Answer:
1) 1,... 2
2) 18
3) n= 3 and I=1
Explanation:
1) when l= 0, its an s-sub-level, and only 1 orbital is possible which can carry only 2-electrons
2) the maximum number of electron is given by 2n^2= 2×3^2= 18
3) in 3p, the coefficient of p is the value of n= 3 and l-value of P is 1
Answer:
2.2×10^8
Explanation:
Cu(OH)2(s)<---------> Cu^2+(aq) + 2OH^-(aq) Ksp=2.2 x 10 ^-20
2H3O^+(aq) + 2OH^-(aq) <-------> 4H2O(l). Kw= 1×10^14
Cu^2+(aq) + 4H2O(l) <--------> [Cu(H2O)4]^2+(aq)
Overall ionic reaction:
Cu(OH)2(s) +2H3O^+(aq) <---------> [Cu(H20)4]^2+(aq)
Equilibrium constant for the reaction: Ksp×Kw= 2.2 x 10 ^-20 × (1/(1×10^-14))^2
Keq= 2.2×10^8
Kw= ion dissociation constant of water
Ionization energy refers to the amount of energy needed to remove an electron from an atom. Ionization energy decreases as we go down a group. Ionization energy increases from left to right across the periodic table.
<h3>What is ionization energy?</h3>
Ionization is the process by which ions are formed by the gain or loss of an electron from an atom or molecule.
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period but decreases along a group.
Smaller is the size of an atom more will be the force of attraction between its protons and electrons. Hence, more amount of energy is required to remove an electron.
Thus, we can conclude that the energy required to remove an electron from a gaseous atom is called ionization energy.
Learn more about the ionization energy here:
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In general chemistry, isotopes are substances that belong to one specific element. So, they all have the same atomic numbers. But they only differ in the mass numbers, or the number of protons and neutrons in the nucleus. In a nutshell, they only differ in the number of neutrons.
For Nickel, there are 5 naturally occurring isotopes. Their identities, masses and relative abundance are listed below
Isotope Abundance Atomic Mass
Ni-58 68.0769% <span>57.9353 amu
Ni-60 </span>26.2231% <span>59.9308 amu
Ni-61 </span>1.1399 % <span>60.9311 amu
Ni-62 </span>3.6345% <span>61.9283 amu
Ni-64 </span>0.9256% <span>63.9280 amu
To determine the average atomic mass of Nickel, the equation would be:
Average atomic mass = </span>∑Abundance×Atomic Mass
Using the equation, the answer would be:
Average atomic mass = 57.9353(68.0769%) + 59.9308(26.2231%) + 60.9311(1.1399%) + 61.9283(3.6345%) + 63.9280(0.9256%)
Average atomic mass = 58.6933 amu
