I'm thinking the answer you are looking for is light energy. With the sunlight there is light energy and heat.
Answer:
Option B is correct. A nuclear alpha decay
Explanation:
Step 1
This equation is a nuclear reaction. So it can be an alpha decay or a beta decay
An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.
During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.
Thus, radium-226 decays through α-particle emission to form radon-222 according to the equation that is showed.
A Beta decay occurs when, in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other.
Option B is correct. A nuclear alpha decay
Answer:
we can do it again and again and again and again and again and again
Answer:
a) 2-bromopyrrole
Explanation:
Our options for this questions are:
a) 2-bromopyrrole
b) 2,3-dibromopyrrole
c) N-bromopyrrole
d) 3-bromopyrrole
To understand how the reaction works we have to start with the <u>resonance structures</u>. (Figure 1), on these structures, we will obtain a n<u>egative charge on carbon 2</u> in the pyrrole ring, therefore on this carbon we can generate an attack to an electrophile.
The second step is to check how the mechanism take place. An <u>electrophile is generated</u> by the 
and 
. This electrophile can be <u>attacked</u> by the negative charge on carbon 2 producing the 2-bromopyrrole. (See figure 2).
I hope it helps!
Answer:
Less
Explanation:
Since [Cu(NH3)4]2+ and [Cu(H2O)6]2+ are Octahedral Complexes the transitions between d-levels explain the majority of the absorbances seen in those chemical compounds. The difference in energy between d-levels is known as ΔOh (ligand-field splitting parameter) and it depends on several factors:
- The nature of the ligand: A spectrochemical series is a list of ligands ordered on ligand strength. With a higher strength the ΔOh will be higher and thus it requires a higher energy light to make the transition.
- The oxidation state of the metal: Higher oxidation states will strength the ΔOh because of the higher electrostatic attraction between the metal and the ligand
A partial spectrochemical series listing of ligands from small Δ to large Δ:
I− < Br− < S2− < Cl− < N3− < F−< NCO− < OH− < C2O42− < H2O < CH3CN < NH3 < NO2− < PPh3 < CN− < CO
Then NH3 makes the ΔOh higher and it requires a higher energy light to make the transition, which means a shorter wavelength.
