The mass of the object can also be determined if the density and volume of an object are known.
If an electron moves up from the first orbit to the higher energy levels, energy will be absorbed by the electron itself and no emission line produced.
But if it moves from the orbits 6,5,4 and 3 to orbit 2, energy will be released by the electron and different emission lines wll be produced.
Number of O atoms : 24
<h3>Further explanation</h3>
Given
C₆H₁₂O₆ compound
Required
Number of atoms
Solution
A molecular formula shows the number of atomic elements in compound.
The empirical formula is the smallest comparison of the atoms
Glucose-C₆H₁₂O₆ is composed of 3 elements, namely C, H, and O.
The number of atoms in a compound can usually be seen from the subscript number after the atom and the reaction coefficient shows the number of molecules
So number of O atoms :
= 4 x 6 = 24 atoms
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
So diffusion is inversely proportional to mass !
so as mass of the particle increases, diffusion decreases !