<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub> would be
welcome.
<sub>2</sub>CrO<sub>4</sub>, is formed when a solution of chromic acid is neutralised by ammonia, and the mixture
evaporated. It yields golden-yellow needles (monoclinic prisms):
<sub>2</sub>CrO<sub>4</sub> =
(NH<sub>4</sub>
<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub> + 2NH<sub>3</sub> + H<sub>2</sub>O"
<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub>. A few recrystallisations would do it.
<sub>2</sub>SO<sub>4</sub> system might work.
K<sub>2</sub>SO<sub>4</sub> is lower solubility than the other three possibilities, but it's not much less than
K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub>, so it may be a challenge to crystallize (for example, it may exhibit a
eutectic sort of behavior as the solvent evaporates).
+ heat, leaching + HCl +
NH<sub>4</sub>Cl might stand a good chance though, because acc. wiki's solubility table
Na<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub> is highly soluble even cold.
<sub>2</sub>CO<sub>3</sub> could
be used for this then..
<sub>2</sub> +
(NH<sub>4</sub>
<sub>2</sub>CO<sub>3</sub> >
PbCO<sub>3</sub> (s) + NH<sub>4</sub>Cr<sub>2</sub>O<sub>7</sub>Quote: Originally posted by Antiswat ![]() |
Quote: Originally posted by plante1999 ![]() |
Quote: Originally posted by plante1999 ![]() |
Quote: Originally posted by blogfast25 ![]() |
Quote: Originally posted by blogfast25 ![]() |
Quote: Originally posted by blogfast25 ![]() |

