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where did 6.022x10^23 came from in step 2 for this question

where did 6.022x10^23 came from in step 2 for this question

Chemistry
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...
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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
Transcribed Image Text: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 :- ?
Transcribed Image Text: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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