Answer:
Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]
Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]
Explanation:
An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:
HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)
And Ka, the acid dissociation constant is:
<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />
When base:
HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)
And kb, base dissociation constant is:
<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>
The trick for this problem is to understand atomic mass: the fact that different atoms have different masses. What we need to do is add up all the atomic masses of the compound and work out the ratio of mass of water to the mass of sodium carbonate. Atomic masses are often given for each atom in the periodic table, but you can look them up on google too.
You can do this by adding up individual atoms for each molecule, or you can shortcut and lookup the molar mass of the compound (i.e.the task already done for you).
The molar mass of water is 18.01g/mole so for 10 moles of water we have a mass of 180.1g.
The molar mass of sodium carbonate is 106g/mole (google).
So the total mass of the sodium carbonate decahydrate compound is 180.1+106 = 286.1g, of which water would make up 180.1g, so the percentage of water is is 180.1/286.1 = 0.629, so we can round this to 63%
:)
Lipids cannot be compressed since there is only a small distance between the molecules when bonded
Answer:
k= 1.925×10^-4 s^-1
1.2 ×10^20 atoms/s
Explanation:
From the information provided;
t1/2=Half life= 1.00 hour or 3600 seconds
Then;
t1/2= 0.693/k
Where k= rate constant
k= 0.693/t1/2 = 0.693/3600
k= 1.925×10^-4 s^-1
Since 1 mole of the nuclide contains 6.02×10^23 atoms
Rate of decay= rate constant × number of atoms
Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms
Rate of decay= 1.2 ×10^20 atoms/s
Answer:
30.6 g of C is formed.
Explanation:
2A + B → C
Average rate of reaction = 2[A]/Δt = [B]/Δt = [C]/Δt
Average rate of reaction = [C]/Δt
Average rate of reaction = 15 g / 9 min
Average rate of reaction = 1.7 g of C / min
Average rate of reaction = [C]/Δt
[C] = Average rate of reaction x Δt
[C] = 1.7 g of C / min x 18 min
[C] = 30.6 g of C