Answer:
solubility is 1.984x10⁻⁹M
Explanation:
When CaCO₃ is in water, the equilibrium that occurs is:
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Kps = [Ca²⁺] [CO₃²⁻] = 4.96x10⁻⁹
If you have a 0.250M solution of Na₂CO₃, [CO₃²⁻] = 0.250M:
[Ca²⁺] [0.250M] = 4.96x10⁻⁹
Assuming you are adding an amount of CaCO₃:
[X] [0.250 + X] = 4.96x10⁻⁹
<em>Where X is the amoun of CaCO₃ you can add, that means, solubility</em>
X² + 0.250X - 4.96x10⁻⁹ = 0
Solving for X:
X = -0.25M → False answer, there is no negative concentrations.
X = 1.984x10⁻⁹M.
That means, <em>solubility is 1.984x10⁻⁹M</em>
Answer: By carrying out series of wet test
Explanation: Dissolve the ionic compound in water. Ionic compounds will ionise and allow the solution to conduct an electric current. Afterwards specific test will be carried out to detect the presence of cations and anions. Another way would be to find the substance's melting point. Ionically bonded compounds have much higher melting points.
The correct answer is the second option. <span>Baking soda fizzing in vinegar is an example of a chemical change. This example is a chemical reaction which undergoes a chemical change since new substances are being formed the fizzing of the system represent that gas is being produced by the reaction.</span>
Answer:
The molecular formula of benzene = 
Explanation:
% of C = 92.3
Molar mass of C = 12.0107 g/mol
% moles of C = 
= 7.6848
% of H = 7.7
Molar mass of H = 1.00784 g/mol
% moles of H = 
= 7.6401
Taking the simplest ratio for C and H as:
7.6848 : 7.6401
= 1 : 1
The empirical formula is = 
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 12+ 1 = 13 g/mol
Molar mass = 78.0 g/mol
So,
Molecular mass = n × Empirical mass
78.0 = n × 13
⇒ n ≅ 6
<u>The molecular formula of benzene = </u>
