STARTING AMOUNTHow many liters of 0.305 M K3PO4 solution are necessary tocompletely react with 187 mL of 0.0184 M NiCl₂ according to thebalanced chemical reaction:X0.3052 K3PO4 (aq) +ADD FACTORx( )1870.001g NiCl₂3 NiCl₂(aq) → Ni3(PO4)2(S) +7.520.003760.01846.022 x 1023 0.00752ANSWERM K3PO4 L K3PO4 mL K3PO4g K3PO4 mL NiCl₂236 KCl(aq)RESET21000 0.01136M NiCl₂ mol K3PO4 mol NiCl₂L NICI ₂

A balanced chemical equation contain equal number of atoms of each element on both sides of the…

STARTING AMOUNT How many liters of 0.305 M K3PO4 solution are necessary to completely react with 187 mL of 0.0184 M NiCl₂ according to the balanced chemical reaction: X 0.305 2 K3PO4 (aq) + ADD FACTOR ‹( ) 187 0.001 g NiCl₂ 3 NiCl₂(aq) → Ni3(PO4)2 (S) + 7.52 0.00376 0.0184 6.022 x 1023 0.00752 ANSWER M K3PO4 L K3PO4 mL K3PO4 g K3PO4 mL NiCl₂ 2 3 6 KCl(aq) RESET 2 1000 0.0113 6 M NiCl₂ mol K3PO4 mol NiCl₂ L NICI ₂

STARTING AMOUNT How many liters of 0.305 M K3PO4 solution are necessary to completely react with 187 mL of 0.0184 M NiCl₂ according to the balanced chemical reaction: X 0.305 2 K3PO4 (aq) + ADD FACTOR ‹( ) 187 0.001 g NiCl₂ 3 NiCl₂(aq) → Ni3(PO4)2 (S) + 7.52 0.00376 0.0184 6.022 x 1023 0.00752 ANSWER M K3PO4 L K3PO4 mL K3PO4 g K3PO4 mL NiCl₂ 2 3 6 KCl(aq) RESET 2 1000 0.0113 6 M NiCl₂ mol K3PO4 mol NiCl₂ L NICI ₂

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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