The total mass of the products is 10.76 g + 204.44 g = 215.20 g.
The masses of all the reactants but one are known so,
215.20 g - 120.00 g - 8.15 g - 75.00 g = 12.05 g
12.05 g is the mass of the unweighed barium nitrate.
Answer:
None of the given options
Explanation:
Let's go case by case:
A. No matter the volume, the concentration of Fe(NO₃)₃ (and thus of [Fe³⁺] as well) is 0.050 M.
B. We can calculate the moles of Fe₂(SO₄)₃:
- 0.020 M * 0.80 L = 0.016 mol Fe₂(SO₄)₃
Given that there are two Fe⁺³ moles per Fe₂(SO₄)₃ mol, in the solution we have 0.032 moles of Fe⁺³. With that information in mind we <u>can calculate [Fe⁺³]</u>:
- 0.032 mol Fe⁺³ / 0.80 L = 0.040 M
C. Analog to case A., the molar concentration of Fe⁺³ is 0.040 M.
D. Similar to cases A and C., [Fe⁺³] = 0.010 M.
Thus none of the given options would have [Fe⁺³] = 0.020 M.
V1 = 2.0 L
T1 = 25.0 oC = 298 K V2 = V1T2 = (2.0 L)(244 K) = 1.6 L
V2 = ? t1(298 K)
T2 = –28.9 oC = 244 K
Given:
P1 = 13.0 atm
T1 = 20 °C
T2 = 102 °C
Required:
P2 of oxygen
Solution:
At constant volume,
we can apply Gay-Lussac’s law of pressure and temperature relationship
P1/T1=P2/T2
(13.0 atm) / (20 °C)
= P2 / (102 °C)
P2 = 66.3 atm
The answer is not in the choices given.
