2. In your lab last week, you added concentrated NH OH to the anhydrous CuSO, ac

Question

2. In your lab last week, you added concentrated NH OH to the anhydrous CuSO, according to the following equation: If you started with a target weight that I assigned of 0.9000 grams of the copper sulfate, how many ml of 10.5 M ammonium hydroxide would you need to fully react to produce the maximum blue color complex? b. In your experiment, why was the concentrated ammonium hydroxide in such a HUGE excess? (8 pts) 3. Give the correct answer with the correct number of significant figures and/or the correct uncertainty: (19 pts) A. 159.0 + 16.54 + 213 = B. 1.29 + 6.456 x 7.7754 = C. 138.14 + 24.1]/ [57.1 - 24.6]- D. [36.2+0.8-31.8 ±0.21 x 2.960.09 E. (99.31 + 0.09)-(49.32 +0.08) + (2.2000.014) x (29.99+0.12) =

Explanation / Answer

(a)
Balanced equation,
CuSO4 (aq.) + 4 NH4OH (aq.) --------> [Cu(NH3)4]2+ + 4 H2O + SO42-
Mass of CuSO4 = 0.9000 g.
Molar mass of CuSO4 = 159.5 g/mol
Moles of CuSO4 = mass / molar mass = 0.9000 / 159.5 = 0.005643 mol
From the balanced equation,
1 mol of CuSO4 needs 4 mol of NH4OH
Then,
0.005643 mol of CuSO4 needs 4 * 0.005643 = 0.02257 mol of NH4OH
Therefore,
Volume of NH4OH needed = mole / molarity = 0.02257 / 10.5 = 0.002150 L = 2.150 mL
3.
(a)
159.0 + 16.54 + 213. = 388.
(b)
1.29 + (6.456 * 7.7754) = 1.29 + 50.20 = 51.49
(c)
[38.14 + 24.1] / [57.1 - 24.6] = 62.2 / 32.5 = 1.91
(d)
[ 36.2 +/- 0.8 - 31.8 +/- 0.2] * 2.96 +/- 0.09 = 4.4 +/-1.0 * 2.96 +/-0.09 = 13. +/- 1.1
(e)
52.19 +/- 0.18 * 29.99 +/- 0.12 = 1565. +/- 0.30

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