Answer:
(a) adding 0.050 mol of HCl
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.
In the buffer:
1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>
1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>
-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-
Thus:
<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.
(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.
(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>
Answer:
fjskeowkcnekvo Dee five votes come vote for dog even r
Answer:
t = 7.58 * 10¹⁹ seconds
Explanation:
First order rate constant is given as,
k = (2.303
/t) log [A₀]
/[Aₙ]
where [A₀] is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>
[A₀] = 615 calories;
[Aₙ] = 615 - 480 = 135 calories
k = 2.00 * 10⁻²⁰ sec⁻¹
substituting the values in the equation of the rate constant;
2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)
(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)
t = 2.303 / 3.037 * 10⁻²⁰
t = 7.58 * 10¹⁹ seconds
