An irregularly shaped stone was lowered into a graduated cylinder holding a volume of water equal to 2.0 ml. The height of the water rose to 7.0 ml. If the mas of the stone was 25 g, what is the density?

The density of the stone whose mass is 25 g and its volume is 5 mL is 5 g/mL.

We have a graduated cylinder holding a volume of water equal to 2 mL. After a stone is added, the height of the water rose to 7 mL. The volume of the stone is equal to the difference in the levels of water.

The stone has a volume of 5 mL and a mass of 25 g. Its density, that is the ratio of the mass to the volume, is:

The density of the stone whose mass is 25 g and its volume is 5 mL is 5 g/mL.The density of the stone is 5 g/cm³.

Mass = 25 g

Volume = 7 mL – 2 mL = 5 mL = 5 cm³

Density = mass/volume = 25 g/5 cm³ = 5 g/cm³

Explanation:

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6