Answer: C2HNO3
Explanation: C = 27.59/12.011 = 2.297
H = 1.15/1.008 = 1.1409
N = 16.09/14.007 = 1.1487
O = 55.17/15.999 = 3.4483
Divide by smallest result:
C = 2
H=1
N=1
O = 3
Empirical formula = C2HNO3
Explanation:
Given -
- An organic compound gives H₂ gas with Na
- On treatment with alkaline iodine it gives yellow ppt.
- On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)
To Find -
- Name the compound and write the reaction involved
Now,
Let A be the organic compound.
Then,
- A + Na → + H₂↑
- A + I₂ → CHI₃ (yellow ppt.)
- A + CrO₃ + H⁺ → C₂H₄O
Now,
Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.
- Functional group of aldehyde = —CHO
And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).
Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.
It means,
We know that 1° alcohol on oxidation gives aldehyde.
Here it gives 2 Carbon aldehyde.
It means,
Here 2 Carbon and 1° alcohol is used.
Now,
Its cleared that Compound A is Ethanol.
Reaction Involved -
- CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
- CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
- CH₃CH₂OH + CrO₃ + H⁺ → CH₃CHO
Because all matter has density and since air has density then air is matter and plus EVERYTHING is made of matter
The formula for pH given the pKa and the concentrations
are:
pH = pKa + log [a–]/[ha]
<span>
Therefore calculating:</span>
3.75 = 3.75 + log [a–]/[ha]
log [a–]/[ha] = 0
[a–]/[ha] = 10^0
<span>[a–]/[ha] = 1</span>
