Answer:
Cu(s) in Cu(NO₃)₂(aq)
Explanation:
The standard reduction potential (E°) is the energy necessary to reduce the atom in a redox reaction. When an atom reduces it gains electrons from other than oxides. As higher is E°, easily it will reduce. The substance that reduces is at the cathode of a cell, where the electrons go to, and the other that oxides are at the anode of the cell.
The standard reduction potentials from Al(s) and Cu(s) are, respectively, -1.66V and +0.15V, so the half-cell of Cu(s) in Cu(NO₃)₂(aq) is the cathode.
Hey there!
Sulphur is the element which has 6 electrons in the outermost shell.
WhereasCalcium,Neon have 2 and chlorine has 7.
Hope this explanation helps you :)
Have a good day :)
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.
Answer:
This reactivity order reflects both the strength of the C–X bond, and the stability of X(–) as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class.
Answer:
Wavelength = 0.06024 m
Explanation:
Equation of energy of wavelength from Einstein's relativity equation is;
E = hc/λ
Where;
h is Planck's constant = 6.626 × 10⁻³⁴ J. s
c is speed of light = 3 × 10⁸
We are given E = 3.3 × 10^(-24) J
Making wavelength λ the subject, we have;
λ = hc/E
Thus;
(6.626 × 10⁻³⁴ × 3 × 10⁸)/(3.3 × 10^(-24)) = 6.024 × 10^(-2) = 0.06024 m
Wavelength = 0.06024 m
