The volume of the new solution is
calculation
by use of the formula
M1V1= M2V2
M1 (molarity 1) = 2.13 M
V1 (volume 1) = 1.24 l
M2 ( molarity 2) = 1.60 M
V2 (volume 2) = ?
by making V2 the subject of the formula
V2 = M1 V1/ M2
V2=( 1.24 x 2.13)÷ 1.60= 1.651 L
Answer:
The answer is given below.
Explanation:
We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.
AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M
HA → H+ + A-
Initial concentration: 0.1 → 3.98 ^10-3 + 0
equilibrium concentration: 0.1(1-α) → 3.98 * 10-3 + 0.1α 0.1α
pKa of chloroacetic acid is 2.9
-log(Ka) = 2.9
Ka = 1.26 * 10-3
From the equation, Ka = [H+] * [A-] / [HA]
1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)
Since α<<1, we assume 1-α = 1
Solving the equation, we have: α = 0.094
Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%
Answer:
In fact, there are actually several ways crude or refined oil may reach the marine environment. ... Use or consumption of oil (which includes operational discharges from ships and discharges from land-based sources): 37% Transportation (accidental spills from ships): 12% Extraction: 3% :))
<span>0.6 = mass/1.2 </span>
<span>mass = 0.6 x 1.2 </span>
<span>= 0.72 g </span>
