Answer:
Barium nitrate or silver nitrate based on the anion our solute contains
Explanation:
I assume the situation is that currently the solute is soluble in water and you wish to make it insoluble.
It really depends on the soluble material you have, however, let's look at some specific cases.
- We have a salt in our solution. Addition of any of the three reagents will produce a double displacement reaction, that is, our cation will be replaced by another cation, either sodium, barium or silver cation.
- According to the solubility rules, all sodium salts are soluble, so sodium nitrate won't precipitate our anion.
- In case our solute contains sulfate, carbonate or phosphate, we may use barium nitrate to precipitate it, as barium sulfate, barium carbonate and barium phosphate are insoluble.
- In case our solute contains chloride, then silver nitrate is the way to go to precipitate it in an insoluble form of AgCl. Similarly, silver would form precipitates with carbonate, phosphate, iodide, bromide and slightly soluble silver sulfate (barium is the choice for sulfate, however).
Answer:
Iodine. The reaction between hot iron and iodine vapor produces gray iron(II) iodide, and is much less vigorous. This reaction, the equation for which is given below, is difficult to carry out because he product is always contaminated with iodine. Iodine is only capable of oxidizing iron to the +2 oxidation state.
Answer:
Molecules with three electron pairs
Explanation:
=>
Answer:6 valence electrons
and
one valence electron
Explanation:
1) D = 13.6 g / mL
2)ethyl alcohol weighs 158g
3)ρ
_copper = 8.9 g 
Explanation:
1)
D = m / V
=306.0 g / 22.5 mL
D= 13.6 g / mL
2)
density = mass / volume
mass = density × volume
=0.789g /ml × 200.0 ml
M=158g
Ethyl alcohol weighs 158g
3)
ρ (density) = Mass / Volume
ρ
_copper = 1896 g / 8.4cm × 5.5cm × 4.6cm
= 1896g / 212.5
ρ
_copper=8.9 g
