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Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
Answer:- The natural abundance of 
is 0.478 or 47.8% and 
is 0.522 or 52.2% .
Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:
Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)
We have been given with atomic masses for and as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.
Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of as n then the abundance of would be 1-n .
Let's plug in the values in the formula:
151.964=150.919860n+152.921243-152.921243n
on keeping similar terms on same side:
negative sign is on both sides so it is canceled:
The abundance of is 0.478 which is 47.8%.
The abundance of is = 
= 0.522 which is 52.2%
Hence, the natural abundance of is 0.478 or 47.8% and is 0.522 or 52.2% .
Water, Sun/Heat, and Carbon Dioxicide/Air
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