An analytical chemist weighs out 0.133 g of an unknown diprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.0900 M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 22.0 mL of NaOH solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits. g mol 0 X

A bucket of water has a volume of 0.473 kL . How many dL is this? STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 47300 1000 0.001 0.01 10 4730 0.0000473 100 0.473 0.1 1 0.00473 473 0.000473 kL dL +

Pl 5 please answer question and make sure the answer is right show all work please thank you.

Part A Solve for x, where M is molar and s is seconds. (2.7 x 103 M-2s¬1)(0.22 M)³ Enter the answer. Include units. Use the exponent key above the answer box to indicate any exponent on your units. View Available Hint(s) HA Value Units Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining

Question 2 of 21 Submit Convert 797 mm Hg to atm STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 1.01325 1 1.05 760 1000 6.06 x 105 101.325 14.70 105 7.97 797 0.98692 1.01325 x 105 0.001 atm kPa psi torr mm Hg Pa bar

Determine the % error for of of dote provided reports that when data a 25.00ml pipette when used u times, the pipette delivered volume of 24.89 ML, 24.98 AAL, M2, 25.03 m2 and 25.01 me Report your answer to 2 si do not include figures and a unit. significant figu 57900 PERS

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Trial #1 – Read the following burette readings...and add to the Table Initial: Final: Pickling Vinegar Conc NaOH (M) 0.1005 Trial #1 Trial #2 Trail #3 Trial #4 Volume Vinegar (mL) 2.00 2.00 2.00 2.00 Click or tap here to enter text. Initial Volume NaOH (mL) 0.22 1.02 0.44 Click or tap here to enter text. Final Volume (mL) 24.20 25.00 24.45 Click or tap Click or tap Click or tap here to here to enter enter text. here to enter Click or tap here to enter text. Volume NaOH added (mL) text. text.

Mean, standard deviation, Coefficient variable, and 95% confidence limits of HCl molarity's *0.08883M * 0.08743M * 0.08790M

Data Sheet UNKNOWN NUMBER: 1 49.83 4575 8.08 12.47 I1.031 1.44 1. Mass of test tube + naphthalene 2. Mass of empty test tube %3D 3. Mass of naphthalene 4. Mass of vial + Unknown %3D 5. Mass of vial %3D 6. Mass of Unknown Cooling curve data Pure Naphthalene Naphthalene + Unknown Temp Time Temp Time Temp Time Temp Time 79 0.5 T4-9 5 20:7 5:30 78:2 020 749 S:30 85.2 30 84.5 B15 1:30 70.9 6 6.70 72 30 14.9 6 14.9 6:30 75.6 1:08C 74 9 75:52 20.0 69.8 7. 69.4 7:30 68.9 68:3 76:5. Imit 74 75.02.9 749,3. 28.2 2 76 Y 25.0 22. 3. . 2:10 "NS 130 3:30 24.0 9 23-4 9:30 74.9.4.30 23.5 8:)0 679 9 67.2 9.30 . 21.8 71.4 4 330 74.9 94.9 u १:१० :30 67.2 10 20.9 S b6.5 10.30 75-5 750 10:50 66.0 1 7. Freezing point of pure Naphthalene from cooling curve = 8. Freezing point of solution 9. AT 10. Molality of solution, m 11. Molar mass of Unknown 12. Include graphs of cooling curves CALCULATIONS