I want to find acetic acid - sodium hydroxide solution enthalpy. They gave 1.0M acetic acid...
I want to find acetic acid - sodium hydroxide solution enthalpy. They gave 1.0M acetic acid and 1.0M sodium hydroxide have -55.0 kj/mol
And they also gave 70ml of 1.0M acetic acide combine with 50ml 1.0M sodium hydroxide. I want to find the mole of this solution to use -55.0 kj/mol. How can I calculate the mole of acetic acid and sodium hydroxide solution?
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I want to find acetic acid - sodium hydroxide solution enthalpy.
They gave 1.0M acetic acid...
so i know that acetic acid is the solution and i googled the molar mass for...
so i know that acetic acid is
the solution and i googled the molar mass for it which is 60.052,
and the mass was 59.81 so i did 59.81/60.052 which gave me the
answer 0.99597. Why is it wrong ? i dont understand, thats how you
find moles right ? mass/molar mass.
Part A initial temperature of 20.1C. and the final The aluminum cup inside your calorimeter weighs 37.55 g. You add 59.81 g of 1.0 M acetic acid solution...
In a one step reaction of neutralization of solid sodium hydroxide with and acid the enthalpy...
In a one step reaction of neutralization of solid sodium hydroxide with and acid the enthalpy was measured as −123.95kJ/mol. Theoretical value is -98.800 KJ/mol What is the percent error? Round your answer up to integer percents:
Find the enthalpy of neutralization of 47.2 mL 1.00M aqueous sodium hydroxide with 40.3,mL 1.00M hydrochloric acid...
Find the enthalpy of neutralization of 47.2 mL 1.00M aqueous sodium hydroxide with 40.3,mL 1.00M hydrochloric acid if after mixing the two solutions the temperature went up 6.6 C6.6∘C. Report the amount of the limiting reagent: Report the enthalpy of reaction if the heat capacity of the styrofoam cup is 57.5 J/K in kJ/mol
19.25 grams of acetic acid solution (density = 1.00g/mL) required 18.05mL of 0.875M sodium hydroxide solution...
19.25 grams of acetic acid solution (density = 1.00g/mL) required 18.05mL of 0.875M sodium hydroxide solution to reach the endpoint of a titration. Using this data, the acetic acid would have a concentration of
A 50.0 mL solution of 0.805 M acetic acid, a monoprotic acid, is completely neutralized by...
A 50.0 mL solution of 0.805 M acetic acid, a monoprotic acid, is completely neutralized by the addition of 23.55 mL of sodium hydroxide. What is the concentration of the initial sodium hydroxide solution? What is the final concentration of the sodium ions in mol/L? In the reaction in question 1, the initial temperature of each solution is 22.1°C. The temperature after mixing is 29.4°C. How much heat is liberated by this neutralization reaction? What is the heat of neutralization...
Calculate the pH of a solution containing 0.10 mole of acetic acid, to which is added...
Calculate the pH of a solution containing 0.10 mole of acetic acid, to which is added 0.03 mole of sodium hydroxide (Ka = 1.8 x 10-5).
20. We combine 500 cm 1.2 mol/dm of acetic acid and 500 cm 0.5 mol/dm of...
20. We combine 500 cm 1.2 mol/dm of acetic acid and 500 cm 0.5 mol/dm of sodium hydroxide solution. What is the pH of the created buffer solution? (4.59) 21. How will the pH of the buffer solution change in the previous problem if we add 40 cm 1 mol/dm of hydrochloric acid, or 50 cm 1 mol/dm² of sodium hydroxide solution? (4.70, 4.46)
Find Pka value and equivalence point for each :) Titration of Acetic Acid with Sodium Hydroxide...
Find Pka value and equivalence point for each :)
Titration of Acetic Acid with Sodium Hydroxide PH 0 0.1 0.2 0.3 04 Volume NaOH added 0.5 0.6 Titration of Selenious Acid with Sodium Hydroxide PH + NO R 0.1 0.2 0.3 0.4 Volume NaOH added (L) 0.5 0.6 Titration of Selenious Acid with Sodium Hydroxide PH 0.1 0.5 0.6 0.2 0.3 0.4 Volume NaOH added (L)
3. One liter of buffer solution was prepared by mixing 0.1 mole of acetic acid CH3COOH...
3. One liter of buffer solution was prepared by mixing 0.1 mole of acetic acid CH3COOH and 0.05 mole of sodium acetate CH3COONa. Calculate a. pH of that solution b. How much of a strong base, say NaOH, in mol/L needs to be added to that solution to change its pH to 6.0? Notes and useful data: For acetic acid pK4.75 For carbonic acid pKa 6.3 and pKa 10.3 Sodium acetate CH3COONa dissociates entirely to Na'CH3COO
An acetic acid/ sodium acetate buffer solution similar to the one you made in the lab...
An acetic acid/ sodium acetate buffer solution similar to the one you made in the lab was prepared using the following components: 3.46 g of NaC2H3O2∙3H2O (FW. 136 g/mol) 9.0 mL of 3.0 M HC2H3O2 55.0 mL of water If you take half of this solution and add 2 mL of 1.00 M HCl to it, then what is the pH of this new solution?
so i know that acetic acid is
the solution and i googled the molar mass for it which is 60.052,
and the mass was 59.81 so i did 59.81/60.052 which gave me the
answer 0.99597. Why is it wrong ? i dont understand, thats how you
find moles right ? mass/molar mass.
Part A initial temperature of 20.1C. and the final The aluminum cup inside your calorimeter weighs 37.55 g. You add 59.81 g of 1.0 M acetic acid solution...
20. We combine 500 cm 1.2 mol/dm of acetic acid and 500 cm 0.5 mol/dm of sodium hydroxide solution. What is the pH of the created buffer solution? (4.59) 21. How will the pH of the buffer solution change in the previous problem if we add 40 cm 1 mol/dm of hydrochloric acid, or 50 cm 1 mol/dm² of sodium hydroxide solution? (4.70, 4.46)
Find Pka value and equivalence point for each :)
Titration of Acetic Acid with Sodium Hydroxide PH 0 0.1 0.2 0.3 04 Volume NaOH added 0.5 0.6 Titration of Selenious Acid with Sodium Hydroxide PH + NO R 0.1 0.2 0.3 0.4 Volume NaOH added (L) 0.5 0.6 Titration of Selenious Acid with Sodium Hydroxide PH 0.1 0.5 0.6 0.2 0.3 0.4 Volume NaOH added (L)
3. One liter of buffer solution was prepared by mixing 0.1 mole of acetic acid CH3COOH and 0.05 mole of sodium acetate CH3COONa. Calculate a. pH of that solution b. How much of a strong base, say NaOH, in mol/L needs to be added to that solution to change its pH to 6.0? Notes and useful data: For acetic acid pK4.75 For carbonic acid pKa 6.3 and pKa 10.3 Sodium acetate CH3COONa dissociates entirely to Na'CH3COO
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