States of Matter

Aluminium is more electropositive than Cr, so it displaced chromium from Cr? O?
Cr? O? + Al - (Δ)-> Al? O? + Cr

M = (a³ * d * N_A) / Z = (3.608 * 10? )³ * 8.92 * 6.022 * 10²³) / 4
M = (46.96 * 10? ²? * 8.92 * 6.022 * 10²³) / 4 = 63 g/mole
the closest answer is choice (1)

P = (nRT) / V = (2 * 0.0831 * 300) / 10 = 4.986 bar

p? = p? x? is not a correct form of Dalton's law of partial pressures.

Mole of CH? = 6.4 / 16 = 0.4 and mole of CO? = 8.8 / 44 = 0.2
Total mole = (0.4 + 0.2) = 0.6 mole of a non-reacting mixture of gas
Using Ideal Gas Law; P = nRT / V
P = (0.6 * 8.314 * 300) / 10 = 149.65 kPa
Ans = 150 (Rounded off)

Partial Pressure of O? = K? * solubility (K? = Henry's constant)
Solubility = PO? / K? = 20 / (8.0 * 10? ) = 2.5 * 10? = 25 * 10? M
Ans = 25

For ideal gas
PM = dRT
d = [PM]/R * 1/T
So graph between dV ST is not straight line

Kindly go through the solution 

 $\mathrm{}{V}_{rms}>V_{av}>V_{mp}$

Theory based.                                                                                                                                                                             

W = 4.75 g

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

  $P=\frac{740}{760}atm$          

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

Applying : ${\left(n_1+n_2\right)}_{initial}={\left(n_1+n_2\right)}_{final}$  

Assuming the system attains a final temperature of T (Such that 300 < T < 60)

(Heat lost by N₂ of container l) = (Heat gained by N₂ of container II)

$n_{1}Cm\left(300-T\right)=n_{2}Cm\left(T-60\right)$  

$\left(\frac{2.8}{28}\right)\left(300-T\right)=\frac{0.2}{28}\left(T-60\right)$  

14 (300 – T) = T – 60

$P=\left(\frac{3}{28}\right)*8.31*\frac{284}{3*10^{-3}}*10^{-5}bar=0.84287bar$  

$P=84.28*10^{-2}bar$

$\approx 84*10^{-2}bar$