Question

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

Answer 1

`V_(c)=V_{H_(2)O}+V_(Pb)+ V_(Fe)\\\\ V_{H_(2)O}=367mL\\\\\\ V_(Pb):\\ m=5g\\ d=11,34(g)/(mL) \ \ \ \Rightarrow \ \ \ V=(m)/(d)=(5g)/(11,34(g)/(cm^(3)))\approx0,441mL\\\\\\ V_(Fe):\\ m=3,2g\\ d=7,874(g)/(mL) \ \ \ \Rightarrow \ \ \ V=(m)/(d)=(3,2g)/(7,874(g)/(mL))\approx0,406mL\\\\\\ V_(c)=367mL + 0,441mL+0,406mL=367,847mL`