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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

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