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Chemistry: The Molecular Science

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter3: Chemical Reactions

Section: Chapter Questions

Problem 153QRT

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I would just like to ask if where did 6.022x10^23 came from in step 2 for this question?

ll SMART
10:08 PM
99% 4
Question
Lead (II) nitrate is reacted with sodium iodide.
Calculate the number of Pb2+ ions required to
form 0.87 g of precipitate. Write the balanced
molecular, total ionic, and net ionic equations of
the reaction.
Atomic Mass:
Pb: 207.2 g/mol
N: 14.0067 g/mol
O: 15.999 g/mol
Na: 22.989769 g/mol
I: 126.90447 g/mol
Note: Use scientific notation with the format
(n)e(x) wheren= number and x = exponent
(Example: 6.022e23 for 6.022 x 1023)
Expert Answer O
Step1

ll SMART
10:08 PM
99% D
Expert Answer O
Step1
a)
Step-1
Lead 11) nitrate
Pb(NOglz
Sodim
Todide
NaI
foom
of
Precibitate =0.87 g
mass
Step2
b)
reaction
Precipitation
+ 2NaNOg caq
+ 2 NáI
Molar
magg of PbI, =[207.2 + 126.90447 x2) 2
461.008
8imo)
: motes
O. 87 g
-3
= 1.881X İo moles
PbI =
46).008 3mo)
23
= 1.887 x1o°× 6.022×1o
:-
ber of Iona o
1.136 e 21
Anyower
Step3
c)
Net
Ionic
equation
written
in
3 steps :-
?

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