1
$\begingroup$

I'm doing a biology assignment and need to find out a clear liquid that attacks cellular respiration. It needs to be toxic and either absorbed through the skin or inhaled.

$\endgroup$
  • 2
    $\begingroup$ I'd try asking this on Biology.SE, not here. And nevertheless, you should give candidates that you've thought of already and discarded for one reason or the other. This isn't a "do my homework" site. $\endgroup$ – tschoppi Mar 10 '14 at 21:37
  • 1
    $\begingroup$ As below, but potassium cyanide solution. Liquid HCN can explosively polymerize, doi:10.1021/je60007a031 $\endgroup$ – Uncle Al Mar 11 '14 at 19:21
  • 1
    $\begingroup$ @tschoppi: Or a "plan my murder" site... $\endgroup$ – Aesin Mar 11 '14 at 20:46
2
$\begingroup$

enter image description here

Hydrogen cyanide is a clear, colorless, metabolic poison that can inhibit cellular respiration. It interferes in the uptake of oxygen in blood necessary to perform out cellular respiration. The cardiovasclar and nervous system are affected the most due to limited availble options for energy production.

From the CDC, http://www.cdc.gov/niosh/ershdb/EmergencyResponseCard_29750038.html:

Hydrogen cyanide (AC) is a systemic chemical asphyxiant. It interferes with the normal use of oxygen by nearly every organ of the body. Exposure to hydrogen cyanide (AC) can be rapidly fatal. It has whole-body (systemic) effects, particularly affecting those organ systems most sensitive to low oxygen levels: the central nervous system (brain), the cardiovascular system (heart and blood vessels), and the pulmonary system (lungs). Hydrogen cyanide (AC) is a chemical warfare agent (military designation, AC)

$\endgroup$
1
$\begingroup$

If you do not want the more effective acetylcholinesterase inhibitors, metal carbonyls might be an option: Nickeltetracarbonyl, $\ce{Ni(CO)4}$, is a colourless liquid with a boiling point of 43 °C.

$\endgroup$
  • 1
    $\begingroup$ Liquid death... Oooh long Johnson! $\endgroup$ – tschoppi Mar 11 '14 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.