i dont really know i will find some more information on that in about 5 minutes
90 grams due to laws of conservation of mass. Output mass = input mass. Mass can never be created or destroyed.
Think it would be distillation
Answer:
![m_{C_8H_{18}}=85.67gC_8H_{18}](https://tex.z-dn.net/?f=m_%7BC_8H_%7B18%7D%7D%3D85.67gC_8H_%7B18%7D)
Explanation:
Hello there!
In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:
![m_{C_8H_{18}}=300.0gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molC_8H_{18}}{25molO_2}*\frac{114.22gC_8H_{18}}{1molC_8H_{18}} \\\\m_{C_8H_{18}}=85.67gC_8H_{18}](https://tex.z-dn.net/?f=m_%7BC_8H_%7B18%7D%7D%3D300.0gO_2%2A%5Cfrac%7B1molO_2%7D%7B32.00gO_2%7D%2A%5Cfrac%7B2molC_8H_%7B18%7D%7D%7B25molO_2%7D%2A%5Cfrac%7B114.22gC_8H_%7B18%7D%7D%7B1molC_8H_%7B18%7D%7D%20%20%20%5C%5C%5C%5Cm_%7BC_8H_%7B18%7D%7D%3D85.67gC_8H_%7B18%7D)
Regards!
Your answer is C. Both gasoline and litter would need to be physically separated from the water, because neither bonds with the water.