63/29 Cu (copper-63)
65/29 Cu (copper-65)
24/12 Mg (magnesium 24)
25/12 Mg (magnesium 25)
26/12 Mg (magnesium 26)
6/3 Li (lithium 6)
7/3 Li (lithium 7)
The top number is the mass number and you find that by adding the number of neutrons and atomic number
The bottom number is the atomic number and it’s the number or protons
Answer:
The Law of Conservation of Energy states that energy cannot be created or destroyed. In other words, the total energy of a system remains constant. This is an important concept to remember when dealing with energy problems. The two basic forms of energy that we will focus on are kinetic energy and potential energy.
Explanation:
In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.
Im bad at these questions hope it helps and have a good day.
Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.
Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.
The reaction is as shown in the image.
The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid 
is shown in the image.
m-CPBA (meta-perchlorobenzoic acid) is a peracid and forms epoxides on reacting with alkenes.
Answer:
A) oxidizing agent is SO2
B) NaClO is the oxidizing agent
Explanation:
A) This is a redox reaction in which oxidation and reduction occur simultaneously.
Thus, in 2H2S(g) + SO2(g) -> 2H2O(l) + 3S(s);
H2S is reduced as follows;
H2S → S + 2H+ + 2e−
We can see that SO2 has been reduced while H2S gets oxidized since it has changed state from - 2 to 0 . Thus sulphur dioxide is the oxidizing agent.
B) SO2(g) + H2O(l) + NaClO(aq) -> NaCl(aq) + H2SO4(aq)
In this, SO2 undergoes oxidation and NaClO is the oxidizing agent
