This problem is providing us with the molality of a solution of calcium iodide as 0.01 m. So the most likely van't Hoff factor is required and theoretically found to be 3 due to the following:
<h3>Van't Hoff factor:</h3>
In chemistry, the correct characterization of solutions also imply the identification of the ions it will release in aqueous solution. For that reason, the van't Hoff factor gives us an idea of this number, according to the formula the solute has got.
In such a way, for calcium iodide, we write its ionization equation as shown below:
Assuming it is able to ionize due to the low molality, because if it was higher, then it won't ionize. Hence, since we have three moles of ion products, one Ca²⁺ and two I⁻, we can conclude the van't Hoff factor would be 3, although calculations may lead to a different, yet close result.
Learn more about the van't Hoff factor: brainly.com/question/23764376
Explanation:
The balanced equation of the reaction is given as;
Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)
1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?
From the reaction;
2 mol of HBr produces 1 mol of MgBr2
Converting to masses using;
Mass = Number of moles * Molar mass
Molar mass of HBr = 80.91 g/mol
Molar mass of MgBr2 = 184.113 g/mol
This means;
(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2
18.3g would produce x
161.82 = 184.113
18.3 = x
x = (184.113 * 18.3 ) / 161.82 = 20.8 g
2. How many moles of H2O will be produced from 18.3 grams of HBr?
Converting the mass to mol;
Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol
From the reaction;
2 mol of HBr produces 2 mol of H2O
0.226 mol would produce x
2 =2
0.226 = x
x = 0.226 * 2 / 2 = 0.226 mol
3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?
From the reaction;
2 mol of HBr reacts with 1 mol of Mg(OH)2
18.3g of HBr = 0.226 mol
2 = 1
0.226 = x
x = 0.226 * 1 /2
x = 0.113 mol
<span>1 ml of water weighs 1 gram so 1 liter (1000 ml) weighs 1000 grams. A 3% solution (3% = 0.03) of hydrogen peroxide (w/v) would contain 1000 grams x 0.03 or 30 grams. The chemical formula of hydrogen peroxide is H2O2 and a mole weighs 34.0147 grams/mole. So 30 grams of H2O2 divided by 34.0147 grams/mole equals 0.88 moles of H2O2. The concentration of a 3% (w/v) hydrogen peroxide solution therefore contains 30 grams of H202 (or 0.88 moles of H202) per in a liter of water (or 1000 grams H20) would thus be 0.88 moles H2O2 per liter (0.88 moles H2O2/l) .</span>
Answer:
A beaker
Step-by-step explanation:
Specifically, I would use a 250 mL graduated beaker.
A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.
You don't need precisely 100 mL solution.
If the beaker is graduated, you can easily measure 100 mL of the stock solution.
Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).
To squeeze a gas into a dmaller place is to compress it.
