Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 4 - Reactions in Aqueous Solutions - Questions & Problems: 4.78

Answer

Please see the work below.

Work Step by Step

We know that: $moles\space of \space first\space solution\space Ca(NO_3)_2=0.0462L\times 0.568M=0.0262$ $moles\space of\space sodium \space Ca(NO_3)_2=0.0805L\times 1.396M=0.112$ $total\space volume=V_1+V_2=0.1267L$ $new\space concentration=\frac{0.138}{0.1267}=1.09M$
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