<span>1 drop is approximately 0.05mL. Since 0.500L of 0.550M NH4Cl contains 0.275mol of substance (calculated by using c=n/V formula), equal amount of substance of NH3 is needed to neutralize this solution (since pH of 7 is neutral solution). Thus, we need 0.0275L of NH3, i.e. around 550 drops.</span>
Answer:
208 g/mol and 87 g/mol
Explanation:
Number of protons = 82
Atomic mass = 126 + 82 = 208 g/mol
For ^{87}Sr
Number of neutrons = 49
Number of protons = 38
Atomic Mass = 49 + 38 = 87 g/mol
Answer:
0.075 M
Explanation:
5Fe²⁺(aq) + MnO₄⁻(aq) + 8H⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(ℓ)
Using the moles of Fe²⁺ that reacted, we can <em>calculate the reacting moles of MnO₄⁻</em>:
0.0450 mol Fe⁺² *
= 0.0090 mol MnO₄⁻
Now we divide the moles of MnO₄⁻ by the volume in order to <u>calculate the molarity of the solution</u>, keeping in mind that 120.0 mL = 0.120 L.
0.0090 mol MnO₄⁻ / 0.120 L = 0.075 M