I think the best answer from the choices listed above is option B. The best graduated cylinder for this case is the glass cylinder. A metal cylinder is not possible because you cannot measure really since metals are not transparent. A plastic cylinder cannot also be used since the substance should be heated for the experiment.
Answer:
The correct option is acetic acid
Explanation:
Distillation is the process of separating a mixture of substances based on differences in boiling points. During distillation, the compound with the lowest/least boiling point is distilled and collected first and then the one with the next least boiling point and it goes on like that.
From the explanation above, <u>acetic acid has the least boiling point (in the organic layer) with 118°C and thus will distill first</u>. This is then followed by isopentyl alcohol (130°C) and then isopentyl acetate (142°C) and finally sulfuric acid water (290°C).
To find pH, use the following formula ---> pH= - log [H+]
so first we need to calculate the [H+] concentration using the OH concentration. to do this, we need to use this formula--> 1.0x10-14= [H+] X [OH-], so we solve for H+ and plug in
[H+]= 1.0X10-14/[OH-]---> 1.0 x 10-14/ 1.0 x 10-4= 1.0 x 10-10
now that we have the H+ concentration, we can solve of pH
pH= -log (1.0x10-10)= 10
answer is A
The original question is to find the pH and the pOH of 0.023 M of perchloric acid.
Answer:
pH = 1.638
pOH = 12.362
Explanation:
1- getting the pH:
pH can be calculated using the following rule:
pH = -log[H+]
Since the given solution is an acid, this means that [H+] is the same as the concentration of the solution.
This means that:
[H+] = 0.023
Substitute in the above equation to get the pH as follows:
pH = -log[0.023]
pH = 1.638
2- getting the pOH:
We know that:
pH + pOH = 14
We have calculated that pH = 1.638.
Substitute in the above equation to get the pOH as follows:
pOH + 1.638 = 14
pOH = 14 - 1.638
pOH = 12.362
Hope this helps :)
Answer: A. It can identify the elements in the sample.
Explanation: on edge
