how many litres of hydrogen is present in 12 grams of hydrogen​

From the calculation, the equilibrium partial pressure of IBr is 6.5 * 10^-4 atm while that of I2 and Br2 is 0.002 atm.

We must set up the ICE table as shown hence;

              2 IBr(g) < ------> I2(g) + Br2(g)

I              0.00465            0        0

C                  - 2x                + x      + x

E            0.00465 - 2x       + x      + x

Kp = pI2. pBr2/pIBr^2

pI2 = pBr2 = x

0.748 = x^2/  0.00465 - 2x

0.748 (0.00465 - 2x) = x^2

3.5 * 10^-3 - 1.496x = x^2

x^2 + 1.496x  - 3.5 * 10^-3 = 0

x=0.002 atm

Hence;

For IBr =   0.00465 - 2(0.002) = 6.5 * 10^-4 atm

For I2 and Br2 = 0.002 atm

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The number of photons given off will be 50 photons.

To find the answer, we need to know about the plank's equation.

                     [tex]E=-13.6(\frac{1}{(n_f)^2}- \frac{1}{(n_i)^2})eV\\[/tex]

                     [tex]n_f=4\\n_i=2\\N=50[/tex]

                  [tex]E=-13.6(\frac{1}{(4)^2}- \frac{1}{(2)^2})eV\\\\E=-13.6*-0.188=2.55eV[/tex]

                [tex]E=50*2.55eV=127.5eV[/tex]

                            E=nhf

                    [tex]n=\frac{E}{h*f} =\frac{127.5*1.67*10^{-19}J}{(6.63*10^-34)Js} \\[/tex]

                       [tex]\frac{1}{wave length}=R_H(\frac{1}{(n_i)^2}- \frac{1}{(n_f)^2})\\\\1/wv= 1.1*10^5(\frac{1}{(2)^2}- \frac{1}{(4)^2})=20625cm^{-1}.\\wavelength=486nm.[/tex]

                 [tex]f=\frac{c}{wavelength} =\frac{3*10^8}{486*10^{-9}} =6.172*10^{14}s{-1}[/tex]

                       [tex]n=\frac{E}{h*f} =\frac{127.5*1.67*10^{-19}J}{(6.63*10^-34)Js*6.17*10^{14} s^{-1}} \\\\n=52.05 photons[/tex]

Thus, we can conclude that, the number of photons given off will be 50 photons.

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When traveling through a smoke-filled area, a person must be able to have distance of at least 4 feet in order to safely reach a fire exit. That is option C.

Smoke-filled area is an enclosed area that is filled with smoke gas that is usually made up of carbon monoxide.

The smoke-filled area is usually caused by

When there is smoke-filled area die to fire out break, there are some guidelines to follow to safely reach the fire exit early enough.

The importance of these guidelines include the following:

The affected individual should get down and crawl while taking short breath through the nose because cleaner air is nearest to the floor.

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Caprylic acid is a saturated molecule with eight electrons and the COOH functional group.

The initial element and primary component of stars, hydrogen, makes up nine out of every ten atoms in the cosmos. Helium makes up the other 10% of all atoms. Already 100 percent of that. On a cosmic scale, the remaining elements of the periodic table—Elements 3 through 118, lithium through ununoctium—barely register.

Hydrogen may be utilized to heat our homes and places of work in addition to powering vehicles and producing energy. To reach net zero, it will be crucial to reduce our carbon emissions significantly.

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Answer:

Before air is liquefied, water vapor and carbon dioxide are removed, because these substances solidify when cooled and would clog the pipes of the air liquefaction plant.

For any spontaneous process, universe entropy intensifies is known as the second law of thermodynamics.

Entropy is defined as the degree of randomness or disorderliness of a system.

The entropy of a system generally increases for any spontaneous process.

This is according to the second law of thermodynamics.

In conclusion, the entropy of a system is the a measure of randomness of the system.

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The ester formed from the reaction of acetic acid and 1-propanol is O3, which is propyl acetate.

The chemical compound ethyl acetate, also known as ethyl ethanoate or EtOAc or EA, has the formula C4H8O2 and is an organic compound. This clear liquid is used to decaffeinate tea and coffee as well as in glues and nail paint removers. It has a distinctive sweet scent akin to pear drops. The ester of ethanol and acetic acid, ethyl acetate is produced in huge quantities for use as a solvent.

Due to its affordability, low toxicity, and pleasant smell, ethanol acetate is mostly employed as a solvent and diluent. It is frequently utilized, for instance, to clean circuit boards and in some nail polish removers (acetone is also used). This solvent is used to decaffeinate tea leaves and coffee beans. Additionally, it serves as a hardener or activator in paints. Fruits, colognes, and confections all contain ethyl acetate. In perfumes, it swiftly evaporates, leaving the fragrance on the skin.

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Answer:

nah what questionExplanation:

Answer:

we are given that mass of lithium four spade that is equal to 77.7 grams

Your answer would be C, 4.8%.

Answer: it is either 34x or it is 0 im not sure how you do the math or what your question is asking exactly sorry

Explanation:

Mass of 3.40 x 10^35 formula units of KCL is calculated to be 4.20 * 10^13.

In chemistry, formula unit is the empirical formula of ionic or covalent network solid compound which is used as an independent entity for stoichiometric calculations. It is also the lowest whole number ratio of ions represented in ionic compound.

Convert molecules into moles.  

6.02 x 1023 molecules = 1 mole

1 = (1 mole)/(6.02x1023 molecules)

(3.40x10^35 molecules KCL)*(1 mole KCL)/(6.02x1023 molecules KCL)

5.647 *10^11 moles Na2SO4

Molar mass of KCL = 74.55 g/mol

Mass of KCL=  5.647 *10^11  * 74.55

Mass of KCL = 4.20 * 10^13

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The percent yield of carbon dioxide will be 49.0 %.

First, let's look at the equation of the reaction:

[tex]2C_8H_1_8 + 25O_2 -- > 16CO_2 + 18H_2O[/tex]

The mole ratio of octane to oxygen is 2:25.

Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol

Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol

Thus, octane is limiting.

Mole ratio of octane to carbon dioxide = 2:16.

Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol

Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams

Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %

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Answer:

2H2(g) + O2 -> 2H2O

Explanation:

According to the law of conservation of mass, mass cannot be created nor destroyed in a chemical formula.

The only chemical reaction of the five options that follows this law is

2H2(g) + O2 -> 2H2O because the mass of the compounds stays the same before the reaction (arrow) and after.

Before reaction (reactants) we have:

2 (H2) = 4H

1 (O2) = 2O

After reaction takes place (products) we have:

2 (H2O) = 4H and 2O

and so mass is conserved.

Answer: its the 3rd option

Explanation: on edge hope this helps

The order of decreasing bond length is; C- P > C - N > C=N. Option B

A chemical bond is formed when two compounds combine together. We know that the strength of a bond is shown by the bond length and the bond order. Clearly, bonds that have shorter bond lengths tend to be stronger than longer bonds that have a longer bond distance.

This explains why triple bonds are shorter and stronger bonds. This stems for the high s contribution (50%) in the formation of the bond. Therefore, the order of decreasing bond lengths is single bonds > Double bonds > triple bonds.

Now applying this background to the case in point, we can see that the decreasing order of bond energy must have to do with increasing bond order and decreasing bond distance.

Hence, the order of decreasing bond length is; C- P > C - N > C=N.

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Answer: 421.264 grams.

Explanation: This problem is essentially testing you on your knowledge of percent mass except with one extra step.

We are given that there are 2kg or 2000 grams of ore. This ore is %15.1 Antimony. We can multiply 2000 grams by .151 to find the amount of Antimony which is 302 grams. We don't want the amount of Antimony though, we want the amount of Stibnite. To find the amount of Stibite, we can multiply the amount of Antimony by 1/(the percent mass of Antimony in 1 molecule of Stibnite. To find the percent mass we divided 2(molar mass of Sb) by the molar mass of one molecule of Stibnite. This calculation is 2(121.67)/339.69. This is the percent mass of Antimony in Stibnite. We take the inverse because we want to find out how much Stibite there is per amount of Antimony. We do 1/(((2(121.67))/339.69) and multiply that by 302 to find the total amount of Sb2S3.

Hope this helps!

2756.1 liters of fluorine gas is needed to produce 919 liters of sulfur hexafluoride.

The volume of SF₆ = 919 L

Pressure = 2 atm

Temperature = 273.15 k

The volume of fluorine required =?

Balance chemical equation:

   S (s) + F₂(g)    →  SF₆(g)

First of all, we will calculate the moles of SF₆.

 PV = nRT

 n = PV/RT

 n = 2. atm× 919L / 0.0821 L. atm. mol⁻¹. K⁻¹  × 273.15 K

 n = 1838 atm. L/ 22.43 L. atm. mol⁻¹

 n = 81.9 mol

81.9 moles of SF₆ will produce.

Now we will compare the moles of SF₆ and fluorine from the balanced chemical equation.

  SF₆:   F

  1     :    3

 81.9  :   3/1 × 81.9  = 245.7 moles

Now we will calculate the volume of fluorine.

PV = nRT

V  = nRT / P

V= 245.7 mol × 0.0821 L. atm. mol⁻¹. K⁻¹  × 273.15 K / 2 atm

 V = 5512.2 / 2

 V = 2756.1 L

2756.1 liters of fluorine gas are needed to produce 919 liters of sulfur hexafluoride.

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Answer:

Observation 2 is a result of silver ions changing their oxidation state.

Explanation:

Silver is reduced and forms silver precipitate. Observation 1 is a result of coppe being oxidized and forming copper nitrate

The concentration of solution in mass percent is 60%.

Mass percentage refer to the percentage of solute present in solution.

The concentration of substance can be expressed in mass percent.

So, we can write,

Mass percent = mass of solute / mass of solution x 100.

Mass of solution = mass of solute + solvent.

Here, sucrose is the solute and water is solvent.

Mass of solute is 1.5Kg and mass of solvent is 1Kg.

Mass of solution = 1.5 + 1 = 2.5 Kg.

Mass percent = 1.5/2.5x100

So, Concentration of solution in mass percent = 60%.

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When the soil is saturated in a gently sloping area, any additional rainfall in the area will most likely become surface run off.

Super saturation occurs with a chemical solution when the concentration of a solute exceeds the concentration specified by the value equilibrium solubility.

Additional water into the soil will cause the soil to be super saturated and eventually runs off due to the steepness of the area.

Thus, when the soil is saturated in a gently sloping area, any additional rainfall in the area will most likely become surface run off.

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The balanced redox equation of the reaction is given below:

The oxidizing agent is MnO₄ while the reducing agent is I⁻.

Redox equations are equations in which oxidation and reduction reactions occur together.

Redox reactions can take place in alkaline or acidic mediums.

The balanced redox equation of the reaction is given below:

The oxidizing agent is MnO₄ while the reducing agent is I⁻

In conclusion, a balanced redox equation is one in which the atoms and the change in oxidation state is equal on both sides of the reaction.

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a) Orbital designation of F₂⁺ is shown below:

b) σ(2p) is occupied by at least one electron in F₂⁺

c) π*(2p) is occupied by only one electron in F₂⁺

The molecular orbital diagram can be used to show the bonds inside a diatomic molecule. The magnetic characteristics of a molecule and their ionization-related changes can be determined using MO diagrams. The number of bonds shared between the two atoms, or the bond order of the molecule, is likewise shown by them.

The structure A represents an anti-bonding MO of side-wise overlapping of 2p-orbital. Thus designation of this orbital diagram is: π*(2p).

The structure B represents a bonding MO of head-on overlapping of 2p-orbital. Thus designation of this orbital diagram is: σ(2p)

The structure C represents a bonding MO of side-wise overlapping of 2p-orbital. Thus designation of this orbital diagram is: π(2p)

The structure D represents an anti-bonding MO of head-on overlapping of 2p-orbital. Thus designation of this orbital diagram is: σ*(2p)

a) Thus, orbital designation of F₂⁺ is shown below:

b) σ(2p) is occupied by at least one electron in F₂⁺

c) π*(2p) is occupied by only one electron in F₂⁺

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The complete question is as follows:

The molecular orbitals depicted below are derived from 2p atomic orbitals in F₂⁺, (a) Give the orbital designations. (b) Which is occupied by at least one electron in F₂⁺? (c) Which is occupied by only one electron in F₂⁺?

"hydrogen bonds between the orange and red strands are not shown but are implied by the model" - True

When a hydrogen atom bound to a highly electronegative atom is present close to another electronegative atom with a single pair of electrons, a hydrogen bond forms between the two molecules. This unique sort of dipole-dipole attraction is known as a hydrogen bond. Between molecules, there are intermolecular forces.

The unequal distribution of the electron between a hydrogen atom and a negatively charged atom is the source of hydrogen bonding. While a stable electron pair requires two electrons, hydrogen in a bond still only has one electron.

Thus, the given statement is true.

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The balanced equation will be [tex]2NH_3 + 3I_2 -- > N_2I_6 + 3H_2[/tex]

They are chemical equations that obey the law of conservation of atoms.

In other words, they are equations in which the number of atoms before and after reactions are the same.

Thus, the balanced equation for the reaction will be [tex]2NH_3 + 3I_2 -- > N_2I_6 + 3H_2[/tex]

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The rate constant of the second order reaction is 0.137 M-1s-1.

For the second order reaction we can write;

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

[A]o = initial concentration

[A] = final concentration

k = rate constant

t = time

Now;

1/0.319 = 13.5k + 1/ 0.740

1/0.319 - 1/0.740 = 13.5k

3.13 - 1.35 = 13k

k = 3.13 - 1.35/13

k = 0.137 M-1s-1

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C₃H₈ (g) + 5 O₂ (g) → 3 CO₂ (g) + 4 H₂O oxygen atoms are being oxidized and propane are being reduced.

The term redox reaction is defined as when oxidation and reduction takes place simultaneously.

The loss of electron is called as oxidation, it increases the oxidation state. The gain of electron is called as reduction, it decreases the oxidation state.

The oxidation state of the oxygen is decreased hence it gets reduced while that of propane oxidation state is increased hence it gets oxidized.

Thus, C₃H₈ (g) + 5 O₂ (g) → 3 CO₂ (g) + 4 H₂O oxygen atoms are being oxidized and propane are being reduced.

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Answer:

A.) Change only the coefficients

Explanation:

An equation is balanced when there is an equal quantity of each type of element on both sides of a reaction. When balancing an equation, the only way to manipulate the amounts of each element is by changing the coefficient values. The coefficients alter the amount of each molecule in the reaction.

The subscripts cannot be altered. If you were to change the subscripts, you would be altering the amount of atoms in a particular molecule.

Answer:

When an alkane undergoes a halogenation reaction, the majority of the products will add the halogen to the most substituted and most stable carbon. In this case, this would likely be the third carbon (middle).

Key:

Leftmost molecule = C₅H₁₂

Rightmost molecule = C₅H₁₁Cl

*Not pictured = HCl (another product)

The general formula for many YBCO compounds, also known as Y123, is YBa2Cu3O7x. However, materials with other Y: Ba: There are Cu ratios like YBa2Cu4Oy (Y124) and Y2Ba4Cu7Oy (Y247).

A compound is a substance composed of two or more distinct chemical elements arranged in a predetermined ratio in chemistry. Chemical bonds that are difficult to break form when the elements combine and react with one another. Atomic electron sharing or exchanging results in the formation of these bonds.

compound Include in the list Share When a problem is compounded, something is added to make it worse, like pouring water on a grease fire. The combination is a compound word; A compound is a mixture or combination of at least two things.

A substance in science that is made by chemically joining two or more distinct elements. Table salt (NaCl), which is made from the elements sodium and chloride, and water (H2O), which is made from the elements hydrogen and oxygen, are examples of compounds.

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The Wittig reaction is a chemical reaction used to synthesize alkenes from aldehydes or ketones using a phosphonium ylide (also known as a Wittig reagent) as the nucleophile.

When an aldehyde or ketone reacts chemically with a triphenyl phosphonium ylide (often referred to as a Wittig reagent), an alkene and triphenylphosphine oxide are produced. This reaction is known as the Wittig reaction or Wittig olefination.

The synthesis of the Wittig reagent from Ph3P and an alkyl bromide can be outlined as follows:

1. Start with Ph3P (triphenylphosphine) and an alkyl bromide (R-Br) in a suitable solvent, such as dichloromethane (CH2Cl2).

2. Add a strong base, such as n-butyllithium (n-BuLi) or potassium tert-butoxide (t-BuOK), to the mixture. This will deprotonate Ph3P to give the phosphonium cation Ph3P+.

3. The alkyl bromide will act as a nucleophile and will react with the Ph3P+ cation to give the phosphonium ylide intermediate: Ph3P+-R-Br.

4. The intermediate will then lose a molecule of Br- to give the final Wittig reagent which is a phosphonium ylide : Ph3P=CHR

5. This reagent can be used as a nucleophile in the Wittig reaction to form an alkene.

Note that this reaction is not suitable for all types of Alkyl Bromides, and some may not be able to form the ylide intermediate or may form by products.

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Answer:

[tex]V_2 = \frac{5V}{8}[/tex]

Explanation:

I am assuming you are saying what is the final volume of the gas

Known :

Initial volume (V1) = V

Initial temperature (T1) = T

Final temperature (T2) = 5/4 T

Initial pressure (P1) = P

Final pressure (P2) = 2P

Wanted: Final volume (V2)

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\frac{PV}{T} = \frac{(2P)V_2}{(5/4)T}\\\frac{V}{1} = \frac{(2)V_2}{5/4}\\5/4V = 2V_2\\\\V_2 = \frac{5V}{8}[/tex]