I believe B is the correct answer
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
Answer:
Measure 20mL of the stock
Explanation:
All we need to do is to find the volume of the stock solution needed. This is illustrated below:
C1 = 2M
V1 =?
C2 = 0.400 M
V2 = 100mL
C1V1 = C2V2
2 x V1 = 0.4 x 100
Divide both side by 2
V1 = ( 0.4 x 100) /2
V1 = 20mL
Therefore, to prepare the solution, we will measure 20mL of the stock solution and make it up to the 100mL mark.
It is an act or instance of viewing or noting a fact, or an occurrence for some scientific or other special purpose
A - its condensation and gas particles have a higher kinetic energy