<span>A) This solution was not basic when it was heated in part 3. ( in part 3 i convertedCu(OH)2 to CuO).
Incorrectly low, because not all copper compounds will precipitate out
B) A slightly blue solution was decanted from Cu in part V. (in part 5 i reduced Cu(H20)6 ions with zink)
Incorrectly low, because some copper were thrown away
C) In part 5 the water in the beaker boiled away, exposing the evaporating dish to excess heat (same as above).
incorrectly high, because other compounds might be present as well </span>
Answer:
d. 60.8 L
Explanation:
Step 1: Given data
- Heat absorbed (Q): 53.1 J
- External pressure (P): 0.677 atm
- Final volume (V2): 63.2 L
- Change in the internal energy (ΔU): -108.3 J
Step 2: Calculate the work (W) done by the system
We will use the following expression.
ΔU = Q + W
W = ΔU - Q
W = -108.3 J - 53.1 J = -161.4 J
Step 3: Convert W to atm.L
We will use the conversion factor 1 atm.L = 101.325 J.
-161.4 J × 1 atm.L/101.325 J = -1.593 atm.L
Step 4: Calculate the initial volume
First, we will use the following expression.
W = - P × ΔV
ΔV = - W / P
ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L
The initial volume is:
V2 = V1 + ΔV
V1 = V2 - ΔV
V1 = 63.2 L - 2.35 L = 60.8 L
Answer:
b
Explanation:
An acid-base titration is an experimental procedure used to determined the unknown concentration of an acid or base by precisely neutralizing it with an acid or base of known concentration. ... It is filled with a solution of strong acid (or base) of known concentration.
Answer:
2AlCl3 + Ca3N2 - 2AlN+ 3CaCl2
