QUESTION 1: How much KCl03reactant will be needed to produce 105.8g of tetraphosphorus decaoxide (P4010)? G: 105.8g P4010 MM: 1 mol P4010 = 287.88g P,010 1 mol KC103 = 122.45g KC103 MR:3 mol P,010 %3D 10тol кс10; СHOICES: A. 15.01g KC103 В. 150.01g КCI03 С. 1.501g КCI0, D. 0.150g KC103
QUESTION 1: How much KCl03reactant will be needed to produce 105.8g of tetraphosphorus decaoxide (P4010)? G: 105.8g P4010 MM: 1 mol P4010 = 287.88g P,010 1 mol KC103 = 122.45g KC103 MR:3 mol P,010 %3D 10тol кс10; СHOICES: A. 15.01g KC103 В. 150.01g КCI03 С. 1.501g КCI0, D. 0.150g KC103
Chemistry
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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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