Solve mixture problems. George wishes to increase the percent of acid in 50m / of a 15% acid solution in water to 25% acid, how much pure acid must you add?

x = amount of pure acid to be added = 6.67ml

Step-by-step explanation:

George wishes to increase the percent of acid in 50ml of a 15% acid solution in water to 25% acid, how much pure acid must you add?

let

x = amount of pure acid to be added

15% of 50 + x = 25% of 50 + x

(0.15 * 50) + x = 0.25(50 + x)

7.5 + x = 12.5 + 0.25x

Collect like terms

x - 0.25x = 12.5 - 7.5

0.75x = 5

x = 5/0.75

x = 6.6666666666666

Approximately,

x = 6.67

x = amount of pure acid to be added = 6.67ml

Step-by-step explanation:

-8.3 + 9.2 - 4.4 + 3.7

= -8.3 + 9.2 + (-4.4) + 3.7            {Additive inverse}

= -8.3 + (-4.4) + 9.2 + 3.7            {Commutative Property}

= [ -8.3 + (-4.4] + [ 9.2 + 3.7]       {Associative property}

=  - 12.7 + 12.9

= 0.2