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How does the entropy change in the reaction 2C2H2(g) plus 5O2(g) 4CO2(g) plus 2H2O(l)?
In the reaction 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l), the entropy decreases. This is because the reactants consist of gaseous molecules, which have higher entropy due to their greater freedom of movement, while the products include liquid water, which has lower entropy. Additionally, there is a reduction in the number of gas molecules from 7 (2 C2H2 + 5 O2) to 4 (4 CO2), further contributing to the decrease in disorder. Overall, the transition from gas to liquid and the reduction in the number of gas molecules results in a net decrease in entropy.
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What is the complete ionic equation for the reaction 2kohaq h2so4aq 2h2ol k2so4aq?
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.
2H2O2l-- 2H2Ol O2g A. Given the standard free energy of formation values shown below calculate the free energy for this reaction. triangleGf H2O2 -120.4 kjmol?
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}).
What is the net ionic equation for the reaction 2kohaq h2so4aq 2h2ol k2so4aq?
OH- 2H+ -> 2H2O(I) ywwww :)
What is the complete ionic equation for the reaction 2kohaq h2so4aq 2h2ol k2so4aq?
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.
2H2O2l-- 2H2Ol O2g A. Given the standard free energy of formation values shown below calculate the free energy for this reaction. triangleGf H2O2 -120.4 kjmol?
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}).
Why is sodium stored in kerosene and not water?
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.
The neutralization of h2so4aq using naohaq is represented by the following equation 2naohaq h2so4aq -- na2so4aq 2h2ol if 30.40 ml of 0.500 m naohaq is neutralized by 22.02 ml of?
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.
