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:
0.038 g of reactant
Explanation:
Data given:
Heat release for each gram of reactant consumption = 36.2 kJ/g
mass of reactant that release 1360 J of heat = ?
Solution:
As 36.2 kJ of heat release per gram of reactant consumption so first we will convert KJ to J
As we know
1 KJ = 1000 J
So
36.2 kJ = 36.2 x 1000 = 36200 J
So it means that in chemical reaction 36200 J of heat release for each gram of reactant consumed so how much mass of reactant will be consumed if 1360 J heat will release
Apply unity formula
36200 J of heat release ≅ 1 gram of reactant
1360 J of heat release ≅ X gram of reactant
Do cross multiplication
X gram of reactant = 1 g x 1360 J / 36200 J
X gram of reactant = 0.038 g
So 0.038 g of reactant will produce 1360 J of heat.
Light does not travel at a constant speed in a vacuum, compared to in air, because the light is being absorbed by atoms and molecules in the air. But light does travel at a constant speed in a vacuum.
So I agree with A
All that talk about moving forward is irrelevant (I think)
Answer:
= 74.4 grams / mole. Ernest Z. The reaction will produce 15.3 g of KCl
Explanation:
B
Please help
Sorry I need 20 characters to submit this Answer that’s why I’m adding more words
