Dissociation of para-bromobenzoic acid can be represented as:
4-BrC₆H₄COOH----->4-BrC₆H₄COO⁻ + H⁺
Dissociation constant of this acid can be calculated as:
Ka={[BrC₆H₄COO⁻][H⁺]}/[4-BrC₆H₄COOH]
as[4-BrC₆H₄COOH=Concentration of para- bromobenzoic acid=0.35 M
And as 1.69% of this acid dissociates to form [BrC₆H₄COO⁻] and [H⁺], so amount of these ions formed will be:
[BrC₆H₄COO⁻]=[H⁺]=1.69×0.35=0.59
So Now Ka=(0.59×0.59)/0.35
=0.99.
The state of matter is liquid.
Answer: Percent composition is calculated from a molecular formula by dividing the mass of a single element in one mole of a compound by the mass of one mole of the entire compound. This value is presented as a percentage.
Explanation: I hope that helped!
Answer:
2.58g
Explanation:
First calculate the moles of butane used in the reaction
moles = mass÷molar mass
= 0.85÷58
= 0.0147
According to the stoichiometric ratio:
C4H10 : CO2 = 2:8
moles of CO2 =(8÷2)×0.0147
=0.0586 moles
mass of CO2 = 0.0586×44
= 2.58g
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