The ionic radius increases.
Each time you move down to the next Period, you are adding an extra shell of electrons.
Each shell is further from the nucleus than the one before it, so the .
The image below illustrates the increase in ionic radius as you go down a Group.
Melting point is dependent on the intermolecular forces which means the bonds between the molecules of bromine as it is a simple molecular structure
Answer:C is most likely a metalloid
Explanation:A is a metal while B is a non metal. This is because when we look at the properties of metals, they are very lustrous and have a high conductivity. On the other hand, non metals are the opposite but metalloids have properties the are inbetween ie, they react with either an acid or base, and have a medium electrical conductivity unlike metals(high) anc non metals(low).
Answer:
NH3(aq)
Explanation:
Gold III hydroxide is an inorganic compound also known as auric acid. It can be dehydrated at about 140°C to yield gold III oxide. Gold III hydroxide is found to form precipitates in alkaline solutions hence it is not soluble in calcium hydroxide.
However, gold III hydroxide forms an inorganic complex with ammonia which makes the insoluble gold III hydroxide to dissolve in ammonia solution. The equation of this complex formation is shown below;
Au(OH)3(s) + 4 NH3(aq) -------> [Au(NH3)4]^3+(aq) + 3OH^-(aq)
Hence the formation of a tetra amine complex of gold III will lead to the dissolution of gold III hydroxide solid in aqueous ammonia.
Answer:
D. [CO₂]
Explanation:
Let's consider the following equation at equilibrium.
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [CO₂]
