Answer : The correct option is, (c) use of a mobile and a stationary phase.
Explanation :
Chromatography : It is a separation process or technique of a mixture in which a mixture is distributed between the two phases at different rates, one of which is stationary phase and another is mobile phase.
Mobile phase : The mixture is dissolved in a solution is known as mobile phase.
Stationary phase : It is an adsorbent medium and It is a solid, liquid or gel that remains immovable when a liquid or a gas moves over the surface of adsorbent. It remains stationary.
Hence, a characteristic feature of any form of chromatography is the use of a mobile and a stationary phase.
Answer:
Total Ionic equation:
H⁺(aq) + NO₃⁻ (aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + NO₃⁻ (aq)
Explanation:
Chemical equation:
HNO₃ + NaOH → NaNO₃ + H₂O
Balanced chemical equation:
HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)
Total Ionic equation:
H⁺(aq) + NO₃⁻ (aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + NO₃⁻ (aq)
Net ionic equation:
H⁺(aq) + OH⁻(aq) → H₂O(l)
The NO₃⁻ (aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation
Answer:
Since they are in a geostationary orbit, the GOES satellites provide continuous monitoring of the Earth's surface. They are able to constantly monitor the life cycle of significant weather such as hurricanes, tornadoes, flash floods, and hail storms.
Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06