To solve this problem, we can use the method of weighted averages or the mixture equation. Let $x$ represent the volume (in mL) of the 17% acid solution, and $35 - x$ represent the volume (in mL) of the 63% acid solution.

We want to produce 35 mL of a 33% acid solution. The equation for the mixture will be:

\text{(Amount of acid from the 17% solution)} + \text{(Amount of acid from the 63% solution)} = \text{(Amount of acid in the final 33% solution)}

This can be written as:

$0.17x + 0.63(35 - x) = 0.33(35)$

Step-by-step solution:

1. Expand both sides of the equation: $0.17x + 0.63(35) - 0.63x = 0.33(35)$ $0.17x + 22.05 - 0.63x = 11.55$

2. Combine like terms: $(0.17x - 0.63x) + 22.05 = 11.55$ $-0.46x + 22.05 = 11.55$

3. Isolate $x$: $-0.46x = 11.55 - 22.05$ $-0.46x = -10.50$

4. Solve for $x$: $x = \frac{-10.50}{-0.46}$ $x = 22.83 \text{ mL}$

So, you need to mix approximately 22.83 mL of the 17% acid solution with 12.17 mL of the 63% acid solution (since $35 - 22.83 = 12.17$) to get 35 mL of a 33% acid solution.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. What happens if we change the target concentration to 40%?
2. How much of each solution is needed to make 50 mL of a 33% acid solution?
3. How would you set up the equation if you had three different solutions to mix?
4. What is the maximum concentration you can achieve by mixing the two solutions?
5. Can you solve for the volumes if the total quantity is not specified?

Tip: Always check if the total volume and concentration match your initial conditions by plugging the solution back into the original equation.