Question
Asked Sep 12, 2019
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A 30.0‐g aluminum sphere and a 40.0‐g iron sphere are both added to 65.7 mL of water contained in a graduated cylinder. What is the new water level, in milliliters, in the cylinder?

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Expert Answer

Step 1

Volume of 30 g aluminum sphere and 40.0 g iron sphere can be determined using its densities. The densities of aluminium and iron are 2.70 g/cc and 7.87 g/cc respectively.

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30.0g-11.1cm3=11.1mL 2.70g/cm3 mass cm1mL A1 density 40.0g 7.87 g/cm mass -5,08 cm3 =5.08 mL VFe density

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Step 2

Hence, volume occupied by aluminum and iron are 11.1 ml and 5.08 ml...

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Initial volume volume of iron + volume of aluminium Final Volume Final Volume = 65.7+ 11.1 + 5.08 Final Volume = 81.8 ml

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