Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

What volume of a 0.295 M nitric acid solution is required to neutralize 24.8 mL of a 0.107 M potassium hydroxide solution

Sagot :

Medunno13

Answer: 9.00 mL

Explanation:

[tex]M_{A}V_{A}=M_{B}V_{B}\\(0.295)V_{A}=(0.107)(24.8)\\V_{A}=\frac{(0.107)(24.8)}{(0.295)} \approx \boxed{9.00 \text{ mL}}[/tex]

View image Medunno13
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.