For each reaction below, identify the atom oxidized, the atom reduced, the oxidizing agent, and thereducing agent. ### Chemical ReactionsHere are some examples of chemical reactions involving various compounds, often used in educational settings to illustrate balancing equations and identifying reaction types:34. **Reaction involving Potassium Dichromate and Sulfur**: - Equation: \( 2K_2Cr_2O_7 + 2H_2O + 3S \rightarrow 3SO_2 + 4KOH + 2Cr_2O_3 \) - This reaction demonstrates the redox reaction where potassium dichromate reacts with sulfur.35. **Reaction involving Potassium Chlorate and Sucrose**: - Equation: \( 8KClO_3 + C_{12}H_{22}O_{11} \rightarrow 8KCl + 11H_2O + 12CO_2 \) - This is a decomposition reaction paired with the oxidation of sugar, where potassium chlorate is the oxidizing agent.36. **Reaction involving Oxalic Acid and Potassium Permanganate**: - Equation: \( 3H_2C_2O_4 + 2KMnO_4 \rightarrow 6CO_2 + 2K_2O + Mn_2O_3 + 3H_2O \) - This showcases a redox reaction where oxalic acid reduces potassium permanganate resulting in manganese(III) oxide.37. **Another Reaction involving Oxalic Acid and Potassium Permanganate**: - Equation: \( 4H_2C_2O_4 + 2KMnO_4 \rightarrow 8CO_2 + K_2O + Mn_2O_3 + 4H_2O \) - Similar to the previous, it involves the reduction of permanganate ions by oxalic acid, producing carbon dioxide and manganese(III) oxide.### Notes:These examples are useful in understanding stoichiometry and the conservation of mass in chemical equations. Each reaction involves balancing the number of atoms for each element on both sides of the equation to adhere to the law of conservation of mass.

“Hey, since there are multiple questions posted, we will answer first question. If you want any specific question to be answered then please submit that question only or specify the question number in your message.” 34. Given: 2K2Cr2O7+2H2O+3S→3SO2+4KOH+2Cr2O3 First determine the oxidation number for each atom present in the given reaction. Generally, the oxidation number of atom in its pure elemental form is 0. The oxidation number of O in its combined form is -2 and the oxidation number of H in its combined form is +1.The oxidation number of K in its combined form is +1. 2K+12CrX2O-27+2H+12O-2+3S0→3SYO2-2+4K+1O-2H+1+2CrZ2O-23 Oxidation number value of x, y and z are determined as follows, K2Cr2O7⇒1×+2 +2x+7×-2=02+2x-14=0x=+6 SO2⇒1×y+2×-2=0y=+4Cr2O3⇒2z+3-2=0z=+3 2K+12Cr+62O-27+2H+12O-2+3S0→3S+4O2-2+4K+1O-2H+1+2Cr+32O-23Now observing the above oxidation number value it is clear that S is oxidized since oxidation number value is increased and Cr in K2Cr2O7 is reduced since its oxidation number value is decreased. Therefore, the reducing agent is S and the oxidizing agent is K2Cr2O7.