will someone please help me i am desperate? 1.Calculate the volume of 1.78 M sul
ID: 683053 • Letter: W
Question
will someone please help me i am desperate? 1.Calculate the volume of 1.78M sulfuric acid thatwould be needed to neutralize 62.4 mL of a 2.09 M aqueousammonia solution. The equation for the reaction is?2.Calculate the molarity of an acetic acid solution if 85.15 mL ofthe solution is needed to neutralize 128 mL of 1.35 Msodium hydroxide. The equation for the reaction is?
will someone please help me i am desperate? 1.Calculate the volume of 1.78M sulfuric acid thatwould be needed to neutralize 62.4 mL of a 2.09 M aqueousammonia solution. The equation for the reaction is?
2.Calculate the molarity of an acetic acid solution if 85.15 mL ofthe solution is needed to neutralize 128 mL of 1.35 Msodium hydroxide. The equation for the reaction is?
Explanation / Answer
use molarity as your conversion factor and make sure youbalacne the equation correctly problem #1 H2SO4 + NH3 H2SO4 + NH3 the ratio of H2SO4 to NH3 is 1:1 since the ratio is 1:1 you can use M1V1=M2V2 (you can only usethis shortcut if the ration is 1:1) (1.78)(x) = (2.09)(62.4ml) x = 73.27 ml ( if they ask for liters than just convert) problem #2 balance the equation first HCH3COOH + NaOH NaCH3COOH + H2O again the ratio of HCH3COOH to NaOH is 1:1 (x)(85.15ml) = (1.35)(128) x = 2.03 hope this helped. Don't forget to give Karma please. If youhave any more questions feel free to send me a messageRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.