The horizontal center of mass will be located at the horizontal center of mass will be located on the center line. This is further explained below.
<h3>What is Location?</h3>
Generally, a location or place that is inhabited, open for occupation or distinguished by some distinctive feature: situation The setting of the property contributes significantly to its overall allure.
In conclusion, The center line will serve as the location for the horizontal center of mass, which will be found at the same location as the center line.
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Note: The matrix referred to in the question is: ![M = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]](https://tex.z-dn.net/?f=M%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2F2%261%2F3%260%5C%5C1%2F2%261%2F3%260%5C%5C0%261%2F3%261%5Cend%7Barray%7D%5Cright%5D)
Answer:
a) [5/18, 5/18, 4/9]'
Explanation:
The adjacency matrix is ![M = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]](https://tex.z-dn.net/?f=M%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2F2%261%2F3%260%5C%5C1%2F2%261%2F3%260%5C%5C0%261%2F3%261%5Cend%7Barray%7D%5Cright%5D)
To start the power iteration, let us start with an initial non zero approximation,
![X_o = \left[\begin{array}{ccc}1\\1\\1\end{array}\right]](https://tex.z-dn.net/?f=X_o%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C1%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
To get the rank vector for the first Iteration:

![X_1 = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]\left[\begin{array}{ccc}1\\1\\1\end{array}\right] \\\\X_1 = \left[\begin{array}{ccc}5/6\\5/6\\4/3\end{array}\right]\\](https://tex.z-dn.net/?f=X_1%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2F2%261%2F3%260%5C%5C1%2F2%261%2F3%260%5C%5C0%261%2F3%261%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C1%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5CX_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2F6%5C%5C5%2F6%5C%5C4%2F3%5Cend%7Barray%7D%5Cright%5D%5C%5C)
Multiplying the above matrix by 1/3
![X_1 = \left[\begin{array}{ccc}5/18\\5/18\\4/9\end{array}\right]](https://tex.z-dn.net/?f=X_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2F18%5C%5C5%2F18%5C%5C4%2F9%5Cend%7Barray%7D%5Cright%5D)
Answer:
Aqueous carbon dioxide, CO2 (aq), reacts with water forming carbonic acid, H2CO3 (aq). Carbonic acid may loose protons to form bicarbonate, HCO3- , and carbonate, CO32-. In this case the proton is liberated to the water, decreasing pH. The complex chemical equilibria are described using two acid equilibrium equations.
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