What happens if excess carbon dioxide is not removed from the atmosphere?

A) a decrease in the amount of fossil fuels availabe

B) it may lead to global warming

C) more dead plants and animals not decomposing before getting buried

D) increase in the amount of fossils fuels available

please help fast I need to bring my grade up I'll give 30 points.

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:

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 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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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]

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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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)

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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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 isotope that is more abundant, given the data is isotope Li7

Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]

6.94 = [6.02A% + 702 - 7.02A%] / 100

Cross multiply

6.02A% + 702 - 7.02A% = 6.94 × 100

6.02A% + 702 - 7.02A% = 694

Collect like terms

6.02A% - 7.02A% = 694 - 702

-A% = -8

A% = 8%

Thus,

Abundance of B = (100 - A)%

Abundance of B = (100 - 8)%

Abundance of B = 92%

SUMMARY

From the above, isotope Li7 is more abundant.

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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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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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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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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⁺?

Answer:

[tex]\huge\boxed{\sf Ribosomes}[/tex]

Explanation:

[tex]\rule[225]{225}{2}[/tex]

Answer:

A

Explanation:

In a simple perspective, an ion of an element is an element +- [tex]n[/tex] amount of electrons.

We know that electrons have atomic mass, so we can mark off D

We know that with more or fewer electrons, the electron configuration would change

However, the number of protons won't change in the ion of an element, this is because an ion is only the change in electrons, not protons or neutrons (the particles that make up a nucleus of an atom)

Answer:

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

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!

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

If U-235 decays into Cs-135 and 4 neutrons, the other nuclide that will be produced is Rb-96 (option D).

A radioactive decay is the process by which an unstable large nuclei emit subatomic particles and disintegrate into one or more smaller nuclei.

According to this question, a radioactive material Uranium- 235 undergoes radioactive decay into Cs- 135 and 4 neutrons (1/0n).

This means that the mass of the products we have is 135 + 4 = 139.

The mass of the nuclide left must be 235 - 139 = 96, hence, the other nuclide that will be produced is Rb-96.

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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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Answer:

C.) 109.9 g/mol

Explanation:

The molar mass is the sum of each element's atomic weight times their quantity. There is one atom of each element unless denoted by subscripts.

Atomic Mass (Li): 6.9410 g/mol

Atomic Mass (S): 32.065 g/mol

Atomic Mass (O): 15.998 g/mol

Molar Mass (Li₂SO₄): 2(6.9410 g/mol) + 32.065 g/mol + 4(15.998 g/mol)

Molar Mass (Li₂SO₄): 109.939 g/mol

Considering the definition of mole fraction, the mole fraction of O₂ in the mixture is 0.434.

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

In other words, the mole fraction expresses the concentration of solute in a solution as the ratio of moles of substance to total moles of solution:

[tex]mole fraction=\frac{moles of substance}{moles of solution}[/tex]

In this case, you know a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O).

So, the total moles of the solution can be calculated as:

Total moles = moles of oxygen (O₂) + moles of nitrous oxide (N₂O)

Then:

Total moles= 4.60 moles + 6 moles

Total moles= 10.60 moles

Finally, the more fraction of O₂ can be calculated as follow:

[tex]Mole fraction of O_{2} =\frac{moles of O_{2}}{total moles}[/tex]

[tex]Mole fraction of O_{2} =\frac{4.60 moles}{10.6o moles}[/tex]

Solving:

Mole fraction O₂ = 0.434

Finally, the mole fraction of O₂ in the mixture is 0.434.

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Answer:

Mole Fraction (H₂O)  =  0.6303

Mole Fraction (C₂H₅OH)  =  0.3697

Explanation:

(Step 1)

Calculate the mole value of each substance using their molar masses.

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

200.0 g H₂O            1 mole
---------------------  x  ------------------  =  11.10 moles H₂O
                                 18.014 g

Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol

Molar Mass (C₂H₅OH): 46.068 g/mol

300.0 g C₂H₅OH              1 mole
----------------------------  x  --------------------  =  6.512 moles C₂H₅OH
                                         46.068 g

(Step 2)

Using the mole fraction ratio, calculate the mole fraction of each substance.

                                            moles solute
Mole Fraction  =  ------------------------------------------------
                               moles solute + moles solvent

                                                  11.10 moles H₂O
Mole Fraction  =  -------------------------------------------------------------
                               11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (H₂O)  =  0.6303

                                             6.512 moles C₂H₅OH
Mole Fraction  =  -------------------------------------------------------------
                               11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (C₂H₅OH)  =  0.3697

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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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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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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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.

Your answer would be C, 4.8%.

Answer:

nah what questionExplanation: