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Ronch [10]
3 years ago
12

Help plz:))) I’ll mark u Brainliest

Chemistry
1 answer:
loris [4]3 years ago
3 0

Answer:

We report an unusual case of mercury vapor poisoning from using a heated tobacco product. The suspect had added grains of mercury into 20 cigarettes in a pack. When a 36-year-old Japanese man inserted one of these cigarettes into the battery powered holder, it was heated to a temperature of 350 °C, and he inhaled vaporized mercury. After using 14 of the cigarettes over 16 h, he noticed he had flu-like symptoms so he visited the hospital. Although no physical abnormalities were revealed, 99 μg/L of mercury was detected in his serum sample. His general condition improved gradually and his whole blood mercury level had decreased to 38 μg/L 5 days later. When the remaining six cigarettes in the pack were examined, many metallic grains weighing a total of 1.57 g were observed. Energy dispersive X-ray fluorescence spectrometry confirmed the grains as elemental mercury. Accordingly, the victim was diagnosed with mercury poisoning. Because the mercury was incorporated into cigarettes, an unusual and novel intoxication occurred through the heating of the tobacco product. Both medical and forensic scientific examination confirmed this event as attempted murder.

Explanation:

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lilavasa [31]

1.66 M is the concentration of the chemist's working solution.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dilute solution of this reactant.

The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.

We want to know the concentration of this diluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:

n_1= n_2  (1)

and we also know that:

n = M x V_2

If we replace this expression in (1) we have:

M_1 x V_1= M_2 x V_2

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M_2:

4.93 x 210 =  620 xM_2

M_2 = 1.66 M

This is the concentration of the solution prepared.

Learn more about molarity here:

brainly.com/question/19517011

#SPJ1

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