Answer:
H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)
Explanation:
H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq) + SO²⁻₄(aq) + H₂O(l)
A careful observation of the equation above, shows that the equation is already balanced.
To obtain the net ionic equation, we simply cancel Mg²⁺ from both side of the equation as shown below:
H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)
Among formic acid (HCOOH ) and sulfuric acid (H₂SO₄), formic acid is the weak acid. Acidic strength of any acid is the tendency of that acid to loose proton. Among these two acids formic acid has a pKa value of 3.74 greater than that of sulfuric acid i.e. -10. Remember! Greater the pKa value of acid weaker is that acid and vice versa. Below I have drawn the Ionization of both acids to corresponding conjugate bases and protons. The structures below with charges are drawn in order to explain the reason for strength. As it is seen in charged structure of formic acid, there is one positive charge on carbon next to oxygen carrying proton. The electron density is shifted toward carbon as it is electron deficient and demands more electron hence, attracting electron density from oxygen and making the oxygen hydrogen bond more polar. While, in case of sulfuric acid it is depicted that Sulfur attached to oxygen containing proton has 2+ charge, means more electron deficient as compared to carbon of formic acid, hence, more electron demanding and strongly attracting electrons from oxygen and making the oxygen hydrogen bond very polar and highly ionizable.
Answer:
During chemical reactions, matter is neither
created nor destroyed; it just changes form.
Explanation:
Answer:
1.12g/mol
Explanation:
The freezing point depression of a solvent for the addition of a solute follows the equation:
ΔT = Kf*m*i
<em>Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)</em>
<em>ΔT = 13.4°C - (-12.82) = 26.22°C</em>
<em>m is molality of the solution</em>
<em>Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)</em>
<em>And i is Van't Hoff factor (1 for all solutes in benzonitrile)</em>
Replacing:
26.22°C = 5.35°Ckgmol⁻¹*m*1
4.90mol/kg = molality of the compound X
As the mass of the solvent is 100g = 0.100kg:
4.9mol/kg * 0.100kg = 0.490moles
There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:
0.551g / 0.490mol
= 1.12g/mol
<em>This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.</em>
