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vichka [17]
3 years ago
4

Fill in the blanks with vocabulary terms.All vocab terms should be in lower case.

Chemistry
1 answer:
zepelin [54]3 years ago
3 0

Answer :

<u>Oxidation </u>is the loss of electrons.

<u>Reduction </u>is the gain of electrons.

The compound that became reduced acts as the <u>oxidizing </u>agent.

The compound that became oxidized acts as the <u>reducing </u>agent.

The measure of a compounds likeliness to gain or lose an electron is its <u>electrochemical potential</u> (E value).

A common electron carrier we will use a lot in this class is <u>NAD⁺</u> when it is in the oxidized state and <u>NADH </u>when it is in the reduced state.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced.

Electrochemical potential : It is defined as the measurement of the potential difference between two half cells.

Electron carrier : The molecules that are capable of accepting 1 or 2 electrons from one molecule and donating to another molecule in the process of electron transport.

There are two important electron carriers:

Nicotinamide adenine dinucleotide (NAD⁺ in its oxidized form and NADH in its reduced form).

Flavin adenine dinucleotide (FAD)

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How many moles of molecules are in 40g of H2O?
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Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid
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Answer: a) The K_a of acetic acid at 25^0C is 1.82\times 10^{-5}

b) The percent dissociation for the solution is 4.27\times 10^{-3}

Explanation:

CH_3COOH\rightarrow CH_3COO^-H^+

 cM              0             0

c-c\alpha        c\alpha          c\alpha

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.10 M and \alpha = ?

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2.87=-log[H^+]  

[H^+]=1.35\times 10^{-3}M

[CH_3COO^-]=1.35\times 10^{-3}M

[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M

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\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}

5 0
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