This is a chemical reaction equation
H2SO4(aq) + Ca(OH)2(s) --> CaSO4(s) + 2H2O(l)
OH- 2H+ -> 2H2O(I) ywwww :)
Because it will react violently in water producing hydrogen gas in a highly exothermic reaction which often causes ignition of the hydrogen gas liberated: 2Nas + 2H2Ol --> H2,g + 2(NaOH)aq The kerosene used for storage must be importantly water free, or anhydrous, otherwise this reaction will also occur but more slowly. This is often exploited to remove trace water from organic solvents for chemical synthesis.
OH- 2H+ -> 2H2O(I) ywwww :)
Aqueous Sulfuric Acid
To find the molarity of the sulfuric acid (H2SO4), we first calculate the number of moles of NaOH used: 30.40 mL * 0.500 mol/L = 15.20 mmol NaOH. Since the mole ratio between NaOH and H2SO4 is 2:1, 15.20 mmol of NaOH would neutralize 7.60 mmol of H2SO4. Now we can find the molarity of H2SO4 using its volume: 7.60 mmol / 22.02 mL = 0.345 M H2SO4.
To write the complete ionic equation for the reaction between potassium hydroxide (KOH) and sulfuric acid (H₂SO₄), we first recognize that KOH dissociates into K⁺ and OH⁻ ions, while H₂SO₄ dissociates into 2 H⁺ and SO₄²⁻ ions. The complete ionic equation is: 2 K⁺(aq) + 2 OH⁻(aq) + 2 H⁺(aq) + SO₄²⁻(aq) → 2 H₂O(l) + K₂SO₄(aq). This shows the ions involved in the reaction and the products formed.
2H+ + SO42- + Ca2+ + 2I- CaSO4 + 2H+ + 2I
Because it will react violently in water producing hydrogen gas in a highly exothermic reaction which often causes ignition of the hydrogen gas liberated: 2Nas + 2H2Ol --> H2,g + 2(NaOH)aq The kerosene used for storage must be importantly water free, or anhydrous, otherwise this reaction will also occur but more slowly. This is often exploited to remove trace water from organic solvents for chemical synthesis.
To calculate the standard free energy change (ΔG) for the reaction (2 \text{H}_2\text{O}_2 (l) \rightarrow 2 \text{H}_2\text{O} (l) + \text{O}_2 (g)), we can use the formula: [ \Delta G = \sum \Delta G_f \text{(products)} - \sum \Delta G_f \text{(reactants)} ] Given that (\Delta G_f \text{(H}_2\text{O}) = -237.13 , \text{kJ/mol}) (standard value) and (\Delta G_f \text{(O}_2) = 0 , \text{kJ/mol}), we can substitute: [ \Delta G = [2(-237.13) + 0] - [2(-120.4)] ] [ \Delta G = -474.26 + 240.8 = -233.46 , \text{kJ} ] Thus, the free energy change for the reaction is approximately (-233.46 , \text{kJ}).