Liquid? maybe, its really inbetween if you get what i mean
<h3>
The supplements that are minerals are</h3>
- calcium
- sodium
- iron
- zinc
<u><em> Explanation</em></u>
- calcium and sodium are major minerals which are required by the body for
calcium- needed for muscle,hearing bone and for the support of synthesis and function of cells
sodium- is needed to control blood pressure and also for proper muscle and nerve function
- Zinc and iron are required in trace and both are needed for good health
Answer:
23.8 L
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the volume in liters of a 0.0380M potassium iodide solution that contains 150 g of potassium iodide. Be sure your answer has the correct number of significant digits.</em>
<em />
The molar mass of potassium iodide is 166.00 g/mol. The moles corresponding to 150 grams are:
150 g × (1 mol/166.00 g) = 0.904 mol
0.904 moles of potassium iodide are contained in an unknown volume of a 0.0380 mol/L potassium iodide solution. The volume is:
0.904 mol × (1 L/0.0380 mol) = 23.8 L
Answer:
295.7 mL of 24% trichloroacetic acid (tca) is needed .
Explanation:
Let the volume of 24% trichloroacetic acid solution be x
Volume of required 10% trichloroacetic acid solution =8 bottles of 3 ounces
= 24 ounces = 709.68 mL
(1 ounces = 29.57 mL)
Amount of trichloroacetic acid in 24% solution of x volume of solution will be equal to amount of trichloroacetic acid in 10% solution of volume 709.68 mL.
x = 295.7 mL
295.7 mL of 24% trichloroacetic acid (tca) is needed .
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
