A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup (see sketch at right).

Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to 2 significant digits.

You'll need to know that the density of fresh water at the temperature of the sea around the sailor is 0.999/gcm3. You'll also want to remember Archimedes' Principle, that objects float when they displace a mass of water equal to their own mass.

Answer:

0.0630

Explanation:

The molar mass of urea = 60 g/mol

we all know that:

[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]

Then; the number of moles of urea

= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]

= 0.0667 mol

Similarly; the number of moles of methanol

= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]

= 0.9988 mol

The total number of moles = (0.0667 + 0.9988) mol

= 1.0655 mol

Finally,the mole fraction of urea  [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]

[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]

= 0.0630

Answer:

When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed

Explanation:

A neutralization reaction is one in which an acid (or acidic oxide) reacts with a base (or basic oxide). In the reaction a salt is formed and in most cases water is formed. A Salt is an ionic compound formed by the union of ions and cations through ionic bonds.

In the reactions of a strong acid (those substances that completely dissociate) with a strong base (they dissociate completely, giving up all their OH-), the complete neutralization of the species is carried out:

2 HBr (aq) + Ca(OH)₂ (s) → CaBr₂ (aq) + 2 H₂O (l)

The reaction is already balanced, complying with the law of conservation of matter. This law states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reactants must be equal to the number of atoms present in the products.

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), 2 moles of water H₂O are formed.

On the other hand, Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. Avogadro's number applies to any substance.

Then you can apply the following rule of three: if 1 mole of H₂O contains 6.023*10²³ molecules, 2 moles of H₂O, how many molecules does it contain?

[tex]amount of molecules=\frac{2moles*6.023*10^{23}molecules }{1 mole}[/tex]

amount of molecules= 1.2046*10²⁴ molecules

When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed

Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

Answer:

Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!

Explanation:

The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.  

Answer:

Explanation:

we know that

ΔH=m C ΔT

where ΔH is the change in enthalpy (j)

m is the mass of the given substance which is water in this case

ΔT IS the change in temperature and c is the specific heat constant  

we know that given mass=2.9 g

ΔT=T2-T1 =98.9 °C-23.9°C=75°C

specific heat constant for water is 4.18 j/g°C

therefore ΔH=2.9 g*4.18 j/g°C*75°C

ΔH=909.15 j

Answer:

0.032 moles

Explanation:

no of moles =

[tex] \frac{mass \: in \: grams}{relative \: molecular \: mass} [/tex]

=

[tex] \frac{2.4}{40 + 32 + 2} [/tex]

= 0.032

Calcium hydroxide reacted with nitric acid the total number of moles will be 0.032 moles.

A mole is Avogadro's number of particles, which is exactly 6.02214076×1023.

The mole is widely used in chemistry as a convenient way to express amounts of reactants and products of chemical reactions. For example, the chemical equation 2H2 + O2 → 2H2O can be interpreted to mean that for each 2 mol dihydrogen (H2) and 1 mol dioxygen (O2) that react 2 mol of water (H2O) form.

Number of moles = Mass of substance / Mass of one mole Number of moles

mass of substance = 2.4g

molar mass of calcium hydroxide is  (1 ×40.078g/mol Ca) +(2 × 15.999g/mol O) + (2 × 1.008g/mol H) = 74.092 g/mol Ca (OH)2

substituting the value,

number of moles = 2.4 / 74.029

                          = 0.032 moles

Therefore, moles of calcium hydroxide will be 0.032 moles

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Answer:

Ka = 6.15x10⁻⁷

Explanation:

Ka is defined as dissociation constant in the equilibrium of a weak acid with water. The general reaction is:

HA(aq) + H₂O(l) ⇆ H₃O⁺(aq) + A⁻(aq)

And Ka is defined as the ratio between molar concentrations in equilibrium of products over reactants as follows:

Ka = [H₃O⁺] [A⁻] / [HA]

You don't take water in the equilibrium beacuse is a pure liquid

Replacing with the concentrations of the problem:

Ka = [H₃O⁺] [A⁻] / [HA]

Ka = [4.00x10⁻⁴] [4.00x10⁻⁴] / [0.260]

Answer:

1.05 V

Explanation:

Since;

E°cell= E°cathode- E°anode

E°cathode= -0.40 V

E°anode= -1.45 V

E°cell= -0.40-(-1.45) = 1.05 V

Equation of the process;

2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)

n= 8 electrons transferred

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]

Since log 1=0

Ecell= E°cell= 1.05 V

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are in equilibrium

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are the actual concentrations

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

Where X is defined as the reaction coordinate

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

[CrO₄²⁻] = 0.135M

Answer:

a) Monomers: monomers are unit molecules, that can react together with other monomers, to form a long chain molecule called a polymer. Th polymer formed can also be in a three dimensional network. The process of this conversion of monomers to polymers is called polymerization.

b) Repeating unit: A repeating unit is a unit of the polymer formed, whose repetition would produce a long complete polymer chain. A polymer is made up of these repeating links of molecules that form a long chain of molecules.

c) Condensation polymerization: This is a form of condensation reaction, that involves the combination of molecules into polymers with the loss of small molecules such as water or methanol as by products.

d) Cross-linked polymer: This is a polymer formed from a type of bonding of molecules. The bonding is usually in the form of covalent bonds or ionic bonds and the polymers can be either synthetic polymers or natural polymers.  The cross-links leads to an alteration in the physical properties of the polymer.

The definition of following terms are :

a) Monomers:

The monomers are unit atoms, that can respond in conjunction with other monomers, to create a long chain molecule called a polymer.

The polymer shaped can too be in a three dimensional arrange.

b) Repeating unit:

A rehashing unit may be a unit of the polymer shaped, whose reiteration would produce a long total polymer chain.

A polymer is made up of these rehashing joins of atoms that shape a long chain of molecules.

c) Condensation polymerization:

This is often a frame of condensation response, that includes the combination of particles into polymers with the misfortune of little particles such as water or methanol as by products.

d) Cross-linked polymer:

This can be a polymer shaped from a sort of holding of particles.

The cross-links leads to an modification within the physical properties.

Definition is a rhetorical style that uses various techniques to impress upon the reader the meaning of a term, idea, or concept.

Definition may be used for an entire essay but is often used as a rhetorical style within an essay that may mix rhetorical styles.

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Answer:

CO3^2-

Explanation:

In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.

When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.

Answer:

There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value

Explanation:

The radioactive decay follows always first-order kinetics where its general law is:

Ln[A] = -Kt + ln[A]₀

Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.

We can find rate constant from half-life as follows:

Rate constant:

t(1/2) = ln 2 / K

As half-life of Cesium-137 is 30.2 years:

30.2 years = ln 2 / K

K = 0.02295 years⁻¹

Replacing this result and with the given data of the problem:

Ln[A] = -Kt + ln[A]₀

Ln[A] = -0.02295 years⁻¹* t + ln[A]₀

Ln ([A] / [A₀]) = -0.02295 years⁻¹* t

As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:

Ln (0.2) = -0.02295 years⁻¹* t

70.1 years = t

Answer:

True

Explanation:

The percentage by mass of a substance in a solution can be calculated by dividing the mass of the substance dissolved in the solution by the total mass of the solution. This can be expressed mathematically as:

Percentage by mass = mass of substance in solution/mass of solution x 100

In this case;

mass of KBr = 4.34 grams

mass of water = 17.4 grams

mass of solution = mass of KBr + mass of water = 4.34 + 17.4 = 21.74

Percentage by mass of KBr = 4.34/21.74 x 100

                                              = 19.96 %

19.96 is approximately 20%.

Hence, the statement is true.

Answer:

Weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.

Explanation:

The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation:

[tex] pH = pKa + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]  

We have that pH = 10.3 and the Ka is 4.7x10⁻¹¹, so:

[tex] 10.3 = -log(4.7 \cdot 10^{-11}) + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]  

[tex] \frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]  (1)

Also, we know that:

[tex] [Na_{2}CO_{3}] + [NaHCO_{3}] = 0.20 M [/tex]    (2)

From equation (2) we have:

[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] [/tex]   (3)

By entering (3) into (1):

[tex] \frac{0.20 - [NaHCO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]

[tex] 0.94*[NaHCO_{3}] + [NaHCO_{3}] = 0.20 [/tex]

[tex] [NaHCO_{3}] = 0.103 M [/tex]  

Hence, the [Na_{2}CO_{3}] is:

[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] = 0.20 M - 0.103 M = 0.097 M [/tex]  

Now, having the concentrations and knowing the volume of the buffer solution we can find the mass of the sodium carbonate and the sodium bicarbonate, as follows:

[tex]m_{Na_{2}CO_{3}} = C*V*M = 0.097 mol/L*0.050 L*105.99 g/mol = 0.5141 g[/tex]

[tex]m_{NaHCO_{3}} = C*V*M = 0.103 mol/L*0.050 L*84.007 g/mol = 0.4326 g[/tex]

Therefore, to prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3 we need to weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.      

I hope it helps you!

Answer:

1. the siren has a lower pitch to Jane

2. the siren has a higher pitch to John

3. sound different due to moving away from Jane making the sound wave lengths longer and moving toward John making the wave lengths shorter

Explanation:

The Doppler effect expresses that sound is comparative with the spectator or observer. This is demonstrated valid by the model given with Jane and John. To one individual it could sound low and to someone else it could sound high, in light of where they are tuning in from. To John, the police alarm playing is a higher pitch. Be that as it may, to Jane this equivalent alarm is a totally extraordinary pitch and is heard lower than in comparison to the john.

This is a prime case of the Doppler Effect. They sound distinctive on the grounds that the sound is moving far from Jane making the sound frequencies longer and it is advancing toward John making the frequencies shorter. This impacts how the sound is heard by the human ear.

Answer:

7.23 J

Explanation:

Step 1: Given data

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)

Q = 7.23 J

Answer:

PLEASE LOOK INN TO THE FILE YOU WILL GET ANSWER AND ALSO SUMMARY THANKS FOR ASKING QUESTION.

Explanation:

Answer:

2,2,3,3-tetrapropyloxirane

Explanation:

In this case, we have to know first the alkene that will react with the peroxyacid. So:

What do we know about the unknown alkene?

We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If [tex](CH_3CH_2CH_2)_2C=O[/tex] is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.

What is the product with the peroxyacid?

This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)

Answer:

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Explanation:

Hello,

In this case, we can apply the following principles to explain the order:

- The greater the molar mass, the larger the standard molar entropy.

- The greater the molar mass and the structural complexity, the larger the standard molar entropy.

- The greater the structural complexity, the larger the standard molar entropy.

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

This is due to the fact that the greater the molar mass, the larger the standard molar entropy.

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).

Regards.

Answer:12.9e-12g or in short 12.9pg

Explanation:as p=1e-12

Answer:The concentration of the unknown HNO3 solution = 0.01809 M

Explanation:

For the acid-base reaction,  HNO3 + NaOH-----> NaN03 + H20

we have that

C1 V1 = C2 V2

Where ,

C1 = concentration of HNO3=?

V1 = volume of HNO3 = 25.00 mL,

V2 = volume of NaOH = 22.62 mL,  

C2 = concentration of NaOH = 0.02000 M

Therefore ,

25.00 mL x C1 = 22.62 mL x 0.02000 M    

 = (22.62 mL / 25.00 mL) x 0.02000 M = 0.01809 M

The concentration of the unknown HNO3 solution = 0.01809 M

Answer:

2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)

Explanation:

The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)

For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:

Answer:

5000g

Explanation:

mass= density × volume

Since the unit of density here is g/mL, we need to convert the volume to mL.

1L= 1000mL

4L= 4 ×1000 = 4000 mL

Mass of object

= 1.25 ×4000

= 5000g

Answer:

5,000 grams

Explanation:

The mass of an object can be found by multiplying the volume by the density.

mass= volume * density

The density is 1.25 g/mL and the volume is 4 L.

First, we must convert the volume to mL. The density is given in grams per milliliter, but the volume is given in liters.

There are 1,000 mL per L. The volume is 4 L. Therefore, we can multiply 4 and 1,000.

4 * 1,000 = 4,000

The volume is 4,000 mL.

Now, find the mass of the object.

mass= volume * density

volume = 4,000

density= 1.25

mass= 4,000 * 1.25 = 5,000

Add the appropriate units for mass, in this case, grams, or g.

mass= 5,000 g

The mass of the object is 5,000 grams.

Answer:

Pb + 2H2O --> PbO2 + 2H2

Explanation:

Products:

Solid metal; PbO2

Hydrogen; H

Reactants:

Metal; Pb

Steam; H2O

Reactants --> Products

Pb + H2O --> PbO2 + H2

Upon balancing we have;

Pb + 2H2O --> PbO2 + 2H2

Answer:

0.9718M

Explanation:

Rate constant, k =  0.255 M-1s-1

time, t = 8.00 s

Initial concentration, [A]o = 1.33 M

Final concentration, [A] = ?

These quantities are represented by the equation;

1 / [A] = 1 / [A]o + kt

1 / [A] = 1 /1.33 + (0.255 * 8)

1 / [A] = 0.7519 + 2.04

[A] = 1 / 2.7919 = 0.3582 M

How much of NO2 decomposed is obtained from the change in concentration;

Change in concentration = Initial - Final

Change = 1.33 - 0.3582 = 0.9718M

Answer:

8.05 moles

Explanation:

5.50 / 2.954 = x / 4.325

x = 8.05

According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added  so that the balloon expands to 4.325 L.

The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.

In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314  so n=V/R=4.325/8.314=0.520 moles.

Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.

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Answer:

D. The Ca[OH]2 solution may have been unsaturated

Explanation:

The solubility product constant Ksp of any given chemical compound is a term used to describe the equilibrium between a solid and the ions it contains solution. The value of the Ksp indicates the extent to which any compound can dissociate into ions in water. A higher the Ksp, implies more greater solubility of the compound in water.

If the Ksp is more than the value in literature, this false value must have arisen from the fact that the solution was unsaturated hence it appears to be more soluble than it should normally be when saturated.

A 1.89 pints of blood would contain 873 grams of platelets.

To calculate the amount of platelets present in 1.89 pints, it is first necessary to transform this unit of volume into liters:

                                  [tex]1.89 \times 473.2 = 894.3 mL[/tex]

                                         [tex]\frac{894.3}{1000}= 0.84L[/tex]

Now, just calculate the amount of platelets present in 0.84L:

                                    [tex]\frac{1.04\times10^{3}g}{xg}=\frac{1L}{0.84L}[/tex]

                                       x = 873 grams

So, a 1.89 pints of blood would contain 873 grams of platelets.

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Answer:

One carbon atom and the nitrogen atom have nonzero formal charges.

Explanation:

The compound Acetonitrile has sixteen valence electrons as is easily San from its structure. It contains a carbon nitrogen triple bond with a lone pair of electrons on nitrogen. All atoms satisfy the octet rule and there is no hyper valent atom in the molecule.

The formal charge an carbon and nitrogen is calculated as follows;

No. of valence electron on atom - [non bonded electrons + no. of bonds]

Therefore, for carbon and nitrogen, we have;

formal charge on carbon = 4 - (0 + 4) = 0

formal charge on nitrogen = 5 - (2 + 3) = 0

Hence carbon and nitrogen both possess zero formal charges.