Answer:
The use of phenol (carbolic acid) as a wound disinfectant was first practiced by Lister. The correct option is E
Explanation:
Phenol( carbolic acid) is an aromatic organic compound. It has the following properties:
- it's soluble in water
- it's a weak acid and
- highly reactive toward electrophilic aromatic substitution
In Medical field, phenol was first used as an antiseptic by Joseph Lister.Joseph Lister was a student at University College London under Robert Liston, later rising to the rank of Surgeon at Glasgow Royal Infirmary. He believed that patients who passed through surgery, died due to infections caused by microorganisms.He theorized that if germs could be killed or prevented, no infection would occur. Lister reasoned that a chemical could be used to destroy the micro-organisms that cause infection. He made use of the carbolic acid by soaking a clean cloth in them, and placing it over the surgical wounds of patients.
Answer:
subatomic particles are smaller then atoms
Answer:
Zn =⇒ Zn+2(0.10) + 2e- (anode)
Zn+2(?M) + 2e- === Zn(s) (cathode)
Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn
E = E^o -0.0592 log Q; in this case E^o is zero.
E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2
23 mV x 1 volt/1000mv = 0.023 Volts
0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
0.023 = -0.0296 { log 0.10 – log [Zn+2] }
0.023 = -0.0296{ -1 - log[Zn+2] }
0.023 = +0.0296 + 0.0296log[Zn+2]
-0.0066 = 0.0296log[Zn+2]
-0.22= log[Zn+2]
[Zn+2] = 10^-0.22 = 0.603 Molar
Answer:
After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.
Explanation:
