the ram of a pile driver drops onto the top of an iron beam, driving it partway into the ground. the distance that the beam sinks into the ground depends on the

While out camping one spring night,  the moon looks orange and the possible explanation is that the dust in the atmosphere makes the moon appear orange which is denoted as option D.

This is Earth's only natural satellite and it is referred to as a large round object that circles the Earth and that shines at night by reflecting light from the sun.

In some areas, the atmosphere can be filled with air pollution, dust, and even smoke from wildfires which leads to the particles scattering light in the same way described above, leading to an orange or red Moon high in the sky.

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Nonpolar molecules are generally the least soluble in water. Option B is correct.

Water is a polar molecule, meaning that it has a partial positive charge on one end and a partial negative charge on the other end. This polarity allows water to interact with other polar molecules, such as other water molecules or polar solutes, through hydrogen bonding.

Nonpolar molecules, on the other hand, do not have regions of partial positive and negative charges and are not able to form hydrogen bonds with water molecules. As a result, they are generally not soluble in water. Examples of nonpolar molecules include hydrocarbons, such as oil or grease.

Ionic compounds, polar molecules, and hydrophilic compounds are all more likely to be soluble in water than nonpolar molecules because they can interact with water molecules through ionic interactions, dipole-dipole interactions, or hydrogen bonding.

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--The given question is incomplete, the complete question is

"Which of the following is least soluble in water? group of answer choices A) ionic compounds B) nonpolar molecules C) hydrophilic compounds D)polar molecules"--

The atomic number 46 and the letter Pd serve as the symbol for the chemical element palladium.

This rare and brilliant silvery-white metal was first discovered in 1803 by the English chemist William Hyde Wollaston. In homage to the asteroid Pallas, which was given that name by the Greek goddess Athena after killing Pallas he gave it the name Pallas. The elements palladium, platinum, rhodium, ruthenium, iridium, and osmium are all included in the platinum group metals, which represent a group of different materials (PGMs). Palladium is the least dense and has the lowest melting point of the group despite having similar chemical properties.

The majority of palladium is used in catalytic converters for automobiles. It is also used in many jewelry pieces as well as dental crowns and fillings. Gold that has undergone decolorization through alloying with another metal, occasionally palladium, is known as white gold.

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The average reaction rate between 0 and 1500 s is -0.000105 M/s and  the average reaction rate between 500 and 1200 s is -0.0000571 M/s

Part A: To calculate the average reaction rate, we can use the formula:

Average rate = (change in concentration)/(change in time)

The change in concentration of A is [A]t - [A]0, and the change in time is t - 0. Using the given data, we get:

Average rate = ([A]1500 - [A]0)/(1500 - 0) = (0.016 - 0.184)/(1500) = -0.000105 M/s

Therefore, the average reaction rate between 0 and 1500 s is -0.000105 M/s.

Part B: Following the same formula as above, we can calculate the average rate between 500 and 1200 s using the given data:

Average rate = ([A]1200 - [A]500)/(1200 - 500) = (0.031 - 0.069)/(700) = -0.0000571 M/s

Therefore, the average reaction rate between 500 and 1200 s is -0.0000571 M/s.

Part C: The instantaneous rate at t=800 s can be determined by drawing a tangent to the concentration vs. time plot at that point and finding its slope.

From the given data, we can plot [A] vs. time and draw a tangent at t=800 s

The slope of this tangent gives us the instantaneous rate at that point. Using a straight edge or ruler, we can estimate the slope to be approximately -0.0000575 M/s.

Therefore, the instantaneous rate of the reaction at t=800 s is approximately -0.0000575 M/s.

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Rounding to the correct number of significant figures, the molarity of the HCl solution is 0.150 mol/L.

The balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and hydrochloric acid (HCl) is:

Na2CO3 + 2 HCl → 2 NaCl + H2O + CO2 :From the equation, we can see that 2 moles of HCl are required to react with 1 mole of Na2CO3.

First, we need to determine the number of moles of HCl used in the titration:

moles of HCl = volume of HCl (in L) × molarity of HCl

We know the volume of HCl used in the titration is 13.23 mL, or 0.01323 L. We need to find the molarity of the HCl solution to determine the number of moles of HCl used.

To do this, we need to know the mass of Na2CO3 used in the titration and the balanced chemical equation to find the number of moles of Na2CO3:

molar mass of Na2CO3 = 2 × atomic mass of Na + atomic mass of C + 3 × atomic mass of O

= 2 × 22.99 g/mol + 12.01 g/mol + 3 × 16.00 g/mol

= 105.99 g/mol

moles of Na2CO3 = mass of Na2CO3 / molar mass of Na2CO3

= 0.1052 g / 105.99 g/mol

= 0.000992 mol

Since 2 moles of HCl are required to react with 1 mole of Na2CO3, the number of moles of HCl used in the titration is:

moles of HCl = 2 × moles of Na2CO3

= 2 × 0.000992 mol

= 0.001984 mol

Now we can calculate the molarity of the HCl solution:

molarity of HCl = moles of HCl / volume of HCl (in L)

= 0.001984 mol / 0.01323 L

= 0.1499 mol/L

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It is true that the substance with hydrogen-bonded molecules will have a greater boiling point than the substance with non-hydrogen-bonded molecules.

Hydrogen bonds can form in any molecule where an oxygen or nitrogen atom is directly connected to a hydrogen atom. Despite the absence of the HF group in other compounds, hydrogen bonds can still form when hydrogen is linked to fluorine.

An electronegative atom from another molecule, such as nitrogen, oxygen, or fluorine, interacts with a hydrogen atom to form a hydrogen bond. To establish the link, the hydrogen has to build a covalent bond with an additional electronegative atom.

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Only ethanol is among the possibilities given that can dissolve in water. This is because ethanol has an alcoholic functional group that will create a hydrogen bond with the proton in water.

Underground is where one can find the sole free hydrogen on earth. Stars and gas giant planets are where you'll find the majority of hydrogen. Hydrogen atoms are changed into helium atoms in the core of stars due to the intense pressure there.

Incorrect handling of the highly flammable gas utilized in fuel cells, hydrogen, can result in flames and explosions. A worker should always assume that a flame is present if they suspect a hydrogen leak because hydrogen fires are undetectable.

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A vapour pressure of the a fluid at a given temperature is determined by the pressure that the saturated vapour exerts on the liquid surface. The phases of liquid and vapour are in balance.

The equilibrium constants do not change even when the system's pressure varies. Only a temperature change can modify an equilibrium constant. The equilibrium position may change if the pressure is changed.

The equilibrium constant is temperature-dependent and unaffected by the precise ratios of reactants to products, either presence of a catalyst, or the presence of inert substances. Additionally, it is unaffected by the volumes, pressures, and concentrations of the reactants and products.

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The correct question is

A)It remains constant when the temperature increases.

B) It decreases to half its original value if the volume of the gas phase is doubled...

✓ C) It is independent of the volume of the vapor phase.

The quantity of the product obtained from the reaction is generally expressed in terms of the yield of the reaction. The amount obtained actually is called the actual yield.

The amount of the product which is predicted by the stoichiometry is called the theoretical yield. A reaction yield is reported as the percentage of the theoretical amount.

The percentage yield is the ratio of actual yield to the theoretical yield.

% yield = Actual yield / Theoretical yield × 100

a. Molar mass of PbCO₃ = 267.21 g/mol

Molar mass of PbCl₂ = 278.1 g/mol

2.871 g of PbCO₃ gives:

2.871 × 278.1 / 267.21 = 2.988 g  PbCl₂

b. % yield = 2.385 / 2.988 × 100 = 79.82 %

Thus the theoretical yield is 2.988 g and the % yield is 79.82 %.

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The volume of the butane starting material is 24500 mL, or 24.5 liters.

We use the balanced chemical equation for the combustion of a hydrocarbon to determine the number of moles of the hydrocarbon that were used in the experiment:

Hydrocarbon + O₂ → CO₂ + H₂O

We know that 6 moles of CO₂ and 6 moles of H₂O were produced. This means that the hydrocarbon contained 6 carbon atoms  (since each mole of CO₂ produced in the reaction comes from one carbon atom in the hydrocarbon).

Since we also know that the hydrocarbon has less than 5 carbons, this means that it must be either propane (C₃H₈) or butane (C₄H₁₀).

To determine which of these two hydrocarbons was used, we can calculate the theoretical yield of CO₂ for each one and compare it to the actual yield of 6 moles.

For propane, the theoretical yield of CO₂ would be 3 moles, while for butane, it would be 4 moles. Since the actual yield was 6 moles, it is more likely that butane was the hydrocarbon used.

To calculate the volume of the butane starting material, we need to use the ideal gas law:

PV = nRT

Rearranging the equation, we get:

V = nRT/P = (1 mol)(0.0821 L·atm/mol·K)(298 K)/(1 atm) = 24.5 L

Since the volume is in milliliters (mL), we can convert:

24.5 L x 1000 mL/L = 24500 mL

Therefore, the volume of the butane starting material is 24500 mL, or 24.5 liters.

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The electron changes from a higher energy level to a lower energy level, and a photon is emitted.

What is electrons?

Electrons are the smallest and most fundamental particles that make up matter. They are negatively charged particles that exist in the orbits of atoms and molecules and that participate in chemical reactions. Electrons are found in all atoms, and determine the chemical properties of the atom. They are also responsible for electricity and magnetism, and can be used to create electrical current in circuits. Electrons have a very small mass and move around the nucleus of an atom very quickly. In addition, electrons can be excited by certain energies, causing them to move to higher energy levels, where they can then be used to create electrical current.

Therefore, The electron changes from a higher energy level to a lower energy level, and a photon is emitted.

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The equation for net potential energy between two adjacent ions, En, is given by:

En = A/r^n + B/r^n

To calculate the bonding energy E0 in terms of the parameters A, B, and n, we can follow the following procedure:

1)Since the curve of En vs r has a minimum at E0, differentiate En with respect to r, and then put the resultant expression equal to zero. This results in:

dEn/dr = -An/r^(n+1) - Bn/r^(n+1) = 0

2) In order to determine no, the equilibrium interionic spacing, one must solve for r in terms of A, B, and n. This results in:

r0 = (An/Bn)^1/n

3) Determine the expression for E0 by substitution of ro into Equation. This gives us:

E0 = A/r0^n + B/r0^n

= A(Bn/An)^n/n + B(An/Bn)^n/n

= (An^n + Bn^n)/n

a) In order to superimpose the three energies on a single plot versus the distance r up to 1.0 nm, we must first calculate each of the energies at each given distance. The attractive energy EA is given by the equation EA = -5.86 x 10-6 p9 ED. The repulsive energy Er is given by Er = 1.436 EA. The net energy En is the sum of the two energies, En = EA + Er.

We can then plot the three energies on a single graph, with the distance r on the x-axis and the energies on the y-axis.

b) From the graph, we can determine the equilibrium spacing r0 and the magnitude of the bonding energy E, between the two ions. The equilibrium spacing r0 is the distance at which the net energy En is at a minimum, which is approximately 0.45 nm. The magnitude of the bonding energy E, is the difference between the net energy at the equilibrium spacing r0 and the repulsive energy at that distance. In this case, the magnitude of the bonding energy E, is approximately 0.2 eV/K+-Cl pair.

c) We can also mathematically determine the r0 and E, values using the solutions to Problem 2. For the equilibrium spacing r0, we can solve the equation En(r) = 0 for r, which yields r0 = 0.45 nm. For the magnitude of the bonding energy E, we can subtract the repulsive energy Er at the equilibrium spacing r0 from the net energy En at that distance, which yields E, = 0.2 eV/K+-Cl pair. This result matches the value we obtained graphically in part b.

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To determine if the dew point will be reached when the temperature drops, we need to calculate the dew point temperature at the current relative humidity and temperature.

Using the table, we can find that the amount of water vapor needed for saturation at 35°C is 42.1 g/m3. To calculate the dew point temperature, we can use the following formula:

dew point temperature = (243.12 × ln(RH/100) + 17.62 × T / (243.12 + T)) / (17.62 - ln(RH/100) - 243.12 × T / (243.12 + T))

where RH is the relative humidity (80 percent) and T is the temperature in degrees Celsius (35°C).

Plugging in the values, we get:

dew point temperature = (243.12 × ln(80/100) + 17.62 × 35 / (243.12 + 35)) / (17.62 - ln(80/100) - 243.12 × 35 / (243.12 + 35))

dew point temperature = 29.9°C

This means that the air would have to cool to 29.9°C in order for the dew point to be reached and condensation to occur. Since the temperature is only dropping to 25°C later in the evening, the dew point will not be reached and condensation will not occur.

The concentration of HI present at equilibrium is 0.516 M when 0.500 mol of H2 and 0.500 mol of I2 were placed in a 1.00-L container to react.

The balanced chemical equation for the reaction is:

H2 (g) + I2 (g) ⇌ 2HI (g)

The initial molar concentrations are:

[H2] = 0.500 mol/1.00 L = 0.500 M

[I2] = 0.500 mol/1.00 L = 0.500 M

[HI] = 0 M (since there is no HI initially)

Let x be the change in concentration of H2, I2 and HI at equilibrium. Since 2 mol of HI is formed for every 1 mol of H2 or I2 reacted, the equilibrium concentrations can be expressed as:

[H2] = 0.500 - x

[I2] = 0.500 - x

[HI] = 2x

Substitute these expressions into the equilibrium constant expression:

[tex]Kc = [HI]^2 / ([H2] *[I2])[/tex]

[tex]53.3 = (2x)^2 / ((0.500 - x) * (0.500 - x))[/tex]

Solving for x using quadratic formula gives x = 0.258 M.

Therefore, the concentration of HI at equilibrium is:

[HI] = 2x = 2(0.258) = 0.516 M.

So the concentration of HI present at equilibrium is 0.516 M.

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The Part II spectra for Base/Phenolphthalein shows two distinct absorption peaks at 420 nm and 540 nm. These peaks are consistent with the form of the absorption spectrum obtained in Part I of the experiment.

In Part I, we added HCl to the basic solution of phenolphthalein, causing the solution to change from pink to colorless. The color change occurred due to the dissociation of the phenolphthalein molecule into its ionized forms. The ionized forms of phenolphthalein have different absorption spectra than the neutral form. The neutral form absorbs light at a different wavelength than the ionized forms, resulting in a change in the color of the solution. The absorption peaks observed in Part II correspond to the ionized forms of phenolphthalein. The peak at 420 nm is due to the absorption of the hydrogen-bonded form of the ionized molecule, while the peak at 540 nm is due to the non-hydrogen-bonded form. These peaks are consistent with the color changes observed in Part I, as the ionized forms of phenolphthalein have different colors than the neutral form.

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The molarity of p after 10.677 minutes is approximately 1.950 M.

First-order kinetics is a type of reaction kinetics where the rate of the reaction depends on the concentration of only one reactant, raised to the power of one.

If the reaction has first-order kinetics, the rate of the reaction can be described by the following equation:

rate = k[A]

Where

The integrated rate law for a first-order reaction is:

ln([A]t/[A]0) = -kt

Where

To find the concentration of p after 10.677 minutes, we can plug in the values we have:

ln([A]t/[A]0) = -kt

ln([p]t/3.489 M) = -(0.00151 s^-1)(10.677 min)(60 s/min)

ln([p]t/3.489 M) = -1.026

Taking the exponential of both sides gives:

[p]t/3.489 M = e^-1.026

[p]t = (e^-1.026)(3.489 M) = 1.950 M

Therefore, the molarity of p after 10.677 minutes is approximately 1.950 M.

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The differences in electronic properties between azulene and naphthalene can be ascribed to differences in their resonance structures, which influence and charge distribution in the molecule.

Azulene, on the other hand, has an additional ring fused to one of the rings in naphthalene, resulting in a 7-membered ring and a shift in the position of double bonds in the molecule.

 :            :

H2C=CH-      -CH=CH2

  ||   \    /   ||

H2C=CH      CH=CH2

   :     \ /     :

   :      C      :

   :    // \\    :

   :   /    \    :

   :  CH    CH  :

   :   \    /    :

   :    \\ //    :

   :      C      :

   :     / \     :

H2C=CH      CH=CH2

  ||   /    \   ||

H2C=CH-      -CH=CH2

   :            :

One of the azulene resonance structures has one carbon atom with a positive charge and one nitrogen element with a negative charge. Because the double bonds in naphthalene are equally distributed between the two rings.

:            :

H2C=CH-      -CH=CH2

  ||   \    /   ||

H2C=CH      CH=CH2

   :      ||      :

   :      ||      :

   :      C      :

   :      ||      :

   :      C     :

   :      ||      :

H2C=CH-CH=CH-CH=CH2

   :      ||      :

   :      C     :

   :      ||      :

H2C=CH-CH=CH-CH=CH2

  ||   /    \   ||

H2C=CH      CH=CH2

   :            :

This results in the formation of a very strong dipole in the molecule, which adds to its bright blue color. Naphthalene, on the other hand, has two resonance patterns.

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Scottish physician and chemist William Cullen introduced bonding brackets and  affinity arrows to enable chemists to pictorially represent affinity reactions, and develop chemical reaction equations.

William Cullen was a Scottish physician and chemist who lived from 1710 to 1790. He made significant contributions to the field of chemistry, including the development of the theory of "affinity," which helped explain chemical reactions.

Cullen's ideas were important precursors to the modern understanding of chemical bonding and reactions. His use of arrows to represent affinity is similar to the modern use of arrows to represent electron movement in chemical reactions. His use of brackets to represent the structure of molecules is also similar to modern notation.

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The major organic product of the reaction is a disubstituted dihalomethane, with a structure of [tex]CH3-CH(CH3)-CH-I-CH3[/tex]. This compound is optically inactive, and thus a meso compound.

The reaction of [tex]CH3-CH(CH3)-CH=CH2[/tex] with [tex]CH2I2[/tex] upon reacting with Zn(Cu) produces a meso compound, meaning that the molecule contains two or more identical groups attached to opposite sides of a stereocenter. Stereochemistry is not applicable here, so the structure does not need to be drawn with wedge/hash bonds. The reaction produces a racemic mixture, which means that the product is an equal mixture of two stereoisomers that are mirror images of each other. As such, since only one stereoisomer needs to be drawn, the structure of [tex]CH3-CH(CH3)-CH-I-CH3[/tex] is sufficient to represent the major organic product.

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This radiation can be emitted in the form of positively charged alpha particles, negatively charged beta particles, gamma rays, or x-rays, as explained below.

Gamma rays are a type of electromagnetic radiation that results from a redistribution of electric charge within a nucleus. Gamma rays are essentially very energetic X rays ; the distinction between the two is not based on their intrinsic nature but rather on their origins.

chemist Paul Villard Gamma radiation is one of the three types of natural radioactivity discovered by Becquerel in 1896. Gamma rays were first observed in 1900 by the French chemist Paul Villard when he was investigating radiation from radium .

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Same number of protons but different numbers of neutrons is the correct answer.

What are isotopes ?

Isotopes are variants of a chemical element that have the same number of protons in their atomic nucleus but differ in the number of neutrons. Isotopes of the same element have the same atomic number, but different atomic masses. For example, carbon-12, carbon-13, and carbon-14 are three isotopes of the element carbon.Isotopes can have different physical and chemical properties due to differences in their atomic mass. For example, isotopes of an element can have different boiling points, melting points, and densities. This can be important in a wide range of fields, including chemistry, geology, and physics.Some isotopes are unstable and undergo radioactive decay, which can result in the emission of alpha particles, beta particles, or gamma rays. This property of certain isotopes has important applications in nuclear medicine, nuclear power generation, and radiocarbon dating.

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Although cannabinol and methyl alcohol are both alcohols, cannabinol is only slightly soluble in methyl alcohol at room temperature due to the reason that methyl alcohol has a very short chain of carbon  while cannabinol is a pure hydrocarbon due to the reason that it has a hydroxyl functional group.

Generally, solubility is defined as the maximum amount of a substance that will dissolve in a given amount of solvent at a specified temperature. Solubility is a characteristic property of a specific solute–solvent combination, and different substances have greatly differing solubilities.

Cannabinol and methyl alcohol are both alcohols, cannabinol is only slightly soluble in methyl alcohol at room temperature due to the reason that methyl alcohol has a very short chain of carbon while cannabinol is a pure hydrocarbon due to the reason that it has a hydroxyl functional group.

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A satellite terminates one revolution of a planet in one hour. At the end of one hour, the satellite has moved 2.0 x 10⁷ meters and is only 10 meters away from its starting point. After one hour, the average velocity of the satellite is 2.7 × 10 ⁻² m/s.

Alteration in position divided by the time of travel defines the average velocity.

Given:

Distance = 2.0 x 10⁷ meters

Time = 1 hour

Average speed is calculated as:

Average speed = distance covered by the object ÷ Time to cover the distance

= 2.0 x 10⁷ ÷ 1

= 2.0 x 10⁷ meters ÷ hour

= 5,555 and 5/9 meters/second.

Now,

The term velocity is defined as the ratio of displacement to the time for the displacement.

Velocity = Displacement / Time

= 10 meters/hour

= 0.002777 meters/second

= 2.7 × 10 ⁻² m/s

Thus, 2.7 × 10 ⁻² m/s is the satellite's average velocity after one hour.

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An element is a substance composed of atoms with the same number of protons in the nucleus and cannot be broken down into simpler substances by chemical means.

Elements are the building blocks of matter and are represented by unique symbols on the periodic table. Each element has a unique set of properties, such as atomic number, atomic mass, and electron configuration. Elements can combine to form compounds, which are made up of two or more different elements.

The study of elements and their properties is an important part of chemistry, and understanding the behavior of elements is critical for understanding chemical reactions and the properties of different materials.

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If sodium-24 has a half-life of 14.96 hours then rate constant K is 0.0463 hr²−1 Rounded to two significant figures, the answer is (a) 4.63 × 10²−2 hr²−1.

The first-order rate equation is given by:

ln(N_t/N_0) = -kt

where N_0 is the initial amount of the substance, N_t is the amount of substance at time t, k is the rate constant, and ln is the natural logarithm.

The half-life of sodium-24 is given as 14.96 hours, which means that after 14.96 hours, the amount of sodium-24 will be reduced to half of its initial amount. Therefore, we can write:

N_t = (1/2)N_0

Substituting this into the first-order rate equation, we get:

ln(1/2) = -k(14.96)

Solving for k, we get:

k = [ln(1/2)]/14.96

k ≈ 0.0463 hr²−1

Therefore, the rate constant for the radioactive decay of sodium-24 is approximately 0.0463 hr²−1. Rounded to two significant figures, the answer is (a) 4.63 × 10²−2 hr²−1.

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Answer:

Explanation:

 240          4              236

Pu   ---->        He   +        U

  94            2                 92

Two observations that can indicate that an organic liquid is dry when using the drying agent are:

Organic liquid is liquid that contains volatile organic compounds (often shorten to VOCs), such as crude oils and petroleum distillates. It is used in many stuff, ranging from paints, lacquers, glues, and adhesives; even in the production of dyes, textiles, and pharmaceuticals.

Drying agents are things that absorb moisture from their vicinity. They can be used to remove trace amounts of water from an organic solution. Some examples of drying agents are sodium metal or calcium hydride.

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The reaction of diethyl ether and oxygen to form a hydroperoxide can be represented as follows:

O2 + C2H5–O–C2H5 → C2H5–O–O–C2H5 + H2O

The curved fishhook arrows show the movement of electrons in this step of the reaction mechanism. The initiation of the reaction is by a radical, specifically a radical of the ether molecule. This radical abstracts an electron from the oxygen molecule, forming a new radical which reacts with the ether molecule, resulting in the formation of the hydroperoxide molecule and the release of water. The arrows indicate the movement of electrons from the oxygen molecule to the ether molecule.

O2 → •O2

•O2 + C2H5–O–C2H5 → C2H5–O–O–C2H5 + H2O

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For the given chemical equation, 8 grams of HNO₃ will produce 0.76 grams of water.

The chemical reaction is:

S + 6 HNO₃ → H₂SO₄ + 6 NO₂ + 2 H₂O

From the table we know that:

molar mass H = 1

molar mass N = 14

molar mass O = 16

Molar mass of water (H₂O) = 2 x molar mass H + molar mass O

                                            = 2 + 16 = 18 gram/mole

Molar mass of HNO₃ = 1 + 14 + 3 x 16 =  63 gram/mole

8 gram HNO₃  is equivalent to 8/63 moles

6 moles HNO₃ produce 2 moles water

8/63 moles HNO₃ will produce (2/6) x (8/63) moles water

(2/6) x (8/63) moles H₂O is equivalent to (2/6) x (8/63) x 18 gram, which equals 0.76 gram water.

Your question is incomplete, most likely it was:

S + 6 HNO₃ → H₂SO₄ + 6 NO₂ + 2 H₂O

In the above equation how many grams of water can be made when 8 grams of hno3 are consumed? use the following molar masses

Element   Molar Mass

Hydrogen 1

Nitrogen        14

Sulfur        32

Oxygen         16

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