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:
Answer:
B. The student chose the correct tile, but needs to flip the tile to make the units cancel
Explanation:
Based on the reaction:
2AgNO₃(aq) + Cu(s) → 2Ag(s) + Cu(NO₃)₂ (aq)
<em>2 moles of AgNO₃ react per mole of Cu producing 2 moles of Ag and 1 mole of Cu(NO₃)₂</em>
Thus, if you want to produce 6.75moles of Cu(NO₃)₂ you need:
= 13.50 moles of AgNO₃ are needed
Thus, if you analize the tile shown by the student:
<em>B. The student chose the correct tile, but needs to flip the tile to make the units cancel</em>
Answer:
For n=3 and l=1=p
It is 3p-orbital.
Magnetic quantum number m
l
have values from -l to +l and total of 2l+1 values.
Forl=1, m
l
values are:
m
l
=−1,0,1 for l=1; total m
l
values =3= Number of orbitals
Each orbital can occupy maximum of two electron
Number of electrons =2×3=6
Thus 6 electrons will show same quantum number values of n=3 and l=1.
Number of elements with last electron in 3p orbitals = 6
The reactants are aluminum and iron nitrate.
