问一道数学题~

Yvonne320

题目是这样的。。The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentracion of chemical B is increased by 100 percent. which is the closest percentage in the concentration of chemical A required to keep the reaction rate unchaged?

答案是 40 % increase

bull11389

rate=K(一个系数)*(A成分的平方)/ (B成分）。现在分母上乘以了2，分

子也要变2倍，rate才能保持不变。比较选项，找出根号下（1+x%）最接近2

的选项，1.40的平方为1.96，应该是最接近2的吧~~

liuqiaoqi77

根据题意列式子：
M=  K.A（平方）/B  
B增加100% 则M'=K.A(平方）/2B
M'=M
则分子变成K.(根号2.A )平方
则A的浓度变成----根号2*A----约等于1.4A
增加40%

Yvonne320

谢谢啊。。我懂了