Answer:
2.3 * 10^-5
Explanation:
Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.
Hence;
Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols
From the question;
1.2 * 10^-3 mols dissolves in 250 mL
x moles dissolves in 1000mL
x = 1.2 * 10^-3 mols * 1000/250
x= 4.8 * 10^-3 moldm^-3
CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)
Hence Ksp = [Ca^2+] [C2O4^2-]
Where;
[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3
Ksp = (4.8 * 10^-3)^2
Ksp = 2.3 * 10^-5
Determining the identity of substances is a critical part of chemistry because once the substance's identity is known, we can predict its behavior and understand the scenarios that it is involved in better.
For example, consider an industrial pipe where fouling (scaling) is occurring. If the compounds present in the scales are identified, steps may be taken to prevent and remove the scaling. This is one of many examples where identifying chemical substances is of high importance.
Answer:- 0.273 kg
Solution:- A double replacement reaction takes place. The balanced equation is:
We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.
= 63.8 g aluminum nitrate
From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.
We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.
= 
sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.
= 0.273 kg
So, 0.273 kg of 35% m/m sodium chlorate solution are required.
Mass of ammonium sulfate = 660.7 g
<h3>Further explanation</h3>
Given
3.01 x 10²⁴ molecules of ammonium sulfate
Required
mass
Solution
The mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
mol ammonium sulfate (NH₄)₂SO₄ :
n = N : No
n = 3.01 x 10²⁴ : 6.02 x 10²³
n = 5
mass ammonium sulfate :
= mol x MW
= 5 x 132,14 g/mol
= 660.7 g
