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ctheowl

  • 2 years ago

Can anyone please help me? You need a 20% alcohol solution. On hand, you have a 400 mL of a 15% alcohol mixture. How much pure alcohol will you need to add to obtain the desired solution?

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  1. princeofwetlot
    • 2 years ago
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    you can set upthe equation as: x mL of 100% alchohol + 400 mL of 15% alcohol = (x + 400) mL of 20% alcohol

  2. satellite73
    • 2 years ago
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    you have 15% of 400 ml = \(.15\times 400=60\) ml of alcohol at present. if you add say \(x\) ml of alcohol you will have \(60+x\) ml of alcohol and \(400+x\) ml of solution you want \(400+x\) ml to be 20% alcohol, so set \[60+x=.20(400+x)\] and solve for \(x\)

  3. satellite73
    • 2 years ago
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    probably easiest to multiply by 10 on both sides to get \[600+10x=2(400+x)\] \[600+10x=800+2x\] \[8x=200\] etc

  4. ctheowl
    • 2 years ago
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    THANK YOU VERY MUCH!

  5. satellite73
    • 2 years ago
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    yw hope steps are clear

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