Answer:
2.2 x 10²² molecules.
Explanation:
- Firstly, we need to calculate the no. of moles in (6.0 g) sodium phosphate:
<em>no. of moles = mass/molar mass </em>= (6.0 g)/(163.94 g/mol) = <em>0.0366 mol.</em>
- <em>It is known that every mole of a molecule contains Avogadro's number (6.022 x 10²³) of molecules.</em>
<em />
<u><em>using cross multiplication:</em></u>
1.0 mole of sodium phosphate contains → 6.022 x 10²³ molecules.
0.0366 mole of sodium phosphate contains → ??? molecules.
<em>∴ The no. of molecules in 6.0 g of sodium phosphate</em> = (6.022 x 10²³ molecules)(0.0366 mole)/(1.0 mole) = <em>2.2 x 10²² molecules.</em>
Answer: The Answer is 18.7ml.
Explanation: Solved in the attached picture.
Answer:
6.022 x 10²³; it is a conversion factor between moles and number of particles
Explanation:
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole of hydrogen = 6.022 × 10²³ atoms of hydrogen
238 g of uranium = 1 mole of uranium = 6.022 × 10²³ atoms of uranium
By taking ions:
62 g of NO⁻₃ = 1 mole of NO⁻₃ = 6.022 × 10²³ ions of NO⁻₃
96 g of SO₄²⁻ = 1 mole of SO₄²⁻ = 6.022 × 10²³ ions of SO₄²⁻
Answer:
N- 1s2 2s2 2p3
Mg- 1s2 2s2 2p6 3s2
O- 1s2 2s2 2p4
F- 1s2 2s2 2p5
Al-1s2 2s2 2p6 3s2 3p1
Explanation:
Order of decreasing atomic radius
Mg,Al, N,O,F
Order of increasing ionization energy
Mg,Al, N,O,F
Reason:
Atomic radius decreases with increase in nonmetallic character. Looking at the electronic configurations, as effective nuclear charge increases, the atom becomes smaller and the attractive force between the nucleus and the outermost electrons increases. Hence, the radius of the atom decreases and ionization energy increases. Note that the addition of more orbital electrons implies addition of more nuclear charge since the both must exactly balance for the atom to remain electrically neutral. The more the electrons in the outermost shell, the higher the first ionization energy.
