Answered: Q3. 1.5g of 0.1M NaCl has been prepared…

Q3. 1.5g of 0.1M NaCl has been prepared in 1L and only 25 of NaCl solution has been used to titrate it against 0.5M silver nitrate AgNO, calculate the% yield of NaCl [ molar mass of NaCl=58.5g/mol].

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

A solution of citric acid (H.CeHsO,) with a known concentration of 0.200 M H.CcHsO, is titrated with a 0,750 | M NAOH solution. How many mL of NaOH are required to reach the third equivalence point with a starting volume of 25.0 mL H:CCH:O, according to the following balanced chemical equation: H.CcHsO, + 3 3 H:0 NaOH – Na:CeHsO, + STARTING AMOUNT ADD FACTOR ANSWER RESET 1 3 1000 6.67 15.0 20.0 0,020 25.0 11.25 2.22 0.001 0.750 2.00 x 10 0.200 + g H.CcHsO, M H.CcHsO, mL NaOH mol NaOH L NaOH mol H.CHSO, LH.CeHsO, 11:33 PM 68°F Mostly cloudy 10/5/2021 P Type here to search prt sc delete tg tvo insert 144 トト」米 esc home @ 23 backspace 3. 4 6. 7 8 Y P Q tab
7. The separate samples of nitric and acetic acids are both titrated with a 0.100 M solution of NaOH(aq). (Y) 25.0mL of 1.0 M HNO3(aq) (Z) 25.0 mL of 1.0 M CH3COOH(aq) Determine whether each of the following statements concerning this titration is true or false. (A) A larger volume of NaOH(aq) is needed to reach the equivalence point in the titration of HNO3 (B) The pH at the equivalence point in the HNO3 titration will be lower than the pH at the equivalence point in the CH3COOH titration (C) Phenolphthalein would be a suitable indicator for both titrations Group of answer choices a.) A) False B) True C) True b.) A) False B) False C) True c.) A) False B) True C) False d.) A) True B)True C)True
Q/ Mohr's method uses the potassium chromate indicator and this indicator makes the equivalent point not identical to the end point. A sample weighing 5.592 g of BaCl2 is dissolved in 500 mL of water. Calculate the percentage of chloride in the sample when 25 mL of the sample is titrated and requires 33.47 mL of 0.1098 N AGNO. SOLUTION:
A solution of 25.0 mL of 0.40 M H2SO4 is titrated with 0.50 M of LIOH. H₂SO4 + 2LIOH →>> Li₂SO4 + 2H₂O What volume of LiOH is required to neutralize the acid completely? Vol LiOH = [?] mL Hint: Be careful of the mole ratios! Vol LiOH (mL) Enter
A solid sample containing some Fe2+ion had a total mass of .9791 g. It required 18.2 ml of 0.02304 M KMnO4 to titrate the Fe2+ in the dissolved sample to a pink endpoint. How many moles of Fe2+ were in the the sample? How many grams of irion were present in the sample and what was the mass percent iron in the sample?
What weight of potassium hydrogen phthalate will require 30. mL of a 0.30M NaOH solution to reach the equivalence point? KHP MM 204.22 g/mol .CO;K' .CO;K' NaOH + H2O + *CO,H COƠ,Na'
A solution of malonic acid (H2C3H2O4) with a known concentration of 0.500 M H2C3H2O4 is tirated with a 0.255 M KOH solution. How many mL of KOH are required to reach the second equivalence point with a starting volume of 60.0 mL H2C3H2O4, according to the following balanced chemical equation: H2C3H2O4 + 2 KOH---> K2C3H2O4 + 2 H2O
Caculate the volume (mL) of 1.0 M potassium carbonate (K₂CO₃) needed to prepare 5.0g of the COPPER pigment. This is where the balanced chemical equation come in. You should be able to relate the moles of pigment to the moles of K₂CO₃, and from that calculate the volume of 1.0 M K₂CO₃ solution needed. Please report your value as XX.X mL (one deicmal point and no units).