Nindow Help Lab 11- Lab Practi. Lab 11-Lab Practi.. 125% make all needed measure

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Nindow Help Lab 11- Lab Practi. Lab 11-Lab Practi.. 125% make all needed measurements and include space for those measurements in your data table. Solve the following in your notebook - you will want to write a plan in your prelab notes for how to solve these prior to entering the laboratory. This plan should be included as part of your prelab in the calculation notes section of the prelab. (3 points) Calculate the pressure (P), considering the following conditions: A 10.0 mL sample of your acid is vaporized in a 20.0 L container at the room temperature in the lab on the day the titration is conducted. Determine the gas pressure created by just the acid (ignore the vaporized water). Hint: use the ideal gas law . (3 points) Consider the following: To prepare your acid solution. (of which you just determined the Molarity). a concentrated stock solution was used. 0.0200 L of this concentrated acid solution was diluted to 1.00 L. Calculate the Molarity of the concentrated stock solution (3 points) Assume that the specific heat of your acid solution is the same as that of water. Calculate the heat gained by your acid solution (in Joules) when it is heated from the room temperature at the time the experiment was conducted to a temperature of 35.0 °C. Use the data from your second acid sample (your second titration trial) Page 2 of 2

Explanation / Answer

1)

According ideal gas equation,

P1V1 = P2V2 at constant temperature and constant number of moles

P1 = initial pressure of the gas = P1

V1 = initial volume of the gas = 10 * 10*-3 L

P2 final pressure of the gas = P2

V2 = final volume of the gas = 20 L

P1*10*10^-3 = P2 * 20

P2 = 5*10^-4 * P1

The final pressure will be 5*10^-4 times of initial prsseure.

2)

according to molarity of solution, M1V1 = M2V2

M1 = concentration of stock solution

V1 = volume of stock solution

M2 = concentration of final solution

V2 = volume of the final solution

M1*0.02 = M2*1

M1 = M2.0.02

M1 = M2 * 5 * 10^-3

Concentration of stock solution is equal to 5*10^-3 times of the final solution concentration.

3)

Water specific heat is given by 4.184 J/g 0C

the acid specific heat is also = 4.184 J/g 0C

Q = m * s * dT

Q = heat energy gained

m = mass of the acid

s = specific heat capacity

dT = change in temperature

Q = m * 4.184 * (35 - 25) ( room temperature = 25 0C)

Q = m * 41.84 J

Heat energy gained by the acid is equal to 41.84 times of the mass of the acid.

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