Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:
n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M
C.
centi- is essentially 10^2 of one meter.
If you had 100m, multiplying 100 by 10^2 (or 100) would give you 10000 cm.
Answer: 69.152% → 63^Cu
30.848% → 65^Cu
Explanation:
As you know, the average atomic mass of an element is determined by taking the weighted average of the atomic masses of its naturally occurring isotopes.
Simply put, an element's naturally occurring isotopes will contribute to the average atomic mass of the element proportionally to their abundance.
Explanation:
The mass number goes on top, and the atomic number goes on bottom.
Therefore, the answer is C:
40
K
19
Answer:
I should use a volumetric flask.
Explanation:
If the accuracy of the concentration is important, we need to use a volumetric flask.
