A. 100ml of 0.5M acetic avid is titration with 20ml of 0.5M KOH. What is the pH of the acid before titration? (ka acetic acid is 1.8 x 10^-5)

B. What is the pH of the solution after titration?

calculate the wavelengths of the following: A) a muon (a subatomic particle with a mass of 1.884x10^-25g) traveling at 2.30x10^2 m/s B) an electron (m=9.10939x10^-28g) moving at 3.80x10^6 m/s in an electron microscope

What is a possible procedure I can use to determine the direct solubility of KHT following these guidelines?

1. Based on the principle of making a saturated solution, KHT is added until excess solid is formed. The room temperature needs to be recorded and kept to avoid solubility being changed by temperature duing preparation.

2. The excess KHT solid should be filtered.

3. Method of titration using a standard soltuion of NaOH.

4. Finally, how will you test the temperature effect on solubility of KHT?

Using electron configurations, shielding, and effective nuclear charge, explain why Mg has a higher ionization energy than Na and Al.

The line spectrum of helium has many more lines in it than the line spectrum of hydrogen. Draw orbital energy diagrams for both H and He, showing all orbitals up through n = 3. Use these diagrams to explain why helium’s line spectrum is so much more complex. They should differ in more than just the number of electrons!

1.) Acetic acid is a weak monoprotic acid with Ka=1.8*10^-5. In an acid base titration, 100 mL of 0.100M acetic acid is titrated with 0.100M NaOH. What is the pH of the solution:

a.)Before any NaOH is added

b.)Before addition of 15.0mL of 0.100M NaOH

c.)At the half-equivalence point

d.)After addition of total of 65.0mL of 0.100M NaOH

e.)At equivalence point

f.)After addition of a total of 125.0mL of 0.100M NaOH

The half-equivalence point and the equivalence point should not be noted when measuring pH changes during a titration. True or False ?

How many grams of each of the following per 1000g of water in your car radiator are needed to give equal protection against freezing down to -10.0°C? (a) Methyl alcohol (CH3OH) b.p is 65°C, (b) ethyl alcohol (C2H5OH) b.p is 78°C, (c) ethylene glycol (C2H4(OH)2) b.p is 197°C. In spite of higher cost, what advantages does ethylene glycol have over the other alcohols as an all-winter antifreeze?

Which one of the following will act as a buffer solution?

• A.

  25 ml of 0.1 M NH₄Cl solution mixed with 12.5 mL of 0.1 M NaCl solution

• B.

  25 mL of 0.1 M NH₄Cl solution mixed with 25 mL of 0.1 M NaOH solution

• C.

  25 mL of 0.1 M HCl solution mixed with 12.5 mL of 0.1 M NaCl solution

• D.

  12.5 mL of 0.1 M acetic acid solution mixed with 25 mL of 0.1 M NaOH solution

• E.

  25 mL of 0.1 M acetic acid solution mixed with 12.5 mL of 0.1 M NaOH solution

clear solution would be greatly appreciated :)!

will rate, thank you!

The Winkler method for dissolved oxygen in water is based on the rapid oxidation of solid Mn(OH)₂ to Mn(OH)₃ in alkaline medium. When acidified, the Mn (III) readily releases iodine from iodide. A 240.-mL water sample, in a stoppered vessel, was treated with 1.10 mL of a concentrated solution of NaI and NaOH and 1.10 mL of a manganese(II) solution. Oxidation of the Mn(OH)₂ was complete in about 1 min. The precipitates were then dissolved by addition of 2.20 mL of concentrated H₂SO₄, whereupon an amount of iodine equivalent to the Mn(OH)₃ (and hence to the dissolved O₂) was liberated. A 24.5-mL aliquot (of the 244 mL) was titrated with 14.2 mL of 0.00861 M thiosulfate. Calculate the mass in milligrams of O₂ per milliliter sample. Assume that the concentrated reagents are O₂ free and take their dilutions of the sample into account.

14a. Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC₈H₄O₄. This is often written in shorthand notation as KHP.
How many grams of KHP are needed to exactly neutralize 22.3 mL of a 0.504 M calcium hydroxide solution ?
_______grams KHC₈H₄O₄

_14b. Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC₈H₄O₄. This is often written in shorthand notation as KHP.
What volume of a 0.561 M barium hydroxide solution is needed to exactly neutralize 5.17 grams of KHP ?
_____mL barium hydroxide

14c. Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC₈H₄O₄. This is often written in shorthand notation as KHP.
If 2.62 grams of KHP are needed to exactly neutralize 25.9 mL of a potassium hydroxide solution, what is the concentration of the base solution ?
_______M potassium hydroxide

A sheet of gold weighing 12.0 g and at a temperature of 20.0°C is placed flat on a sheet of iron weighing 21.5 g and at a temperature of 59.4°C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.

A 34.51−g stainless steel ball bearing at 104.89 °C is placed in a constant-pressure calorimeter containing 116.9 g of water at 22.47°C. If the specific heat of the ball bearing is 0.474 J / (g ·°C)calculate the final temperature of both the water and steel when they equilibrate. Assume the calorimeter to have negligible heat capacity.

A quantity of 2.00 ×10² mL of 0.641 M HCl is mixed with 2.00 ×10² mL of 0.321 M Ba(OH)₂ in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and Ba(OH)₂ solutions is the same at 21.87°C. For the process below, the heat of neutralization is −56.2 kJ/mol. What is the final temperature of the mixed solutions?

H⁺(aq) + OH⁻(aq) → H₂O(l)

A) Calculate the mass defect of the helium nucleus 52He. The mass of neutral 52He is given by MHe=5.012225amu. Answer in amu

B) Calculate the binding energy E of the helium nucleus 52He (1eV=1.602×10−19J). Answer in Mev

C) Calculate the binding energy per nucleon of the helium nucleus 52He. Answer in MeV

Hydrazine, N 2 H 4 , reacts with oxygen to form nitrogen gas and water. N 2 H 4 ( aq ) + O 2 ( g ) ⟶ N 2 ( g ) + 2 H 2 O ( l ) If 2.45 g of N 2 H 4 reacts with excess oxygen and produces 0.650 L of N 2 , at 295 K and 1.00 atm, what is the percent yield of the reaction?