Conservation of mass can be checked in an experiment . There are three steps to do it in a best way:
1. Weigh all the equipment and materials required in the experiment before the experiment.
2. Avoid spillage and evaporation during the experiment.
3. Weigh all the equipment and materials after the experiment.
If the mass is conserved then weight from step 1 is equal to weight from step 3.
Answer:
1.60.
Explanation:
- The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.
- The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.
<em>Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.</em>
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∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.
∵ pH = - log[H⁺]
<em>∴ pH = - log[H⁺] </em>= - log(0.025) = <em>1.602 ≅ 1.60.</em>
Anode- oxidization
Cathode-reduction
Answer: XF8
Explanation:
Empirical Formular shows the simplest ratio of elements in a compound.
Xe = 46.3% F = 53.7%
Divide the percentage composition of each element by the atomic mass.
Xe = 46.3/ 131.3 F= 53.7/ 19
= 0.353( approx) = 2.826 (approx)
Divide through with the smallest of the answers gotten in previous step.
Xe = 0.353 / 0.353 F = 2.826/ 0.353
= 1 = 8.0
Empirical formular = XF8
