How many mL of 10% and 50% solutions must be combined to prepare 8 oz of 30% solution?

A. 100 mL of 50% solution; 140 mL of 10% solution
B. 240 mL of 50% solution; 240 mL of 10% solution
C. 140 mL of 50% solution; 100 mL of 10% solution
D. 120 mL of 50% solution; 120 mL of 10% solution

Question

05-02-2025
Chemistry

High School

We appreciate your visit to How many mL of 10 and 50 solutions must be combined to prepare 8 oz of 30 solution A 100 mL of 50 solution 140. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

How many mL of 10% and 50% solutions must be combined to prepare 8 oz of 30% solution?

A. 100 mL of 50% solution; 140 mL of 10% solution
B. 240 mL of 50% solution; 240 mL of 10% solution
C. 140 mL of 50% solution; 100 mL of 10% solution
D. 120 mL of 50% solution; 120 mL of 10% solution

Answer :

Other Questions

A 3.00-kg object has a velocity of [tex]6.00 \hat{i} + 2.00 \hat{j}[/tex]

Elige la subclase del género lírico que más te haya llamado la

Which statement is true with regard to keystone species? A. Keystone species

When stacking rails in a single hydro hoist way with a rise

A patient had a procedure done and she has a catheter. She

A garden has a perimeter of 120 feet. The difference between the

1. How would you describe the universe to someone not in our

Garbage located in a garbage can inside the curtilage can be searched

What color would you report seeing if a researcher projects the longest

Window unit air conditioners are designed primarily to supply cool air to:

riyapandey3400 riyapandey3400 · Answer 1 · 2023-09-22T16:59:56-04:00

Final answer:

To prepare 8 oz of 30% solution, you should combine 140 mL of a 10% solution with 100 mL of a 50% solution. The correct answer is a. 100 mL of 50% solution; 140 mL of 10% solution.

Explanation:

To solve this problem, we can use the concept of mixtures. Let x be the volume (in mL) of the 10% solution and y be the volume (in mL) of the 50% solution.

We can set up two equations based on the information given:

x + y = 8 (since we need a total of 8 oz of solution)

0.10x + 0.50y = 0.30(8)

Solving these equations, we find that 140 mL of the 10% solution and 100 mL of the 50% solution need to be combined to prepare 8 oz of 30% solution.

To answer the student's question regarding how many mL of 10% and 50% solutions must be combined to prepare 8 oz of a 30% solution, we need to use the concept of mixtures in algebra. First, we must convert the volume from ounces to milliliters, since the rest of the problem uses the metric system.

There are approximately 29.5735 mL in 1 oz, so 8 oz is approximately 236.588 mL (8 x 29.5735 = 236.588). Let's call the volume of the 10% solution 'V10' and the volume of the 50% solution 'V50'. The equation is then:

0.10V10 + 0.50V50 = 0.30(236.588)

We must also take into account that the total volume of the mixed solution is the sum of V10 and V50:

V10 + V50 = 236.588

How many mL of 10% and 50% solutions must be combined to prepare 8 oz of 30% solution?A. 100 mL of 50% solution; 140 mL of 10% solutionB. 240 mL of 50% solution; 240 mL of 10% solutionC. 140 mL of 50% solution; 100 mL of 10% solutionD. 120 mL of 50% solution; 120 mL of 10% solution

Answer :

Final answer:

Explanation:

Other Questions

How many mL of 10% and 50% solutions must be combined to prepare 8 oz of 30% solution?

A. 100 mL of 50% solution; 140 mL of 10% solution
B. 240 mL of 50% solution; 240 mL of 10% solution
C. 140 mL of 50% solution; 100 mL of 10% solution
D. 120 mL of 50% solution; 120 mL of 10% solution