Homework Answers
It is evident from the graph that at the equivalence point, the mass of H2O2 solution is 3.0g.
Since the solution is 3% m/m, Hence the amount of H2O2 in the solution at the equivalence point is
= 0.03*3.0
=0.09g of H2O2
moles of H2O2 = given mass/ molar mass of H2O2
= 0.09g/34.0147gmol-1
=2.646*10-3 moles of H2O2
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Approximate the number of moles of hydrogen peroxoide at equivalence point in ghe graph, supposing 3.00% m/m solution...
thjs is the question this is the data these are the background information please help thank...
thjs is the question
this is the data
these
are the background information
please help thank you
4. Approximate the number of moles of hydrogen peroxide at the equivalence point in the graph in the introduction, supposing a 3.00% m/m solution. Thus the densities will be- Trial Mass(g) 0.448 0.450 3 Density(g/ml) 0.448 g/ 0.400 ml = 1.12 g/ml 0.450 g/ 0.400 ml = 1.125 g/ml 0.437 g/ 0.400 ml = 1.0925 g/m 0.442 g/ 0.400 ml = 1.105 g/ml...
If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of...
If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the equivalence point is 5.125 g, how many moles of hydrogen peroxide molecules are present?
Please answer #3! You will be using 25.00 mL of 3.00% by mass solution ({mass H_2O_2/...
Please answer #3!
You will be using 25.00 mL of 3.00% by mass solution ({mass H_2O_2/ mass solution}*100) of hydrogen peroxide. Assume the density of this solution is 1.000 g / mL. If all hydrogen peroxide decomposes to water and oxygen according to the reaction 2H_2O_2 (aq) 2H_2O (I) + O_2(g) What is the total volume of the oxygen gas generated if the temperature is 24 degree C and the pressure is 770 Torr (760 Torr = 1 atm) ?...
In Part A, you explored the relationship between molarity, the number of moles, and solution volume....
In Part A, you explored the relationship between molarity, the number of moles, and solution volume. However, when you prepare a solution, you are unable to directly measure the number of moles of solute. Instead, you can mass the solute and convert between the number of moles and mass using the molar mass of a substance as a conversion factor. The molar mass of a substance can be calculated based on its molecular formula; otherwise, it can be calculated from...
pH at equivalence is 8.40 It's a tritration with a breaker filled with 0.1894 M CH3COOH...
pH at equivalence is 8.40
It's a tritration with a breaker filled with 0.1894 M CH3COOH and a buret with 0.2006 M NaOH. | added 9.44 mL of NaOH to the CH3COOH to find the equivalence point. Find the Ka of CH3COOH 9. Use your data to estimate Ka of the weak acid. Show all working (give your answer to thee sig figs) Hint: Estimation of K The pH at the equivalence point can be used to estimate the K....
In Part A, you explored the relationship between molarity, the number of moles, and solution volume....
In Part A, you explored the relationship between molarity, the number of moles, and solution volume. However, when you prepare a solution, you are unable to directly measure the number of moles of solute. Instead, you can mass the solute and convert between the number of moles and mass using the molar mass of a substance as a conversion factor. The molar mass of a substance can be calculated based on its molecular formula; otherwise, it can be calculated from...
How do you graph and find equivalence point I don't understand UPDATE: This is all the...
How do you graph and find equivalence point I don't
understand
UPDATE: This is all the information I got
Conductivity measurements • Enter 12 data points • All 12 points are needed. • Make sure your minimum value is included, with at least 5 values before and after that point. . Your graph is expected to show a "V" shape. • Use at least 2 s.f. for the conductivity data. Total Volume of H2SO.(a) Dispensed Conductivity Units AS/cm Data Point...
12. Molarity, M, is defined as A. moles of solute dissolved in 1 mol of solvent....
12. Molarity, M, is defined as A. moles of solute dissolved in 1 mol of solvent. B. moles of solute dissolved in 1 kg of solvent. C. moles of solute dissolved in 1 L of solvent. D. moles of solute dissolved in 1 L of solution. E. moles of solute dissolved in the solution. 13. What volume of 2.50 M NaOH (40.00 g/mol) contains 0.100 mole of NaOH? A. 0.250 L D. 0.250 mL B. 40.0 mL E. 0.0400 mL...
Mass potassium hydrogen phthalate: .459 g Concentration NaOH solution: .10 M Strong Base titrated...
Mass potassium hydrogen phthalate: .459 g
Concentration NaOH solution: .10 M
Strong Base titrated into weak acid
1. ml base added at equivalence point?
2. ml base added at half-neutralization point 3. moles acid left unreacted at half-neutralization point 3 a. moles salt present at half-neutralization point b. moles of anion of salt present at half-neutralization point 5. ratio of moles of anion to moles of unreacted acid at half-neutralization point 6. . pKa of potassium hydrogen phthalate b. explanation...
2)250.0 mL of a 0.375 M solution of lactic acid (HC:H:03) is titrated with a solution...
2)250.0 mL of a 0.375 M solution of lactic acid (HC:H:03) is titrated with a solution of 0.75 M NaOH. The K, for lactic acid is 1.38 x 104. (22 pts total) a) what is the pH before addition of NaOH? (Hint: what is the relevant equilibrium?) (4 pts) b) what is the total number of moles of acid? (2 pts) c) what volume of the NaOH solution is required to reach the equivalence point? (2 pts) d) what is...
thjs is the question
this is the data
these
are the background information
please help thank you
4. Approximate the number of moles of hydrogen peroxide at the equivalence point in the graph in the introduction, supposing a 3.00% m/m solution. Thus the densities will be- Trial Mass(g) 0.448 0.450 3 Density(g/ml) 0.448 g/ 0.400 ml = 1.12 g/ml 0.450 g/ 0.400 ml = 1.125 g/ml 0.437 g/ 0.400 ml = 1.0925 g/m 0.442 g/ 0.400 ml = 1.105 g/ml...
Please answer #3!
You will be using 25.00 mL of 3.00% by mass solution ({mass H_2O_2/ mass solution}*100) of hydrogen peroxide. Assume the density of this solution is 1.000 g / mL. If all hydrogen peroxide decomposes to water and oxygen according to the reaction 2H_2O_2 (aq) 2H_2O (I) + O_2(g) What is the total volume of the oxygen gas generated if the temperature is 24 degree C and the pressure is 770 Torr (760 Torr = 1 atm) ?...
pH at equivalence is 8.40
It's a tritration with a breaker filled with 0.1894 M CH3COOH and a buret with 0.2006 M NaOH. | added 9.44 mL of NaOH to the CH3COOH to find the equivalence point. Find the Ka of CH3COOH 9. Use your data to estimate Ka of the weak acid. Show all working (give your answer to thee sig figs) Hint: Estimation of K The pH at the equivalence point can be used to estimate the K....
In Part A, you explored the relationship between molarity, the number of moles, and solution volume. However, when you prepare a solution, you are unable to directly measure the number of moles of solute. Instead, you can mass the solute and convert between the number of moles and mass using the molar mass of a substance as a conversion factor. The molar mass of a substance can be calculated based on its molecular formula; otherwise, it can be calculated from...
How do you graph and find equivalence point I don't
understand
UPDATE: This is all the information I got
Conductivity measurements • Enter 12 data points • All 12 points are needed. • Make sure your minimum value is included, with at least 5 values before and after that point. . Your graph is expected to show a "V" shape. • Use at least 2 s.f. for the conductivity data. Total Volume of H2SO.(a) Dispensed Conductivity Units AS/cm Data Point...
12. Molarity, M, is defined as A. moles of solute dissolved in 1 mol of solvent. B. moles of solute dissolved in 1 kg of solvent. C. moles of solute dissolved in 1 L of solvent. D. moles of solute dissolved in 1 L of solution. E. moles of solute dissolved in the solution. 13. What volume of 2.50 M NaOH (40.00 g/mol) contains 0.100 mole of NaOH? A. 0.250 L D. 0.250 mL B. 40.0 mL E. 0.0400 mL...
Mass potassium hydrogen phthalate: .459 g
Concentration NaOH solution: .10 M
Strong Base titrated into weak acid
1. ml base added at equivalence point?
2. ml base added at half-neutralization point 3. moles acid left unreacted at half-neutralization point 3 a. moles salt present at half-neutralization point b. moles of anion of salt present at half-neutralization point 5. ratio of moles of anion to moles of unreacted acid at half-neutralization point 6. . pKa of potassium hydrogen phthalate b. explanation...
2)250.0 mL of a 0.375 M solution of lactic acid (HC:H:03) is titrated with a solution of 0.75 M NaOH. The K, for lactic acid is 1.38 x 104. (22 pts total) a) what is the pH before addition of NaOH? (Hint: what is the relevant equilibrium?) (4 pts) b) what is the total number of moles of acid? (2 pts) c) what volume of the NaOH solution is required to reach the equivalence point? (2 pts) d) what is...
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