Question

an electrolyte solution has an average current density of 1 ampere per square decimeter (\\(\frac{a}{dm^{2}}\\)).
what is the current density of the solution in \\(\frac{a}{m^{2}}\\)?
\\(\square\frac{a}{m^{2}}\\)

Explanation:

Step1: 单位换算关系

因为\(1m = 10dm\)，所以\(1m^{2}=(10dm)^{2}=100dm^{2}\)。

Step2: 计算电流密度

已知电流密度为\(1\frac{A}{dm^{2}}\)，将\(dm^{2}\)换算为\(m^{2}\)，根据\(1m^{2}=100dm^{2}\)，则\(1\frac{A}{dm^{2}}=\frac{1A}{\frac{1}{100}m^{2}}\)。
根据分数除法法则\(a\div\frac{b}{c}=a\times\frac{c}{b}\)（\(a = 1A\)，\(b = 1\)，\(c = 100\)），可得\(\frac{1A}{\frac{1}{100}m^{2}}=1\times100\frac{A}{m^{2}}\)。

Answer:

\(100\)