<h2>DIFFERENCE BETWEEN CRYSTALLINE AND AMORPHOUS SOLIDS :</h2><h2><em><u> Amorphous solids do not have definite melting points but melt over a wide range of temperature because of the irregular shape. Crystalline solids, on the other hand, have a sharp melting point.</u></em></h2>
<u>Answer</u>:
Charlie adjust the coefficients to make the number of atoms in the reactants the same as the number of atoms in the
<u>Explanation</u>:
Stoichiometric coefficients are the number that appears before the symbol in each compound or element in the equation of a chemical reaction. By rule, it is positive for products and negative for reactants. Stoichiometric coefficients help in describing the stoichiometry of a chemical reaction.
According to "law of conservation of mass" that states "total reactant's mass equals the overall mass of obatined from the product". Thus, Charlie would adjust coefficients to equalise the number of atom in reactants and products.
Answer: The Ionization energy of Barium is lower and hence it can lose its electrons easily.
Explanation:
Ionization energy is defined as the energy required to remove an electron from an isolated gaseous atom. It is represented as 
This energy will be higher for fully filled and half-filled electronic configuration than partially filled electronic configuration. This is so because half filled and fully filled configurations are stable.
Magnesium is the 12th element of the periodic table having electronic configuration of = ![[Ne]3s^2](https://tex.z-dn.net/?f=%5BNe%5D3s%5E2)
Barium is the 56th element of the periodic table having electronic configuration of ![[Xe]6s^2](https://tex.z-dn.net/?f=%5BXe%5D6s%5E2)
Both magnesium and barium belong to the same group as they have similar valence electrons. But as Barium is larger in size, the valence electrons lie farther from the nucleus and lesser energy is required to remove the electron and thus ionization energy will be lower for barium and thus will be more reactive.
AgNO3 + NaCl ---> AgCl(s)
+ NaNO3
find moles of each
mmol = M*V
mmol = 55*1 = 55 mmol of Ag+
mmol = 25 * 1.06 = 26.5 mmol
of Cl-
then limitn greactant is Cl
we can produce up to 26.5
mmol of Agcl
<span>mass = mol*MW = (26.5 * 10^-3)
* 143.32 = 3.7980 g of AgCl</span>
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