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