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Setler [38]
1 year ago
7

The question is on the pic, thanks :)

Chemistry
1 answer:
Inessa05 [86]1 year ago
8 0
It’s the BOA not the dog or kangaroo
You might be interested in
Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.
Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0
2 years ago
Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are and . Does the cell po
Free_Kalibri [48]

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

6 0
2 years ago
Consider the general reversible reaction. Lower A upper A plus lower B upper B double-headed arrow lower C upper C plus Lower d
dimulka [17.4K]

Answer: K e q equals StartFraction StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D over StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b EndFraction.

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients. It is represented by the symbol K_{eq}

The balanced chemical reaction is:

aA+bB\rightleftharpoons cC+dD     

The expression for K_{eq} is written as:

K_{eq}=\frac{[C]^c\times [D]^d}{[A]^a\times [B]^b}

Thus the correct option is K e q equals StartFraction StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D over StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b EndFraction.

3 0
2 years ago
A tiny sample of an aqueous solution of two substances R and P is sketched below as if it was under an imaginary microscope so p
PilotLPTM [1.2K]

Answer:

R = 6, P = 6

Explanation:

Answer

Number of R molecules = 6

Number of P molecules = 6

Explanation

R(aq) <-------> P(aq)

K = [P]/[R]

at equillibrium

[ P ] = 10+x

[R] = 2-x

10 + x /(2-x) = 1

10 + x = 2- x

2x = -8

x = -4

Therefore,

[ R] = 2 - ( -4) = 6

[ P ] = 10 + ( -4) = 6

7 0
2 years ago
A student made a graph to show the chemical equilibrium position of a reaction.
pogonyaev

Answer:

Concentration, because the amounts of reactants and products remain constant after equilibrium is reached.

Explanation:

The rate of reaction refers to the amount of reactants converted or products formed per unit time.

As the reaction progresses, reactions are converted into products. This continues until equilibrium is attained in a closed system.

When equilibrium is attained, the rate of forward reaction is equal to the rate of reverse reaction, hence the concentration of reactants and products in the system remain fairly constant over time.

When deducing the rate of reaction, concentration of the specie of interest is plotted on the y-axis against time on the x-axis.

8 0
2 years ago
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