Measure the brightness of a star through two filters and compare the ratio of red to blue light.
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
Answer: sodium-11 iron-26 sulfur-16
Explanation:
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87
