Answer:
Explanation:
it is very important because with out it we will die
The mole<span> is the </span>unit of measurement<span> in the </span>International System of Units<span> (SI) for </span>amount of substance<span>. It is defined as the </span>amount<span> of a </span>chemical substance<span> that contains as many representative particles, e.g., </span>atoms<span>, </span>molecules<span>, </span>ions<span>, </span>electrons<span>, or </span>photons<span>, as there are atoms in 12 </span>grams<span> of </span>carbon-12<span> (</span>12<span>C), the </span>isotope<span> of </span>carbon<span> with </span>relative atomic mass<span> 12 by definition.
so to solve the moles, divide the mass with molar mass
moles = 4177 g / </span><span>133.34 g/mol
moles = 31.33 moles</span>
<u>Answer:</u> The correct option is A) They have fixed energy values.
<u>Explanation:</u>
Electron is one of the sub-atomic particle present around the nucleus of an atom which is negatively charged.
In an atomic model, it is assumed that the electron revolves around the nucleus in discrete orbits having fixed energy levels.
These electrons when jumping from one energy level to another, some amount of radiation is either emitted or absorbed.
These fixed energy levels are given by the Bohr model and thus, the electrons are quantized.
Hence, the correct option is A) They have fixed energy values.
The compound : C₄₀H₄₄N₄O
<h3>Further explanation</h3>
The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components
The principle of determining empirical formula
• Determine the mass ratio of the constituent elements of the compound.
• Determine the mole ratio by dividing the percentage by the atomic mass
The mol ratio of composition : C : H : N : O
