If 83.6 grams of H2 and 257 grams of N2 react, how many grams of ammonia will be produced?

True. For these before leaving the ER undergo a process called N-terminal signal peptide cleavage. During this process, the signal peptide, which is an amino acid sequence that helps guide the protein into the ER, is cleaved, resulting in a smaller polypeptide.

Polypeptides are composed of a variety of different amino acids that are linked together in a sequence. This sequence of amino acids determines the structure and function of the protein that they make up. Polypeptides are essential components of cells, playing a role in many essential processes, including:

They are found in the cytoplasm and nucleus of cells, as well as in extracellular fluid. Polypeptides are also important components of:

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The energy of an electron in a hydrogen atom with an orbit of n = 3 is -5.45 x 10⁻¹⁹ J.

To calculate the energy of an electron in a hydrogen atom with an orbit of n=3, we know that the value of k is given as k = 2.18 × 10⁻¹⁸ J. We can use the Rydberg formula to calculate the energy of an electron in the hydrogen atom. The Rydberg formula states that:

1/wavelength = R(1/n1² - 1/n2²)

where R is the Rydberg constant, which is equal to 1.097 x 10⁷ m⁻¹. We can use the formula E = hν to calculate the energy of a photon with frequency ν. Where h is the Planck constant, which is equal to 6.626 x 10⁻³⁴ J s.

The energy of an electron in the hydrogen atom can be calculated using the formula

E = -Rh/n²

where Rh is the Rydberg energy, which is equal to 2.18 x 10⁻¹⁸ J, and n is the principal quantum number. The negative sign indicates that the electron is bound to the nucleus.

Substituting n = 3 and Rh = 2.18 x 10⁻¹⁸ J into the formula gives:

E = - Rh/n²

= - 2.18 × 10⁻¹⁸ J / 3²

= - 5.45 x 10⁻¹⁹ J

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The empirical formula of the compound C₄H₁₀O₁.

The empirical method of a chemical compound is the simplest entire range ratio of atoms present in a compound. The empirical formula of a compound is the most effective whole variety ratio of atoms of each detail in the compound.

Step 1: decide the masses. Step 2: decide the variety of moles by dividing the grams by means of the atomic mass. Step 3: Divide the wide variety of moles of every detail through the smallest quantity of moles. Step 4: Convert numbers to complete numbers.

                                         divide by smallest

64.9% C /12 = 5.4             5.4/ 13.5  = 4

13.5% / 1 H  = 13.5             13.5 /1.35  = 10

21.6% / 16 O  = 1.35          1.5/1.5 = 1

Empherical formula = C₄H₁₀O₁

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Answer: if we lost our core it would be an entire fallout like fallout 76 we would have no way to protect ourselves from the radiation of space except for a select few that actually own astronaut suits tho we would not turn into monsters, sadly :( I wanna be a ghoul

The wavelength associated with the photon is 329 nanometers.

To determine the energy, frequency, and wavelength associated with the excitation of an electron from level n=2 to n=6 in a hydrogen-like ion (with three protons, two neutrons, and one electron), we can use the formula for the energy of a photon:

E = -13.6 eV/n^2

where E is the energy in electron volts (eV) and n is the principal quantum number.

(i) Energy of the photon:

To find the energy of the photon, we calculate the energy difference between the two levels:

ΔE = E_final - E_initial

= (-13.6 eV/6^2) - (-13.6 eV/2^2)

= (-13.6 eV/36) - (-13.6 eV/4)

= -0.3778 eV

The energy of the photon required to cause excitation is 0.3778 eV.

(ii) Frequency associated with the photon:

The energy of a photon can be related to its frequency (ν) using the equation:

E = hν

where h is Planck's constant (4.1357 x 10^-15 eV·s).

Substituting the values, we can solve for the frequency:

0.3778 eV = (4.1357 x 10^-15 eV·s) ν

ν = (0.3778 eV) / (4.1357 x 10^-15 eV·s)

ν ≈ 9.125 x 10^14 Hz

The frequency associated with the photon is approximately 9.125 x 10^14 Hz.

(iii) Wavelength associated with the photon:

The wavelength (λ) of the photon can be determined using the equation: c = νλ

where c is the speed of light (3.0 x 10^8 m/s).

Substituting the values, we can solve for the wavelength:

(3.0 x 10^8 m/s) = (9.125 x 10^14 Hz) λ

λ = (3.0 x 10^8 m/s) / (9.125 x 10^14 Hz)

λ ≈ 3.29 x 10^-7 m

Converting the wavelength to nanometers:

λ ≈ 329 nm

The wavelength associated with the photon is approximately 329 nanometers.

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A and B are correct. Crubbers are expensive flue gas desulfurization devices used to remove 90% or more of SO2 pollution from industrial emissions.

Crubbers are indeed flue gas desulfurization devices used as pollution control equipment to remove SO2 (sulfur dioxide) from industrial exhaust gases. They are designed to reduce SO2 emissions to levels that are 90 percent or more below baseline levels. The term "baseline levels" refers to the initial levels of SO2 emissions before the implementation of the flue gas desulfurization system.

Crubbers, or flue gas desulfurization systems, work by utilizing various chemical processes to react with and remove sulfur dioxide from the flue gas. This helps mitigate the harmful effects of SO2 emissions on the environment and human health. However, it's important to note that while rubbers are effective in reducing SO2 pollution, they can be expensive equipment to install and maintain due to their complex design and operation. Therefore, both options A and B, which state that clubbers are flue gas desulfurization devices and expensive equipment, are correct.

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The reaction of 1° or 2° alcohols with PBr₃ or SOCl₂/N(C₂H₅)₃ leads to the formation of alkyl bromides or alkyl chlorides, respectively, through the corresponding SN₂ reaction mechanism.

The reaction mechanism involves the formation of an oxonium or ammonium intermediate followed by the attack of the halide ion on the carbon atom bearing the hydroxyl group.

In the case of PBr₃, the reaction proceeds as follows:

R-OH + PBr₃ → R-Br + H₃PO₃

In the case of SOCl₂/N(C₂H₅)₃, the reaction proceeds as follows:

R-OH + SOCl₂/N(C₂H₅)₃ → R-Cl + SO₂/N(C₂H₅)₃H

where R is an alkyl group.

It is worth noting that the reaction with PBr₃ is more efficient and selective than with SOCl₂/N(C₂H₅)₃. Additionally, 1° alcohols react faster than 2° alcohols in both cases.

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Answer:

angular momentum

Explanation:

The hybridization schemes corresponding to each electron geometry:

a. Linear: The hybridization scheme for linear electron geometry is sp hybridization.
b. Trigonal planar: The hybridization scheme for trigonal planar electron geometry is sp2 hybridization.
c. Tetrahedral: The hybridization scheme for tetrahedral electron geometry is sp3 hybridization.
d. Trigonal bipyramidal: The hybridization scheme for trigonal bipyramidal electron geometry is sp3d hybridization.
e. Octahedral: The hybridization scheme for octahedral electron geometry is sp3d2 hybridization.

In each case, the hybridization scheme is determined by the combination of s, p, and d orbitals required to accommodate the electron geometry.

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a. Linear: sp, b. Trigonal planar: sp², c. Tetrahedral: sp³, d. Trigonal bipyramidal: sp³d, e. Octahedral: sp³d². These hybridization schemes describe the arrangement of orbitals around the central atom in each respective electron geometry.

a. The hybridization scheme for a linear electron geometry is sp.

b. The hybridization scheme for a trigonal planar electron geometry is sp².

c. The hybridization scheme for a tetrahedral electron geometry is sp³.

d. The hybridization scheme for a trigonal bipyramidal electron geometry is sp³d.

e. The hybridization scheme for an octahedral electron geometry is sp³d².

In the case of linear electron geometry (a), the central atom is surrounded by two electron groups, resulting in a linear arrangement. The atom undergoes sp hybridization, where one s orbital and one p orbital hybridize to form two sp hybrid orbitals.

For trigonal planar electron geometry (b), the central atom is surrounded by three electron groups, forming a planar arrangement. The atom undergoes sp² hybridization, where one s orbital and two p orbitals hybridize to form three sp² hybrid orbitals.

In tetrahedral electron geometry (c), the central atom is surrounded by four electron groups, resulting in a three-dimensional arrangement. The atom undergoes sp³ hybridization, where one s orbital and three p orbitals hybridize to form four sp³ hybrid orbitals.

For trigonal bipyramidal electron geometry (d), the central atom is surrounded by five electron groups, forming a complex arrangement. The atom undergoes sp³d hybridization, where one s orbital, three p orbitals, and one d orbital hybridize to form five sp³d hybrid orbitals.

In octahedral electron geometry (e), the central atom is surrounded by six electron groups, resulting in a symmetrical arrangement. The atom undergoes sp³d² hybridization, where one s orbital, three p orbitals, and two d orbitals hybridize to form six sp³d² hybrid orbitals.

Therefore, a. Linear: sp, b. Trigonal planar: sp², c. Tetrahedral: sp³, d. Trigonal bipyramidal: sp³d, e. Octahedral: sp³d². These hybridizations correspond to the electron geometries and describe the arrangement of orbitals around the central atom.

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Answer:

Carbon

Explanation:

Steel is an alloy of iron and carbon. Stainless steels are steels containing at least 10.5% chromium, less than 1.2% carbon and other alloying elements

Answer: B

Explanation:Steel is an alloy of iron and carbon.

Answer: 12

Explanation: 18mL 14L(1.6x10-2) 23ml 12L

1 Milliliter (mL) is equal to 0.001 liter (L)

1.6x10-2= 14  you see my point?

18 and 23 are both mL so bye bye 14 bigger then 12 so there

According to the question the q(gas) in J for this compression process is 0J.

Gas is a state of matter in which particles are spread out and have enough energy to move around freely. Gas is composed of molecules in constant motion and takes the shape and volume of its container. Gas can be either naturally occurring or man-made and is found in the atmosphere. Examples of naturally occurring gases include oxygen, nitrogen, and carbon dioxide. Man-made gases include helium, chlorine, and hydrogen. Gas is often used as a source of energy and is burned to produce heat, which can be used to power machines and vehicles. Gas is also used in many industries, such as in the production of chemicals and plastics.

In this case, n = 0.100 moles,
\(C_v\) = (3/2)R = (3/2)(8.314 J/mol K) = 12.471 J/mol K, and
T₁ = 298.0 K,
T2 = 298.0 K.
Therefore, q(gas)
= nCv (T₂- T₁)
= 0.100 mol × 12.471 J/mol K × (298.0 K - 298.0 K)
= 0 J.

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Answer:

they stayed alive?

Explanation:

easy

Answer:

yur moms been too mars

Explanation:

The process of photosynthesis requires water and air in addition to grow ,obtain their energy from the sun  and supply oxygen to the environment .

It is defined as a process by which plants and other photosynthetic organisms convert the light energy in to chemical energy  through the process of cellular respiration.

Some of the energy which is converted  is stored in molecules of carbohydrates like sugar and starches  which are made up of from carbon dioxide and water . Photosynthetic organisms which can perform photosynthesis  are algae and cyanobacteria. Photosynthesis is largely responsible for producing and maintaining the content of oxygen in earth's atmosphere.

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A complex is a compound that is composed of a central metal atom/ion and then ligands.   MnO4- is a d^0 specie hence it has no d electron.

A complex is a compound that is composed of a central metal atom/ion and then ligands. These ligands may lead to the color of the complex. However, the color of complexes more commonly result from the d-orbital transition in the metal atom/ion.

Now we know that the neutral electron configuration of Mn is [Ar]. 3d5. 4s2. It then follows that MnO4- is a d^0 specie hence it has no d electron.

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The correct answer is B. Metalloids

Explanation:

In this periodic table, the green boxes include elements such as Boron (B), Arsenic (As), and Antimony (Sb). All these elements are known in chemistry as semimetals or metalloids. This classification is because these elements cannot be completely classified into metals or non-metals. Moreover, they all display similar properties including some level of conductivity, malleability, and solid-state. According to this, the correct answer is B.

Answer:

B. Metalloids

Explanation:

In 1911, astronomers Hertzsprung and Russell independently plotted the spectral class (temperature) against the luminosity (energy emited) for known stars. They discovered that the stars were grouped together in different regions on the graph. This graph is now known as the Hertzsprung- Russell Diagram (see attached image).

The HR diagram shows us that there are different types of stars and that stars evolve in different ways depending on their initial mass. This can tell us what reactions are occurring in the stars' cores.

Stars follow a distinct path as seen on the diagram, and evolve in the following ways:

- Main Sequence > red giant > planetary nebula > white dwarf

- Main Sequence > supergiant > supernova > blackhole/neutron star

- Main Sequence > white dwarf

Temperature: surface temperature of stars

Absolute magnitude: measure of the luminosity or brightness of a star

Luminosity: the relative magnitude, relative to the magnitude of our sun

Spectral Class: temperature group of stars. categorised into OBAFGKM. Stars on the left the the hottest.

Main Sequence: Majority of stars lie in the main sequence, including our sun. These stars are fusing hydrogen to helium in their cores.

Red giants/Super giants: consists of a small minority of stars found at the top right of the HR diagram. These are very large and luminous, but have a much cooler temperature.

White Dwarfs: consists of a majority of stars, found at the bottom left of the HR diagram. These have very low luminosity, despite relatively high surface temp. and undergo fusion. These will not evolve anymore and will continue until all energy is used up in its core, and die out.

Blue Giants: rare, short-lived stars, and very luminous, hot, bright, and massive. These are found in the top left of HR diagrams, and are fusing heavier elements in their cores. They don't last long and will quickly evolve into white dwarfs.

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When the hybridization occur molecules have a linear arrangement of the atoms with a bond angle of 180 degree.

Hybrid orbitals are formed by combining the valence orbital on an atom. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it. Two such regions imply sp hybridization, three, sp2 hybridization, four, sp3 hybridization five, sp3d hybridization and six, sp3d2 hybridization. Pi (π) bonds are formed from unhybridized atomic orbitals that is p or d orbitals. The geometry and bond angle for the molecules can be identified by the VSEPR model from the molecule central atom which is enclosed by the electron pairs.

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Answer:

Chemistry, physics, astronomy and earth science.

Explanation:

\(CH_3CH_2OH\) is a polar compound and has hydrogen bonding. It has a higher boiling point. It will have the lowest vapor pressure among the three compounds.

Vapor pressure is the pressure produced by the vapors of a liquid in a sealed container when the liquid is heated. It is a property of a liquid that describes the tendency of a liquid to evaporate. It is measured in units of pressure, typically in mmHg or torr. The higher the vapor pressure, the more volatile a compound is. So, the decreasing order of vapor pressure for the given compounds is: \(CH_3CH_3 > CH_3CH_2F > CH_3CH_2OH\)

Here, \(CH_3CH_3\) is a non-polar compound while \(CH_3CH_2F\) and \(CH_3CH_2OH\) are polar compounds. The polar compound has hydrogen bonding, which increases the intermolecular forces between molecules. As a result, the polar compound will have a lower vapor pressure than the non-polar compound.

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The value of Y in the formula when 2.8g of silicon react with 3.2g of oxygen, is 2

The numbers in the formula of a compound determine the ratio of moles of each element in the compound.

Given:

2.8g of silicon

3.2g of oxygen

[relative atomic mass of silicon is 28 and of oxygen is 16]

For silicon, n = 2.8 ÷ 28 = 0.1    (n= moles)

For oxygen, n = 3.2 ÷ 016 = 0.2

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.1 moles.

For Silicon = 0.1/0.1 = 1

For Oxygen = 0.2/0.1 = 2

Taking the mole ratio as their subscripts.

The ratio of Si:O is 1:2

SiO\(_{2}\) is the formula of silicon dioxide.

The value of y in the formula SiO\(_{Y}\) is 2 when 2.8g of silicon react with 3.2g of oxygen.

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The ideal gas equation can be used to determine the molar mass of the substance with the help of the moles. The molar mass of the gas having a mass of 0.458 grams is 9.15g/mol.

Ideal gas law gave the relation between the moles, pressure, temperature, volume, and the gas constant. The moles of the ideal hypothetical gas can be used to calculate the molar mass.

An ideal gas equation is given as:

PV =nRT

Given,

Mass = 0.458 gm

Pressure (P) = 0.98 atm

Volume (V) = 1.2 L

Temperature (T) = 287 K

Gas constant (R) = 0.0821

Substituting values in ideal gas as:

0.98 × 1.2 = n × 0.0821 × 287

n = 1.18 ÷ 23.56

n = 0.05moles

Now, moles are used to calculate molar mass as:

molar mass = mass ÷ moles

= 0.458 ÷ 0.05

= 9.15g/mol

Therefore, 9.15g/mol is the molar mass of the unknown gas.

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Answer:

Inactive ingredients are components of a drug product that do not increase or affect the therapeutic action of the active ingredient, which is usually the active drug. Inactive ingredients are added during the manufacturing process of pharmaceutical products such as tablets, capsules, suppositories, and injections

Electrostatic attraction between metal cations and delocalized electrons produces metallic bonds. Many of the physical characteristics of metals, including conductivity and malleability, are explained by the type of metallic bonding that exists.

The substance in which the atoms are held together by metallic bonding is A. Cr (Chromium). Metallic bonding occurs between metal atoms, where the valence electrons are shared by all the atoms in a lattice structure, creating a strong bond. Cr is a transition metal and its atoms have a partially filled d orbital, which allows them to share their valence electrons and form metallic bonds.

B. Si (Silicon) is a non-metal and forms covalent bonds, where atoms share electrons with each other to form a stable molecule.

C. S8 (Sulfur) is a molecular substance where eight sulfur atoms are covalently bonded together in a ring structure, with weak van der Waals forces holding the molecules together.

D. CO2 (Carbon dioxide) is a molecular substance where one carbon atom is covalently bonded to two oxygen atoms, with the bonds formed by sharing electrons between atoms.

E. Br2 (Bromine) is a molecular substance where two bromine atoms are covalently bonded together, with the bonds formed by sharing electrons between atoms. Br2150 is not a known substance.

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Answer: One formula unit of NaCl consists of one cation, whose chemical symbol is \(Na^+\) and one anion whose chemical symbol is \(Cl^-\)

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

The cation is formed by the metal sodium which forms  \(Na^+\) and the anion is formed by non metal chlorine which forms \(Cl^-\).

For a formula unit of sodium chloride,  the charges have to be balanced , thus the valencies of ions are exchanged and the neutral compound result. Thus

\(Na^+\) and  \(Cl^-\) combine to form neutral \(NaCl\)

70.9-gram sample of \(Cl_{2}\) gas will occupy Opton C. 22.4 liters at STP.

To determine the volume occupied by the sample of \(Cl_{2}\) (g) at STP, we can use the ideal gas law equation, PV = nRT

where P = pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature.

At STP (Standard Temperature and Pressure), the pressure is 1 atmosphere (atm) and the temperature is 273.15 Kelvin (K).

First, calculate the number of moles of \(Cl_{2}\) (g) using its molar mass. The molar mass  \(Cl_{2}\) is 70.9 grams/mol.

Number of moles (n) = mass (m) / molar mass (M)

n = 70.9 g / 70.9 g/mol

n = 1 mol

Now, we can calculate the volume using the ideal gas law:

V = (nRT) / P

V = (1 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

V ≈ 22.4 L

Therefore, the correct answer is C. 22.4 L.

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