Assume that a compound is a cyclic, planar, completely conjugated ring. Which number of p electrons would make it aromatic?
a) 0 p electrons
b) 2 p electrons
c) 3 p electrons
d) 4 p electrons
e) 32 p electrons

Answers

Answer 1

Answer:

option b is correct

2 p electron makes aromatic

Explanation:

An aromatic compound which is cyclic, planar and has a complete conjugate ring must have (4n + 2)pi electrons(Huckel's rule)

Huckel's Standard (4n+2 rule): For a compound to be an aromatic, a particle must have a specific number of pi (electrons with pi bonds, or lone pairs inside p orbitals) inside a shut loop of parallel, adjoining p orbitals. The pi electron tally is characterized by the arrangement of numbers created from 4n+2 where n = zero or any positive whole number (i..e, n = 0, 1, 2, and so forth.). The most widely recognized case in six pi electrons (n = 1) which is found for example in benzene, pyrrole, furan, and pyridine.

where n is the number of pi electrons

where n = 0

(4n +2) pi electrons = 2pi electrons

attached is an example of aromatic which is cyclic, planar and a complete conjugate ring

Assume That A Compound Is A Cyclic, Planar, Completely Conjugated Ring. Which Number Of P Electrons Would
Answer 2

Answer:

2 p electrons.

Explanation:

For any compound to be considered an aromatic compound it must be cyclic,flat, conjugated and it must obey Huckel's rule that states an aromatic compound must have 4n + 2 pi electrons in it's p orbitals for it to be an aromatic compound.

n can represent an integer from 2, 6,10, 14,........

The lone pair is actually in a pure 2p orbital perpendicular to the ring, which means they count as π electrons.


Related Questions

How many moles of aqueous magnesium ions and chloride ions are formed when 0.250 mol of magnesium chloride dissolves in water

Answers

Answer:

0.250 mol Mg²⁺

0.500 mol Cl⁻

Explanation:

Magnesium chloride (MgCl₂) dissociates into ions according to the following equilibrium:

MgCl₂  ⇒  Mg²⁺ + 2 Cl⁻

1 mol      1 mol   2 mol

1 mol of Mg²⁺ and 2 moles of Cl⁻ are formed per mole of MgCl₂.  If we have 0.250 mol of MgCl₂, the following amounts of ions will be formed:

0.250 mol MgCl₂ x 1 mol Mg²⁺/mol MgCl₂= 0.250 mol Mg²⁺

0.250 mol MgCl₂ x 2 mol Cl⁻/mol MgCl₂= 0.500 mol Cl⁻

Answer:

HEY THE ANSWER ABOVE ME IS RIGHT!! i defientely misclicked my rating :/

5/5 all the way.

Explanation:

Sulfuric acid is commonly used as an electrolyte in car batteries. Suppose you spill some on your garage floor. Before cleaning it up, you wisely decide to neutralize it with sodium bicarbonate (baking soda) from your kitchen. The reaction of sodium bicarbonate and sulfuric acid is

Answers

Answer:

The mass of NaHCO3 required is 235.22 g

Explanation:

*******

Continuation of Question:

2NaHCO3(s) + H2SO4(aq)  →  Na2SO4(aq) + 2CO2(g) + 2H2O(l)

You estimate that your acid spill contains about 1.4 mol H2SO4. What mass of NaHCO3 do you need to neutralize the acid?

********\

The question requires us to calculate the mass of NaHCO3  to neutralize the acid.

From the balanced chemical equation;

1 mol of H2SO4 requires 2 mol of NaHCO3

1.4 would require x?

Upon solving for x we have;

x = 1.4 * 2 = 2.8 mol of NaHCO3

The relationship between mass and number of moles is given as;

Mass = Number of moles * Molar mass

Mass = 2.8 mol * 84.007 g/mol

Mass =  235.22 g

Answer all to the best of your ability please

Answers

Answer:

2–Ethyl–3–methlypentanal.

Explanation:

To name the compound given in the question above, we must observe the following:

1. The functional group of the compound is Alkanal i.e Aldehyde,

—CHO and it is located at carbon 1.

Note: the aldehyde functional group is always at carbon 1 and there will be no need to state it's position in the compound.

2. The longest continuous carbon chain is 5 i.e pentane. But the presence of the functional group will replace the –e at the end of pentane with –al, making the name to the pentanal.

3. The substituents attached are:

a. Ethyl, CH2CH3 at carbon 2.

b. Methyl, CH3 at carbon 3.

4. Combine the above to get the name of the compound.

Therefore, the name of the compound is:

2–Ethyl–3–methlypentanal.

Question 14 of 25
What type of reaction is BaCl2 + Na,504 → 2NaCl + Baso,?
A. Single-replacement
B. Synthesis
C. Double-replacement
D. Decomposition

Answers

double displacement

bcoz each of the reactants combines with other reactants to obtain the product

What is the empirical formula for the compound: C8H8S2?

Answers

Answer:

Empirical formula = C4H4S

Explanation:

The subscripts in a formula determine the ratio of the moles of each element in the compound. To convert this formula to the empirical formula, divide each subscript by 2. This is similar to reducing a fraction to its lowest denominator.

Which of the following is an alkaline earth metal?
A. Silicon (Si)
B. Magnesium (Mg)
C. Carbon (C)
D. Aluminum (AI)

Answers

Answer:

B

Explanation:

The alkaline earth metals are the elements located in Group 2. The only element out of our choices that is in Group 2 is Magnesium.

Magnesium (Mg) is an alkaline earth metal. So, option (B) is the correct answer.

What are Alkaline Earth Metals?

Alkaline earth metals are metals with two valence electrons in their outermost shell.

Group 2A (or IIA) elements:
- Beryllium (Be)
- Magnesium (Mg)
- Calcium (Ca)
- Strontium (Sr)
- Barium (Ba)
- Radium (Ra)


The six alkaline earth metals are barium, radium, strontium, magnesium, and calcium. By giving their outermost electrons, which boosts their stability, they acquire the electron configuration of noble gases. The electrons are transported to an electronegative atom, giving the alkaline earth metals a positive charge.

The second column of the periodic table contains alkaline earth metals, which are very reactive metals. Every single object in the world is made using these metals. Sulfates of these metals may be found all throughout nature.

Minerals come in a variety of forms, some of which include gypsum, calcium sulfate, epsomite, magnesium sulfate, barite, and barium sulfate.

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A chemist adds of a mercury(I) chloride solution to a reaction flask. Calculate the micromoles of mercury(I) chloride the chemist has added to the flask.

Answers

Answer:

3.383x10⁻³ micromoles of HgCl

Explanation:

The chemist adds 170mL of a 1.99x10⁻⁵mmol/L Mercury (I) chloride, HgCl.

The solution contains 1.99x10⁻⁵milimoles of HgCl in 1L. That means in 170mL = 0.170L there are:

0.170L × (1.99x10⁻⁵milimoles HgCl / L) = 3.383x10⁻⁶ milimoles of HgCl.

Now, in 1milimole you have 1000 micromoles. That means in 3.383x10⁻⁶ milimoles of HgCl you have:

3.383x10⁻⁶ milimoles of HgCl ₓ (1000micromoles / 1milimole) =

3.383x10⁻³ micromoles of HgCl

Some metal oxides, such as Sc2O3, do not react with pure water, but they do react when the solution becomes either acidic or basic. Do you expect Sc2O3 to react when the solution becomes acidic or when it becomes basic?

Write a balanced chemical equation to support your answer.

Answers

Answer:

[tex]Sc_2O_3[/tex] reacts with an acidic solution

Explanation:

Scandium Oxide [tex]Sc_2O_3[/tex] is a basic metal oxide which therefore reacts with acidic solution. An oxide is  a compound that contains only two elements, one of which is oxygen .

The objective of this question is to Write a balanced chemical equation to support your answer.

The chemical equation to support the reaction of [tex]Sc_2O_3[/tex] with acidic solution is as follows:

Assuming the acidic solution to be HCl

[tex]\mathbf{Sc_2O_3_{(s)} + 6 HCl_{(aq)} ----> 2 ScCl_{3(aq)} + 3H_2O_{(l)}}[/tex]

The ionic equation :

[tex]\mathbf{Sc_2O_{3(s)} + 6H^+_{(aq)} ---> 2Sc^{3+}_{(aq)} + 3H_2O_{(l)}}[/tex]

True or False: Adding 4.18 joules to water will increase the temperature more than adding 1 calorie to water.

Answers

Answer:

Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C. If you look at the specific heat graph shown below, you will see that 4.18 is an unusually large value.

A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?

Answers

Answer:

V2 = 17371.43ml

Explanation:

We use Boyles laws

since temperature is constant

P1V1=P2V2

760 x 400 = 17.5 x V2

304000 = 17.5 x V2

V2 = 304000/17.5

V2 = 17371.43ml

The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at  760 torrs will be 18 ml.

What is vapor pressure?

The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.

The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -

P1 V1 / T1 = P2 V2 / T1

here, P = pressure

       T = temperature

        V = volume

substituting the value in the equation,

400 ×760 / 20 = 17.5× V / 20

V = 400× 760 / 20 × 17.5 / 20

V = 18 ml

Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.

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Given a fixed amount of gas in a rigid container (no change in volume), what pressure will the gas exert if the pressure is initially 1.50 atm at 22.0oC, and the temperature is changed to 11.0oC?
A. 301 atm
B. 1.56 atm
C. 0.750 atm
D. 1.44 atm
E. 3.00 atm

Answers

Answer:

The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)

Explanation:

Gay Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move more rapidly. Then the number of collisions against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

[tex]\frac{P}{T}=k[/tex]

Where P = pressure, T = temperature, K = Constant

You have a gas that is at a pressure P1 and at a temperature T1. When the temperature varies to a new T2 value, then the pressure will change to P2, and then:

[tex]\frac{P1}{T1}=\frac{P2}{T2}[/tex]

In this case:

P1= 1.50 atmT1= 22 °C= 295 °K (being 0°C= 273 °K)P2= ?T2= 11 °C= 284 K

Replacing:

[tex]\frac{1.5 atm}{295 K}=\frac{P2}{284 K}[/tex]

Solving:

[tex]P2= 284 K*\frac{1.5 atm}{295 K}[/tex]

P2=1.44 atm

The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)

Explain why o-vanillin does not fully protonate p-toluidine. Reference appropriate pKa values and include a balanced chemical reaction and an appropriate reaction arrow in your answer.

Answers

Answer:

Here's what I get  

Explanation:

pKₐ of o-vanillin = 7.81; pKₐ of p-toluidine = 4.44

The higher the pKₐ, the weaker the acid.

Thus, o-vanillin is the weaker acid and has a stronger conjugate base.

The conjugate acid of p-toluidine is the stronger and has the weaker conjugate base.

The equation for the equilibrium is

H-OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₂ ⇌ ⁻OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₃⁺

    weaker acid              weaker base          stronger  base        stronger acid

The reaction between the stronger acid and the stronger base pushes the position of equilibrium to the left.

Thus, o-vanillin does not fully protonate p-toluidine.

 

O-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base; hence, o-vanillin does not fully protonate p-toluidine.

The pKa is defined as the negative logarithm of Ka. The dissociation constant of an acid Ka shows the extent of dissociation of an acid in solution. The higher the pKa, the lower the Ka and the weaker the acid.

The pKₐ of o-vanillin is 7.81 while the pKₐ of p-toluidine is 4.44. This means that  o-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base. Hence, o-vanillin does not fully protonate p-toluidine.

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All the following are oxidation–reduction reactions except:________
a. H2(g) + F2(g) → 2HF(g).
b. Ca(s) + H2(g) → CaH2(s).
c. 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g).
d. 6Li(s) + N2(g) → 2Li3N(s).
e. Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g).

Answers

Answer:

e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)

Explanation:

All the following are oxidation–reduction reactions except:________

a. H₂(g) + F₂(g) → 2HF(g).  Redox. H is oxidized and F is reduced.

b. Ca(s) + H₂(g) → CaH₂(s).  Redox. Ca is oxidized and H is reduced.

c. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g).  Redox. K is oxidized and H is reduced.

d. 6Li(s) + N₂(g) → 2Li₃N(s).  Redox. Li is oxidized and N is reduced.

e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g). Not redox. All the elements have the same oxidation number

The reaction Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g) is not a redox reaction.

Redox reactions are those reactions in which there is a change in the oxidation number of species from left to right in the reaction. A specie is oxidized leading to increase in oxidation number while another specie is reduced leading to decrease in oxidation number.

The reaction in which there is no change in oxidation number of species from left to right is the reaction; Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g).

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Calculate the amount of heat energy, in kJ, needed to increase the temperature of 200g of water by 300 C.

Answers

Answer:

Q = 246 kJ

Explanation:

It is given that,

Mass of water, m = 200 g

Let initial temperature, [tex]T_i=5^{\circ}[/tex]

Final temperature of water, [tex]T_f=300^{\circ} C[/tex]

We know that the specific heat capacity of water, [tex]c=4.18\ J/g-^{\circ} C[/tex]

So, the heat energy needed to raise the temperature is given by :

[tex]Q=mc\Delta T\\\\Q=200\times 4.18\times (300-5)\\\\Q=246620\ J[/tex]

or

Q = 246 kJ

So, the heat energy of 246 kJ is needed.

Sample gas has a volume of 3.40 L at 10°C what will be its volume at 100°C pressure remaining constant

Answers

Answer:

V2 = 4.48L

Explanation:

using charles law

V1/T1=V2/T2

3.4/283=V2/373

0.012=V2/373

V2= 0.012 x 373

V2 = 4.48L

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of copper(II) cation in the solution. You can assume the volume of the solution doesn't change when the copper(II) acetate is dissolved in it. Round your answer to significant digit.

Answers

Answer:

Molarity Cu²⁺ = 0.423M Cu²⁺

Explanation:

40.8g of copper (II) acetate into 200mL of a 0.700M sodium chromate

The reaction of copper acetate with sodium chromate occurs as follows:

Cu(CH₃COO)₂(aq) + Na₂CrO₄(aq) → CuCrO₄(s) + 2CH₃COONa

In water, the Copper(II) acetate dissociates in Cu²⁺ cation.

To know final molarity of Cu²⁺ we need to calculate the moles of Cu²⁺ that don't react with chromate ion, thus:

Moles of 40.8g of copper(II) acetate (Molar mass: 181.63g/mol) are:

40.8g × (1mol / 181.63g) = 0.2246 moles of Copper(II) acetate

Moles of sodium chromate are:

0.200L ₓ (0.700mol / L) = 0.140 moles of sodium chromate.

As 1 mole of Copper(II) acetate reacts per mole of sodium chromate, moles of Copper(II) acetate = Moles of Cu²⁺ that remains after the reaction are:

0.2246mol - 0.140moles = 0.0846 moles of Cu²⁺

Molarity is ratio between moles of solute (Moles Cu²⁺) and volume in liters of solution (200mL = 0.200L):

Molarity Cu²⁺ = 0.0846 moles / 0.200L

Molarity Cu²⁺ = 0.423M Cu²⁺

question attached ! asap please

Answers

Answer:

A. copper is highly water soluble. It will turn into 5 different hydrates as it absorbs more and more water.

b. Glycerol is easily soluble in water, due to the ability of the polyol groups to form hydrogen bonds with water molecules

c. octane is considered to be non-polar, it will not be soluble in water, since water is a polar solvent. This will happen because octane (hydrocarbons in general) contains neither ionic groups, nor polar functional groups that can interact with water molecules.

d. Nitric acid decomposes into water, nitrogen dioxide, and oxygen, forming a brownish yellow solution.

e. Barium carbonate is a white powder. It is insoluble in water and soluble in most acids

Explanation:

The following reaction is part of the electron transport chain. Complete the reaction and identify which species is reduced. The abbreviation Q represents coenzyme Q. Use the appropriate abbreviation for the product.
FADH2+Q→
The reactant that is reduced is: _____

Answers

Answer:

[tex]FADH_2+Q --> FAD + QH_2[/tex]

The reactant that is reduced is Q.

Explanation:

The complete equation for the reaction is such that:

[tex]FADH_2+Q --> FAD + QH_2[/tex]

Two molecules of H atom is lost from [tex]FADH_2[/tex] and the H atoms are gained by the coenzyme Q. Consequently,  [tex]FADH_2[/tex] becomes FAD while Q becomes [tex]QH_2[/tex].

From the definition of oxidation as loss of hydrogen and reduction as the addition of hydrogen, it can be concluded that the FADH2 that lost hydrogen is a reactant that is oxidized while the coenzyme Q that gained hydrogen is a reactant that is reduced in the reaction.

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.

Answers

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

[tex]\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }[/tex]

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

[tex]\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}[/tex]

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH has been added

Answers

Answer:

The correct answer is 1.60.

Explanation:

Based on the given question, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be determined by using the formula,  

Moles = volume * concentration of HCl

= 25/1000*0.10 = 0.0025 moles

Similarly the moles of NaOH added will be determined by using the formula,  

Moles of NaOH added = volume * concentration of NaOH

= 15/1000 * 0.10 = 0.0015 moles

The reaction taking place in the given case is,  

HCl + NaOH = NaCl + H2O

Now the moles of excess H+ = moles of excess HCl

= 0.0025 - 0.0015 = 0.001 moles

Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L

[H+] = moles of H+/total volume

= 0.001 / 0.040 = 0.025 M

pH = -log[H+]

= -log[0.025]  

= 1.60

The pH after 15 ml of NaOH has been volume is 1.60.

Calculation of Concentration of HCl Moles

It is based on the given question that is, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be specified by using the formula,

Moles is = volume * concentration of HCl

Then is = 25/1000*0.10 = 0.0025 moles

Besides, the moles of NaOH added will be determined by using the formula,

When the Moles of NaOH added is = volume * concentration of NaOH

= 15/1000 * 0.10 = 0.0015 moles

When The reaction taking place in the given case is,

HCl + NaOH = NaCl + H2O

Now the moles of excess H+ = moles of excess HCl

= 0.0025 - 0.0015 = 0.001 moles

Then It Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L

[H+] = moles of H+/total volume

After that = 0.001 / 0.040 = 0.025 M

pH = -log[H+]

Then = -log[0.025]

Therefore, = 1.60

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1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the actual average molarity you measured in the standardization. What do you notice

Answers

Answer:

0.1M solution of NaOH

Explanation:

1 mole of NaOH - 40g

? moles - 1 g = 1/40 = 0.025 moles.

Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume

= 0.025/0.25

= 0.1M.

Which of the following solutions would have the highest pH? Assume that they are all 0.10 M in acid at 25°C. The acid is followed by its Ka value.

a. HCHO2, 1.8 x 10-4
b. HF, 3.5 x 10-4
c. HClO2, 1.1 x 10-2
d. HCN, 4.9 x 10-10
e. HNO2, 4.6 x 10-4

Answers

Answer:

[tex]HCN~~Ka=4.9x10^-^1^0[/tex]

Explanation:

In this case, we have to remember the relationship between the Ka value and the pH. We can use the general reaction for any acid with his Ka value expression:

[tex]HA~->~H^+~+~A^-[/tex]    [tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]

In the Ka expression, we have a proportional relationship between Ka and the concentration of [tex]H^+[/tex]. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher

So, if we have to find the higher pH value we need to search the smaller Ka value in this case [tex]HCN~~Ka=4.9x10^-^1^0[/tex].

I hope helps!

HCN has the highest pH among all the acids listed in the question.

The Ka is called the acid dissociation constant. It shows the extent to which an acid is ionized in water. The pH shows the hydrogen ion concentration of water. The higher the Ka, the higher the hydrogen ion concentration and the lower the pH.

Hence, HCN has the lowest Ka and the lowest hydrogen ion concentration. Therefore, HCN has the highest pH among all the acids listed in the question.

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If D+2 would react with E-1, what do you predict to be the formula?

Answers

Answer:

DE2

Explanation: for every one D+2 you need two E-1 because +2=-2

Calculate the heat absorbed by a sample of water that has a mass of 45.00 g when the temperature increases from 21.0oC to 38.5 oC. (s=4.184 J/g.o C)

Answers

Answer:

The heat absorbed by the sample of water is 3,294.9 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

Q=?m= 45 gc= 4.184 [tex]\frac{J}{g*C}[/tex]ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 C

Replacing:

Q= 4.184 [tex]\frac{J}{g*C}[/tex] * 45 g* 17.5 C

Solving:

Q=3,294.9 J

The heat absorbed by the sample of water is 3,294.9 J

For the reaction system, 2SO2(g) + O2(g) <--> 2SO3(g), the equilibrium concentrations are: SO3: 0.120M SO2: 0.860M O2: 0.330M Calculate the value of Kc for this reaction.

Answers

Answer:

0.0590 M⁻¹

Explanation:

Kc represents the equilibrium constant. It is given as;

Kc = [products] / [reactants]

For the reaction; 2SO2(g) + O2(g) <--> 2SO3

Products = SO3

Reactants = SO2 and O2

Kc is given as;

Kc = [SO3]² / [SO2]² [O2]

Kc = 0.120² / (0.860)² (0.330)

Kc = 0.0144 / 0.2440 = 0.0590 M⁻¹

Determine which complex of the electron transport chain (respiratory chain) each phrase describes. (Coenzyme Q is also called ubiquinone or ubiquinol, depending on whether it is in oxidized or reduced form.)
Complex I:
Complex II:
Complex III:
Complex IV:
Here are the choices that need to be put in the correct complex:
1) NADH-ubiquinone(NADH-coenzyme Q oxidoreductase)
2) Coenzyme Q-cytochrome c oxidoreductase
3) Electron transfer from succinate to ubiquinone (coenzyme Q)
4) Electron transfer from cytochrome c to O2
5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)
6) Cytochrome c oxidase
7) Electron transfer from ubiquinol (QH2) to cytochrom c
8) Electron transfer from NADH to ubiquinone (coenzyme Q)

Answers

Answer:

Complex I:  (1) NADH-ubiquinone(NADH-coenzyme Q oxidoreductase), (8) Electron transfer from NADH to ubiquinone (coenzyme Q)

Complex II:  (3) Electron transfer from succinate to ubiquinone (coenzyme Q) (5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)

Complex III:  (2) Coenzyme Q-cytochrome c oxidoreductase, (7) Electron transfer from ubiquinol (QH2) to cytochrome c

Complex IV: (4) Electron transfer from cytochrome c to O2, (6) Cytochrome c oxidase

Explanation:

The electron transport chain (ETC) in the mitochondria provides a pathway by which electrons are transferred from NADH and FADH₂ through a series of membrane-bound carriers to  molecular oxygen reducing it to water.

The electron transport chain electron carriers are organized into four complexes, Complexes I - IV.

Complex I : It is also called NADH:ubiquinone reductase. It transfers electrons from NADH to ubiquinone (also known as coenzyme Q)

Complex II : It is also called succinate dehydrogenase. It functions to tranfer electrons from succinate to FAD and then to ubiquinone.

Complex III : It is also called ubiquinone:cytochrome c oxidoreductase. It functions to transfer electrons from ubiquinol (reduced ubiquinone) to cytochrome c.

Complex IV : It is also called cytochrome oxidase. It functions to transfer electrons from cytochrome c to molecular oxygen reducing it to water.

The electron transporter chain is a series of enzymatic reactions to produce and store energy for the organism’s correct functioning. Complex I: 1 and 8. Complex II: 3 and 5. Complex III: 2 and 7. Complex IV: 4 and 6.

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Electron transporter chain

The electron transporter chain is located in the internal mitochondrial membrane. It constitutes a series of enzymatic reactions to release and save energy for the organism’s correct functioning.  

Along the chain, there are four proteinic complexes in the membrane, I, II, III, and IV, that contain the electrons transporters and the enzymes necessary to catalyze the electrons' transference from one complex to the other.  

Different redox reactions occur to pass electrons along the chain.  

Released energy creates a proton concentration gradient used to synthesize ATP.  

1)  NADH provides electrons to the first complex, Complex I (NADH-  

   ubiquinone or NADH-coenzyme Q oxidoreductase).

From there, electrons go to the coenzyme Q (Ubiquinone) that carries them to complex II and III. Meanwhile, complex I pomp four protons to the intermembrane space.  

2) Complex II (succinate-dehydrogenase) receives electrons from CoQ and also receives electrons from FADH2. Electrons are sent from complex II to ubiquinone Q that carries these electrons to complex III.

3) Complex III (Cytochrome C-reductase) receives electrons from ubiquinone Q and pomps protons to the intermembrane space.

Electrons are transferred to Cytochrome c.  

Electrons travel from cytochrome c to complex IV.

4) Complex IV (Cytochrome C-oxidase)  is the last complex that pomps protons to the intermembrane space. It takes electrons from cytochrome C and sends them to oxygen.

5) Electrons are sent to O₂ molecules, which also receive protons in the matrix to create water molecules.

Four electrons are needed to produce two water molecules from one O₂ molecule.  

The proton gradient is used to produce ATP molecules.

Now, we can join the complexes with the phrases.

Complex I:

1) NADH-ubiquinone (NADH-coenzyme Q oxidoreductase)

8) Electron transfer from NADH to ubiquinone (coenzyme Q)

Complex II:

3) Electron transfer from succinate to ubiquinone (coenzyme Q)

5) Succinate-coenzyme Q Oxidoreductase (succinate dehydrogenase)

Complex III:

2) Coenzyme Q - cytochrome c oxidoreductase

7) Electron transfer from ubiquinol (QH₂) to cytochrom c

Complex IV:

6) Cytochrome C oxidase

4) Electron transfer from cytochrome c to O₂

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Assume that a nickel weighs exactly 5.038650 g for the sets of weights listed below obtained by a single weighing on the balance below

Answers

Answer:

afshkkyfugutuiryfyi

oxygen get stable configuration by ____________two electrons



please give the answer as fast as you can


please ​

Answers

Answer:

gaining two electrons

Explanation:

electron configuration

2:6

so add two to 6 to get stable 2:8

Gaining electrons easy

If the SN2 reaction of an aromatic alcohol with an alkyl halide, like the synthesis of nerolin, is successful, what changes would be seen in the IR spectrum for the product compared to the starting material

Answers

Answer:

O-H stretch signal at 3300 cm-1

Explanation:

In this question, we can start with the reaction mechanism for the synthesis of Nerolin. We have to start with naphthalen-2-ol adding NaOH we can produce the alkoxide. Then this alkoxide can react by an Sn2 reaction with bromomethane to produce Nerolin (see figure 1).

In the starting molecule (naphthalen-2-ol) we have an "OH" group. Therefore we will have an O-H stretch signal around 3300 cm^-1. The alcohol signals are very broad and very intense, so this will be the main signal for the initial molecule. In the final product, we dont have the "OH" therefore this signal will disappear (see figure 2).

I hope it helps!

Calculate the mass percent of .485g of H, which reacts with O to form 2.32g H2O?

Answers

Answer:

53.1% of hydrogen reacts

Explanation:

The mixture of 2 atoms of H with 1 atom of O produce 1 molecule of H₂.

The mass of hydrogen in 2.32g of H₂O could be obtained using molar mass of H₂O (18.01g/mol) and molar mass of hydrogen (1.01g/mol) as follows:

Moles H₂O: 2.32g H₂O × (1mole / 18.01g) = 0.1288 moles of water

1 mole of H₂O contains 2 moles of H, moles of hydrogen in 0.1288 moles of water are:

0.1288 moles H₂O × (2 moles H / 1 mole H₂O) = 0.2576 moles of H

In mass:

0.2576 moles H × (1.01g/ mol H) = 0.260g H you have in the formed water

As before reaction you had 0.485g of H and just 0.260g reacted, mass percent is:

(Mass that reacts / Mass added) × 100

(0.260g / 0.485g) × 100 =

53.1% of hydrogen reacts

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