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Musya8 [376]
3 years ago
11

Explain how a redox reaction involves electrons in the same way that acid-base reaction involve protons

Chemistry
1 answer:
lana [24]3 years ago
8 0
Redox reaction stands for reduction/oxidation reaction whereas acid-base reaction involves acid and base in its reaction. 
Acid-base reaction involves the transfer of protons whereas redox reaction involves the transfer of electrons. Thus in both the case, a charged species is transferred from one molecule to another. 
In case of redox reaction, oxidation involves loss of electrons and reduction involves gain of electrons. Thus, one substance gets oxidized (acts as reducing agent and donates electrons) while other gets reduced (acts as oxidizing agent and gains electrons).
Similarly, in acid-base reaction one substance accepts proton while other releases proton. Thus, one substance is called conjugate acid (the base that accepts protons) and other is called conjugate base (acid that donates protons). 
Thus reducer and oxidizer are analogous terms for acid and base respectively.
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A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a
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<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

We are given:

Mass percentage of CH_4 = 20 %

So, mole fraction of CH_4 = 0.2

Mass percentage of C_2H_4 = 30 %

So, mole fraction of C_2H_4 = 0.3

Mass percentage of C_2H_2 = 35 %

So, mole fraction of C_2H_2 = 0.35

Mass percentage of C_2H_2O = 15 %

So, mole fraction of C_2H_2O = 0.15

We know that:

Molar mass of CH_4 = 16 g/mol

Molar mass of C_2H_4 = 28 g/mol

Molar mass of C_2H_2 = 26 g/mol

Molar mass of C_2H_2O = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}

where,

\chi_i = mole fractions of i-th species

m_i = molar masses of i-th species

n_i = number of observations

Putting values in above equation:

\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}

\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75

Hence, the correct answer is Option d.

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