Complete and balance the equations for the following reactions. a. \(\operatorname{Cs}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow\) b. \(\mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \rightarrow\)

First, we need to write the products formed when solid cesium (Cs) reacts with liquid water (H2O). Cesium is an alkali metal, and these metals react with water to produce a metal hydroxide and hydrogen gas. In this case, it's cesium hydroxide (CsOH) and hydrogen gas (H2). Now that we have the products written, we can balance the equation. There is one cesium atom on each side of the equation, but we have two hydrogen atoms in water, one hydrogen in cesium hydroxide, and two hydrogen atoms in hydrogen gas. To balance the hydrogen atoms, we need to place a coefficient of '2' in front of CsOH, giving us: Cs(s) + H2O(l) → 2CsOH(s) + H2(g) All elements are now balanced, with two hydrogen atoms and one cesium atom on both sides.

First, we need to write the product formed when solid sodium (Na) reacts with chlorine gas (Cl2). Sodium is an alkali metal, and when reacting with a halogen, it forms a salt (ionic compound). In this case, it's sodium chloride (NaCl). Before balancing the equation, the unbalanced equation is: Na(s) + Cl2(g) → NaCl(s) Now, we can balance the equations. There is one sodium atom on each side of the equation. For the chlorine atoms, we have two chlorine atoms in chlorine gas and one chlorine atom in sodium chloride. We need to balance the chlorine atoms by placing a coefficient of '2' in front of NaCl, giving us: 2Na(s) + Cl2(g) → 2NaCl(s) All elements are now balanced, with two sodium atoms and two chlorine atoms on both sides.