Here are the solutions to your chemistry questions: 20. Thermal decomposition of Sodium Hydrogen Carbonate (NaHCO$_3$) a) Sodium Hydrogen Carbonate (NaHCO$_3$) is an ionic compound composed of a sodium ion and a hydrogen carbonate ion. The two radicals (polyatomic ions) present or closely related in this context are: 1. Hydrogen Carbonate radical (also known as Bicarbonate radical): $\boxed{\text{HCO}_3^-}$ 2. Carbonate radical: $\boxed{\text{CO}_3^{2-}}$ (This radical is formed during the decomposition of sodium hydrogen carbonate). b) The thermal decomposition of Sodium Hydrogen Carbonate produces Sodium Carbonate, Water, and Carbon (IV) oxide. Step 1: Write the unbalanced chemical equation with state symbols. $$ \text{NaHCO}_3(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g) $$ Step 2: Balance the sodium atoms. There is 1 Na on the left and 2 Na on the right. Place a coefficient of 2 in front of NaHCO$_3$. $$ 2\text{NaHCO}_3(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g) $$ Step 3: Verify the balance of all other atoms: Sodium (Na): Left = $2 \times 1 = 2$, Right = $1 \times 2 = 2$ (Balanced) Hydrogen (H): Left = $2 \times 1 = 2$, Right = $1 \times 2 = 2$ (Balanced) Carbon (C): Left = $2 \times 1 = 2$, Right = $1 \times 1 (\text{in Na}_2\text{CO}_3) + 1 \times 1 (\text{in CO}_2) = 2$ (Balanced) Oxygen (O): Left = $2 \times 3 = 6$, Right = $1 \times 3 (\text{in Na}_2\text{CO}_3) + 1 \times 1 (\text{in H}_2\text{O}) + 1 \times 2 (\text{in CO}_2) = 3 + 1 + 2 = 6$ (Balanced) The balanced chemical equation for this thermal decomposition is: $$ \boxed{2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)} $$ 21. Doping Silicon a) Silicon (Si) has an atomic number Z=14, meaning a neutral Silicon atom has 14 electrons. Step 1: Fill the atomic orbitals in order of increasing energy. 1s orbital: 2 electrons 2s orbital: 2 electrons 2p orbital: 6 electrons 3s orbital: 2 electrons 3p orbital: The remaining $14 - (2+2+6+2) = 2$ electrons. The s, p configuration for a neutral Silicon atom is: $$ \boxed{1s^2 2s^2 2p^6 3s^2 3p^2} $$ b) The element used for doping has "exactly one more electron in its p-orbital than Silicon." Step 1: Silicon's outermost p-orbital configuration is $3p^2$. Step 2: An element with one more electron in its p-orbital would have a $3p^3$ configuration. This means it is in the same period (Period 3) but one group to the right of Silicon (Group 14). Step 3: The element in Period 3, Group 15 is Phosphorus (P). Step 4: Verify the electron configuration for Phosphorus (P, Z=15). 1s orbital: 2 electrons 2s orbital: 2 electrons 2p orbital: 6 electrons 3s orbital: 2 electrons 3p orbital: The remaining $15 - (2+2+6+2) = 3$ electrons. The element used for doping is Phosphorus (P). Its s, p configuration is: $$ \boxed{1s^2 2s^2 2p^6 3s^2 3p^3} $$