The following equation shows the equilibrium in an aqueous solution of ammonia:NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Which of the follo
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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.
That's the answer on that picture
