Chemistry

Mass of CuSO₄. 5H₂O = 80 g

Volume of solution = 5L

Molar mass of CuSO₄.5H₂O = 249.54g/ml

$Concentration=\frac{moles}{volumeofsolution}$          

$\begin{array}{l}=\frac{0.32}{5}=0.064M\\ =64*10^{-3}M\end{array}$                

Ammonium salt in rain drop resulting wet deposition

$N{H}_4^{+}?Salt+H_{2}O?N{H}_{4}OH$

Oxides of N & S settle down on ground as dry deposition (SO₂).

$Zn{\left(OH\right)}_2?Z{n}^{2+}+2O{H}^{-}$  

S                           -            -

NaOHNa⁺ + OH^--

0.1 M    -            -

$\left[Z{n}^{2+}\right]=S$                

Here ; 2 * 10^-20 = S (0.1)²

So; S = 2 * 10^-18 M

$M{n}^{+2}\to groupIV\left(M{n}^{+2},C{o}^{+2},Z{n}^{+2},N{i}^{2+}\right)$

$A{s}^{+3}\to groupIIB\left(A{s}^{+3},A{s}^{+5},S{b}^{+3},S{b}^{+5},S{n}^{+2},S{n}^{+4}\right)$

$\begin{array}{l}C{u}^{+2}\to groupIIA\left(C{u}^{+2},P{b}^{+2},H{g}^{+2},C{d}^{+2},B{i}^{+3}\right)\\ A{\mathcal{l}}^{+3}\to groupIII\left(F{e}^{+3},A{\mathcal{l}}^{+3},C{r}^{+3}\right)\end{array}$

$\left[Ag{\left(N{H}_3\right)}_2\right]\left[Ag{\left(CN\right)}_2\right]contains{\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}and{\left[Ag{\left(CN\right)}_2\right]}^{-}$

 ln ${\left[Ag{\left(N{H}_3\right)}_2\right]}^{+},$  oxidation state of Ag = +1

$ln{\left[Ag{\left(CN\right)}_2\right]}^{-}$   , oxidation state Ag = +1

Hence; sum of oxidation state = 2

Mass of empty cylinder = 14.8 kg

Mass of cylinder when full = 29 kg

Mass of gas in cylinder when filled; W₁ = 29 – 14.8 = 14.2 kg

Mass of gas in cylinder after using, W₂ = 23 – 14.8 = 8.2 kg

Initial pressure ; P₁ = 3.47 atm

Final pressure ; P₂ =?

Using

  $PV=\frac{W}{M}RT$

$P_{1}V=\frac{W_1}{M}RT..........\left(I\right)$              

$P_{2}V=\frac{W_2}{M}RT...........\left(II\right)$                 

$\frac{P_1}{P_2}=\frac{W_1}{W_2}$                

$\frac{3.47}{P_2}=\frac{14.2}{8.2}$                

P₂ = atm

Glyptal, Dacron & PHBV are polyesters.

Novalac is copolymer of phenol & formaldehyde but not polyesters.

Electrophilic addition of bromine to an alkene is anti-addition, in which cis-alkene gives two enantiomers and trans – alkene gives meso form

Here; trans-but-2-ene will give meso products

(a) $\Delta G=\Delta G°+RT\mathcal{l}nQ$

$\Delta G°=\Delta H°-T\Delta S°..............\left(1\right)$                      

At equilibrium $\Delta G=0,Q=K_{eq}$

  $\therefore \Delta G°=-RT\mathcal{l}n{K}_{eq}.................\left(2\right)$                   

$\therefore \Delta H°-T\Delta S°=-RT\mathcal{l}n{K}_{eq}$                      

$\mathcal{l}nKeq=-\left(\frac{\Delta H°-T\Delta S°}{RT}\right)$         

(b) $W_{rev}=-nRT\mathcal{l}n\left(\frac{V_f}{V_i}\right)$  

(c) $\Delta G=-T\Delta S_{Total}$ (at constant P)

$\therefore \frac{\Delta G}{\Delta S_{Total}}=-T$        

(d) $\Delta G°=-RT\mathcal{l}nKeq$

$\therefore Keq=e^{\left(-\Delta G°/RT\right)}$  

Electrophilic addition of bromine to an alkene is anti-addition, in which cis-alkene gives two enantiomers and trans – alkene gives meso form

Here; trans-but-2-ene will give meso products