The density does not change because it is still the same liquid as before
<h3>The <u>
complete question</u>
would be:</h3>
In a common laboratory experiment in general chemistry, students are asked to determine the relative amounts of benzoic acid and charcoal in a solid mixture. Benzoic acid is relatively soluble in hot water, but charcoal is not. Devise a method for separating the two components in the mixture.
<h3>The <u>
solution</u> for that would be:</h3>
First you put the mixture in hot water and allow the benzoic acid to separate. Next funnel it and then see what's on the funnel paper. Lastly, heat the leftover water and see what's left after it evaporates.
<h3><u>What are laboratory experiments ?</u></h3>
- Using controlled surroundings to test theories, researchers may develop laboratory experiments.
- Laboratories are rooms or specially constructed facilities within buildings that are often found on college and university campuses and are utilized for academic research.
- It's crucial to differentiate between experiments and other kinds of research investigations carried out in lab settings.
- Even if a research study takes place in a lab, it is not always an experiment.
- In order to evaluate causal conclusions regarding the links between independent and dependent variables, laboratory studies, like all genuine experimental designs, use procedures of random assignment of participants and control groups.
To view more about experiments, refer to:
brainly.com/question/17143317
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This correct option is C.
10^-2 = 0.001
8.7 * 10^-2 = 8.7 * 0.001 = 0.0087
Thus, 8.7 * 10^-2 is equivalent to 0.0087.
Answer:
100 mL
Explanation:
The reaction that takes place is:
- CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
First we <u>convert 500 mg of CaCO₃ into mmoles</u>, using its <em>molar mass</em>:
- 500 mg ÷ 100 mg/mmol = 5 mmol CaCO₃
Then we <u>convert 5 mmoles of CaCO₃ into HCl mmoles</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 5 mmol CaCO₃ *
= 10 mmol HCl
Finally we <u>calculate the volume of a 0.10 M HCl solution (such as stomach acid) that would contain 10 mmoles</u>:
- 10 mmol / 0.10 M = 100 mL
