Answer:
12.6.
Explanation:
- We should calculate the no. of millimoles of KOH and HCl:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
- It is clear that the no. of millimoles of KOH is higher than that of HCl:
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
Answer:
Explanation:
MW of NaOH = 40 g/mol
MW of KCl = 74.55 g/mp;
250 mL = .25 L
100g NaOH / 40 g = 25 mol
100g KCl/ 74.55g = 1.34 mol
Molarity of NaOH: 25/.25 = 100M
Molarity of KCl: 1.34/.25 = 5.36 M
Answer:
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Explanation:
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Answer: They digest excess or worn out organelles, food particles, and engulfed viruses or bacteria.
Explanation:
