A chemist on a field trip forgot to pack the 40 Liters of 76% copper solution she needs. She has access to as much 30% and 85% copper solutions as she needs. How much of each should she use to get 40 Liters of 76% copper solution? Round to the nearest hundredth if necessary.

This is a classic mixture problem. We can use algebra to find the amount of each solution (30% and 85%) needed to make 40 liters of a 76% copper solution.

Let:
- \( x \) = the amount of 30% copper solution (in liters),
- \( y \) = the amount of 85% copper solution (in liters).

We have two important conditions:
1. The total amount of solution should be 40 liters:
   \[
   x + y = 40
   \]
2. The total amount of copper in the mixture must match the copper concentration in the final 76% solution. This gives us:
   \[
   0.30x + 0.85y = 0.76 \times 40
   \]

### Step 1: Solve for \( y \) in terms of \( x \)
From the first equation, we can express \( y \) as:
\[
y = 40 - x
\]

### Step 2: Substitute \( y \) in the second equation
Substitute \( y = 40 - x \) into the second equation:
\[
0.30x + 0.85(40 - x) = 30.4
\]

### Step 3: Simplify the equation
Distribute the 0.85 on the left-hand side:
\[
0.30x + 34 - 0.85x = 30.4
\]
Combine like terms:
\[
-0.55x + 34 = 30.4
\]
Subtract 34 from both sides:
\[
-0.55x = -3.6
\]
Solve for \( x \):
\[
x = \frac{-3.6}{-0.55} \approx 6.55
\]

### Step 4: Find \( y \)
Since \( y = 40 - x \), we substitute \( x = 6.55 \):
\[
y = 40 - 6.55 = 33.45
\]

### Final Answer:
The chemist should mix approximately **6.55 liters of the 30% copper solution** and **33.45 liters of the 85% copper solution** to get 40 liters of a 76% copper solution.

Would you like more details on this process, or do you have any questions?

Here are some related questions to consider:
1. What is the general approach to solving mixture problems in chemistry?
2. How do you derive equations for percentage-based mixtures in word problems?
3. Can you apply this method to mixtures of more than two solutions?
4. How do changes in percentage concentrations affect the volume needed for mixtures?
5. How do you solve for non-linear mixture problems?

**Tip:** Always check your final solution by substituting back into the original conditions to ensure accuracy.

A chemist on a field trip forgot to pack the 40 Liters of 76% copper solution she needs. She has access to as much 30% and 85% copper solutions as she needs. How much of each should she use to get 40 Liters of 76% copper solution?

Solve a mixture problem involving 40 liters of 76% copper solution, using algebraic equations to find how much of 30% and 85% copper solutions are needed. This step-by-step solution is ideal for students in Grades 9-12.

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