Respuesta :
Answer: [tex]Ce(IO_3)_4[/tex] = 0.123g , [tex]RaSO_4[/tex] = [tex]2.1*10^-^4g[/tex] and [tex](NH_4)2SeO_4[/tex] = 96 g
Explanations: The question asks about the solubility of each of the compounds in grams per 100 mL of water. density of water is 1 gram per mL. So, 100 mL of water would be same as 100 g water.
In solubility charts, the solubilities are also shown in grams of compound per 100 grams of water.
Solubility of [tex]Ce(IO_3)_4[/tex] is 0.123 g, solubility of [tex]RaSO_4[/tex] is 0.00021 g and the solubility of [tex](NH_4)2SeO_4[/tex] is 96 g per 100 g of water.
Since the question also asks about grams of each soluble in 100 mL that is 100 g of water, the grams of each salts will be same as their above solubilities that is, [tex]Ce(IO_3)_4[/tex] is 0.123 g, [tex]RaSO_4[/tex] is 0.00021 g and the solubility of [tex](NH_4)2SeO_4[/tex] is 96 g.
The solubility of the given compounds in grams per 100 mL, as obtained
online are presented as follows;
[tex]\begin{array}{|l|l|c|l|c|} \mathbf{\underline{Compound}} & \mathbf{\underline{Formula}}&\underline{0^{\circ}C}&\underline{20^{\circ}}&\\&&&&\\Cerium(IV)iodate&Ce(IO_3)_4&&&0.015\\Radium \ sulfate&RaSO_4&&2.1 \times 10^{-4}&\\Ammonium \ selenate&(NH_4)_2SeO_4&96&115&\end{array}\right][/tex]
The details of the solubility of the substances in the above table of values,
are;
- 0.015 gram of cerium(IV) iodate, Ce(IO₃)₄ will dissolve in 100 mL of water.
- 2.1 × 10⁻⁴ gram of radium sulfate, RaSO₄, will dissolve in 100 mL of water at 20°C.
- 96 grams of ammonium selenate, (NH₄)₂SeO₄, will dissolve in 100 mL of water at 0°C, and 115 grams of (NH₄)₂SeO₄, will dissolve in 100 mL of water at 20°C.
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