Answer:
Acid(BSA) = CH₃COOH
Base (BSB) = H₂O
Conjugate base (CB) = CH₃COO⁻
Conjugate acid (CA) = H₃O⁺
Explanation:
Equation of reaction;
CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺
Hello,
From my understanding of the question, we are required to identify the
1) Acid
2) Base
3) conjugate acid
4) conjugate base in the reaction
Acid (BSA) = CH₃COOH
Base (BSB) = H₂O
CA = conjugate acid = H₃O⁺
CB = conjugate base = CH₃COO⁻
The dissociation equation will be
NH4OH ---> NH4+ + OH-
Initial 0.006 0 0
Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053
Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053
Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053
Ka = 1.78 X 10^-5
Just look it up on goog^le or a chart
Answer:
a. True
Explanation:
Alkanes are chains of carbon atoms surrounded by hydrogen atoms. TRUE.
Alkanes are hydrocarbons, that is, they are organic compounds formed only by carbon and hydrogen. In alkanes, carbon atoms are bonded to each other through single covalent bonds and they are also bonded to hydrogen atoms through the same type of bonds. Alkanes have the general formula CnH2n+2.
Answer: -
Concentration of PbI₂ = 1.5 x 10⁻³ M
PbI₂ dissociates in water as
PbI₂ ⇄ Pb²⁺ + 2 I⁻
So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.
Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =
= 1.5 x 10⁻³ x 2 M
= 3 x 10⁻³ M
PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.
[Pb²⁺] = 1.5 x 10⁻³ M
So solubility product for PbI₂
Ksp = [Pb²⁺] x [ I⁻]²
=1.5 x 10⁻³ x (3 x 10⁻³)²
= 4.5 x 10⁻⁹
