Question

solve the problem. 5) in a chemistry class, 7 liters of a 4% silver iodide solution must be mixed with a 10% solution to get a 6% solution. how many liters of the 10% solution are needed? a) 4.5 l b) 7.0 l c) 2.5 l d) 3.5 l

Explanation:

Step1: Set up the equation

Let $x$ be the volume of the 10% solution. The amount of silver - iodide in the 4% solution is $0.04\times7$ liters, the amount of silver - iodide in the 10% solution is $0.1x$ liters, and the amount of silver - iodide in the final 6% solution is $0.06\times(7 + x)$ liters. Using the fact that the sum of the amounts of silver - iodide in the two initial solutions equals the amount of silver - iodide in the final solution, we get the equation $0.04\times7+0.1x=0.06\times(7 + x)$.

Step2: Expand and simplify the equation

First, calculate $0.04\times7 = 0.28$ and $0.06\times(7 + x)=0.42+0.06x$. So the equation becomes $0.28 + 0.1x=0.42+0.06x$.

Step3: Solve for $x$

Subtract $0.06x$ from both sides: $0.28 + 0.1x-0.06x=0.42+0.06x - 0.06x$, which simplifies to $0.28 + 0.04x=0.42$. Then subtract 0.28 from both sides: $0.04x=0.42 - 0.28$, so $0.04x=0.14$. Divide both sides by 0.04: $x=\frac{0.14}{0.04}=3.5$.

Answer:

D. 3.5 L