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A 5.0 gram sample of lead and a 3.2 g sample of iron are placed into 367 mL of water. What will be the new volume level of water in units of mL? The density of lead is 11.34 g/cc and the density of iron is 7.874 g/mL. Round your answer to three significant figures. Do not enter “mL” as part of your answer.




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Answer :

Density=mass / volume     ⇒   volume=mass / density

1)We compute the volume of 5 g of lead

Data:
Density= 11.34 g / cm³ 
mass= 5 g
1 ml=1 cm³

volume= 5 g / (11.34 g/ cc)=0.441 cm³=0.441 ml

2) we compute the volume of 3.2 g of iron.
Data:
density= 7.874 g/ ml
mass=3.2 g

volume=3.2 g / (7.874 g/ml)=0.406 ml

3) we compute the total volume ( lead + iron) of the metals.
Total volume of the metals = 0.441 ml + 0.406 ml=0.847 ml

4) we compute the new volume level of water :
New volume level of water=367 ml + 0.847 ml=367.847 ml

Answer: The new volume level of water in units of ml is 367.847 

Answer: 368

Explanation: The question asks you to round your answer to 3 sig figs.

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