Answer:
<u>only sodium chloride compounds</u> MOST LIKELY products of the reaction between sodium metal and chlorine gas.
Correct answer - C
Explanation:
A solid sodium metal reacts with gaseous form of chlorine gas to form a solid sodium chloride.
This formation of sodium chloride arises due to transfer of one valence electron from sodium to chlorine atom and the both atoms get stable octet electronic configuration.
As a result, an ionic compound solid sodium chloride is formed.
The chemical reaction of formation sodium chloride is as follows.

Therefore, <u>only sodium chloride compounds</u> MOST LIKELY products of the reaction between sodium metal and chlorine gas.
Answer: D) helium.
Explanation:
Nuclear fission is a process which involves the conversion of a heavier nuclei into two or more small and stable nuclei along with the release of energy.

Nuclear fusion is a process which involves the conversion of two small nuclei to form a heavy nuclei along with release of energy.
Example: 
Thus when deuterium and tritium , the two isotopes of hydrogen are fused, a heavier nuclei helium is being formed from two smaller nuclei releasing a neutron.
739 is the number of atoms of O in 92.3 moles of Cr3(PO4)2.
Explanation:
Molecular formula given is = Cr3(PO4)2
number of moles of the compound is 92.3 moles
number of 0 atoms in 92.3 moles =?
From the chemical formula 1 mole of the compound has 8 atoms of oxygen
So, it can be written as
1 mole Cr3(PO4)2 has 8 atoms of oxygen
92.3 moles of Cr3(PO4)2 has x atoms of oxygen
= 
x = 8 x 92.3
x = 738.4 atoms
There will be 739 oxygen atoms in the 92.3 moles of Cr3(PO4)2.
Cryosurgery. You automatically know this because 'cryo' means ice or something cold, thus you can assume that it is that
Answer : The maximum amount of nickel(II) cyanide is 
Explanation :
The solubility equilibrium reaction will be:

Initial conc. 0.220 0
At eqm. (0.220+s) 2s
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ni^{2+}][CN^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BNi%5E%7B2%2B%7D%5D%5BCN%5E-%5D%5E2)
Now put all the given values in this expression, we get:


Therefore, the maximum amount of nickel(II) cyanide is 