Answer:
Released
Explanation:
When particles that attract each other come together, energy is usually released. The combination of the two particles is expected to result in a lower energy system. This lower energy system will be more stable than the different individual particles.
Hence, as this lower energy system is formed, the excess energy originally possessed by the particles is evolved hence energy is released when particles that attract each other are allowed to come together.
Number of neutrons = Atomic mass - Atomic number (number of protons)
Hope that helps!
Answer:
Follows this order: B=> A => C.
Explanation:
NB: kindly check the attachment for the diagram of compounds A, B and C.
Elution is a very important concept in chromatography separation techniques. It deals with the use of eluent in the removal of an adsobate from an adsorbent. The principle behind Elution is just about how polar the solvent is.
So, in this question Compound B will go with the Elution first because of its polarity. Compound B has lesser polarity as compared to Compounds A and B.
Compound A will then elutes second because of its polarity too as resonance increases its polarity.
Last, compound C elutes because it has the highest polarity which is caused by electronegative atoms.
Answer:

Explanation:
Hello.
In this case, given the volume (1cm³=1mL) and density of the bromine we are to firstly compute the mass since it will allow us to compute the representative particles:


Next, since the mass of one mole of diatomic bromine is 159.82 g (one bromine weights 78.91), we can next compute the moles in that sample:

Finally, via the Avogadro's number we can compute the representative particles of bromine as follows:

Best regards.