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Kaylis [27]
3 years ago
7

The MSDS for chloroform indicates that it is a clear liquid that has a pleasant smell and substantial vapor pressure. People sho

uld avoid inhaling its vapors, and it is sensitive to light. Malik needs 10 mls of chloroform for an experiment.
According to this information, how should he safely pour the chloroform?

He should locate the chloroform stored in a transparent container in chemical storage and pour directly into his beaker from that location.

He should locate the chloroform stored in a transparent container in chemical storage and should take it to the fume hood to pour.

He should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.

He should locate the chloroform stored in a dark container in chemical storage and pour directly into his beaker from that location.
Physics
1 answer:
Norma-Jean [14]3 years ago
8 0

Answer:

The correct option is the third option

Explanation:

Firstly, it must be noted that chemicals/reagents that are sensitive to sunlight are stored in dark/amber container in the laboratory. Hence, the chloroform can only be found in an amber/dark bottle.

Also, reagents/chemicals that release poisonous/offensive gases are handled in the fume cupboard in the laboratory. Thus, If Malik is going to pour the chloroform, he should pour it in a fume cupboard to avoid inhaling it because of the toxicity of it's vapor.

From the above explanation, it can be deduced that <u>Malik should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour</u>.

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An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then
gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

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So, the velocity can be obtained by integrating this expression:

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The velocity is, by definition: v=\frac{dx}{dt}, so

dx=1.76tdt\\x=1.76\frac{t^{2}}{2}.

Do x=11 in order to find the time spent.

11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s

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