Answer:
a )Li
b)O
c)F
Explanation:
a) Li-1s^2 2s^1
F-1s^2 2s^2 2p^5
it is easy to pull out e- from 2p orbit than 2s because 2s orbit is close to nucleus.Therefore Li have high ionisation enthalpy
b)oxygen ion is larger than Na because o have fewer proton
c)F because it requires only 1e to achieve stable noble gas configuration.Therefore to achieve stable nobke gas electonic configuration it accept 1e.
Answer:
0.11%
Explanation:
Without mincing words, let us dive straight into the solution to the question/problem. The first step to solve this question is to write out the chemical reaction, that is the reaction showing the dissociation of acetic acid.
CH3COOH <=======================================> CH3COO⁻ + H⁺
Initially, the amount present in the acetic acid which is = 12M, the concentration for CH3COO⁻ and H⁺ is 0 respectively.
At equilibrium, the amount present in the acetic acid which is = 12 - x, the concentration for CH3COO⁻ = x and H⁺ = x respectively. Note that the ka for acetic acid = 1.8 × 10⁻⁵.
1.8 × 10⁻⁵ = x²/ 14 - x. Therefore, x = 0.0158 M.
The next thing to do is to calculate for the percentage of dissociation, this can be done as given below:
percentage of dissociation = x/14 × 100. Recall that the value that we got for x = 0.0158 M. Hence, the percentage of dissociation = 0.0158 M/ 14m × 100 = 0.11%
Answer:
space tools are the one ur llooking for :)
The number of protons in the nucleus of the atom is equal to the atomic number (Z). The number of electrons in a neutral atom is equal to the number of protons.
C
because the mass never changes, it is always equal on both sides.
