Most of the carbon is put away in sedimentary carbonates and kerogens, with the rest being spread between the sea, the air, biomass, for example, plants and creatures, and petroleum products.
<u>Explanation</u>:
- The carbon cycle is the procedure where carbon goes from the surrounding into living beings and to the Earth and then again goes into the air. Plants take carbon dioxide from the air and use it for food preparation. Creatures at that point eat the nourishment and carbon is put away in their bodies or discharged as CO2 through the breath.
- Most of the carbon is put away in sedimentary carbonates and kerogens, with the rest being spread between the sea, the air, biomass, for example, plants and creatures, and petroleum products. This is known as carbon storage.
- For instance, carbon, a fundamental component in natural particles, is preserved as it is moved from inorganic carbon in a biological system to natural atoms in living life forms of the biological system and back as inorganic carbon to the earth.
The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment
Answer:
Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z
Explanation:
The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.
The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.
Compound 2 (2,5-dimethylhexane) structure shows that the cleavage of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.
Answer:
i) increase
ii) decrease
iii) remain the same
iv) No, because it dissociates completely.
Explanation:
On a 10-fold dilution of a weak acid, the pH will increase because the concentration of hydrogen ions will decrease thereby increasing the pH to close to that of water.
On a 10-fold dilution of a weak base, the pH will decrease due to the removal of hydroxide ions from the solution. This results in the solution having a H closer to that of water.
If one adds a very small amount of strong base to a buffered solution, the pH will remain constant because a buffer solution acts to withstand any change to its pH on the addition of small quantities of either an acid or a base.
A buffer solution cannot be made with a strong acid because thy undergo complete dissociation. Therefore, any small addition of base or acid will result in very large changes in the pH of the solution. A buffer solution is made with a weak acid and its conjugate base or a weak base and its conjugate acid.
System to surroundings since energy is released in an exothermic reaction
