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

4.6875 pounds

step-by-step explanation:

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2