Class 12th

∴ % of S in the compound
= (32/233) * (mass of BaSO? / mass of compound) * 100 = (32 * 0.35 * 100) / (233 * 0.25) = 19.227 ≈ 19.23

Kindly go through the solution

Ans. (8.00)

Kindly go through the solution

Ans. (8.40 to 8.57)

Factual
(A) Magnalium contain Al and Mg
(B) Bell metal ⇒ Cu and Sn
(C) Gun metal ⇒ Cu, Sn and Zn
(D) Chrome steel ⇒ Fe and Cr.

$M={\chi }_{m}H$ and ${}_{}\frac{}{}$ ${\chi }_m=\frac{C}{T}($ Curie law $$ $)$ ,

$$ $M=\frac{CH}{T}$  or $\frac{M_1}{M_2}=\frac{C{H}_1/T_1}{C{H}_2/T_2}=\left(\frac{H_1}{H_2}\right)\left(\frac{T_2}{T_1}\right)$ $\frac{{}_{}}{{}_{}}\frac{{}_{}{}_{}}{{}_{}{}_{}}\left(\frac{{}_{}}{{}_{}}\right)\left(\frac{{}_{}}{{}_{}}\right)$

$M_2=M_1\left(\frac{H_2}{H_1}\right)\left(\frac{T_1}{T_2}\right)=12(\mathrm{A}/\mathrm{m})\left[\frac{0.2T}{0.6T}\right]\left[\frac{4\mathrm{K}}{16\mathrm{K}}\right]$

$=1\mathrm{A}/\mathrm{m}$

x = 4

y = 6

z = 6

Ans. (3)

$C_v=\frac{3R}{2}$  

AC  polytropic P $\propto$  V

$\Rightarrow P.V^{-1}=constant$            

$C_{AC}=C_V+\frac{R}{1-\left(-1\right)}=2R$

AB isobaric

$C_{AB}=C_P=\frac{5R}{2}$       

  $\therefore \frac{Q_{AC}}{Q_{AB}}=\frac{C_{AC}}{C_{AB}}\left[?\Delta Tsameforboth\right]$

$=\frac{2R}{\left(\frac{5R}{2}\right)}=\frac{4}{5}=0.8$           

(i)  give NO₂   (ii) NH₄NO₃         (iii) gives NO₂     (iv) Give NO₂      (v) give NO₂       (vi) Given N₂O

(vii) Gives NO₂       (viii) Gives NO₂  (ix) Gives NO

σ bonded organometallic compound ⇒ M – C
σ-bond
and in π – bonded organo metallic compound
M – C
π bond
In ferrocene, there is π-bond

Let V ml is taken

  $\frac{5*V+1*150}{\left(V+150\right)+0.9}=2.5$            

5V + 150 = 2.25 V + 337.5

$V=\frac{187.5}{2.75}=68.18ml$