when a nucleus of 235u undergoes fission, it breaks into two smaller, more tightly bound fragments. part a calculate the binding energy per nucleon for 235u

The pKa of Ph3P+CH2C(O)Ph, also known as Wittig reagent, is approximately 17.

This is due to the presence of a phosphonium group (Ph3P+) which is a strong electron-withdrawing group. This results in the stabilization of the carbanion formed during the reaction, which in turn leads to a higher pKa value.

The higher pKa value of the Wittig reagent is essential for its reactivity in organic synthesis reactions, where it is commonly used for the synthesis of alkenes from aldehydes or ketones. T

he high pKa allows for the deprotonation of the reagent and subsequent formation of the ylide intermediate, which can then react with the electrophilic carbonyl compound to form the desired alkene product.

Overall, the pKa of Ph3P+CH2C(O)Ph plays a crucial role in its reactivity and usefulness in organic synthesis.

The pKa value is a measure of the acidity of a compound. In this case, you are asking about the pKa of Ph3P+CH2C(O)Ph, which is a phosphonium ylide compound. These compounds are widely used in organic chemistry, particularly in the Wittig reaction, where they react with carbonyl compounds to form alkenes.

Phosphonium ylides are typically non-acidic and possess a pKa value much higher than those of typical acidic compounds. Their basic nature arises from the positive charge on the phosphorus atom and the negative charge on the carbon atom, leading to a highly polarized and reactive molecule. The pKa value is not typically reported for these compounds because they do not display the traditional acidic behavior you would expect from molecules with low pKa values.

Instead of focusing on the pKa of Ph3P+CH2C(O)Ph, it's more helpful to understand its reactivity and the chemical transformations it can undergo, such as the Wittig reaction. This information will be more relevant when using this compound in organic synthesis or other related applications.

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The electron-pushing mechanism for the decarboxylation reaction involves the addition of three curved arrows. The reaction starts with a carboxylic acid, which undergoes a proton transfer to form a carboxylate anion. The carboxylate anion then experiences nucleophilic attack by a base, leading to the formation of a cyclic intermediate. The cyclic intermediate undergoes ring opening, resulting in the expulsion of carbon dioxide and the formation of a new bond. The electron-pushing mechanism helps illustrate the movement of electrons throughout the reaction, highlighting the formation and breaking of bonds.

In the decarboxylation reaction, we start with a carboxylic acid, which is represented by the structure: C(O)OH. The first step involves the proton transfer, where a curved arrow is drawn from one of the lone pairs on the oxygen of the carboxylic acid to the hydrogen atom attached to the same oxygen. This proton transfer leads to the formation of a carboxylate anion, depicted as C(O)O-.

The second step involves nucleophilic attack by a base. Draw a curved arrow from the lone pair on the oxygen of the carboxylate anion to the carbon atom adjacent to the carbonyl group. This movement of electrons results in the formation of a cyclic intermediate. The cyclic intermediate is represented by a ring structure formed by the interaction between the carbon atom and the oxygen atom of the carboxylate group.

The third step involves ring opening. Draw a curved arrow from the carbon atom of the cyclic intermediate to the oxygen atom of the carboxylate group. This movement of electrons breaks the bond between the carbon atom and the oxygen atom, resulting in the expulsion of carbon dioxide (CO2). Simultaneously, a new bond is formed between the carbon atom and the neighboring atom (not specified in the given question).

The overall reaction can be summarized as the decarboxylation of a carboxylic acid, leading to the removal of a carbon dioxide molecule and the formation of a new bond. The electron-pushing mechanism helps visualize the flow of electrons during the reaction, highlighting the formation and breaking of bonds at each step.

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In the final state volume occupied by gas is 7.96 L

To solve this problem, we can use the combined gas law:

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

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

First, we need to convert the initial and final temperatures to Kelvin:

\(T1 = 29.0\°C + 273.15 = 302.15 K\\T2 = 91.0\°C + 273.15 = 364.15 K\)

Now we can plug in the values:

\((3.17\ atm * 6.46 L) / (302.15 K) = (5.80\ atm * V2) / (364.15 K)\)

Solving for V2, we get:

\(V2 = (3.17 atm * 6.46 L * 364.15 K) / (5.80 atm * 302.15 K)\\V2 = 7.96 L\)

Therefore, the gas occupies a volume of 7.96 L in its final state.

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Answer:

A- Before the mechanic left, the molecules were moving away from each other. When she returned, they were moving around each other.

Explanation:

Brainliest?

Answer:

true

Explanation:

cause thereare both in liquid form

The solutes in an aqueous solution of KF in order of increasing concentration are : Lowest concentration: K+ (aq),H+ (aq),OH- (aq),F- (aq),Highest concentration: HF (aq)

In an aqueous solution of KF, K+ ions come from the dissociation of KF, but KF is a strong electrolyte and dissociates almost completely, so the concentration of K+ ions is relatively high. H+ ions are present in water due to the self-ionization of water, but their concentration is relatively low. OH- ions are also present due to the self-ionization of water, but their concentration is lower than that of H+ ions. F- ions come from the dissociation of KF, so their concentration is higher than that of OH- ions. HF is a weak acid that partially dissociates in water, resulting in a higher concentration of HF compared to the other ions in the solution.

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To determine the number of moles of hydrogen needed to crack one mole of the long saturated hydrocarbon chain (C40H82), we can analyze the reactants and products involved in the cracking reaction.

The cracking reaction is given as: C40H82 -> 3 C8H18 + n C2H6. From the equation, we can see that one mole of the long hydrocarbon chain (C40H82) produces three moles of octane (C8H18) and n moles of ethane (C2H6). Since the cracking process involves breaking the carbon-carbon bonds and forming new carbon-hydrogen bonds, the number of hydrogen atoms in the products should remain the same as in the reactant.

The long hydrocarbon chain (C40H82) contains 82 hydrogen atoms, and the products, 3 moles of octane (C8H18), contain (3 moles) * (18 hydrogen atoms/mole) = 54 hydrogen atoms. Therefore, the number of moles of hydrogen needed for cracking one mole of the long hydrocarbon chain can be calculated as: Number of moles of hydrogen = 82 - 54 = 28 moles. Hence, 28 moles of hydrogen are required to crack one mole of the long saturated hydrocarbon chain (C40H82).

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Answer:

Proteins are made up of 20 amino acids. How an Isotope Technique Helps Determine Protein Quality. However be produced by the body and therefore are participated in the labelling process.

Explanation:

Answer: A central nervous system stimulant

Explanation: arguably the most frequently ingested pharmacologically active substance in the world.

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The Correct option is :

\( \large \boxed{\mathrm{B.}}\)

The particles diffuse from the region of higher concentration to the region of lower concentration.

for example :

during Inhalation, oxygen is in higher concentration in our lungs but in lower concentration blood capillaries, so oxygen diffuses into the blood and carbon dioxide which is in higher concentration in blood diffuses out from the blood capillaries.

_____________________________

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A student measures the molar solubility of nickel(II) cyanide in a water solution to be 2. 00×10-8 M. The Ksp of nickel(II) cyanide is \(8 \times10^{-24}\)

The solubility product constant, Ksp, serves as the equilibrium constant for solids that dissolve in water. The level of solute dissolution in solution is what it stands for. The more soluble a material is, the higher its Ksp value.

The following equation may be used to get the solubility product constant (Ksp) for nickel(II) cyanide from the molar solubility:

\(Ni(CN)_2(s)\) ⇌ \(Ni_2+(aq) + 2CN^-(aq)\)

\(Ksp = [Ni_2^+][CN^-]^2\)

As per the given information,  

the molar solubility of nickel(II) cyanide is given as \(2.00\times 10^{-8}\)M, the concentrations of \(Ni_2^+\) and \(CN^-\) ions are also \(2.00 \times 10^{-8}\) M.

We have to find the Ksp of nickel(II) cyanide.

Hence, the Ksp can be calculated as:

Ksp = \((2.00\times10^{-8})(2.00\times10^{-8})^{2}\)

Ksp =\(8.00\times10^{-24}\)

Hence, the Ksp of nickel(II) cyanide is \(8.00\times10^{-24}\)

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a. The attractive forces between molecules increase in the transition from a solid to a liquid and then decrease from a liquid to a gas.

b. The substance with the strongest attractive forces is eicosane (lubricant) due to its solid state at a relatively high melting point. The substance with the weakest attractive forces is natural gas because it exists as a gas at a very low boiling point.

c. Methane exhibits London dispersion forces, octane exhibits London dispersion forces, and eicosane exhibits London dispersion forces.

a. When a substance changes from a solid to a liquid, the attractive forces between molecules weaken. In a solid, the molecules are tightly packed and held together by strong intermolecular forces, such as London dispersion forces, dipole-dipole interactions, or hydrogen bonding. As the solid absorbs heat, the molecules gain energy, and the intermolecular forces weaken, allowing the substance to transition into a liquid state. In this liquid state, the molecules have more freedom to move and slide past each other.

Similarly, when a substance changes from a liquid to a gas, the attractive forces between molecules further decrease. As the liquid absorbs more heat, the molecules gain even more energy, leading to an increase in their kinetic energy. The intermolecular forces become weaker, allowing the molecules to overcome these forces and transition into a gaseous state. In the gas phase, the molecules are relatively far apart and move freely, exhibiting minimal intermolecular interactions.

b. From the properties provided, we can determine the strength of attractive forces. Eicosane (lubricant) has the strongest attractive forces because it exists as a solid at a relatively high melting point (MP: 37°C). The solid state indicates strong intermolecular forces that hold the molecules together. Octane (C8H18) in gasoline is a liquid at room temperature and exhibits weaker attractive forces compared to eicosane. Natural gas, composed mainly of methane (CH4), exists as a gas at a very low boiling point (BP: -161°C), indicating the weakest attractive forces among the three compounds.

c. Methane, octane, and eicosane all exhibit London dispersion forces as their primary intermolecular force. London dispersion forces are temporary and induced by temporary fluctuations in electron density within a molecule, resulting in temporary dipoles. These temporary dipoles induce dipoles in neighboring molecules, leading to attractive forces between them. The strength of London dispersion forces increases with the size and shape of the molecules involved. As the size of a molecule increases, the number of electrons and the surface area available for temporary dipoles also increase, enhancing the strength of London dispersion forces.

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when sodium carbonate reacts with a dilute acid carbondioxide is produces

Answer:  carbon dioxide

Explanation:  a salt, carbon dioxide, and water are formed in this reaction

Answer:

a. 5 8/100 is equivalent to $ 5.08.

Explanation:

i hope this helps :)

      An ion is an atom or collection of atoms where the number of electrons and the number of protons are different. A positive ion, also known as a cation, is a particle that exists when the number of electrons exceeds the number of protons.

      Because they have an equal number of protons and electrons, atoms are neutral particles. Ions are electrically charged particles that can be created by either taking electrons away from neutral atoms to form positive ions or adding electrons to neutral atoms to produce negative ions.

    Depending on whether the number of electrons in an atom is greater or less than the number of protons in the atom, an atom can develop a positive charge or a negative charge. An atom is referred to as an ION when it is drawn to another atom due to an imbalance in its electron and proton numbers.

    Any element that has a variable ratio of protons to electrons, producing either a positively or negatively charged atom, is said to contain an ion. It is relatively easy to determine whether an element is an ion.

       To become a noble gas configuration, elements create ions in order to become more stable. They lose or gain electrons to create ions, which results in a fully filled outer shell configuration and subsequently stability.

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Answer:

The 50 kg object will have a smaller acceleration.

Explanation:

The two forces exerted on the objects are equal and F=ma, so a greater mass means a smaller acceleration while a greater acceleration means a smaller mass.

Answer:

92.01 g/mol

Explanation:

So first you need to find the empirical formula by the percents. That would be, assuming that you have 100 grams of the the sample, divide each quantity of each element found by its respective molar mass.

30.4 g of N ÷ 14 g/mol N= 2.17 mol of N

69.6 g of O ÷ 16g/mol= 4.35 mol of O

You can establish now the empirical formula.

N2.17O4.35,

but since you can't have a decimal subscript, you divide each subscript by the minimum subscript

NO2

So then you're said that the molecular formula derived from that empirical formula has 2 nitrogen, so you multiply all the subscripts, by 2:

N2O4

-Dinitrogen Tetraoxide

-Nitrogen oxide (IV)

Then all you have to do is find the molecular mass of the compound using the periodic table and what you obtain is the molar mass.

remember: molecular mass is correspondent to molar mass.

Answer:

Organism: An organism refers to a living thing that has an organised structure, can react to stimuli, reproduce, grow, adapt, and maintain homeostasis.

Example : sea worms, jellyfish, sea anemones, shellfish, squid, octopus and starfish.

Population: A population is a distinct group of individuals, whether that group comprises a nation or a group of people with a common characteristic.

Example:  Risso´s dolphin population, Bottlenose dolphin population, Killer whale population.

Community: an interacting group of various species in a common location.

Example :  1) A  group of dolphin, 2)group of sea anemones and 3) clownfish or 4) group of whales.

Marine ecosystems include: the abyssal plain, polar regions such as the Antarctic and Arctic, coral reefs, the deep sea, kelp forests, mangroves, the open ocean, rocky shores, salt marshes and sandy shores.

The example of an organism, population, and community for a marine ecosystem are small microorganisms like algae the marine plants like hydra to the largest mammal whale these all are marine.

This ecosystem consist of aquatic environments with high level of salts dissolved in it also consist of deep sea ocean and marine costal.

The examples are coral reef, including the fishes like whale and turtles, sea horse many aquatic plants and the microorganism like amoeba. ocean is covered by plants in the 71% of the area.

Therefore, example of an organism, population, and community for a marine ecosystem are small microorganisms like algae the marine plants like hydra to the largest mammal whale these all are marine.

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Answer:

idc

Explanation:

idc looooooooooool

The concentration of A after 30 seconds when the given reaction obeys the rate law rate = k[A]²[B].

We use the initial concentration of A and B and the rate constant of the reaction to find the rates at these concentrations. Using the integrated rate law for a second-order reaction, we find the concentration of A after 30 seconds to be 0.0934 M.

Given reaction obeys the rate law, rate=k[A]²[B].

Here, the initial concentration of A= 0.10 M,

initial concentration of B = 0.05 M, and

rate constant, k = 2.0 × 10⁻⁴ M⁻¹s⁻¹

We have to find the concentration of A, after 30 seconds.

To find the concentration of A, we need to know the rate at 0.10 M and 0.05 M. Therefore, we have to calculate the rates at these concentrations.

rate1 = k[A]²[B]

= (2.0 × 10⁻⁴ M⁻¹s⁻¹)(0.10 M)²(0.05 M)

= 1.0 × 10⁻⁷ M/srate2

= k[A]²[B] = (2.0 × 10⁻⁴ M⁻¹s⁻¹)(0.09 M)²(0.04 M)

= 6.48 × 10⁻⁸ M/s

Using the integrated rate law for a second-order reaction: [A] = [A]₀ - kt where [A]₀ = initial concentration of A, k = rate constant, and t = time in seconds.

We know [A]₀ = 0.10 M and k = 2.0 × 10⁻⁴ M⁻¹s⁻¹.

Substituting the values in the above equation, we get: [A] = [A]₀ - kt= 0.10 M - (2.0 × 10⁻⁴ M⁻¹s⁻¹)(30 s)≈ 0.0934 M

Therefore, the concentration of A in the vessel after 30 seconds is 0.0934 M.

This question requires us to calculate the concentration of A after 30 seconds when the given reaction obeys the rate law rate = k[A]²[B].

We are given the initial concentration of A and B and the rate constant of the reaction. To find the concentration of A after 30 seconds, we need to calculate the rates at the initial concentrations of A and B.

Using the integrated rate law for a second-order reaction, we can find the concentration of A at any given time. We substitute the given values in the formula and solve for [A]. We get the concentration of A as 0.0934 M after 30 seconds. This calculation is based on the assumption that no other reaction is important.

The concentration of A after 30 seconds when the given reaction obeys the rate law rate = k[A]²[B]. We use the initial concentration of A and B and the rate constant of the reaction to find the rates at these concentrations. Using the integrated rate law for a second-order reaction, we find the concentration of A after 30 seconds to be 0.0934 M. This calculation assumes that no other reaction is important.

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A buffer solution is defined as a solution that resists a change in pH when a small amount of acid or base is added to it. the buffer capacity of the solution will prevent the pH from changing too much.

The buffer solution containing nitrous acid, HNO2, and potassium nitrite, KNO2, will experience the following changes when a small amount of HClO4 is added to it: The HClO4 added to the buffer solution will react with the potassium nitrite, KNO2, to form the salt, KClO4.T

he HNO2 will be converted to nitric acid, HNO3, by the HClO4.The HNO3 formed in the previous step will react with the potassium nitrite, KNO2, to form nitric oxide, NO, and potassium nitrate, KNO3.The net effect of adding HClO4 to the buffer solution containing nitrous acid, HNO2, and potassium nitrite, KNO2, will be to shift the buffer solution to a more acidic pH range.

However, the buffer capacity of the solution will prevent the pH from changing too much.

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The speed of light in terms of km/day is \(2.59\)×\(10^{10}\) km/day.

The speed of light is \(3.0\)×\(10^{8}\) \(m/s\).

1  \(m/s\) = 86.4 km/day

by using this  \(3.0\)×\(10^{8}\) \(m/s\) can be converted to  \(2.59\)×\(10^{10}\) km/day.

The speed of light in a vacuum, commonly abbreviated as c, is one universal physical constant that is important in many fields of physics. The actual speed of light is 299,792,458 meters per second, or c.

In physics class, Einstein had already learnt what a light beam was: a collection of oscillating electric and magnetic fields moving at the observed speed of light, or 186,000 miles per second.

Thus, somewhat strangely, de Rham claims that the only thing that can travel faster than the speed of light is light itself, but only when it is not in the vacuum of space. It is interesting to note that light will never travel faster than 186,282 miles per second regardless of the medium.

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The correct statement regarding the number of valence electrons to include in the Lewis structure for PCl3 is: Each Cl atom has 7 valence electrons, and the P atom has 5 valence electrons, so the Lewis structure for PCl3 will include 26 electrons.

This is because each Cl atom contributes 7 valence electrons and the P atom contributes 5 valence electrons, resulting in a total of 26 valence electrons for the molecule. The bonding domains in the Lewis structure show the arrangement of these valence electrons around the central P atom in PCl3. Phosphorus (P) has 5 valence electrons, and each chlorine atom (Cl) has 7 valence electrons. The Lewis structure for PCl3 includes a single bond between the P atom and each Cl atom, with three lone pairs of electrons on each Cl atom. Therefore, the total number of valence electrons to include in the Lewis structure is:1 × 5 (valence electrons for P) + 3 × 7 (valence electrons for each Cl) = 26 valence electrons.

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Malonate is a competitive inhibitor of succinate dehydrogenase and a structural analog of succinate.

1. The intermediate has the structure -OOC-CH₂-CH₂-COO- (succinate 2)

2. Succinate builds up to 3 when succinate dehydrogenase is inhibited.

3. The accumulation of the reaction's substrate occurs when any reaction in a pathway is inhibited. The accumulation of this substrate alters the effective ΔG of the preceding reaction, changing it in turn for all subsequent steps in the pathway.

The pathway's net rate slows and eventually becomes almost nonexistent. In the case of the citric acid cycle, stopping production of the primary product, NADH, stops electron transport and oxygen consumption, which is the end acceptor of NADH-derived electrons.

4. Since malonate is a serious inhibitor, the expansion of a lot of succinate will beat the hindrance.

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The molarity of copper sulfate in the diluted solution if 4 ml of 0.0250 m cuso4 is diluted to 10.0 ml with pure water is 0.02M.

Molarity is defined as the ratio of moles of solute to the volume of solution.

Mathematically,

Molarity = moles of solute / volume of solution

Given,

Initial volume of CuSO4 = 4 ml = 4/1000 = 0.004 l

Molarity of CuSO4 = 0.0250 m

By substituting all the values, we get

0.025 = moles/ 0.004

moles = 0.025 × 0.004 = 0.000100 mol

Moles = 0.0001 mol.

Total volume = 0.004 l + 10 ml

= 0.004 + 0.001 = 0.005 l

Moles of CuSO4 = 0.0001 mol

Molarity = 0.0001 / 0.005 = 0.02 M

Thus, we concluded that the molarity of copper sulfate in the diluted solution if 4 ml of 0.0250 m cuso4 is diluted to 10.0 ml with pure water is 0.02M.

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Assessment of a spontaneously perfusing patient's ETCO2 with small capnographic waveforms and a reading of 22 mm Hg indicates respiratory acidosis.

End-tidal carbon dioxide (ETCO2) is a measurement of the partial pressure of carbon dioxide (CO2) at the end of exhalation. It provides valuable information about a patient's ventilation and the balance between CO2 production and elimination. In this scenario, small capnographic waveforms and an ETCO2 reading of 22 mm Hg suggest respiratory acidosis.

Respiratory acidosis occurs when there is an increase in the partial pressure of CO2 in the blood, leading to an imbalance in the body's acid-base status. It is typically caused by hypoventilation or inadequate removal of CO2 from the lungs. In this case, the small capnographic waveforms indicate a reduced amount of CO2 being exhaled during each breath.

A reading of 22 mm Hg suggests an elevated level of CO2 in the arterial blood. Normal ETCO2 levels range from 35 to 45 mm Hg. Therefore, a reading of 22 mm Hg indicates a lower than normal level of CO2 elimination, which contributes to an accumulation of CO2 and the development of respiratory acidosis.

Respiratory acidosis can result in symptoms such as confusion, drowsiness, shortness of breath, and potentially lead to serious complications if left untreated. Prompt assessment and appropriate interventions, such as improving ventilation or addressing the underlying cause, are necessary to restore acid-base balance and optimize patient outcomes.

the assessment of ETCO2, the interpretation of capnographic waveforms, and the diagnosis and management of respiratory acidosis for a comprehensive understanding of this clinical scenario.

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Mass of potassium bromide must be dissolved to prepare 100 ml of a 0.164 m aqueous solution is 1.95 gram

It is given that

volume of aqueous solution = 100 ml

molarity of solution = 0.164 m

We have to find mass of potassium bromide

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per liters of a solution. Molarity is also known as the molar concentration of a solution

Molarity = (given mass/molar mass) x (1/volume of solution)

0.164 = (mass/119) x 1000/100

0.164 = mass/119 x 10

0.164 x 119/10 = mass

mass = 1.95 gram

Hence, Mass of potassium bromide must be dissolved to prepare 100 ml of a 0.164 m aqueous solution is 1.95 gram

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Explanation:

The reaction of aqueous solutions of iron(II) sulfate and ammonium phosphate leads to the formation of solid precipitate of iron phosphate and an aqueous solution of ammonium sulphate.

Molecular equation :

3

F

e

(

S

O

4

)

(

a

q

)

+

2

(

N

H

4

)

3

P

O

4

(

a

q

)

→

F

e

3

(

P

O

4

)

2

(

s

)

+

3

(

N

H

4

)

2

S

O

4

(

a

q

)

Total ionic equation :

3

F

e

+

2

(

a

)

+

3

(

S

O

4

)

2

−

(

a

q

)

+

6

(

N

H

4

)

+

(

a

q

)

+

2

P

O

3

−

4

(

a

q

)

→

F

e

3

(

P

O

4

)

2

(

s

)

+

6

(

N

H

4

)

+

(

a

q

)

+

3

S

O

2

−

4

(

a

q

)

Net ionic equation :

By canceling the spectator ions from total ionic equation, we get

3

F

e

+

2

(

a

q

)

+

2

P

O

3

−

4

(

a

q

)

→

F

e

3

(

P

O

4

)

2

(

s

)

Answer:

gas and liquid

Explanation:

right?