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Viefleur [7K]
3 years ago
9

It is necessary to have a 40% antifreeze solution in the radiator of a certain car. the radiator now has 70 liters of 20% soluti

on. how many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?
Chemistry
2 answers:
Akimi4 [234]3 years ago
6 0
The answer is : 17.5 liters drained and replaced by 17.5 liters of 100% solution. 
x = amount drained and replaced
(70-x) = remaining amount of 20% solution
<span>.20(70-x) + 1.00(x) = .40(70)
14 - .2x + 1x = 28
1x - .2x = 28 - 14
</span><span>.8x = 14
</span><span>x = 14/.8
x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)

</span>
Snowcat [4.5K]3 years ago
5 0

Answer:

17.5 litres removed and 17.5 litres of pure antifreeze added

Explanation:

Let k equal the amount of the solution to be removed

amount of antifreeze to be added

0.2(70 - x) + x = 0.4(70)

14 - 0.2x +x = 28

0.8x = 28 -14

x = 14/0.8 = 17.5

x = 17.5 litres removed and 17.5 litres of pure antifreeze added

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