Decomposition is a part of the carbon cycle that occurs slowly hence movement of carbon dioxide into the atmosphere when bacteria decomposes dead matter is a slow part of the carbon cycle.
<h3>What is the carbon cycle?</h3>
The carbon cycle is part of the biogeochemical cycles that exist in nature. It refers to the movement of carbon in the ecosystem. The carbon cycle cuts across the air, the land and the water bodies.
The process in the carbon cycle that occurs slowly among the options is the movement of carbon dioxide into the atmosphere when bacteria decomposes dead matter.
Learn more about carbon cycle: brainly.com/question/1627609
Explanation:
According to Bohr's postulates, the electron in the present in the lower energy level can absorb energy and exits to higher energy level. Also, when this electron returns back to its orbit, it emits some energy.
Since the hydrogen consists of 1 electron and 1 proton. The lowest energy configuration of the hydrogen is when n =1 or, when the electron is present in the K-shell or the ground state.
The possible transition for the electron given in the question is :
n = 2, 3 and 4
The schematic diagram of the hydrogen atom consisting of these four quantum levels in which the electron can jump (Absorption) and comeback to from these energy levels (emission) .
Answer:
Molecular formula
Explanation:
Molecular formula in the first place is required to understand which compound we have. We then should refer to the periodic table and find the molecular weight for each atom. Adding individual molecular weights together would yield the molar mass of a compound.
Then, dividing the total molar mass of a specific atom by the molar mass of a compound and converting into percentage will provide us with the percentage of that specific atom.
E. g., calculate the percent composition of water:
- molecular formula is ;
- calculate its molar mass: [tex]M = 2M_H + M_O = 2\cdot 1.00784 g/mol + 16.00 g/mol = 18.016 g/mol;
- find the percentage of hydrogen: [tex]\omega_H = \frac{2\cdot 1.00784 g/mol}{18.016 g/mol}\cdot 100 \% = 11.19 %;
- find the percentage of oxygen: [tex]\omega_O = \frac{16.00 g/mol}{18.016 g/mol}\cdot 100 \% = 88.81 %.