Answer: a. Carbon monoxide
Explanation:
The carbon monoxide gas competes with the oxygen in the blood to occupy the binding affinity of the hemoglobin thus the blood changes it's color to cherry red.
In postmortem lividity, the body becomes discolored due to the effect of flow of blood from the interstitial tissues and suspended under the force of gravity. This can be seen on the dependent parts of the body and the position of body after death. The color of the postmortem lividity depends on the color of the hemoglobin.
In case of carbon monoxide poisoning the color of postmortem lividity appears to be cherry red as color of the hemoglobin is cherry red. Thus cherry red or dark pinkish patches appear in the corpse as lividity.
They are all things you can do to elements on the periodic table?
Answer:
The specie which is oxidized is:- 
The specie which is reduced is:- 
Explanation:
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
For the given chemical reaction:
The half cell reactions for the above reaction follows:
Oxidation half reaction: 
Reduction half reaction: 
Thus, the specie which is oxidized is:- 
The specie which is reduced is:- 
Answer: Option (d) is the correct answer.
Explanation:
Electronegativity value of hydrogen is 2.2.
Electronegativity value of chlorine is 3.16.
Electronegativity value of carbon is 2.55.
Electronegativity value of oxygen is 3.44.
Electronegativity value of nitrogen is 3.04.
Electronegativity value of sodium is 0.93.
Electronegativity value of iodine is 2.66.
Therefore, calculate the electronegativity difference between the bonded atoms as follows.
- Electronegativity difference of HCl = Electronegativity value of chlorine - electronegativity value of hydrogen
= 3.16 - 2.2
= 0.96
- Electronegativity difference of CO = Electronegativity value of oxygen - electronegativity value of carbon
= 3.44 - 2.55
= 0.89
- Electronegativity difference of
= Electronegativity value of nitrogen - electronegativity value of nitrogen
= 3.04 - 3.04
= 0
- Electronegativity difference of NaI = Electronegativity value of iodine - electronegativity value of sodium
= 2.66 - 0.93
= 1.73
So, we can see that highest electronegativity difference is 1.73 and it is shown by NaI molecule.
Thus, we can conclude that a group 1 alkali metal bonded to iodide, such as NaI has the greatest electronegativity difference between the bonded atoms.
Answer:
The answer to your question is Aluminum
Explanation:
Number of clues
1.- If this element has 3 rings in its Bohr model, we are looking for and element located in the third period of the periodic table.
For example Sodium, Magnesium, Aluminum, Silicate, Phosphorus, Sulfur, Chlorine and, Argon.
2.- It makes three bonds to become stable, then we are looking for and element located in the third group like
Boron, Aluminum, Gallium, Indium, etc
Conclusion
The element that has both characteristics is Aluminum