Chemistry

Sphalarite           ZnS

Calamine            ZnCO₃

Galena                PbS

Siderite               FeCO₃

Acidic oxide => Cl₂O₇

Neutral oxide => N₂O, NO

Basic oxide => Na₂O

Amphoteric oxide => As₂O₃

Emulsions gets separated into two layers on standing.

For stabilization of emulsion. Emulsifying agents added into it but not electrolyte.

$\underset{\_}{\begin{array}{l}A\left(g\right)?B\left(g\right)+\frac{1}{2}C\left(g\right)\\ t=0amoles00\\ -a\alpha moles+a\alpha moles+\frac{a\alpha }{2}moles\end{array}}$

Eq.   a(1 - α)          aα           (aα/2)

Moles           moles       moles

Total no. of moles at equilibrium

= nA + nB + nC

$=a\left(1-\alpha \right)+a\alpha +\frac{a\alpha }{2}=a\left[1+\frac{\alpha }{2}\right]$

${\left(P_A\right)}_{eq}={\left(x_A\right)}_{eq}*P_{eq}=\frac{a\left(1-\alpha \right)}{\left(1+\alpha /2\right)}P=\frac{1-\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_B\right)}_{eq}={\left(x_B\right)}_{eq}*P_{eq}=\frac{a\alpha }{a\left(1+\alpha /2\right)}P=\frac{\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_C\right)}_{eq}={\left(x_C\right)}_{eq}*P_{eq}=\frac{a\alpha /2}{a\left(1+\alpha /2\right)}P=\frac{\alpha /2}{\left(1+\alpha /2\right)}P$

$K_P=\frac{{\left(P_B\right)}_{eq}*{\left(P_C\right)}_{eq}^{1/2}}{{\left(P_A\right)}_{eq}}=\frac{{\left(\alpha \right)}^{3/2}{P}^{1/2}}{\left(1-\alpha \right){\left(2+\alpha \right)}^{1/2}}$

${\left[PtC{l}_4\right]}^{2-}\to Pt$ has dsp² hybridization

BrF₅ -> Br has sp³d² hybridization

PCl₅ -> P has sp³d hybridization

[Co (NH₃)₆]³⁺ -> Co has d²sp³ hybridization.

Degenerate orbitals must have same value of energy

Orbitals with same n and   values are degenerate orbitals.

Mass of C₁₅H₃₀ = volume * Density

= 1000 $m\mathcal{l}$  * 0.756 gm/ $m\mathcal{l}$  

C₁₅H₃₀ + 22.5 O₂   $\stackrel{}{\to }$ 15CO₂ + 15H₂O

No. of moles of C₁₅H₃₀ = $\frac{756}{210}$ moles

No. of moles of O₂ required =   $\left(22.5*\frac{756}{210}\right)moles$

Mass of O₂ required = 22.5 *   $\frac{756}{210}*32=2592gm$

No. of moles of CO₂ liberated = 15 *   $\left(\frac{756}{210}\right)$ moles

Mass of O₂ liberated =   $15*\frac{756}{210}*44=2376gm$

n-factor of KMnO₄ in acidic medium = 5

n-factor of Mohr's salt = 1

meq of KMnO₄ = meq of Mohr's salt

0.01 * 5 * V = 0.05 * 1 * 20

Volume of KMnO₄ used, V = 20 mL

So; Volume of KMnO₄ left in burette = 50 -20mL

= 30 mL

Moles of C = moles of CO₂

=  $\frac{0.793}{44}mol$

Mass of C =  $\frac{0.793}{44}*12g$

= 0.261g

Moles of H = 2 *  $\frac{0.442}{18}mol$

Mass of H =  $2*\frac{0.442}{18}*1g$

Total mass of compound = 0.492g (given)

So; mass of O = (0.492 – 0.216 – 0.049) g

= 0.227g

% of O = $\frac{0.227}{0.492}*100$  

= 46.14%

the nearest integer = 46

Bromination through free radical mechanism occurs at allylic carbon.