Hemoglobin has a much greater affinity for carbon monoxide than oxygen. In a hyperbaric chamber (containing high levels of oxygen) can treat carbon monoxide poisoning, by displacing carbon monoxide from Hemoglobin competitively.
Hemoglobin has a much greater affinity for carbon monoxide than oxygen. This is because, a coordinate bond is formed with Carbon monoxide and Haem structure of the hemoglobin.
Carbon monoxide with Hemoglobin is called as Carboxy haemoglobin.
Presence of oxygen displaces the Carbon monoxide with Hemoglobin that is formed due to poisoning.
Hyperbaric chamber is a chamber which contains pure oxygen in a chamber. The atmospheric pressure is kept about three to four times than the normal, such that the replacement of Carbon monoxide from Haem can occur as fast as possible since this reduces the half life of the Carboxy haemoglobin.
It is advisable not to treat Carbon monoxide poisoning yourself.
Hyperbaric oxygen is used to treat the following conditions as well:
- Infections
- Wounds
- Air bubble is blood
Learn more about Carbon Monoxide here, brainly.com/question/11313918
#SPJ4
b. Na2HPO4 + NaH2PO4.
A <em>buffer </em>is a solution of a weak acid and its conjugate base. The weak acid is H2PO4^(-) and its conjugate base is HPO4^(2-).
All the other options are incorrect because they consist of only a single component.
Answer:
a. electrophilic aromatic substitution
b. nucleophilic aromatic substitution
c. nucleophilic aromatic substitution
d. electrophilic aromatic substitution
e. nucleophilic aromatic substitution
f. electrophilic aromatic substitution
Explanation:
Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).
A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).
Answer is: 3,4
Chemical reaction: HNO₂ + NaOH → NaNO₂ + H₂O.
c₀(HNO₂) = 1,2 M = 1,2 mol/dm³.
c₀(NaNO₂) = 0,8 M = 0,8 mol/dm³.
V₀(HNO₂) = V₀(NaNO₂) = 1 dm³ = 1 L.
c₀(NaOH) = 0,5 M = 0,5 mol/dm³.
n₀(HNO₂)= 1,2 mol/dm³ · 1 dm³ = 1,2 mol.
n₀(NaNO₂) = 0,8 mol/dm³ · 1 dm³ = 0,8 mol.
V(NaOH) = 400 mL · 0,001 dm³/mL = 0,4 dm³.
n₀(NaOH) = c₀(NaOH) · V₀(NaOH).
n₀(NaOH) = 0,5 mol/dm³ · 0,4 dm³ = 0,2 mol.
n(HNO₂) = 1,2 mol - 0,2 mol = 1 mol.
n(NaNO₂) = 0,8 mol + 0,2 mol = 1 mol.
c(HNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
c(NaNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
pH = pKa + log (c(HNO₂) / c(NaNO₂)).
pH = 3,4 + log (0,714 mol/dm³ / 0,714 mol/dm³) = 3,4.
