A46.9 gram sample of a substance has a volume of about 3.5 centimeters3. it is solid at a room temperature of 23ºc. out of the four substances whose properties are given, which is the most likely identity of this substance? substance density (g/cm3) melting point (°c) boiling point (°c) molybdenum 10.28 2,623 4,639 mercury 13.53 -39 357 hafnium 13.31 2,233 4,603 lead 11.34 327 1,749 a. molybdenum b. mercury c. hafnium d. lead

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution :  Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5

First, find the Density of given substance.

Formula used :    

Now,put all the values in this formula, we get

= 13.4 g/

So, we conclude that the density of given substance (13.4 g/) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

in theory, limestone is composed principally of the compound calcium carbonate - caco3, but in reality this compound is deposited in conditions involving a number of other compounds as impurities so it ends up being a mixture of many compounds.

hope this : -)

tony (lordcaos066)

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.