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3 years ago

Do free electrons appear anywhere in the balanced equation for a redox reaction?

2 answers:

7 0

Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.

*Note: Overall redox reactions do not show free electrons because they cancel each other out from both sides of the half-reaction equations.

Andru [333]3 years ago

4 0

Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.

Explanation:

A half reaction is either the chemical reaction or reduction reaction part of an oxidoreduction reaction. A half reaction is obtained by considering the amendment in chemical reaction states of individual substances concerned within the oxidoreduction reaction. Half-reactions are usually used as a way of leveling oxidoreduction reactions.The half-reaction on the anode, wherever chemical reaction happens, is Zn(s) = Zn2+ (aq) + (2e-).

The metal loses 2 electrons to create Zn2+. The half-reaction on the cathode wherever reduction happens is Cu2+ (aq) + 2e- = Cu(s).

Here, the copper ions gain electrons and become solid copper.

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Which of these statements is true about the temperature of a substance?

Svetradugi [14.3K]

Answer:

(c)

Explanation:

4 0

3 years ago

Please help me please i will mark you brainliest. In the nucleus of an atom, there are 17 protons and 18 neutrons. What is the a

prohojiy [21]

Answer:

i can only help with 2 :( atimic mass is: 35.453 and number of electrons is:17

Explanation:

7 0

2 years ago

A hydrocarbon contains 85.7% carbon and the remainder

Viefleur [7K]

Answer:

CH₂ ; 67.1 %

Explanation:

To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation

Assume 100 grams of the compound.

# mol C = 85.7 g / 12.01 g/mol = 7.14 mol

# mol H = 14.3 g / 1.008 g/mol = 14.19 mol

The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C

So the empirical formula is CH₂

For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.

We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )

1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol

Theoretical yield from balanced chemical equation:

0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆

Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g

% yield = 0.295 g/ 0.440 g x 100 = 67.1 %

6 0

3 years ago

Why do you require an acid catalyst to make an ester? Why not just mix acid and alcohol? Describe an alternate method of making

djverab [1.8K]

Answer:Acid catalyst is needed to increase the electrophilicity of Carbonyl group of Carboxylic acid as alcohol is a weak nucleophile.

Alternatively esters can be synthesised by converting carboxylic acid into acyl chloride using thionyl chloride(SOCl_{2} and then further treating acyl chloride with alcohol.

Carboxylic acid and esters can be easily distinguished on the basis of IR as carboxylic acid would contain a broad intense peak in 2500-3200cm_{-1} corresponding to OH stretching frequency whereas esters would not contain any such broad intense peak.

Alcohol and esters can also be distinguished using IR as alcohols would contain a broad intense peak at around 3200-3600cm_{-1}

Explanation: For the synthesis of esters using alcohol and carboxylic acid we need to add a little amount of acid in the reaction . The acid used here increases the electrophilicity of carbonyl carbon and hence makes it easier for a weaker nucleophile like alcohol to attack the carbonyl carbon of acid.

The oxygen of the carbonyl group is protonated using the acidic proton which leads to the generation of positive charge on the oxygen. The positive charge generated is delocalised over the whole acid molecule and hence the electrophilicity of carbonyl group is increased. Kindly refer attachment for the structures.

If we simply mix the acid and alcohol then no appreciable reaction would take place between them and ester formation would not take place because the carboxylic acid in that case is not a good electrophile whereas alcohol is also not a very strong nucleophile which can attack the carbonyl group.

Alternatively we can use thionyl chloride or any other reagent which can convert the carboxylic acid into acyl chloride. Acyl chloride is very elctrophilic and alcohol can very easily attack the acyl chloride and esters could be synthesized.

The carboxylic acid and ester can very easily be distinguished on the basis of broad intense OH stretching frequency peak at around 2500-3200cm_{-1} . The broad intense OH stretching frequency peak is present in carboxylic acids as they contain OH groups and absent in case of esters .

Likewise esters and alcohols can also be distinguished on the basis IR spectra as alcohols will have broad intense spectra at around 3200-3600cm_{-1}corresponding to OH stretching frequency whereas esters will not have any such peak. Rather esters would be having a Carbonyl stretching frequency at around 1720-1760

4 0

3 years ago

Examine the diagram that shows the levels of organization of an organ system.

romanna [79]

Answer:

lol

Explanation:

8 0

3 years ago