125 mile *1gallon/35 mi = 135/35 = (27/7) gallon gasoline
27/7 gallon * 1 L/0.264 gallon = 14.6 L gasoline
14.6 L gasoline * 2.5kg CO2/1L gasoline= 36.5 kg CO2
36.5 kg CO2 * 1lb/0.454 kg = 80.4 lb
Answer: 80.4 lb CO2
22.3 is the answer to this question
The structure of compound A would be solid that is dense enough for antimicrobial form
For the answer to the question above asking, h<span>ow many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
The answer to your question is the the third one among the given choices which is 6.8 mol.
</span><span>moles glucose = 1.5 x 4.5 = 6.8 </span>