Step 1: Identifying Compound (A). When \( \text{FeCr}_2\text{O}_4 \) is fused with \( \text{Na}_2\text{CO}_3 \) in the presence of air, it forms a yellow solution of sodium chromate (\( \text{Na}_2\text{CrO}_4 \)). Thus, \( A = \text{Na}_2\text{CrO}_4 \).
Step 2: Identifying Compound (B). Upon acidification, \( \text{Na}_2\text{CrO}_4 \) is converted to potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)), which is orange in color. Thus, \( B = \text{K}_2\text{Cr}_2\text{O}_7 \).
Step 3: Identifying Compound (C). The reaction of \( \text{K}_2\text{Cr}_2\text{O}_7 \) with KCl leads to the formation of orange-colored \( \text{Cr}_2\text{O}_7^{2-} \) ions. Thus, \( C \) is an orange-colored chromium compound. Thus, \( C = \text{K}_2\text{Cr}_2\text{O}_7 \). \bigskip
Step 4: Identifying Compound (D). An acidified solution of \( C \) oxidizes sodium sulfite (\( \text{Na}_2\text{SO}_3 \)) to sodium sulfate (\( \text{Na}_2\text{SO}_4 \)). This is a typical redox reaction where chromium compounds act as oxidizing agents. Thus, \( D = \text{Na}_2\text{SO}_4 \). So, the identities are:
\( A = \text{Na}_2\text{CrO}_4 \)
\( B = \text{K}_2\text{Cr}_2\text{O}_7 \)
\( C = \text{K}_2\text{Cr}_2\text{O}_7 \)
\( D = \text{Na}_2\text{SO}_4 \)
A school is organizing a debate competition with participants as speakers and judges. $ S = \{S_1, S_2, S_3, S_4\} $ where $ S = \{S_1, S_2, S_3, S_4\} $ represents the set of speakers. The judges are represented by the set: $ J = \{J_1, J_2, J_3\} $ where $ J = \{J_1, J_2, J_3\} $ represents the set of judges. Each speaker can be assigned only one judge. Let $ R $ be a relation from set $ S $ to $ J $ defined as: $ R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y, x \in S, y \in J\} $.