# rohanali16593

Lv9
5 Followers
6 Following
0 Helped

Published269

### Subjects

Architecture1History1English3Mechanical Engineering2Sociology1Psychology1Algebra6Computer Science4Accounting59Calculus15Biology14Statistics84Physics24Finance2Chemistry52
Answer: a. Net amount raised = no.of shares * share price = 6540000*20.55 = 13...
Answer:Sorry bro its too hard for m to ans Sorry
Answer:Sorry bro its too hard for m to ans Sorry
Sorry bro its too hard for m to ans Sorry
Answer:Sorry bro its too hard for m to answr Sorry

I am working on this lab but can't seem to figure out how tofinish it.

1. For the NEW bottle of hydrogenperoxide, record and calculate the following:

 a Volumeof potassium permanganate used in each titration (mL) 50mL b Thevolume of potassium permanganate required to titrate 10 mL of thenew hydrogen peroxide (mL) 18.04mL c How many moles of potassiumpermanganate were required to titrate 10mL of the new hydrogenperoxide (moles) 0.013mol d Given the stoichiometry of thereaction, how many moles of hydrogen peroxide were in the 10mL ofthe solution (moles) e How many grams ofH2O2 were in the hydrogen peroxide solution(MW of H2O2 is 34.01) (grams) f 1 mL ofwater weighs 1 g. Using this information and the numberof grams of H2O2 produced in the reaction, calculate the percentmass of H2O2 in the NEW solution. (% mass = (mass H2O2 /mass water) * 100%)

2. Repeat the calculations for the OLD bottle ofH2O2:

 a Volumeof potassium permanganate used in each titration (mL): 50mL b Thevolume of potassium permanganate required to titrate 10 mL of theold hydrogen peroxide (mL) 12.60mL c How many moles of potassiumpermanganate were required to titrate 10mL of the old hydrogenperoxide (moles) 0.013mol d Given the stoichiometry of thereaction, how many moles of hydrogen peroxide were in the 10mL ofthe solution (moles) e How many grams ofH2O2 were in the hydrogen peroxide solution(MW of H2O2 is 34.01) (grams) f 1 mL ofwater weighs 1 g. Using this information and the numberof grams of H2O2 produced in the reaction, calculate the percentmass of H2O2 in the OLD solution. (% mass = (mass H2O2 /mass water) * 100%)