These ions are disjoint by the charge on the ion into four dissimilar tables and listed alphabetically within each table. Each polyatomic ion, has it called, chemical, formula, two dimensional drawing, and three dimensional representation are given.
The three dimensional buildings are drawn as CPK models. CPK structures represent the atoms as sphere, where the radius of the sphere is equal to the van der waals radius of the atom; these buildings give a measure up the volume of the polyatomic atom.
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Answer:</h3>
Al- [Ne] 3s²3p¹
As- [Ar] 4s²3d¹⁰ 4p³
Explanation:
- Electron configuration of an element shows the arrangement of electrons in the energy levels or orbitals in the atom.
- Noble-gas configuration involves use of noble gases to write the configuration of other elements.
- This is done by identifying the atomic number of the element and then identifying the noble gas that comes before that particular element on the periodic table.
- Aluminium: The atomic number of Al is 13. The noble gas before Aluminium is Neon which has 10 electrons. Therefore the remaining 3 electrons fills up the 3s and 3p sub orbitals.
- Thus, the noble-gas configuration of Al is [Ne] 3s²3p¹
2. Arsenic, Atomic number is 33
- Noble gas before Arsenic is Ar,. Argon has 17 electrons, then the remaining electrons fills up the 4s, 3d and 4p sub-orbitals.
- Thus, the noble-gas configuration of As is [Ar] 4s²3d¹⁰ 4p³
Answer:
molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x 1 mole CH4 = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.
Explanation:
not to certain if this is right or not.. but hope it helps!
The rate constant of the reaction K we can get it from this formula:
K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B
So when after 10 s and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) = 16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14
