Answer:
0,13 moles of acid are produced
Explanation:
The reaction of the problem is:
PCl₅ + 4H₂O → 5HCl + H₃PO₄
Based on the reaction, 1 mole of PCl₅ produces 6 moles of acid (5 moles of HCl + 1 mole of H₃PO₄).
The molecular mass of PCl₅ is:
1P = 30,97g/mol + 5Cl = 5×35,45g/mol = <em>208,24 g/mol</em>
That means 4,5g of PCl₅ are:
4,5g PCl₅×(1mol / 208,24g) = 0,0216 moles of PCl₅. As 1 mole of PCl₅ produces 6 moles of acid, 0,0216 moles of PCl₅ produce:
0,0216 moles PCl₅ × (6 moles acid / mole of PCl₅) =
<em>0,13 moles of acid are produced</em>
I hope it helps!
Answer:
tepo jds fnsdmv dfjv dfj pvjdf ifjv dfjv sdijghdfuiExplanation:
I belive it is synaptic cleft
A protective layer composed of overlapping cells, like fish scales or roof tiles, but facing downwards. The outer cuticle holds your hair in your hair follicle by means of a Velcro-like bond. It also minimizes the movement of water (moisture) in and out of the underlying cortex.
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
