Simon has collected three samples from the coral reef where he observes marine life. He must determine whether each one is a pur
2 answers:
Answer: Sample A is a pure substance, Sample B is a homogeneous mixture and Sample C is a heterogeneous mixture.
Explanation:
Homogeneous mixture : In these mixture, all the component are uniformly distributed throughout the mixture.For example: sugar-water solution.
Heterogeneous mixture : In these mixture, all the component are not uniformly distributed throughout the mixture.For example: sand-water solution.
Pure substance : These are the substance made up only one type of atom or molecule.They have fixed value of boiling and melting points. For example: diamond, pure sugar etc.
Sample A is a clear liquid which evaporates at with no change in appearance. Hence, pure substance.
Sample B is a clear blue liquid which boils at and leaves blue crystals behind with change in appearance. Hence, homogeneous mixture
Sample C is a opaque, whitish liquid which boils at and leaves white crystals behind with dust particles settled in the bottom. Hence, heterogeneous mixture.
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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