Answer is: 5.22·10²² atoms of Iodine.
m(CaI₂) = 12.75 g; mass of calcium iodide.
M(CaI₂) = 293.9 g/mol; molar mass of calcium iodide.
n(CaI₂) = m(CaI₂) ÷ M(CaI₂).
n(CaI₂) = 12.75 g ÷ 293.9 g/mol.
n(CaI₂) = 0.043 mol; amount of calcium iodide.
In one molecule of calcium iodide, there are two iodine atoms
n(I) = 2 · n(CaI₂).
n(I) = 0.086 mol; amount of iodine atoms.
Na = 6.022·10²³ 1/mol; Avogadro number.
N(I) = n(I) · Na.
N(I) = 0.086 mol · 6.022·10²³ 1/mol.
N(I) = 5.22·10²²; number of iodine atoms.
Electronegativity
The tendency of an atom to pull electrons towards itself is called electronegativity.
It is the tendency of an atom to attract bonding pair of electrons towards itself. These atoms which are more electronegative are able to bear a negative charge and be stable.
Unlike electropositive elements which tend to lose electrons , electronegative elements hold on tightly to the electrons.
The electronegativity is measured using the Pauling Scale where the most electronegative element of the periodic table, Fluorine , is given a value of 4 and the rest of the elements have values lower than 4 according to the trends followed by their groups and periods.
The least electronegative element of the periodic table , Cesium has a value of 0.7 on the Pauling Scale.
Electronegativity increases when we move from left to right in the periodic table and it decreases (in general) when we move down the group.
(To know more about Electronegativity: https://brainly.in/question/5742635 )
Answer:
The answer to your question is:
Explanation:
Data
carbon 7.3% = 7.3g
hydrogen 4.5% = 4.5g
oxygen 36.4% = 36.4 g
nitrogen 31.8% = 31.8 g
Now
For carbon
12 g --------------------1 mol
7.3 g ------------- x
x = 7.3/12 = 0.608 mol
For hydrogen
1 g -------------------- 1 mol
4.5 g ------------------ x
x = 4.5 mol
For oxygen
16 g ------------------- 1 mol
36.4 g ---------------- x
x = 2.28 mol
For nitrogen
14 g ---------------- 1 mol
31.8 g --------------- x
x = 2.27 mol
Now divide by the lowest result, the is 0.608 from carbon
carbon 0.608/0.608 = 1
hydrogen 4.5/ 0.608 = 7.4
oxygen 2.28/0.608 = 3.75
nitrogen 2.27/0.608 = 3.73
Empirical formula = CH₇O₄N₄
Answer:
Limitations of Rutherford Atomic Model
Although the Rutherford atomic model was based on experimental observations it failed to explain certain things. Rutherford proposed that the electrons revolve around the nucleus in fixed paths called orbits. ... Ultimately the electrons would collapse in the nucleus.
