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
Cations from smallest to largest
Li⁺ ,Na⁺, K⁺ (from Periodic Table, the bigger number of period, the bigger size, of atom, so the bigger size of cation)
1) LiF smaller cation then KF
1,036 <span>853
</span><span>The lattice energy increases as cations get smaller, as shown by LiF and KF.
</span><span>I think this one should be correct answer, because the compared substances have also the same anion, and we can compare cations in them.
2) The same cation Li , so wrong statement.
3)</span>The same cation Na , so wrong statement.
4) NaCl smaller cation then KF
786 853
Answer:
V = 38.48 L
Explanation:
Given that,
No. of moles = 1.5 mol
Pressure, P = 700 torr
Temperature, T = 15°C = 288 K
We need to find the volume of the gas. The ideal gas equation is given by :
, R = L.Torr.K⁻¹.mol⁻¹
So, the required volume is equal to 38.48 L.
Answer:
Favorite Answer
1.0 x10^-14 = (1.0 x 10^-13) (x)
x = 1.0 x 10^-1 = 0.1 M (this is the [OH-])
pOH = -log 0.1 = 1.0
Explanation:
I hope this helps :) sorry if not :(
Answer:
Explanation:
According to the Law of Conservation of Mass, the mass of the products must equal the mass of the reactants.
- mass products = mass reactants
In this problem, the reaction is:
- The reactants are iron and oxygen. We know the mass of the iron sample is 10 grams.
- The product is ferric oxide. The mass of the ferric oxide sample is 18.2 grams.
We want to find how many grams of oxygen reacted. We have to get the oxygen by itself. 10 is being added to oxygen. The inverse of addition is subtraction. Subtract 10 from both sides of the equation.
<u>8.2 grams of oxygen </u>reacted with 10 grams of iron to form 18.2 grams of ferric oxide.
