Asolution of 10% sulfuric acid and one of 25% sulfuric acid are to be used to make 500 ml of 20% sulfuric acid. how many ml of each solution needs to be used in the mixture? (round answers to the nearest ml)

Step-by-step explanation:

If x is the volume of the 10% sulfuric acid, and y is the volume of 25% sulfuric acid, then:

x + y = 500

0.10 x + 0.25 y = 0.20 (500)

Solve the system of equations with substitution or elimination.  Using substitution:

y = 500 − x

0.10 x + 0.25 (500 − x) = 0.20 (500)

0.10 x + 125 − 0.25 x = 100

25 = 0.15 x

x = 167 mL

y = 333 mL

You need 167 mL of 10% sulfuric acid and 333 mL of 25% sulfuric acid.

Step-by-step explanation:

x ml of 10 % sulfuric acid+(500-x) of 25% sulfuric acid=500 ml of 25% sulfuric acid

4+n=5

step-by-step explanation:

this is because the four is four hours and the n represents the 1 extra hour over the weekend.

b= i'm srry i don't know that one.

4+n=5 n=1 because if vernon plays 5 hours a week then 4+1=5 hours.

hope this : )

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step-by-step explanation:

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