a 35.0g silver object and a 60.0g gold object are both addedto 80.5 ml of water

Question

a 35.0g silver object and a 60.0g gold object are both addedto 80.5 ml of water contained in a graduated cylinder. what is thenew water level in the cylinder? When density of silver is 10.5 g/ml and density of gold is 19.3 g/ml """ i actually got this question from the mastering chemistrywebsite I'm doing HW on that website for my class"""" a 35.0g silver object and a 60.0g gold object are both addedto 80.5 ml of water contained in a graduated cylinder. what is thenew water level in the cylinder? When density of silver is 10.5 g/ml and density of gold is 19.3 g/ml """ i actually got this question from the mastering chemistrywebsite I'm doing HW on that website for my class""""

Explanation / Answer

The new water level is the new volume of water which is the sum ofthe initial volume of water and the volumes of silver &gold. To find the volume of silver and gold, use the equation density =mass/volume => volume = mass/density VAg= 35.0 g Ag / 10.5 g/mL = 3.33 mL Ag VAu= 60.0 g Au / 19.3 g/mL = 3.11 mL Au So the total volume is 80.5 mL + 3.33 mL + 3.11 mL = 86.9 mL (exactsig figs) Hope this helps!

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