Carbon monoxide is a gas that is 43% carbon. carbon dioxide only 27% carbon. suppose that the environmental protection agency (epa) analyzes exhaust from cars, a mixture of carbon monoxide and carbon dioxide. if the epa finds that a 1600 mg sample of exhaust gas has 32% carbon, how many mg of the sample were carbon monoxide, and how many were carbon dioxide?

500 mg Carbon Monoxide

1100 mg Carbon Dioxide

Step-by-step explanation:

Weight of total sample = 1600 mg

Percentage of Carbon in Carbon Monoxide = 43%

Percentage of Carbon in Carbon Dioxide = 27%

Percentage of carbon in Mixture = 32%

Let amount of Carbon Monoxide = mg

Amount of Carbon Dioxide in mixture = (1600-) mg

Amount of carbon in mg of Carbon Monoxide =

Amount of carbon in (1600-) mg of Carbon Monoxide =

We know that the sample of 1600 mg contains carbon because of Carbon Monoxide and Carbon Dioxide present in it.

Adding (1) and (2) and equating to 32% of 1600 mg:

Amount of Carbon Monoxide in sample, = 500 mg

Amount of Carbon Dioxide in sample = (1600 - ) = (1600 - 500) = 1100 mg

Hence, answer is

500 mg Carbon Monoxide

1100 mg Carbon Dioxide

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: