First identify which is being oxidized and reduced. In this case, the Mg is being oxidized and the Hg is being reduced.
Mg --> Mg+2
<span>Hg+2 --> Hg+1
</span>
Then you have to balance each half reaction first with electrons before adding them together in one equation

⇒

and

⇒
and then combine them together to form

⇒

It isn't necessary to keep the electrons but its essential to know how many there are in order to know how many are in the equation in order to calculate the reaction energy. Note: A<span>dd H+ and H2O to balance the H's and O's in acidic solution if needed.</span>
Water is produce bases and says
Answer:
There’s a particular way of writing what’s in a molecule called a chemical formula. The chemical formulae for all the elements that form each molecule and uses a small number to the bottom right of an element’s symbol to stand for the number of atoms of that element. For example, the chemical formula for water is H 2 O.
Explanation:
The top of a wave is called the crest.
Hope I helped! :D
Answer: 32.94 g
Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.
for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm
V = 15.0 L
T = 350.8 + 273 = 623.8 K
For krypton, 
n = 0.146 moles
for chlorine, 
n = 0.439
From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.
Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.
So, the amount of product formed is calculated from moles of krypton.
Molar mass of krypton tetrachloride is 225.61 gram per mol.
There is 1:1 mol ratio between krypton and krypton tetrachloride.

= 32.94 g of 
So, 32.94 g of the product will form.