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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   $\frac{85000}{23000}\times 3500$

= 12934.78 mg

= 12.93478 gm

$\cong 13$          

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  ${E^o}_{Cell}={E^o}_{A{g}^{+}/Ag}-{E^o}_{Z{n}^{+2}/Zn}=0.8+0.76=1.56V$

Anode :  $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$    

Cathode :  2Ag⁺(aq) + 2e^- -> 2Ag(s)

Zn(s) + 2Ag⁺(aq) -> Zn²⁺ (aq) + 2Ag(s)

  $E_{cell}={E^o}_{Cell}-\frac{0.0591}{n}log\frac{\left[Z{n}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}=1.56-\frac{0.0591}{2}log\left(\frac{0.1}{10^{-4}}\right)$          

x = 147

Cis (optically active)                          Trans (optically inactive)

d & l form (2)                                                    (1)

So, total 3 isomerism.

The TS EAMCET mock test, provides candidates with an actual exam like feel. When students access the mock test, they can - 

• Solve it online by referring to the questions and know the types of questions asked in the exam.
• There is a timer also just like the exam centre which helps students to keep track of the time.
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• Mock tests are the best way to prepare for the TS EAMCET exam.

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

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

$=6.76\cong 7$          

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The TS EAMCET mock test is just like a practice session which can accessed via any third party edutech platform. It provides candidates with an actual exam-like feel. To ensure they prepare for the exam effectively.

$f\left(x\right)=a{x}^2+bx+c$  

g (x) = px + q

  $f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$

$?8x^2?2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

  $9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

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

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

$f\left(x\right)=2co{s}^{-1}x+4co{t}^{-1}x-3x^2-2x+10\forall x\in \left[-1,1\right]$

$\Rightarrow f\text{'}\left(x\right)=-\frac{2}{\sqrt[]{1-x^2}}-\frac{4}{1+x^2}-6x-2<0\forall x\in \left[-1,1\right]$

So, f (x) is decreasing function and range of f (x) is

$\left[f\left(1\right),f\left(-1\right)\right],$ which is $\left[\pi +5,5\pi +9\right]$

Now 4a – b = 4 (p + 5) - (5p + 9) = 11 - “π”

To prepare for the TS EAMCET exam, candidates can refer to the NCERT books, however, they can also take help from supplementary books, but over burdening oneself with a lot of books is not recommended.

$f\left(x\right)=a{x}^2+bx+c$  

g (x) = px + q

  $f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$

$\Rightarrow 8x^2-2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

  $9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

Fraction of molecules having enough energy to form product  =   $e^{-Ea/RT}$

 Fraction of molecules having enough energy to form product =   $e^{-\frac{80.9\times 10^3}{8.314\times 700}}$

$=e^{-13.8}\cong e^{-14}$

So, x = 14

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

$\Delta T_f=K_f\times m\times i$  

0.93 = 1.86 × 1 × i

i =  $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  

$i=1+\left(\frac{1}{n}-1\right)$

$\therefore n=2$            

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

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