Agraduated cylinder contains 63.0 ml of water. a piece of gold, which has a density of 19.3 g/ cm3, is added to the water and the volume goes up to 64.5 ml. calculate the mass in grams of the gold that was added to the water. explain how you got your answer.

29.0 g

Explanation:

The mass of the piece of gold is given by:

m = dV

where

m is the mass

d is the density

V is the volume of the piece of gold

The density of gold is

d = 19.3 g/cm^3

while the volume of the sample is equal to the volume of displaced water, so

V = 64.5 mL - 63.0 mL = 1.5 mL

And since

1 mL = 1 cm^3

the volume is

V = 1.5 cm^3

So the mass of the piece of gold is:

m = (19.3 g/cm^3)(1.5 cm^3)=29.0 g

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.