Agraduated cylinder contains 155 ml of water. a 15.0-g piece of iron (density = 7.86 g/ml) and a 20.0-g piece of lead (density = 11.3 g/ml) are added. what is the new water level, in milliliters, in the cylinder?

Initial volume of water is 155 mL. Convert the mass of iron and lead into volume using density values as follows:

d=\frac{m}{V}

Here, d is density, m is mass and V is volume of objects

volume of  piece of iron can be calculated as:

V=\frac{m}{d}=\frac{15 g}{7.86 g/mL}=1.90 mL

Similarly, volume of lead can be calculated as:

V=\frac{m}{d}=\frac{20 g}{11.3g/mL}=1.77 mL

To calculate the final volume of water, volume of both iron and lead should be added to the initial volume of water.

V_{final}=V_{initial}+V_{iron}+V_{lead}=155 mL+1.90 mL+1.77mL=158.67 mL

Thus, new water level in milliliters will be 158.67 mL.

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