Answer:
0.00011 [gr].
Step-by-step explanation:
1. one mol of CO2 weights 12+32=44 [gr];
2. part of C in CO2 is 12/44=3/11, then
3. the required mass of C in 0.0004 [gr] is 0.0004*3/11≈0.0001(09) [gr].
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