Answer:
Im sorry i though this was a joke i don't know I'm sorry:
The reason that some of the elements of period three and beyond are steady in spite of not sticking to the octet rule is due to the fact of possessing the tendency of forming large size, and a tendency of making more than four bonds. For example, sulfur, it belongs to period 3 and is big enough to hold six fluorine atoms as can be seen in the molecule SF₆, while the second period of an element like nitrogen may not be big to comprise 6 fluorine atoms.
The existence of unoccupied d orbitals are accessible for bonding for period 3 elements and beyond, the size plays a prime function than the tendency to produce more bonds. Hence, the suggestion of the second friend is correct.
Answer:
3Mg(s) +2P(s) -------> Mg3P2(s) + energy
Keq= [Mg3P2]/[Mg]^3 [P]^2
Explanation:
The equation for the formation of magnesium phosphide from its elements is;
3Mg(s) +2P(s) -------> Mg3P2(s) + energy
Hence we can see that three moles of magnesium atoms combines with two moles of phosphorus atoms to yield one mole of magnesium phosphide. The equation written above is the balanced chemical reaction equation for the formation of the magnesium phosphide.
The equilibrium expression for the reaction K(eq) will be given by;
Keq= [Mg3P2]/[Mg]^3 [P]^2
<span>Answer:
K because it is metal and typically forms ionic bonds. Ar is also unlikely to form any bonds because it has a full outer shell of electrons, but it can form covalent bonds.</span>
Explanation:
Because molarity is mol/L, we'll have to convert 17g to mol.
After obtaining the mol, we'll divide that by the volume to obtain Molarity.
