Answer:
We could do two 1:50 dilutions and one 1:4 dilutions.
Explanation:
Hi there!
A solution that is 1000 ug/ ml (or 1000 mg / l) is 1000 ppm.
Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.
Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.
We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):
First step (1:50 dilution):
Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).
Step 2 (1:50 dilution):
Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)
Step 3 (1:4 dilution):
Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.
Answer: The correct answer is a) Acetylcholine.
Explanation:
The acetylcholine is a very important neurotransmitter involved in the mediation of synaptic activity between neurons. Among its most important functions we find: expression of mood states, vasodilation, gastrointestinal motility, bronchoconstriction, circular iris muscle contraction and sweat gland secretion.
Therefore the correct answer is a) Acetylcholine.
26g --- 1 mol
56g --- X
X= 56/26 = 2,154 mol
959 ml = 959cm³ = 0,959dm³
C = n/V
C = 2,154/0,959
C = 2,246 mol/dm³
Best Answer: <span>(a)
8.9 x 10^-7 = x^2 / 0.15-x
x = [OH-] = 0.00037 M
pOH = 3.4
pH = 14 - 3.4 = 10.6
(b)
Ka = Kw/Kb = 5.6 x 10^-10 = x^2 / 0.20-x
x = [H+] = 0.000011 M
pH = 5.0</span>
The balanced chemical equation would be as follows:
<span>K2PtCl4(aq) + 2NH3(aq) --> Pt(NH3)2Cl2(s) + 2KCl(aq)
We are given the amount of </span>K2PtCl4 to be used in the reaction. This will be the starting point for our calculations. We do as follows:
65 g K2PtCl4 ( 1 mol / 415.09 g ) ( 1 mol Pt(NH3)2Cl2 / 1 mol K2PtCl ) ( 300.051 g / 1 mol ) = 46.99 g Pt(NH3)2Cl produced