prep数学题求问

senru819

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?100% decrease
50% decrease
40% decrease
40% increase
50% increase


答案是40%increase


这个应该怎么列方程呢？

senru819

哪位大牛支个招呀~万分感谢！

飞扬一直飞

我之前做这道题目也错了……
贴上manhattan的解释
A direct relationship between y and x can be written y = kx.
An inverse relationship between y and x can be written y = k/x.

So in this problem, r = kA^2 / B

Now in our new situation, let's call the new concentration of A as A', and the new concentration of B as B'. In our new situation the rate is the same, so
new rate = old rate
k (A')^2 / B' = k A^2 / B

We know that if B is increased by 100%, that means B is doubled, so B' = 2B.
k (A')^2 / 2B = k A^2 / B

We can multiply both sides by B/k, so
(A')^2 = 2 A^2

This means that A' = A sqrt (2). You should have memorized that:
sqrt (2) = 1.4
sqrt (3) = 1.7
so A' = 1.4 A, or a 40% increase.

senru819

谢谢！