A research chemist wants to make 10.0L of gasoline containing 2.0% of a new experimental additive, one with 5.0% and the other with 6.0% are to be used. if four times as much gasoline without additive as the 5.0% mixture is to be used, how much of each is neede?
sweetysiri92:
is there any one to answer the question
and the answers are 6.4L,1.6L and 2.0L
please do the sum
is it 20% additive??
a research chemist wants to make 10.0L of gasoline containing 2.0% of a new experimentaladditive. Gasoline without additive and two mixtures of gasoline with additive, on ]e with 5.0% and the other with 6.0% are used. if four times as much gasoline without additive as the 5.0% mixture is used, how much of each is needed
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a research chemist wants to make 10.0L of gasoline containing 2.0% of a new experimentaladditive. Gasoline without additive and two mixtures of gasoline with additive, on ]e with 5.0% and the other with 6.0% are used. if four times as much gasoline without additive as the 5.0% mixture is used, how much of each is needed
x+y+z= 10
0*x+0.05*y+0.06*z=0.02*10=0.2
or, 5y+6z=20
further, x=4y
putting this in first equation 5y+z=10
subtracting from the second equation: 6z-z =10
z=2
5y=8
y=1.6
x=6.4
use 6.4, 1.6 and 2 litres respectively of the three solutions. (answer)
x+y+z= 10
0*x+0.05*y+0.06*z=0.02*10=0.2
or, 5y+6z=20
further, x=4y
putting this in first equation 5y+z=10
subtracting from the second equation: 6z-z =10
z=2
5y=8
y=1.6
x=6.4
use 6.4, 1.6 and 2 litres respectively of the three solutions. (answer)
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