Answer:
l = 0 → s = 2 electrons;
l = 1 → p = 6 electrons;
l = 2 → d = 10 electrons;
l= 3 → f = 14 electrons.
Explanation:
For the quantum theory, the probability to find an electron is higher in the space region called orbital. It's impossible to determine where the electron is and his velocity at the same time (uncertainty principle). So, the theory determines four quantum numbers to characterize an electron, so it's easy to identify it:
- n is the principal quantum number and identify the shell where the electron is. It varies from 1 to 7 and is represented by the letters K, L, M, N, O, P, and Q;
- l is the azimuthal quantum number and identify the subshell (or sublevel) where the electron is. It varies from 0 to 3 and is represented by the letters s, p, d, and f;
- ml is the magnetic quantum number, and it represents the orbital. It varies from -l to +l, passing by 0. Each orbital can have at least 2 electrons;
- ms is the spin number and represents the spin of the electrons. It can be +1/2 or - 1/2.
Then, the sublevel s (l= 0) only has 1 orbital (ml = 0) so, it can have ate least 2 electrons; the sublevel p (l= 1) has 3 orbitals (ml = -1, ml= 0, ml = +1), so it can have at least 6 electrons; the sublevel d (l = 2) has 5 orbitals (ml = -2, ml = -1, ml = 0, ml = +1, ml = +2), so it can have at least 10 electrons; and the sublevel f (l = 3) has 7 orbitals (ml = -3, ml = -2, ml = -1, ml = 0, ml = +1, ml = +2, ml = +3), so it can have at least 14 electrons.
There is only one cesium ion on reactant (left hand side) of the reaction equation.
The formation of cesium fluoride involves the combination of cesium ion and fluoride ion. Cesium fluoride is an ionic compound since it is composed of ions.
The reaction between cesium ions and fluoride ions to form cesium fluoride occurs as follows;
Cs^+ + F^- ------> CsF
So, there is only one cesium ion on reactant (left hand side) of the reaction equation.
Learn more: brainly.com/question/1781817
B I think, it has layers that protects the cells nucleus
Answer:
Explanation:
Firstly of all we have to construct the min-heap of the k-sub list and each sub list which is a node in the constructed min-heap.
We have several steps to follows:
Step-1. When we compare the two sub lists, at the starting we can compare their first elements which is actually their minimum elements.
Step-2. The min-heap formation will cost be O(k) time.
Step-3. After the step 1 & step 2 we can run the minimum algorithm which can be extracted from the minimum element in the root list.
Step-4. Then Update the root list in the heap and after that simplify the min-heap as maintained by the new minimum element in the root list.
Step-5. If any root sub-list becomes empty in the step 4 then we can take any leaf sub-list from the root and simplify it.
Step-6. At every Extraction of the element it can take up to O(log k) time.
Hence, We can say that the extract of n element in the total whose
Running time will be O(n log k + k) which can be equal to the O(n log k+ k) (since k < n).
