Answer is: there are 3.011·10²³ atoms of calcium.
n(Ca) = 0.50 mol; amount of substance(calcium).
Na = 6.022·10²³ 1/mol; Avogadro's constant or number.
N(Ca) = n(Ca) · Na.
N(Ca) = 0.50 mol · 6.022·10²³ 1/mol.
N(Ca) = 3.011·10²³; number of calcium atoms.
The mole is an SI unit which measures the number of particles in substance. One mole is equal to <span><span>6.022</span></span>·<span><span><span>10</span></span></span>²³<span> atoms.</span>
Yes, it is a special case of enthalpy of neutralization.
The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.
The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.
(26) All atoms area...<span><span>with the number of protons equaling the number of electrons
</span>(27) </span>The particles that are found in the nucleus of an atom are...<span> protons and neutrons.
(28) </span>As a consequence of the discovery of the nucleus by Rutherford, which model of the atom is thought to be true?...<span>Protons. electrons, and neutrons are evenly distributed throughout the volume of the atom.
(29) </span>The nucleus of an atom is...<span>the central core and is composed of protons and neutrons</span>.
Answer:
C.)One electron in each p orbital
Explanation:
In a P-sublevel with 3 electrons, they should be arranged with one electron going into each p-orbitals.
This is in accordance with the Hund's rule of maximum multiplicity.
The rule states that "electrons go into degenerate orbitals or sub-levels(p,d and f) singly before paring up".
Since the p-orbital is 3-fold degenerate with a capacity to accommodate a maximum number of 6 electrons, given 3 electrons, they will follow the Hund's rule in order to fill the orbitals.
So one electron will go in each p - orbitals easily.
1 mols of Aluminium ion forms 1 mol aluminium phosphate
Molar mass of AlPO_4
Moles of AlPO_4
- 61µg/106
- 0.000061/106
- 5.75×10^{-7}
- 57.5µmol
Moles of Al3+=57.5µmol
