A 2.67 g piece of zinc (density = 7.14 g/mL) is added to a graduated cylinder that contains 12.13 mL H 2 O . What will be the final volume reading on the graduated cylinder, in mL?

Taking into account the definition of density, the final volume reading on the graduated cylinder will be 12.5039 mL.

But first you must know that density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance and the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:  

In this case, you know that:

Replacing in the definition of density:

Solving:

volume= 0.3739 mL

This volume is added to a graduated cylinder that contains 12.13 mL H₂O. Then, the final volume reading on the graduated cylinder is the sum of both volumes:

total volume= 0.3739 mL + 12.13 mL

total volume= 12.5039 mL

In summary, the final volume reading on the graduated cylinder will be 12.5039 mL.

Learn more about density:

12.50 mL

Explanation:

From the question given above, the following data were obtained:

Mass of Zinc = 2.67 g

Density of Zinc = 7.14 g/mL

Volume of water = 12.13 mL

Final volume reading =?

Next, we shall determine the density of the piece of zinc. This can be obtained as follow:

Mass of Zinc = 2.67 g

Density of Zinc = 7.14 g/mL

Volume of Zinc =?

Density = mass / volume

7.14 = 2.67 / volume of Zinc

Cross multiply

7.14 × volume of Zinc = 2.67

Divide both side by 7.14

Volume of Zinc = 2.67 / 7.14

Volume of Zinc = 0.37 mL

Finally, we shall determine the final volume reading of the graduated cylinder. This can be obtained as follow:

Volume of water = 12.13 mL

Volume of Zinc = 0.37 mL

Final volume reading =?

Final volume reading = (Volume of water) + (Volume of Zinc)

Final volume reading = 12.13 + 0.37

Final volume reading = 12.50 mL

Thus, the final volume reading of the graduated cylinder is 12.50 mL

the formula would be mgso4

mgso[subscript]4

fr8t7iot7o87oglgt87ot79p0[90