Answer:
[tex]r_B=9.6M/s[/tex]
Explanation:
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In this case, according to the given chemical reaction, it is possible for us to relate the rate of formation of C and the rate of consumption of B as shown below:
[tex]\frac{r_B}{-4}=\frac{r_C}{2} \\\\[/tex]
Thus, we solve for the rate of change of B as shown below:
[tex]r_B=\frac{-4r_C}{2}\\\\r_B=\frac{-4(4.8M/s)}{2} \\\\r_B=9.6M/s[/tex]
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Based on the equation of the reaction, rate of decrease for [B] when [C] is increasing at 4.8 M/s is 9.6 M/s.
The equation of the reaction is given below:
From the equation of the reaction, 4 moles of B decomposes to give 2 moles of C at any given time.
Given that the rate of increase in [C] is 4.8 M/s, we can derive an equation thus:
rate of [B]/4 = rate of [C]/+2
rate of [B]= rate of [C] × 4/2
rate of [B] = 4.8 M/s × 4/2
rate of [B] = 9.6 M/s.
Therefore, rate of decrease for [B] when [C] is increasing at 4.8 M/s is 9.6 M/s.
Learn more about rate of reaction at: https://brainly.com/question/24795637
