Class 11th

Mass of Na⁺ in 50 ml = 70 * 50 = 3500 mg

23000 mg of Na⁺ is present in 85000 mg of NaNO₃ (1 mole NaNO₃ contains 1 mole Na⁺)

$\therefore 3500$   mg Na⁺ will be present in   $\frac{85000}{23000}*3500$

= 12934.78 mg

= 12.93478 gm

$\cong 13$          

pH =   $\frac{1}{2}\left[p{K}_w+p{K}_a-p{K}_b\right]$

$=\frac{1}{2}\left[14+4.75-5.23\right]$

$=6.76\cong 7$          

Uncertainty in speed = 90   $*\frac{5}{100}=4.5m/s=\Delta V$

  $\Delta x=\frac{h}{4\pi m\Delta V}=\frac{6.63*10^{-34}}{4*3.14*10^{-2}*4.5}=1.173*10^{-33}$            

Fix the unit place, find the chances for the first three digits

unit digit as 1, total ways = 9.10²

unit digit as 2, total ways = 4.5²

unit digit as 3 total ways = 3.4²

unit digit as 4 total ways = 2.3²

unit digit as 5 total ways = 1.2²

unit digit as 6 total ways = 1.2²

unit digit as 7 total ways = 1.2²

unit digit as 8 total ways = 1.2²

unit digit as 9 total ways = 1.2²

$\overline{x}=6=\frac{a+b+8+5+10}{5}\Rightarrow a+b=7$ …… (i)

and

${\sigma }^2=\frac{a^2+b^2+8^2+5^2+10^2}{5}-6^2=6.8$  

⇒a² + b² = 25                  ……. (ii)

From (i) and (ii) (a, b) = (3, 4) or (4, 3)

Now mean deviation about mean

$M=\frac{1}{5}\left(3+2+2+1+4\right)=\frac{12}{5}$  

⇒ 25M = 60

$S\left(g\right)+6F\left(g\right)\to S{F}_6\left(g\right)$  

$\Delta {H^0}_R=\Delta {H^0}_f\left(S{F}_6\right)-\Delta {H^0}_f\left(S\right)-6*\Delta {H^0}_f\left(F\right)$            

= (-1100) - (275) -6 (80) = -1855

$\Delta {H^0}_{S-F}=\frac{1855}{6}=309.16KJ/mol\cong 309$            

sin x = 1 – sin² x

=> sin x = $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$  

draw y = sin x

y = $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.

First common term to both AP's is 9

t₇₈ of $\left(3,6,9,......\right)=78*3=234$  

t₅₉ of $\left(5,9,13,........\right)=5+\left(5-1\right)4=237$  

n^th common term $\le 234$  

9 + (n – 1) 12 $\le$  234

n < $\frac{237}{12}\Rightarrow n=19$  

Now sum of 19 terms with a = 9, d = 12

  $=\frac{19}{2}\left(2.9+\left(19-1\right)\cdot 12\right)=2223$