You have two pure substances that you cannot identify each sample is solid at room temperature describe at least five steps in the appropriate sequence that you would take to be able to identify the substances

Explanation:

If we have two solid samples, in order to identify what they are a series of ordered steps have to be performed.

1) The first thing to do is to observe the sample. If there is color it may indicate the presence of certain anions: for example if the sample is purple, it can be because of the presence of the permanganate ion (MnO₄⁻), if it is yellow it can be chromate ion (CrO₄⁻), if it is orange it can be the dichromate Cr₂O₇²⁻), etcetera. If the color of the sample is white we have no indication whatsoever.

2) Then we can use certain reactants to precipitate the cations of the sample. For example, we can add first HCN 3N to our sample. If there is precipitation, it means that the cations Ag⁺ or Pb²⁺ are present. If not, there are other cations and we must use a different reactant to precipitate them.

3) We then add H₂S to the sample (not adding it per se, but generating it heating thioacetamide with water). If we see a black precipitate, it can be because of the cations Pb²⁺, Bi³⁺ or Cu²⁺. If we see a yellow precipitate, it corresponds to Cd²⁺. If we do not see a precipitate, we need to add other reactant.

4) We add NaOH to the sample. If we see precipitate, it can be because of the the ions Fe³⁺, Ni²⁺, Co²⁺ or Mn²⁺.

5) We observe the color of this precipitate. If it is brown is Fe(OH)₃, if it is green is Ni(OH)₂, if it is pink is Co(OH)₂, and if it is white is Mn(OH)₂.

Se toma la muestra problema o alícuota y se añade HCl 2N. Con este reactivo precipitan los cationes del Grupo I ( Plata (I), Plomo (II) y Mercurio (I)): AgCl, PbCl2 y Hg2Cl2. Sobre el mismo embudo se añade agua de ebullición, quedando en el papel de filtro el AgCl y el Hg2Cl2; el Pb2+ puede identificar añadiendo KI, que origina un precipitado de PbI2 que se disuelve en caliente, que sirve para identificarlo mediante la llamada lluvia de oro.1

Sobre el mismo papel de filtro se añade NH3 2N. En el papel de filtro si existe Hg22+ y se forma una mancha blanca, gris o negro, que es una mezcla de HgClNH2 y Hg0. En la disolución se forman Ag(NH3)2+, que se puede identificar con KI dando un precipitado de AgI amarillo claro.

answerhypothesis;