Answer
See the answer below.
Work Step by Step
a) Group 7, period 5
b) $5s\rightarrow n=5,\ell=0$
$\ell=0\rightarrow m_\ell=0$
c) $E=h\times\nu$
$0.141\times10^6\times1.6022\times10^{-19}\ J =6.626\times10^{-34}\ J.s\times \nu$
$\nu=3.41\times10^{19}\ 1/s$
$c=\lambda\times\nu$
$\lambda=2.998\times10^8\ m/s\div 3.41\times10^{19}\ 1/s=8.79\times10^{-12}\ m$
d) i) $HTcO_4(aq)+NaOH(aq)\rightarrow H_2O(l)+NaTcO_4(aq)$
ii) Number of moles of Tc:
$0.0045\ g\div 98\ g/mol=4.59\times10^{-5}\ mol$
Which is the same as $NaTcO_4$ produced and NaOH required so:
$4.59\times10^{-5}\ mol\times 184.99\ g/mol=8.49\times10^{-3}\ g=8.49\ mg\ NaTcO_4$
$4.59\times10^{-5}\ mol\times 40.00\ g/mol=1.84\times10^{-3}\ g=1.84\ mg\ NaOH$
e) $1.5\times10^{-6}\ mol\times 184.99\ g/mol=2.77\times10^{-4}\ g=0.277\ mg$
$1.5\times10^{-6}\ mol\div10\times10^{-3}\ L=0.00015\ M$
