So in your question that ask to calculate the Ph result of the resulting solution if 26 ml of 0.260 M HCI(aq) is added to the following substance. The the result are the following:
A. The result is pH= 14-pOH
B. There are 10ml of 0.26m HCL excees in this reaction so the answer is log(H)+
Answer:
0.0225 M
Explanation:
In<em><u> one liter</u></em> of this solution, there are 45 mEq of Ca⁺².
1 mEq of Ca⁺² is equal to 20 mg of Ca⁺. We know this by dividing its molar mass (40) by the charge of the ion (2).
Meaning that <em>45 mEq of Ca⁺² is equal to (45 * 20) 900 mg of Ca⁺²</em>.
Now we <u>convert 900 mg of Ca⁺² into moles</u>, using its<em> molar mass</em>:
- 0.9 g Ca⁺² ÷ 40 g/mol = 0.0225 mol Ca⁺²
Finally we calculate the molarity of the solution:
- Molarity = moles / liters
- 0.0225 mol / 1 L = 0.0225 M
Answer:
b
Explanation:
causes onj the scale of 2 3 4 it equals s
