1.00 liter solution contains 0.40 m acetic acid and 0.52 m potassium acetate. if 0.130 moles of calcium hydroxide are added to this system, indicate whether the following statements are true or false. (assume that the volume does not change upon the addition of calcium hydroxide.)

a. the number of moles of ch3cooh will increase.
b. the number of moles of ch3coo- will decrease.
c. the equilibrium concentration of h3o will remain the same.
d. the ph will decrease.
e. the ratio of [ch3cooh] / [ch3coo-] will increase.

The compound methyl butanoate smells like apples. its percent composition is 58.8% c, 9.9% h, and 31.4% o. what’s the empirical formula ?

What kind of ion is contained in salts that produce an acidic solution? a positive ion that attracts a proton from water a positive ion that releases a proton to water a negative ion that attracts a proton from water a negative ion that releases a proton to water

Ineed someone to see if my answers are correct! if any are wrong let me know what the correct answers would be and how to get that answer! 1. how many moles of sodium chloride are in 28 grams od nacl? a. 265 mole naclb. 856 mole naclc. 479 mole of nacld. 1.2 mole nacl < my choice2. 734 grams of lithium sulfate (li2so4) are dissolved to make 2500 ml of solution what is rhe molaratiy? a. 2.67 mb. 4.56 mc. 3.89 m < my choiced. 1.78 m3. how many grams of cacl2 would be dissolved in 3.0 l of a 0.50 m solution of cacl2? a. 250 g cacl2 b. 166.5 g cacl2c. 113.65 g cacl2d. 98 g cacl2 < my choice4. suppose you had 58.44 g of nacl and you dissolved it in exactly 2.00 liters. the molarity if the solution would be 0.5 mtrue < my choicefalse 5. i would need 22g of naoh to make a 3.0 m solution using 250 ml of solvent.true < my choicefalse6. identify the solute: you have a .0195 m solution made from using 6.5 g of solute and 3 l of solvent. identify the solute by solving for molar weight.a. the solute is nacl because the molar weight is 58.43 g/mol < my choiceb. the solute is h2so4 because the molar weight is 98.06 g/molc. the solute is cacl2 because the molar weight is 111.11 g/mol