or dashes.

The covalent compounds above may or may not have a dipole moment while an ionic bond holds compounds such as NaCl. For the covalent compounds, electrons are shared as in water molecule.

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Gasoline is a complex mixture of hydrocarbons, which are compounds composed of carbon and hydrogen. Ethanol, on the other hand, is a different type of compound that contains both carbon, hydrogen, and oxygen. When gasoline is analyzed using gas chromatography (GC), it is possible to separate the different components of gasoline, including ethanol.

GC works by separating the different components in a mixture based on their physical and chemical properties, such as boiling points and polarity. During the process, the mixture is vaporized and passed through a column packed with a stationary phase, which can be a liquid or a solid. As the vaporized components travel through the column, they interact with the stationary phase and are separated based on their properties.
Even if the hydrocarbons in gasoline are not all separated from each other, ethanol can still be separated from them using chromatography because it has different physical and chemical properties than the hydrocarbons. Ethanol has a lower boiling point and is more polar than many of the hydrocarbons in gasoline. These differences allow ethanol to be separated from the other components during the GC analysis.
Standard addition is a technique used in analytical chemistry to determine the concentration of a specific component in a mixture. It involves adding a known amount of the pure component to the sample and analyzing the resulting mixture using chromatography. By comparing the peak areas of the pure component and the component in the mixture, it is possible to determine the concentration of the component in the sample.
In the case of ethanol in gasoline, standard addition can be used to determine which peak in the chromatogram is due to ethanol. A known amount of pure ethanol is added to a sample of gasoline, and the resulting mixture is analyzed using GC. The peak area of the added ethanol is compared to the peak area of the component in the sample, and the concentration of ethanol in the sample can be calculated. This technique allows for accurate and precise measurements of ethanol in gasoline.

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KOH serves as the base in the reaction in the chemical equation H2SO4 + KOH H2O + K+ + HSO4-. Potassium hydrogen sulphate (KHSO4) and water are produced when the strong base KOH reacts with the strong acid H2SO4.

Potassium hydroxide, also known as KOH, is a potent alkali used in a wide range of industrial and laboratory processes. It is a white, odourless solid that dissolves quickly in water and produces a potent alkaline solution. In chemical reactions, KOH is frequently used as a base, especially in the creation of soaps, detergents, and biodiesel. It is also employed in the production of potassium salts, fertilisers, and dyes, as well as an electrolyte in alkaline batteries. KOH is also used in the manufacturing of some pharmaceuticals and cosmetic products, as well as in the food industry as a pH regulator.

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1. The H on the boxed carbon atom in ethyl acetoacetate is highly acidic due to the presence of two electron-withdrawing groups - the ester and keto groups - adjacent to the H atom. These groups pull electron density away from the H atom, making it easier for it to dissociate and form a negatively charged enolate intermediate. This intermediate is important in many reactions involving ethyl acetoacetate, such as the Claisen condensation.

2. The α-unsaturated carbon in trans-chalcone is the carbon adjacent to the carbonyl group, while the β-unsaturated carbon is the carbon on the other side of the double bond. Therefore, in trans-chalcone, the α-unsaturated carbon is the one labeled with a double bond to an oxygen atom, and the β-unsaturated carbon is the one labeled with a double bond to a carbon atom.

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0.831 liters of a 2.18 M solution can be made from 200.0 g K2S. To determine how many liters of a 2.18 M solution can be made from 200.0 g K2S, we first need to calculate the number of moles of K2S in the 200.0 g sample.

The molar mass of K2S is 110.26 g/mol (39.10 g/mol for potassium and 32.07 g/mol for sulfur, each multiplied by 2 for the two potassium atoms and one sulfur atom in K2S).

Using the formula:

moles = mass (in grams) / molar mass

we get:

moles K2S = 200.0 g / 110.26 g/mol = 1.813 mol

Next, we can use the formula for calculating Molarity:

Molarity = moles of solute / liters of solution

Rearranging the formula:

Liters of solution = moles of solute / Molarity

Substituting the values we know:

Liters of solution = 1.813 mol / 2.18 mol/L = 0.831 L

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An equimolar mixture of he and Xe is placed in a container with a pinhole. The mole fraction of helium in the mixture is 0.5

The rate of effusion of a gas is inversely proportional to the square root of its molar mass.

Therefore, the relative rates of effusion of two gases can be determined from the ratio of the square roots of their molar masses.

Given:

Helium (He) has a molar mass of 4.00 g/mol

Xenon (Xe) has a molar mass of 131.29 g/mol.

The ratio of the square roots of their molar masses is:

= 0.301

This means that the rate of effusion of helium is about 0.301 times the rate of effusion of xenon.

The mixture is equimolar, there are equal numbers of moles of helium and xenon in the container.

Therefore, the mole fraction of helium in the mixture is:

Mole fraction of He = number of moles of He / total number of moles

Since there are equal numbers of moles of helium and xenon, the mole fraction of helium is:

Mole fraction of He = 0.5

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The same sample occupy at 300k and 4.5 atm in 3.1 liters.

To find out how many liters the same sample of hydrogen gas occupies at 300K and 4.5 atm, we can use the Combined Gas law formula, which relates the initial and final states of a gas sample:

P1 * V1 / T1 = P2 * V2 / T2
Where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.

The initial conditions are:

P1 = 3.7 atm, V1 = 5.1 L, and T1 = 400K, and the final conditions:

P2 = 4.5 atm and T2 = 300K, we can solve for V2.

Rearrange the equation to solve for V2:

V2 = (P1 * V1 * T2) / (P2 * T1)

Plug in the given values:

V2 = (3.7 atm * 5.1 L * 300K) / (4.5 atm * 400K)

V2 = (5583) / (1800)

V2 ≈ 3.1 L
So, the same sample of hydrogen gas occupies approximately 3.1 liters at 300K and 4.5 atm.

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When heating ammonium nitrate, the reaction releases ammonia gas (NH3). The ammonia gas is alkaline in nature, meaning it is basic and can react with acidic substances.

Red litmus paper is an indicator that changes color in the presence of acids and bases. Initially, red litmus paper is red because it is sensitive to acidic conditions. When exposed to the ammonia gas released during the heating of ammonium nitrate, the gas reacts with the moisture present on the litmus paper's surface. Ammonia gas is basic and can neutralize the acidic properties of the litmus paper. As a result, the red litmus paper turns blue, indicating a basic or alkaline environment. However, as the heating continues and the ammonia gas disperses, the litmus paper gradually loses contact with the alkaline gas and returns to its original acidic state. Consequently, the litmus paper changes back to red, indicating the restoration of the acidic conditions. In summary, the color change of red litmus paper from red to blue and then back to red when heating ammonium nitrate is due to the reaction between the released ammonia gas (which is basic) and the acid-sensitive litmus paper.

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The three-dimensional representation (using wedges and dashed lines) of acetic acid (CH₃COOH) is in the image attached.

The wedge represents a bond coming out of the plane of the paper towards you, and the dashed line represents a bond going into the plane of the paper away from you. In this representation, the hydrogen atoms and the hydroxyl group are both in front of the plane of the paper, while the carbon and oxygen atoms are behind the plane.

The wedge is used to indicate that the hydrogen atom attached to the carbon is closer to you, while the dashed line indicates that the oxygen atom is further away from you. This is just one of many possible ways to represent the three-dimensional structure of acetic acid.

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The false statement is D. If ΔG for the reaction is < 0, then NaI is a stronger reducing agent than HCl.

ΔG represents the change in Gibbs free energy of a system and is related to the spontaneity of a reaction. If ΔG for a reaction is negative, the reaction is spontaneous and can occur without external intervention.

In the given reaction, NaI is oxidized to NaIO3, while HOCl is reduced to HCl. Therefore, NaI is the reducing agent, and HOCl is the oxidizing agent. Option A is true, while option B is also true since chlorine (Cl) in HOCl gains electrons and is reduced to HCl. Option C is also true as NaI loses electrons and undergoes oxidation, making it a reducing agent.

However, option D is false because ΔG cannot be used to determine the relative strength of reducing agents. The strength of a reducing agent is determined by its ability to donate electrons and reduce other species. In this reaction, NaI is a stronger reducing agent than HCl since it has a greater tendency to donate electrons and undergo oxidation.

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The voltage drop across resistor A will be three times greater than the voltage drop across resistor B. Therefore, the correct answer is option C.

Based on Ohm's Law, the current through a resistor is directly proportional to the potential difference across it, and inversely proportional to its resistance. In this circuit, since resistor A has three times the resistance of resistor B, the current through A will be one-third the current through B (option B is incorrect). However, the potential difference across each resistor will depend on the total voltage of the circuit and the individual resistances.

Assuming the voltage across the circuit is constant, the potential difference across A will be three times the potential difference across B (option C is correct). This is because the voltage drop across each resistor is proportional to its resistance, and the voltage drop across resistor A will be three times greater than the voltage drop across resistor B. Therefore, the correct answer is option C.

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The delta g for the phase change CO₂(g) → CO₂(aq) at 25°C is 5.63 kJ/mol.

To determine delta g for the phase change CO₂(g) → CO₂(aq) at 25°C, we need to use the Gibbs free energy equation: delta g = delta h - T delta s.

First, we need to find the enthalpy change (delta h) and entropy change (delta s) for the phase change. The enthalpy change is the heat absorbed or released during the phase change, and the entropy change is the measure of disorder or randomness in the system.

Since CO₂(g) is converting to CO₂(aq), we know that heat is being absorbed (endothermic) and the disorder in the system is increasing (more molecules are able to move around freely in the aqueous solution compared to the gas phase). Therefore, delta h is positive and delta s is also positive.

Next, we need to know the values of delta h and delta s. We can look these values up in a reference table or calculate them using thermodynamic data. For this phase change, the standard enthalpy change is 54.58 kJ/mol and the standard entropy change is 163.7 J/mol*K.

Finally, we can plug these values into the Gibbs free energy equation to solve for delta g:

delta g = (54.58 kJ/mol) - (298 K) * (163.7 J/mol*K)
delta g = 54.58 kJ/mol - 48.95 kJ/mol
delta g = 5.63 kJ/mol

Therefore, the delta g for the phase change CO₂(g) → CO₂(aq) at 25°C is 5.63 kJ/mol.

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The alkene that would give only a ketone with three carbons as a product of oxidative cleavage is propene.

Oxidative cleavage of alkenes involves the breaking of the double bond and the addition of oxygen to form two carbonyl groups. The product formed depends on the position of the double bond and the number of carbon atoms on either side of it.

When propene undergoes oxidative cleavage, it forms a ketone with three carbons, namely acetone, as the only product. This is because propene has a double bond between the second and third carbon atoms, and upon cleavage, it forms two carbonyl groups, one on each end of the double bond, resulting in acetone with three carbons.

In contrast, alkenes with four or more carbon atoms will form a mixture of ketones and aldehydes upon oxidative cleavage.

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Oxygen cylinders stored indoors must be kept at least five feet away from other flammable materials.

This is due to the fact that oxygen is a highly reactive gas that can rapidly accelerate a fire if it comes into contact with flammable materials such as oil, grease, or other combustible substances. Additionally, oxygen cylinders should be stored in a well-ventilated area away from direct sunlight, heat sources, and electrical equipment to prevent the risk of combustion or explosion.

It is important to follow proper safety protocols when handling oxygen cylinders, as any mishandling or improper storage can pose a significant risk to both the individual and the surrounding environment. Therefore, it is crucial to always take the necessary precautions and maintain a safe distance from flammable materials when storing oxygen cylinders indoors.

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Answer:6

Explanation:

The enthalpy change of a reaction can be calculated using the standard enthalpies of formation of the reactants and products, as given by the following equation:

ΔH° = ΣnΔHf°(products) - ΣmΔHf°(reactants)

where ΔHf° is the standard enthalpy of formation of a compound, n and m are the stoichiometric coefficients of the products and reactants, respectively.

For the reaction: OH(g) → H(g) + O(g)

The standard enthalpies of formation of the reactants and products are:

ΔHf°(OH) = 37.2 kJ/mol

ΔHf°(H) = 218.0 kJ/mol

ΔHf°(O) = 249.2 kJ/mol

The stoichiometric coefficients are:

n(H) = 1

n(O) = 1

m(OH) = 1

Using the equation above, we can calculate the standard enthalpy change of the reaction:

ΔH° = [ΔHf°(H) + ΔHf°(O)] - ΔHf°(OH)

    = [(218.0 kJ/mol) + (249.2 kJ/mol)] - (37.2 kJ/mol)

    = 430.0 kJ/mol - 37.2 kJ/mol

    = 392.8 kJ/mol

Therefore, the standard enthalpy change for the reaction OH(g) → H(g) + O(g) is 392.8 kJ/mol.

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When bromine reacts with phenol and decolorizes the orange color, it forms a (c) white precipitate.

When bromine reacts with phenol, it undergoes a substitution reaction and replaces one of the hydrogen atoms in the hydroxyl group of the phenol. This reaction results in the formation of 2,4,6-tribromophenol, which is a white precipitate.

The orange color of phenol is due to the presence of an unsaturated benzene ring, which absorbs visible light in the range of orange color. However, when bromine is added to phenol, it reacts with the benzene ring and changes its structure, which alters its ability to absorb visible light in the orange range. This change in the structure results in the decolorization of the orange color of phenol.

The formation of a white precipitate is due to the fact that 2,4,6-tribromophenol is an insoluble compound, which forms a precipitate in the solution. The white precipitate that is formed is a visual confirmation of the reaction between bromine and phenol.

In summary, when bromine reacts with phenol, it decolorizes the orange color of phenol and forms a white precipitate of 2,4,6-tribromophenol. Therefore, the correct answer to the question is (c) white precipitate.

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The following transition metals would be expected to have the smallest atomic radius is Technetium (Tc) is expected to have the smallest atomic radius among the given transition metals.

As we move across a period from left to right, the atomic radius generally decreases. This is due to an increase in the effective nuclear charge, which pulls the electrons closer to the nucleus.

Comparing the given elements:
A) Yttrium (Y) - Group 3, Period 5
B) Zirconium (Zr) - Group 4, Period 5
C) Niobium (Nb) - Group 5, Period 5
D) Technetium (Tc) - Group 7, Period 5
E) Ruthenium (Ru) - Group 8, Period 5

Since all these elements are in the same period (Period 5), we can simply look for the one that is furthest to the right, as it will have the smallest atomic radius due to the increase in effective nuclear charge.

Your answer: D) Technetium (Tc) is expected to have the smallest atomic radius among the given transition metals.

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If a drain cleaner solution is a strong electrolyte then drain cleaner is highly ionized. The correct option is C).

A strong electrolyte is a substance that completely dissociates into ions in a solution. This means that when a strong electrolyte such as a drain cleaner is dissolved in water, it forms a high concentration of ions in the solution. The high ion concentration makes the solution highly conductive of electricity.

In the case of a drain cleaner, the high concentration of ions allows it to react more effectively with clogs and blockages in drains. Therefore, option A and B can be ruled out as they do not accurately describe the behavior of a strong electrolyte.

Option D is also incorrect as a strong electrolyte is highly ionized, not slightly ionized. Option E is incorrect as one of the options must be true. This leaves us with option C, which correctly describes the behavior of a strong electrolyte. Therefore, the correct option is C.

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The MnO4- ion is undergoing reduction. There are 0.00194 moles of FAS in trial 2, and the molar concentration of KMnO4 solution is 0.0437 M.

a. The MnO4- ion is undergoing reduction.
b. The moles of FAS in trial 2 are 0.00194 moles.
c. The molar concentration of KMnO4 solution is 0.0437 M.
a. In the given reaction, the MnO4- ion gains electrons and its oxidation state reduces from +7 to +2. Hence, it undergoes reduction.
b. To calculate the moles of FAS in trial 2, use the mass and molar mass:
Moles = mass / molar mass = 0.550 g / 284.05 g/mol = 0.00194 moles
c. To find the molar concentration of KMnO4 solution, use the stoichiometry of the balanced equation and volume data from trial 2:
5 moles Fe+2 react with 1 mole MnO4- (from the equation)
0.00194 moles FAS × (1 mole MnO4- / 5 moles Fe+2) = 0.000388 moles MnO4-
Molar concentration = moles MnO4- / volume in liters = 0.000388 moles / 0.0224 L = 0.0437 M

Summary:
The MnO4- ion is undergoing reduction. There are 0.00194 moles of FAS in trial 2, and the molar concentration of KMnO4 solution is 0.0437 M.

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