Formulas of lonic Compounds 2x+3=+6 Al_(3+)^3Al_(2)O_(3)^3x-2=-6 1x+2=+2 CaBr_(2)^2times -1=-2 Ca^2+ CaBr_(2) 2x+1=+2 Na_(2)CO_(3)^1-2=O_(3)^2-

It looks like there are some errors and inconsistencies in the formulas and equations you've provided. Let's break them down and correct them step by step.<br /><br />1. **Equation 1:**<br /> \[<br /> 2x + 3 = +6<br /> \]<br /> Solving for \( x \):<br /> \[<br /> 2x = 6 - 3<br /> \]<br /> \[<br /> 2x = 3<br /> \]<br /> \[<br /> x = \frac{3}{2}<br /> \]<br /><br />2. **Chemical Formula 1:**<br /> \[<br /> Al_{3+}^{3}Al_{2}O_{3}^{3x-2}=-6<br /> \]<br /> This formula seems incorrect and unclear. Typically, chemical formulas should balance and make sense in terms of charges and stoichiometry. Let's assume it's a typo and try to interpret it correctly.<br /><br />3. **Equation 2:**<br /> \[<br /> 1x + 2 = +2<br /> \]<br /> Solving for \( x \):<br /> \[<br /> x = 2 - 2<br /> \]<br /> \[<br /> x = 0<br /> \]<br /><br />4. **Chemical Formula 2:**<br /> \[<br /> CaBr_{2}^{2\times -1=-2}<br /> \]<br /> This seems to be a misinterpretation of the charges. The correct charges for \( CaBr_2 \) should be:<br /> \[<br /> Ca^{2+} \quad \text{and} \quad 2Br^{-}<br /> \]<br /> So, the formula is correct as \( CaBr_2 \).<br /><br />5. **Chemical Formula 3:**<br /> \[<br /> Na_{2}CO_{3}^{1-2}=O_{3}^{2-}<br /> \]<br /> This formula is incorrect. The correct formula for sodium carbonate is \( Na_2CO_3 \), and it does not involve any charges like \( 1-2 \) or \( 2- \).<br /><br />Let's summarize the corrected and clarified information:<br /><br />1. **Solved Equation 1:**<br /> \[<br /> x = \frac{3}{2}<br /> \]<br /><br />2. **Chemical Formulas:**<br /> - \( Al_2O_3 \) (assuming the correct formula for aluminum oxide)<br /> - \( CaBr_2 \) (correct formula for calcium bromide)<br /> - \( Na_2CO_3 \) (correct formula for sodium carbonate)<br /><br />If you have specific questions or need further clarification on any part, please let me know!