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
Answer:
One mole
Explanation:
The balanced chemical equation is
<u>1</u>NaOH + <u>1</u>HNO₃ ⟶ NaNO₃ + H₂O
<u>1</u> mol <u>1</u> mol
The coefficients in front of the formulas tell you the amount of something that reacts with an equivalent amount of something else.
In this reaction, 1 mol NaOH reacts with 1 mol HNO₃.
Actually, that does not happen until the protostar becomes a star when nuclear ignition starts and is maintained. It takes awhile for new star to go through its T-Tauri stage and settle down on the main sequence.
<span>A STAR does not reach hydrostatic equilibrium until it on the main sequence. Otherwise, it would remain a brown dwarf with not enough mass to to maintain nuclear fusion for more than 3,000 to 10,00 years. </span>
Answer:
C contains one N and three I atoms
