A student enters the lab and performs Experiment 14 as written. In the procedure for "Experiment 1" on p. 230 of the lab manual, the student records the concentration of the stock solution of H2O2 to be 0.965 M. The student proceeds to prepare the reaction solution as stated in the lab manual: They add 10.0 mL of 0.10 M potassium iodide and 15.0 mL of distilled water to their flask. They then add 5.0 mL of the stock H2O2 solution and mix the solution to initiate the reaction. Calculate the molarity of H2O2 in the initial reaction solution. Do not include the units with your answer.

Molarity = 0.193

Explanation:

Molarity = Gram mole of solute ÷ liters of the solution

H2O2 is the solute

The mixture of Potassium iodide and water are the solution

STEP 1: FIND THE MOLE OF SOLUTE;

Concentration of solute = 0.965M

Volume of solute = 5.0mL= 0.005L

Concentration= mole ÷ volume

Therefore

Mole = concentration × volume

Mole = 0.965 × 0.005 = 0.004825mol

STEP2: FIND VOLUME OF SOLUTION;

10.0 mL of potassium iodide + 15.0 mL of distilled water

Therefore;

10mL + 15mL= 25mL = 0.025L

STEP 3: FIND THE MOLARITY;

Molarity = 0.004825 ÷ 0.025 = 0.193

Therefore the Molarity of H202 is

0.193

organisms are more likely to be more preserved in mud than sand because mud is moist and wet, while sand is very dry.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"