Answer:How many moles of aluminum are needed to react completely with 1.2 mol of FeO? 2Al(s) + 3FeO(s) = 3Fe(s) + Al2O3(s). 0.8 mols.
Explanation:
Answer:
6.52 × 10^14 Hz
i don't know if that's right tbh
Dry and desert conditions
Answer:
The oxidation number of the metal decreases
2 Al + Fe₂O₃ → Al₂O₃ + 2 FeO
The metal element iron, is reduced from Fe⁺³ in Fe₂O₃ to Fe⁺² in FeO
Explanation:
When an element gains electron, the element becomes reduced, hence when a metal is reduced, the metal gains electrons, which reduces the oxidation number of the metal
An example of a metal being reduced is;
2 Al + Fe₂O₃ → Al₂O₃ + 2 FeO
In the above reaction, the iron (III) oxide is reduced to iron (II) oxide by aluminium metal.
Answer:
6.4 × 10^-10 M
Explanation:
The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).
CaF2 will dissociate as follows:
CaF2 ⇌Ca2+ + 2F-
1 mole of Calcium ion (x)
2 moles of fluorine ion (2x)
NaF will also dissociate as follows:
NaF ⇌ Na+ + F-
Where Na+ = 0.25M
F- = 0.25M
The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M
Ksp = {Ca2+}{F-}^2
Ksp = {x}{0.25}^2
4.0 × 10^-11 = 0.25^2 × x
4.0 × 10^-11 = 0.0625x
x = 4.0 × 10^-11 ÷ 6.25 × 10^-2
x = 4/6.25 × 10^ (-11+2)
x = 0.64 × 10^-9
x = 6.4 × 10^-10
Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M
