Question

a container holds ( x ) liters of a solution that is 20% ( \text{hno}_3 ) and another container holds ( y ) liters of a solution that is 50% ( \text{hno}_3 ). a chemist wants to create 3 liters of a solution that is 40% ( \text{hno}_3 ).

the system below models the amount of liquid from each container the chemist needs to mix.

what are the values of ( a ), ( b ), and ( c )?

\begin{cases}ax + by = 1.20 \\x + y = cend{cases}

( a = ) select

( b = ) select

( c = ) select

Explanation:

Step1: Analyze the volume equation

The total volume of the final solution is 3 liters, so \( x + y = 3 \), thus \( C = 3 \).

Step2: Analyze the acid amount equation

The amount of \( \text{HNO}_3 \) from the first solution is \( 0.20x \) (since it's 20% of \( x \) liters) and from the second is \( 0.50y \) (50% of \( y \) liters). The total acid in the final solution is \( 0.40\times3 = 1.20 \) liters. So the equation for acid is \( 0.20x + 0.50y = 1.20 \), meaning \( A = 0.20 \) and \( B = 0.50 \).

Answer:

\( A = 0.20 \), \( B = 0.50 \), \( C = 3 \)