t we didnt he ad tne sammpleiong enouan +o remove all the H2Chydratičn ur calculatons would beneavierima king n toO5. What mass of CUSO<(H2O)s (Copper II sulfate pentahydrate), when heated and dehydrated,would result in 6.23 g of anhydrous salt?'n" have been too high orC6.2,39X159,69MOI O390:391CUSOU CH20)5159.69 9059.60+90Y MOV9092/77g x159.G9x 61239_X-189 XC6,23+3.631539,86159-TI626309 x 9= 3,6315-x* 54.6.03903うX= 20175(CUSO4 OEABCH20).201670.03903=CUSO4 CH2O)563,5 +32,1t84:16sced Chemistry with Vernier2-3

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Answered: ing n too 5. What mass of CUSO<(H2O)s…

ing n too 5. What mass of CUSO<(H2O)s (Copper II sulfate pentahydrate), when heated and dehydrated, MO! 6.239X159.69 would result in 6.23 g of anhydrous salt? 0739 CUSOU CH20)5 159 69 90

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.8QAP

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Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.

Question

t we didnt he ad tne sammple
iong enouan +o remove all the H2C
hydratičn ur calculatons would be
neavierima king n toO
5. What mass of CUSO<(H2O)s (Copper II sulfate pentahydrate), when heated and dehydrated,
would result in 6.23 g of anhydrous salt?
'n" have been too high or
C6.2,39X159,69
MOI O39
0:391
CUSOU CH20)5
159.69 90
59.60+90
Y MOV
909
2/77g x
159.G9x 61239_X-
189 X
C6,23+3.631539,86159
-TI626309 x 9= 3,6315
-x* 54.6.03903
うX
= 20175
(CUSO4 OEABCH20).20167
0.03903
=CUSO4 CH2O)5
63,5 +32,1t
84:16
sced Chemistry with Vernier
2-3

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