The acid dissociation constant is 1.3 × 10^-3.
<h3>What is acid-dissociation constant?</h3>
The acid-dissociation constant is a constant that shows the extent of dissociation of an acid in solution. We have to set up the reaction equation as shown below;
Let the acid be HA;
HA + H2O ⇄ H3O^+ + A^-
since the pH of the solution is 2.57 then;
[H3O^+] = Antilog(-pH) = Antilog(-2.57) = 2.7 × 10^-3
We can see that; [H3O^+] = [A^-] so;
Ka = (2.7 × 10^-3)^2/(5.5 × 10^–3)
Ka = 1.3 × 10^-3
Learn more about acid-dissociation constant: brainly.com/question/9728159
Rewrite the formula C=5/9(F-32) substituting 23 for C: 23=5/9(F-32), then multiply both sides by the reciprocal of 5/9.
(9/5)*(23)=(9/5)*5/9(F-32)
41.4=F-32; add 32 to both sides.
41.4+32=F-32+32
73.4=F
Answer:
Mole fraction for solute = 0.1, or 10%
Molality = 6.24 mol/kg
Explanation:
22.3% by mass → In 100 g of solution, we have 22.3 g of HCOOH
Mass of solution = 100 g
Mass of solute = 22.3 g
Mass of solvent = 100 g - 22.3g = 77.7 g
Let's convert the mass to moles
22.3 g . 1mol/ 46 g = 0.485 moles
77.7 g. 1mol / 18 g = 4.32 moles
Total moles = 4.32 moles + 0.485 moles = 4.805 moles
Xm for solute = 0.485 / 4.805 = 0.100 → 10%
Molality → mol/ kg → we convert the mass of solvent to kg
77.7 g. 1 kg / 1000g = 0.0777 kg
0.485 mol / 0.0777 kg = 6.24 m
the solid particles take up the intermolecular spaces in the liquid.
Answer:
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-
Explanation:
