Answer:
12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.
Explanation:
For first solution of sulfuric acid :
C₁ = 40% , V₁ = ?
For second solution of sulfuric acid :
C₂ = 10% , V₂ = ?
For the resultant solution of sulfuric acid:
C₃ = 28% , V₃ = 20L
Also,
<u>V₁ + V₂ = V₃ = 20L</u> ......................................(1)
Using
<u>C₁V₁ + C₂V₂ = C₃V₃</u>
<u>40×V₁ + 10×V₂ = 28×20</u>
So,
40V₁ + 10V₂ = 560........................................(2)
Solving 1 and 2 as:
V₂ = 20 - V₁
Applying in 2
40V₁ + 10(20 - V₁) = 560
40V₁ + 200 - 10V₁ = 560
30V₁ = 360
<u>V₁ = 12 L</u>
So,
<u>V₂ = 20 - V₁ = 8L</u>
<u><em>12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.</em></u>
Mixing a base with an acid results in a chemical reaction called neutralization.<span> The result is a perfectly balanced solution of salt and water with a pH of 7 if the acid and base are balanced properly. Depending on the bases and acids used, it can be a dangerous experiment.
Hope I helped you lots!! :3</span>
I rlly just needed the credits bro
Answer: Malleability
Explanation: cuz ... its right ... lol