Answer : Methanal also known as Formaldehyde
is a chemical Aldehyde which contain ( -CHO) group.
Explanation :
In organic chemistry, a carbonyl group is a functional group which contain a carbon atom double-bonded to an oxygen atom i.e, ( C=O).
If carbonyl group is present in a compound then it can be a carboxylic (RCOOH), aldehyde (RCHO), ketone (RCOR'), ester ((RCOOR') or amide (RCONR'R") group.
Here are some functional groups naming according to the<em> IUPAC</em> rules and image also attached,
Carboxylic acid → (RCOOH) → ( name end in 'OIC ACID' )
Aldehyde → (RCOH) → ( name end in 'AL' )
Ketone → (RCOR') → ( name end in 'ONE' )
Ester → (RCOOR') → ( name end in 'ATE' )
Amide → (RCONR'R") → ( name end in 'AMIDE' )
In an aldehyde, atleast one hydrogen atom must be attached to the carbonyl carbon. For an aldehyde, remove ( -e) from alkane name and add ( -al) at the end of the compound.
Methanal is the IUPAC name for Formaldehyde.
Because of the strong attractions between polar water molecules.
Answer:
4.52 x 10¹⁴ cycles/s
Explanation:
From c = f·λ => f = c/λ = (3.0 x 10⁸ m/s)/(6.63 x 10⁻⁷m) = 4.52 x 10¹⁴ cycles/s.
f = frequency = ?
λ = wavelength = 6.63 x 10⁻⁷ meter
c = speed of light in vacuum = 3.0 x 10⁸ meters/s
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days