Answer:
3.75 L
Explanation:
We can solve this problem by using <em>Charles' law</em>, which states:
Where subscript 1 stands for initial volume and temperature and subscript 2 for final volume and temperature, meaning that in this case:
We <u>input the data</u>:
- 2.5 L * 300 K = V₂ * 200 K
And <u>solve for V₂</u>:
Answer:
Explanation:
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
ionisation constant = 1.36 x 10⁻⁴ .
molecular weight of lactic acid = 90 g
moles of acid used = 20 / 90
= .2222
it is dissolved in one litre so molar concentration of lactic acid formed
C = .2222M
Let n be the fraction of moles ionised
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionisation constant Ka
Ka = nC x nC / C - nC
= n²C ( neglecting n in the denominator )
n² x .2222 = 1.36 x 10⁻⁴
n = 2.47 x 10⁻²
nC = 2.47 x 10⁻² x .2222
= 5.5 x 10⁻³
So concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per litre .
A) 2, 3 , 2 simply valence the rest
