A asystem at equilibrium stops
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Given question is incomplete. The complete question is as follows.
Balance the following equation:

Answer: The balanced chemical equation is as follows.

Explanation:
When a chemical equation contains same number of atoms on both reactant and product side then this equation is known as balanced equation.
For example, 
Number of atoms on reactant side:
H = 5
P = 1
O = 6
Ca = 1
Number of atoms on product side:
H = 6
P = 2
O = 9
Ca = 1
In order to balance this equation, we will multiply
by 2 on reactant side and we will multiply
by 2 on product side. Hence, the balanced chemical equation is as follows.

The answer is low frequency and long wavelength
Volume= mass divided by density
V= m/d
55/3.23
= 17.03