Find the value of `x` in the tremolite abestos:
`Ca_(2)Mg_(x)(Si_(4)O_(11))_(2)(OH)_(2)`

Answer

To find the value of \( x \) in the formula of tremolite asbestos, which is given as:

\[ \text{Ca}_2\text{Mg}_x(\text{Si}_4\text{O}_{11})_2(\text{OH})_2 \]

we need to balance the charges of the elements in the compound. Here’s a step-by-step solution:

### Step 1: Identify the charges of each component
- The charge of Calcium (Ca) is \( +2 \).
- The charge of Magnesium (Mg) is \( +2 \).
- The charge of Silicon (Si) in the formula \( \text{Si}_4\text{O}_{11} \) can be calculated as follows:
- Each Si has a charge of \( +4 \), and there are 4 Si atoms, contributing \( 4 \times +4 = +16 \).
- Each O has a charge of \( -2 \), and there are 11 O atoms, contributing \( 11 \times -2 = -22 \).
- Therefore, the total charge of \( \text{Si}_4\text{O}_{11} \) is \( +16 - 22 = -6 \).
- The charge of Hydroxide (OH) is \( -1 \) and there are 2 OH groups, contributing \( 2 \times -1 = -2 \).

### Step 2: Set up the charge balance equation
The total positive charge must equal the total negative charge for the compound to be neutral. Thus, we can write the equation as:

\[ 2 \times (+2) + x \times (+2) + 2 \times (-6) + 2 \times (-1) = 0 \]

### Step 3: Substitute the values into the equation
Substituting the values we have:

\[ 2 \times 2 + x \times 2 + 2 \times (-6) + 2 \times (-1) = 0 \]

This simplifies to:

\[ 4 + 2x - 12 - 2 = 0 \]

### Step 4: Simplify the equation
Combining the constant terms:

\[ 4 - 12 - 2 + 2x = 0 \]

This simplifies to:

\[ -10 + 2x = 0 \]

### Step 5: Solve for \( x \)
Now, isolate \( x \):

\[ 2x = 10 \]

\[ x = \frac{10}{2} = 5 \]

### Conclusion
Thus, the value of \( x \) in tremolite asbestos is:

\[ \boxed{5} \]