Bath Bombs are colvalent bonds because they have both metals and non-metals before and after it makes contact with water.
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:
116.5 g of SO₂ are formed
Explanation:
The reaction is:
S₈(g) + 8O₂(g) → 8SO₂ (g)
Let's identify the moles of sulfur vapor, by the Ideal Gases Law
We convert the 921.4°C to Absolute T° → 921.4°C + 273 = 1194.4 K
5.87 atm . 3.8L = n . 0.082 L.atm/mol.K . 1194.4K
(5.87 atm . 3.8L) / (0.082 L.atm/mol.K . 1194.4K) = n → 0.228 moles of S₈
Ratio is 1:8, 1 mol of sulfur vapor can produce 8 moles of dioxide
Then, 0.228 moles of S₈ must produce (0.228 . 8) /1 = 1.82 moles
We convert the moles to g → 1.82 moles . 64.06 g /1mol = 116.5 g
Answer:
The correct answer is because the molecular structure.
Explanation:
The difficulty of ammonia and methane to be represented on paper is due to the molecular structure. These compounds have a three-dimensional projection with defined angles. Ammonia presents angles of 109.5º between the atom of Nitrogen and those of Oxygen. The ammonia presents 107.8º between the oxygen atoms.
In the methane molecule, there is 109.5º between the hydrogen molecules and the carbon atom. This results in the need for a 3D representation of the molecule.
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