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Lisa [10]
3 years ago
9

Which is a similarity between alcohol fermentation and aerobic respiration?​

Chemistry
2 answers:
Andrew [12]3 years ago
6 0

Both begin with glycolysis.
Both occur in mitochondria.
Both require oxygen to proceed.
Both end with the electron transport chain.
Alexus [3.1K]3 years ago
3 0

I found this information:

Anaerobic respiration begins the same way as fermentation. The first step is still glycolysis, and it still creates 2 ATP from one carbohydrate molecule. However, instead of ending with glycolysis, as fermentation does, anaerobic respiration creates pyruvate and then continues on the same path as aerobic respiration.

After making a molecule called acetyl coenzyme A, it continues to the citric acid cycle. More electron carriers are made and then everything ends up at the electron transport chain. The electron carriers deposit the electrons at the beginning of the chain and then, through a process called chemiosmosis, produce many ATP. For the electron transport chain to continue working, there must be a final electron acceptor. If that acceptor is oxygen, the process is considered aerobic respiration.

Scientists believe that fermentation and anaerobic respiration are older processes than aerobic respiration.

I hope this help. :)

have a good day!

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You’ve been waiting for the bus and your hands become cold. When you get onto the bus and sit down, you put your hands under you
snow_tiger [21]

Explanation:

The <u>First Law of Thermodynamics</u> states that energy cannot be created or destroyed in an isolated system. In other words, energy can be converted from one form into another, but it cannot be created nor destroyed.

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When you put your cold hands under your legs to warm your hands up, the heat energy from your legs is being transferred to your hands through conduction. However, since energy cannot be created, there is no extra heat energy that can instantaneously replace the heat created by your legs.

3 0
2 years ago
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Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---&gt; CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu
Nitella [24]

ΔH = +438 kJ  

We have three equations:  

(I) N₂ + 3H₂ → 2NH₃; Δ<em>H</em> = -92 kJ  

(II) H₂ +½O₂ → H₂O; Δ<em>H</em> = -286 kJ  

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; Δ<em>H</em> = -632 kJ  

From these, we must devise the target equation:  

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; Δ<em>H</em> = ?  

_________________________________

The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.  

When you reverse an equation, you <em>reverse the sign of its ΔH</em>.  

(V) 2NH₃ → N₂ + 3H₂; Δ<em>H</em> = +92 kJ  

Equation (V) has 1N₂ on the right, and that is not in the target equation.  

You need an equation with 1N₂ on the left.  

<em>Reverse Equation (III).</em>  

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; Δ<em>H</em> = +632 kJ  

Equation <em>(VI)</em> has ³/₂O₂ on the right, and that is not in the target equation.  

You need ³/₂O₂ on the left.  

Multiply <em>Equation (II) by three</em>.  

When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.

(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ  

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.  

When you add equations, you add their Δ<em>H</em> values.  

_______________________________________

We get the target equation (IV):  

(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂;                                    ΔH = +  92 kJ  

(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ  

(VII) <u>3H</u>₂ +³/₂<u>O</u>₂ → <u>3</u>H₂O;                             ΔH =   -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O;          ΔH =  +438 kJ  


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2 years ago
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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
3 years ago
I do not understand this question
boyakko [2]

Answer:

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Explanation: why because 37/19 k toward 0+1e +37/18 AR is equivalent to the beta decay and it describes the best beta decay in the formula.

6 0
3 years ago
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