Answer :
- Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as
Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.
The general formula of carboxylic is, 
. According to the IUPAC naming, the carboxylic are named as alkanoic acids.
Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.
The general representation of aldehyde is, 
. According to the IUPAC naming, the aldehyde are named as alkanals.
Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.
The general representation of ketone is, 
. According to the IUPAC naming, the ketone are named as alkanone.
Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.
The general representation of ester is, 
. According to the IUPAC naming, the ester are named as alkyl alkanoate.
Answer:
A
Explanation:
the correct answer is A
This is the sidereal month. It is measured against the background stars. However, the Earth is also orbiting the Sun, and the moon needs extra time to catch up to its previous position relative to the Sun.
4.1 h = 14760 s
<span>t 1/2 = ln 2 / k </span>
<span>k = rate reaction = 4.97 x 10^-5 </span>
<span>ln 0.045 / 0.36 = - 4.97 x 10^-5 t </span>
<span>2.08 = 4.97 x 10^-5 t </span>
<span>t = 41839.9 s = 11 h 37 min 19 s</span>
A) c = 3 x 10^8 m/s
f = 7.15 x 10^14 Hz
c = λ x f (=) λ = 3 x 10^8 / 7.15 x 10^14 = 4.19 x 10^-7 m = 419.6 nm
B) E = h f
H = Planck's constant = 6.63 x 10^-34 J/s
E = 6.63 x 10^-34 x 7.15 x 10^14 = 4.74 x 10^-19 J
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Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.
