One molecule of water contains two atoms of hydrogen and one atom of oxygen, the atomicity of water is three.
The correct option is C.
An atom can be either in the ground state or in an excited state. An atom is said to be in the ground state, if the total energy of its electron can not be lowered by moving one or more electrons into different orbitals. At the ground state, the electrons in the atom have the lowest energy possible and they are stable. On the other hand, an atom is said to be in an excited state, if the energy of its electrons can be lowered by transferring one or more electrons into different orbitals. An atom in an excited state has more energy and is less stable.
In order to find out the number of electrons present in s orbital of Phosphorous we have to write the electronic configuration of phosphorous.
As the Atomic number of Phosphorous is 15 so it will have 15 electrons in neutral state. So,
P = 15 = 1s², 2s², 2p⁶, 3s², 3p³
Now analyzing electronic configuration, we found that it contains 3 s orbitals, i.e. 1s, 2s and 3s, and each s orbital is completely filled with two electrons each.
Result:
Therefore, a total of six electrons are present in all s orbitals of neutral Phosphorous Atom.
Answer:
K8S4O16 or K8(SO4)4 depending on if the SO4 is supposed to represent sulfate or not
Explanation:
Find the molar mass of K2SO4 first:
2K + S + 4O ≈ 174 g/mol
Divide the goal molar mass of 696 by the molar mass of the empirical formula:
696 / 174 = 4
This means you need to multiply everything in the empirical formula by 4:
K2SO4 --> K8S4O16 or K8(SO4)4 depending on if the SO4 is for sulfate or not
