Chemistry

Correctly identifying isotopes and isobars requires knowing both the atomic and mass numbers. Relying on only one is a common error.

• Isotopes: Same element (atomic number), different mass.
• Isobars: Different elements (atomic number), same mass.

Rutherford's atomic model was a breakthrough, but it was flawed. It couldn't explain atomic stability, as orbiting electrons should lose energy and spiral into the nucleus. It also failed to account for the discrete line spectra observed from excited atoms.

α - sulphur & β - sulphur – Diamagnetic, S₂ – form is paramagnetic due to presence of unpaired electron in π* orbital like O₂.

  $\Delta T_f=k_f.m$

$T_f^0-T_f=5.12*\frac{\left(\frac{10}{58}\right)}{\left(\frac{200}{1000}\right)}$

    5.5 – T_f = $\frac{5.12*5*10}{58}$  

  $T_f=1.086°C=\left(1.086\right)°C\cong 1°C$             

$3CH\equiv CH\left(g\right)\to C_6{H}_6\left(\mathcal{l}\right)$       

$\Delta G^0=-RT\mathcal{l}nK.........\left(i\right)$

$\Delta G^0=\sum \Delta {G^0}_P-\sum \Delta {G^0}_R..........\left(ii\right)$    

Equating (i) & (ii)

-2.303 RTlogk = 4.88 * 10⁵

$logK=\frac{-4.88*10^5}{2.303*R*T}=\frac{-488000}{5705.848}=-85.52=-855*10^{-1}$

So; magnitude of log K = 855 * 10^-1

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

$E_1=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}{\left[\frac{1}{H^{+}}\right]}^8$  

[H⁺] = 1M

$E_2=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}+\frac{0.059}{5}log10^{-32}$   

So, difference in E₁ & E₂ is

= 0.3776 V

= 3776 * 10^-4 V

W = 4.75 g

$n=\frac{4.75}{26}$ T = 323 K, R = 0.0826

  $P=\frac{740}{760}atm$          

 V = $\frac{nRT}{P}=\frac{4.75*0.0826*323}{26\left(\frac{740}{760}\right)}=5litre$

$\frac{1.86}{93}=\frac{w}{135}$

W = 2.70 g

Mass produced actual = 2.70 $*\frac{90}{100}=2.43=243*10^{-2}g$

$C_x{H}_y\left(g\right)+\left(x+\frac{y}{4}\right)O_2\left(g\right)\to xC{O}_2\left(g\right)+\frac{y}{2}{H}_{2}O\left(\mathcal{l}\right)$

V 6V - -

- - 4V

Volume of CO2 = 4 * $V_{C_x{H}_y}$ ${}_{{}_{}{}_{}}$

Vx = 4V

x = 4

Volume of O2 = 6 * $V_{C_x{H}_y}$ ${}_{{}_{}{}_{}}$

$V\left(x+\frac{y}{4}\right)=6V$

$4+\frac{y}{4}=6$

y = 8

$Pb{l}_2?P{b}^{2+}+2l^{-}{K}_{sq}=8.0*10^{-9}$

Pb (NO₃)₂ -> Pb²⁺ + ^{$2N{O}_3^{-}$}

0.1 M-

-0.1 M    0.1 M

  $K_{sq}=8*10^{-9}Using{K}_{sq}=\left[P{b}^{2+}\right]{\left[I^{-}\right]}^2$          

$8*10^{-9}=0.1*{\left(2S\right)}^2$

S = 141 * 10^-6 M