Answer:
Explanation:
To separate the a mixture of chalk and potassium chloride, we must not that chalk is calcium carbonate compound, CaCO₃.
Calcium carbonate has low solubility in water. KCl is readily soluble in water and it is also an ionic compound.
To separate a mixture of compounds with various solubility, we can carryout dissolution, filtration and evaporation.
We first pour pure water into the mixture. Water will dissolve the potassium chloride readily.
Then using a filter paper we filter out the suspended chalk particles. Leave the filtrate to then dry and collect it.
The solution filtered should be evaporated to dryness. This will leave the KCl behind from the solution.
Answer:
Im guessing it's C....................................................
<span>34.2 grams
Lookup the atomic weights of the involved elements
Atomic weight potassium = 39.0983
Atomic weight Chlorine = 35.453
Atomic weight Oxygen = 15.999
Molar mass KClO3 = 39.0983 + 35.453 + 3 * 15.999 = 122.5483 g/mol
Moles KClO3 = 87.4 g / 122.5483 g/mol = 0.713188188 mol
The balanced equation for heating KClO3 is
2 KClO3 = 2 KCl + 3 O2
So 2 moles of KClO3 will break down into 3 moles of oxygen molecules.
0.713188188 mol / 2 * 3 = 1.069782282 mols
So we're going to get 1.069782282 moles of oxygen molecules. Since each molecule has 2 atoms, the mass will be
1.069782282 * 2 * 15.999 = 34.23089345 grams
Rounding the results to 3 significant figures gives 34.2 grams</span>
Answer:
0.29mol/L or 0.29moldm⁻³
Explanation:
Given parameters:
Mass of MgSO₄ = 122g
Volume of solution = 3.5L
Molarity is simply the concentration of substances in a solution.
Molarity = number of moles/ Volume
>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.
Number of moles = mass/ molar mass
Molar mass of MgSO₄:
Atomic masses: Mg = 24g
S = 32g
O = 16g
Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol
= (24 + 32 + 64)g/mol
= 120g/mol
Number of moles = 122/120 = 1.02mol
>>>> From the given number of moles we can evaluate the Molarity using this equation:
Molarity = number of moles/ Volume
Molarity of MgSO₄ = 1.02mol/3.5L
= 0.29mol/L
IL = 1dm³
The Molarity of MgSO₄ = 0.29moldm⁻³
