An analytical chemist has a solution containing chloride ion, Cl − . She decides to determine the amount of chloride ion in the solution by titrating 50.0 mL of this solution by 0.100 M AgNO 3 . As a way to indicate the endpoint of the titration, she added 1.00 g of potassium chromate, K 2 CrO 4 (see Figure 17.5). As she slowly added the silver nitrate to the solution, a white precipitate formed. She continued the titration, with more white precipitate forming. Finally, the solution turned red, from another precipitate. The volume of the solution at this point was 60.3 mL. How many moles of chloride ion were there in the original solution? How many moles of chloride ion were there in the final solution? You may make any reasonable approximations.
An analytical chemist has a solution containing chloride ion, Cl − . She decides to determine the amount of chloride ion in the solution by titrating 50.0 mL of this solution by 0.100 M AgNO 3 . As a way to indicate the endpoint of the titration, she added 1.00 g of potassium chromate, K 2 CrO 4 (see Figure 17.5). As she slowly added the silver nitrate to the solution, a white precipitate formed. She continued the titration, with more white precipitate forming. Finally, the solution turned red, from another precipitate. The volume of the solution at this point was 60.3 mL. How many moles of chloride ion were there in the original solution? How many moles of chloride ion were there in the final solution? You may make any reasonable approximations.
An analytical chemist has a solution containing chloride ion, Cl−. She decides to determine the amount of chloride ion in the solution by titrating 50.0 mL of this solution by 0.100 M AgNO3. As a way to indicate the endpoint of the titration, she added 1.00 g of potassium chromate, K2CrO4 (see Figure 17.5). As she slowly added the silver nitrate to the solution, a white precipitate formed. She continued the titration, with more white precipitate forming. Finally, the solution turned red, from another precipitate. The volume of the solution at this point was 60.3 mL. How many moles of chloride ion were there in the original solution? How many moles of chloride ion were there in the final solution? You may make any reasonable approximations.
JON
Determine the bund energy for
UCI (in kJ/mol Hcl) using me
balanced chemical equation and
bund energies listed?
का
(My (9) +36/2(g)-(((3(g) + 3(g)
A Hryn = -330. KJ
bond energy
и-н 432
bond
bond
C-1413
C=C 839 N-H
391
C=O 1010
S-H 363
б-н 467
02 498
N-N
160
N=N
243
418
C-C 341
C-0 358
C=C
C-C 339 N-Br
243
Br-Br
C-Br 274
193
614
(-1 214||(=olin (02) 799
C=N
615
AAL
Determine the bond energy for HCI (
in kJ/mol HCI) using he balanced
cremiculequecticnand bund energles
listed? also c double bond to N is
615, read numbets carefully please!!!!
Determine the bund energy for
UCI (in kJ/mol cl) using me
balanced chemical equation and
bund energies listed?
51
(My (9) +312(g)-73(g) + 3(g)
=-330. KJ
спод
bond energy
Hryn
H-H
bond
band
432
C-1 413
C=C 839 NH
391
C=O 1010
S-1 343
6-H
02 498
N-N
160
467
N=N
C-C
341
CL-
243
418
339 N-Br
243
C-O
358
Br-Br
C=C
C-Br 274
193
614
(-1 216 (=olin (02) 799
C=N
618
Please correct answer and don't used hand raiting
Chapter 17 Solutions
Student Solutions Manual for Ebbing/Gammon's General Chemistry, 11th
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell