Ncert Solutions Chemistry Class 11th

$\Delta U=-742.24kJ/mole\Delta H_{298}=?$

NH₂CN (s) + $\frac{3}{2}$ O₂ (g) ® N₂ (g) + O₂ (g) + H₂O (l)

= $-742.24+\frac{1}{2}*\frac{8.314}{1000}*298$

$-742.24+1.238=-741.002\approx -741kJ/mol$

So, magnitude of $\Delta H$ = 741 kJ/mol.

Angle of Ht – B – Ht > angle of Hb – B - Hb

Bond angle $\propto \mathrm{\%}S-Character\propto \frac{1}{\mathrm{\%}ofP-character}$ $\frac{}{}$

${}_{}$  $\angle H_t$ B Ht is more so % P-Character is less.

Kindly consider the following figure

Blood – Negatively charged colloid.

According to Hardy – Schulze Rule

FeCl₃ is more efficient for blood clotting

Density = $\frac{mass}{volume}$

Cu > Co > Fe > Cr > Zn

$S\left(g\right)+6F\left(g\right)\to S{F}_6\left(g\right)$  

$\Delta {H^0}_R=\Delta {H^0}_f\left(S{F}_6\right)-\Delta {H^0}_f\left(S\right)-6*\Delta {H^0}_f\left(F\right)$            

= (-1100) - (275) -6 (80) = -1855

$\Delta {H^0}_{S-F}=\frac{1855}{6}=309.16KJ/mol\cong 309$            

Kindly go through the solution

S > Se > Te > O

It does not contains 2^nd most abundant element by weight in earth crust because that is Si Calgon -> Na₂ [Na₄ (PO₃)₆]

$\underset{}{\overset{Watersoluble}{\to }}2N{a}^{+}{\left[N{a}_4{\left(P{O}_3\right)}_6\right]}^{2-}$

$\underset{}{\overset{C{a}^{2+}}{\to }}2N{a}^{+}{\left[N{a}_{2}Ca{\left(P{O}_3\right)}_6\right]}^{2-}$

Tritium is radioactive and it decays into He³ during emission of b-radiation