Answer:
Most positive = rubidium
Most negative = fluorine
Explanation:
Electron affinity of an element is the energy released when an electron is attached to a neural atom to form an ion in its gaseous state.
X + e⁻ → X⁻
Electron affinity is similar to electronegativity which is the tendency at which an atom accepts an ion towards itself.
Electron affinity increases across the period and decreases down the group in the periodic table.
In the above option,
Fluorine has the highest electron affinity
Rubidium has the lowest electron affinity
Tellurium and then finally Phosphorus
Helium in this case would have the lowest electron affinity because it has filled orbital and does not require any electron to attain stability. Technically, Helium has the lowest or is expected to have the lowest electron affinity which is close to zero according to quantum mechanics.
Most positive = rubidium
Most negative = fluorine.
You can check periodic table for their exact values
Answer:
Concentration solution A was 0.5225 M
Explanation:
10.00 mL of solution A was diluted to 50.00 mL and yields 50.00 mL of solution B
According to laws of dilution- 
where,
and
are concentration of solution A and B respectively
and
are volumes of solution A and B respectively
Here
= 0.1045 M,
= 50.00 mL and
= 10.00 mL
Hence, 
So, concentration solution A was 0.5225 M
<u>Given:</u>
Concentration of HNO3 = 7.50 M
% dissociation of HNO3 = 33%
<u>To determine:</u>
The Ka of HNO3
<u>Explanation:</u>
Based on the given data
[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M
The dissociation equilibrium is-
HNO3 ↔ H+ + NO3-
I 7.50 0 0
C -2.48 +2.48 +2.48
E 5.02 2.48 2.48
Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23
Ans: Ka for HNO3 = 1.23
Answer:
is larger
Explanation:
, where
is the acid dissociation constant.
For a monoprotic acid e.g. HA,
and ![\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
So, clearly, higher the
value , lower will the the
In this mixture, at equilibrium,
will be constant.
of HF is grater than
of HCN
Hence, ![(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})](https://tex.z-dn.net/?f=%28%5Cfrac%7BF%5E%7B-%7D%7D%7B%5BHF%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HF%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29%3E%28%5Cfrac%7BCN%5E%7B-%7D%7D%7B%5BHCN%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HCN%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29)
So,
is larger