According to the problem statement, we discuss two scenarios:

1. Both Chinese and mathematics are scheduled in the morning. For the arrangement of Chinese and mathematics, there are 2 ways (either can be first). For the remaining 4 subjects, we need to choose 3 to be scheduled in the afternoon. This can be done in $A_4^3$ ways. Thus, there are $2 \times A_4^3 = 2 \times 4 \times 3 \times 2 = 48$ arrangements for this scenario.

2. Chinese and mathematics are scheduled one in the morning and the other in the afternoon.
- If Chinese is in the morning, there are 3 ways to arrange it (as one slot is taken by mathematics). For mathematics in the afternoon, there are 2 ways. We need to choose 3 from the remaining 4 subjects to arrange at other times, which gives us $A_4^3$ possibilities. Therefore, there are $3 \times 2 \times A_4^3 = 3 \times 2 \times 4 \times 3 \times 2 = 144$ ways for this arrangement.
- By symmetry, if mathematics is in the morning and Chinese is in the afternoon, there are also 144 ways.

By combining the scenarios, the total number of different possible arrangements is $48 + 144 + 144 = 336$. Hence, the solution is:

$$\boxed{336}$$

The principle of addition is applied here to combine the scenarios, and the arrangement is computed by considering the given constraints and focusing on elements with restrictions first.