The density of pure silver is 10.5 g/cm3 at 20 °C. If 5.25 g of pure silver pellets is added to a graduated cylinder containing 11.2 mL of water, to what volume level will the water in the cylinder rise?

since the temperature appears to play no role in the experiment, let's ignore it.

5.25g / (10.5g/cm^3) = 0.5cm^3

so, adding .5ml of Ag to 11.2ml of H₂O, that makes the total volume 11.7ml

11.7

11.8 ml

The goal of the problem is to end up with the total volume of the silver pellets and the water, so we need to first convert the grams of pure silver to mL. 1 cm^3 = 1 mL so we need to first convert the 5.25 g/cm3 into cm3. To do this, divide 5.25 g by 10.5 g/cm3. This cancels out the g and leaves you with cm3.

(5.25 g / x cm3) • (1 cm3/10.5 g) = 0.5 cm3

1 cm3 = 1 mL, so 0.5 mL of pure silver pellets are being added to the water, so simply add 0.5 to 11.2 to get 11.7 as the final volume.

D = m/V

10.5g/cm^3 = m/11.2mL (solve for m)
10.5g/cm^3 * 11.2mL = 117.6g (initial mass)
(add the 5.35g to get 122.85g)
D= m/v
10.5g/cm^3 = 122.85g/V (solve for V)
122.85g/10.5g/cm^3 = 11.7mL

i cant help but have a good day!

This thing makes no sense..;-;