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
Mass of H2C2O4 :
mm = 90.04 g/mol
number of moles : 0.0223 moles
m = n * mm
m = 0.0223 * 90.04
m = 2.007 g
hope this helps!.
1. The reactivity among the alkali metals increases as you go down the group due to the decrease in the effective nuclear charge from the increased shielding by the greater number of electrons. The greater the atomic number, the weaker the hold on the valence electron the nucleus has, and the more easily the element can lose the electron. Conversely, the lower the atomic number, the greater pull the nucleus has on the valence electron, and the less readily would the element be able to lose the electron (relatively speaking). Thus, in the first set comprising group I elements, sodium (Na) would be the least likely to lose its valence electron (and, for that matter, its core electrons).
2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.
3. While the alkali and alkaline earth metals would lose electrons to attain a noble gas configuration, the group VIIA halogens, as we have here, would need to gain a valence electron for an full octet. The trends in the group I and II elements are turned on their head for the halogens: The smaller the atomic number, the less shielding, and so the greater the pull by the nucleus to gain a valence electron. And as the atomic number increases (such as when you go down the group), the more shielding there is, the weaker the effective nuclear charge, and the lesser the tendency to gain a valence electron. Bromine (Br) has the largest atomic number among the halogens in this set, so an electron would feel the smallest pull from a bromine atom; bromine would thus be the least likely here to gain a valence electron.
4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.
Answer: Option (5) is the correct answer.
Explanation:
It is known that the ground state electronic configuration of silicon is ![[Ne]3s^{2}3p^{2}](https://tex.z-dn.net/?f=%5BNe%5D3s%5E%7B2%7D3p%5E%7B2%7D)
.
And, we know that when an atom tends to gain an electron then it acquires a negative charge and when an atom tends to lose an electron then it acquires a positive charge.
As 
has a +4 charge which means that it has lost 4 electrons. Hence, the electronic configuration of is 
.
According to the Aufbau principle, in the ground state of an atom or ion the electrons fill atomic orbitals of the lowest energy levels first, before filling the higher energy levels.
As 2p orbital is filled after the filling of 2s orbital.
Therefore, we can conclude that 2p orbital will be occupied by the electrons of highest energy for the ground-state ion.
