Answer:
34.15% is the mass percentage of calcium in the limestone.
Explanation:
Mass of precipitate that is calcium oxalate = 140.2 mg = 0.1402 g
1 mg = 0.001 g
Moles of calcium oxalate = 
1 mole of calcium oxalate have 1 mole of calcium atom.
Then 0.001095 moles of calcium oxalate will have 0.001095 moles of calcium atom.
Mass of 0.001095 moles of calcium :
0.001095 mol × 40 g/mol = 0.04381 g
Mass of sample of limestone = 128.3 mg = 0.1283 g
Percentage of calcium in limestone:
34.15% is the mass percentage of calcium in the limestone.
Answer:
The precipitate will form.
Explanation:
Let's write the equilibrium expression for the solubility product of calcium sulfate:
⇄ 
The solubility product is defined as the product of the free ions raised to the power of their coefficients, in this case:
![K_{sp}=[Ca^{2+}][SO_4^{2-}]=10^{-4.5}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BSO_4%5E%7B2-%7D%5D%3D10%5E%7B-4.5%7D)
Our idea is to find the solubility quotient, Q, and compare it to the K value. A precipitate will only form if Q > K. If Q < K, the precipitate won't form. In this case:
![Q_{sp}=[Ca^{2+}][SO_4^{2-}]=5.00\cdot10^{-2} M\cdot7.00\cdot10^{-3} M=3.5\cdot10^{-4}](https://tex.z-dn.net/?f=Q_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BSO_4%5E%7B2-%7D%5D%3D5.00%5Ccdot10%5E%7B-2%7D%20M%5Ccdot7.00%5Ccdot10%5E%7B-3%7D%20M%3D3.5%5Ccdot10%5E%7B-4%7D)
Now given the K value of:
Notice that:
This means the precipitate will form, as we have an excess of free ions and the equilibrium will shift towards the formation of a precipitate to decrease the amount of free ions.
The purpose of an universal indicator is to test wether a solution is acid or if its a base. It changes colors according to the PH's. It helps a lot in the indication of a chemical reaction because it can say if <span>each component loses or gains protons depending upon the acidity or basicity of the solution being tested.An universal indicator can say if a determined solution proves to be endothermic or exothermic. If the solution is not tested as acid or base then we cannot knwo if there will be an endothermic reaction or an exhotermic one</span>
Answer:
Ka= 2.3 x 10^-11
Explanation:
so Ka x Kb = Kw where Kw=10^-14 at 25 degrees Celsius
plugging in Kb and Kw, we have
Ka x 4.4 x 10^-4 = 10^-14
solving for Ka,
Ka= (10^-14)/(4.4 x 10^-4)
Ka= 2.3 x 10^-11
The balanced equation for the above neutralisation reaction is as follows;
2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O
stoichiometry of KOH to H₂SO₄ is 2:1
neutralisation is the reaction between H⁺ ions and OH⁻ ions to form water which is neutral
number of KOH moles - 1.56 mol
2 mol of KOH require 1 mol of H₂SO₄ for neutralisation
therefore 1.56 mol of KOH require - 1/2 x 1.56 mol = 0.78 mol
0.78 mol of H₂SO₄ are required for neutralisation
