Answer:
300000Pa or 3×10^5 Pa
Explanation:
Since the problem involves only two parameters of volume and pressure, the formula for Boyle's law is suitably used.
Using Boyle's law
P1V1 = P2V2
P1 is the initial pressure = 1.5×10^5Pa
V1 is the initial volume = 0.08m3
P2 is the final pressure (required)
V2 is the final volume = 0.04 m3
From the formula, P2 = P1V1/V2
P2 = 1.5×10^5 × 0.08 ÷ 0.04
= 300000Pa or 3×10^5 Pa.
Answer:
a. 1.78x10⁻³ = Ka
2.75 = pKa
b. It is irrelevant.
Explanation:
a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.
Equilibrium is:
HA ⇄ H⁺ + A⁻
And Ka is defined as:
Ka = [H⁺] [A⁻] / [HA]
The HA reacts with the base, XOH, thus:
HA + XOH → H₂O + A⁻ + X⁺
As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻
That means:
[HA] = [A⁻]
It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:
pH = pKa + log₁₀ [A⁻] / [HA]
Replacing:
2.75 = pKa + log₁₀ [A⁻] / [HA]
As [HA] = [A⁻]
2.75 = pKa + log₁₀ 1
<h3>2.75 = pKa</h3>
Knowing pKa = -log Ka
2.75 = -log Ka
10^-2.75 = Ka
<h3>1.78x10⁻³ = Ka</h3>
b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.
Answer: A volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.
Explanation:
Given: 
= ?, 
= 0.55 M
= 100.0 mL, 
= 2.50 M
Formula used to calculate the volume of KBr is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that a volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.
Answer: B
Explanation: to have a control, and many samples to investigate and cover the differences and anseretics.
