Answer:
it says the answer is variation
Answer:
569K
Explanation:
Q = 3.5kJ = 3500J
mass = 28.2g
∅1 = 20°C = 20 + 273 = 293K
∅2 = x
c = 0.449
Q = mc∆∅
3500 = 28.2×0.449×∆∅
3500 = 12.6618×∆∅
∆∅ = 3500/12.6618
∆∅ = 276.4220
∅2 - ∅1 = 276.4220
∅2 = 276.4220 + ∅1
∅2 = 276.4220 + 293
∅2 = 569.4220K
∅2 = 569K
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
