Answer:
Step-by-step explanation:
[tex]\text{Let}\ x-\text{a volume of pure acid}\\\\p\%=\dfrac{p}{100}\\\\35\%=\dfrac{35}{100}=0.35\\\\60\%=\dfrac{60}{100}=0.6\\\\600mL-\text{the volume of a 35}\%\ \text{ acid solution}\\\\(600+x)mL-\text{the volume of a 60}\%\ \text{ acid solution}\\\\(0.35)(600)-\text{the volume of a pure acid in a 35}\%\ \text{ acid solution}\\\\(0.35)(600)+x-\text{the volume of a pure acid in a 60}\%\ \text{ acid solution}\\\\(0.6)(600+x)-\text{the volume of a pure acid in a 60}\%\ \text{ acid solution}[/tex]
[tex]\text{The equation:}\\\\(0.35)(600)+x=(0.6)(600+x)\qquad\text{use the distributive property}\\\\210+x=(0.6)(600)+(0.6)(x)\\\\210+x=360+0.6x\qquad\text{subtract 210 from both sides}\\\\x=150+0.6x\qquad\text{subtract}\ 0.6x\ \text{from both sides}\\\\0.4x=150\qquad\text{divide both sides by 0.4}\\\\x=375[/tex]
