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:
0.234 M
Explanation:
C- 12.009 x 7
H- 1.001 x 5
N- 14.006
O- 16 x 3
S- 32.059
___________+
183.133 g/mol
= 0.234 M Cancel out the grams mol/L equals molarity. Lowest significant figure is 3
The most abundant carbon isotope is carbon-12.
The relative atomic mass of carbon is 12.011, which is extremely close to 12.0. This means that the masses C-13, and C-14 are practically negligible when contributing to the relative atomic mass of carbon.
the C-12 isotope makes up 98.9% of carbon atoms, C-13 makes up 1.1% of carbon atoms, and C-14 makes up just a trace of carbon atoms as they are found in nature.
Answer: sodium amide undergoes an acid -base reaction
Explanation:
sodium amide is a ionic compound and basically exists as sodium cation and amide anion. Amide anion is highly basic in nature and hence as soon as there is amide anion generated in the solution , Due to its very pronounced acidity it very quickly abstracts the slightly acidic proton available on methanol.
This leads to formation of ammonia and sodium methoxide.
Hence sodium amide reacts with methanol and abstracts its only acidic proton and form ammonia and sodium Methoxide.
Hence the 3rd statement is a corrects statement.
So we cannot use methanol for sodium amide because sodium amide itself would react with methanol and the inherent molecular natur of sodium amide would then change.
The 1st and 2nd statements both are incorrect because both the compounds methanol as well as sodium amide have dipole moments and hence are polar molecules.
The 4th statement is also incorrect as both the molecules have dipole moment and hence there would be ion-dipole forces operating between them.
The following reaction occurs:
NaNH₂+CH₃OH→NH₃+CH₃ONa
Answer: London dispersion forces
