Balance the following equations: (a) $\mathrm{CF}_{4}(l)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CBr}_{4}(l)+\mathrm{F}_{2}(g)$ (b) $\mathrm{Cu}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ (c) $\mathrm{MnO}_{2}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{MnCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$ (d) $\mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$

First, let's identify the elements in the equation: - Carbon (C) - Fluorine (F) - Bromine (Br)

Count the number of atoms for each element on both sides of the equation: - On the left side: 1 C, 4 F, and 2 Br - On the right side: 1 C, 4 Br, and 2 F

Since there are 4 F on the left side and 2 F on the right side, we need to add a coefficient of 2 in front of \(\mathrm{F}_{2}\) on the right side of the equation. Now our equation looks like: \(\mathrm{CF}_{4}(l)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CBr}_{4}(l)+2\mathrm{F}_{2}(g)\) At this point, the equation is balanced. (b) Balancing the equation $\mathrm{Cu}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$:

First, let's identify the elements in the equation: - Copper (Cu) - Hydrogen (H) - Nitrogen (N) - Oxygen (O)

Count the number of atoms for each element on both sides of the equation: - On the left side: 1 Cu, 1 H, 1 N, and 3 O - On the right side: 1 Cu, 2 H, 5 N, and 9 O

Since there are 1 N and 3 O on the left side, and 5 N and 9 O on the right side, we will need to add a coefficient of 8 in front of \(\mathrm{HNO}_{3}\) on the left side of the equation: $\mathrm{Cu}(s)+8\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ Now, we need to add a coefficient of 2 in front of \(\mathrm{NO}_{2}\) and a coefficient of 4 in front of \(\mathrm{H}_{2} \mathrm{O}\) on the right side of the equation to balance the atoms: $\mathrm{Cu}(s)+8\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2\mathrm{NO}_{2}(g)+4\mathrm{H}_{2} \mathrm{O}(l)$ At this point, the equation is balanced. (c) Balancing the equation $\mathrm{MnO}_{2}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{MnCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$:

First, let's identify the elements in the equation: - Manganese (Mn) - Oxygen (O) - Hydrogen (H) - Chlorine (Cl)

Count the number of atoms for each element on both sides of the equation: - On the left side: 1 Mn, 2 O, 1 H, and 1 Cl - On the right side: 1 Mn, 1 O, 2 H, and 3 Cl

Since there are 1 H and 1 Cl on the left side, and 2 H and 3 Cl on the right side, we will add a coefficient of 4 in front of \(\mathrm{HCl}\) on the left side of the equation: $\mathrm{MnO}_{2}(s)+4\mathrm{HCl}(a q) \longrightarrow \mathrm{MnCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$ And after that, place a coefficient of 2 in front of \(\mathrm{H}_{2} \mathrm{O}\) and a coefficient of 1 in front of \(\mathrm{Cl}_{2}(g)\) on the right side of the equation: $\mathrm{MnO}_{2}(s)+4\mathrm{HCl}(a q) \longrightarrow \mathrm{MnCl}_{2}(s)+2\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$ At this point, the equation is balanced. (d) Balancing the equation $\mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$:

First, let's identify the elements in the equation: - Potassium (K) - Hydrogen (H) - Oxygen (O) - Phosphorus (P)

Count the number of atoms for each element on both sides of the equation: - On the left side: 1 K, 4 H, 1 P, and 4 O - On the right side: 3 K, 2 H, 1 P, and 4 O

Since there are 1 K and 4 H on the left side, and 3 K and 2 H on the right side, we will add a coefficient of 3 in front of \(\mathrm{KOH}\) on the left side of the equation: $3\mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ After that, place a coefficient of 3 in front of \(\mathrm{H}_{2} \mathrm{O}\) on the right side of the equation: $3\mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3\mathrm{H}_{2} \mathrm{O}(l)$ At this point, the equation is balanced.