THE MOLECULE HAS A C=C AND AN -OH GROUP, SO IT IS CALLED AN ENE/OL, I.E., AN ENOL. ENOLS CAN BE FORMED ONLY FROM CARBONYL COMPOUNDS WHICH HAVE ALPHA HYDROGENS. THEY CAN BE FORMED BY ACID OR BASE CATALYSIS, AND ONCE FORMED ARE HIGHLY REACTIVE TOWARD ELECTROPHILES, LIKE BROMINE.
Mitosis has 4 steps and meiosis has 5 steps so its 9 steps total here's a photo to help you if you need it
Answer: the pH of the solution is 4.52
Explanation:
Consider the weak acid as Ha, it is dissociated as expressed below
HA H⁺ + A⁻
the Henderson -Haselbach equation can be expressed as;
pH = pKa + log( [A⁻] / [HA])
the weak acid is dissociated into H⁺ and A⁻ ions in the solution.
now the conjugate base of the weak acid HA is
HA(aq) {weak acid} H⁺(aq) + A⁻(aq) {conjugate base}
so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;
pKₐ = -logKₐ
pKₐ = -log ( 7.4×10⁻⁵ )
pKₐ = 4.13
now thw pH is
pH = pKₐ + log( [A⁻] / [HA])
pH = 4.13 + log( [0.540] / [0.220])
pH = 4.13 + 0.3899
pH = 4.5199 = 4.52
Therefore the pH of the solution is 4.52
Organic elements, depends on the context of the question but I suspect that is the answer. The organic elements are H,O,C,N,P,S
