Class 11th

$\frac{1}{4\pi {\epsilon }_0}\frac{{\left|e\right|}^2}{hc}=\frac{1}{4\pi {\epsilon }_0}\frac{{\left|e\right|}^2}{r^2}*\frac{r^2}{\frac{hc}{\lambda }*\lambda }=\frac{Fr*r}{E*\lambda }$

Dimension of 'Fr' and 'E' are same dimensionless

Since equation of the gas process is given so we can convert it into T & V form.

$P{V}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=C$

$\Rightarrow \frac{T}{\sqrt[]{V}}=C$

$\Rightarrow \frac{T_1}{\text{exception 25:}}=\frac{T_2}{}$$\frac{\text{exception 25:}}{}$

$\Rightarrow \frac{T_1}{T_2}=$= $\frac{1}{\sqrt[]{2}}$

As we know that maximum acceleration will act on the particle when it is at extreme position .

$\omega =\frac{\frac{\pi }{6}}{0.1}=\frac{10\pi }{6}$

$f=m{\omega }^{2}x$

$\frac{f}{m}={\omega }^{2}A=\frac{1000{\pi }^2}{36}*0.36={\pi }^2=9.87N$

The early 20th century experiments on electrical discharge through gases eventually led to the discovery of cathode rays (electrons). The major finding was that the characteristics of these cathode rays (electrons) were independent of the material used for the electrodes and the nature of the gas present in the cathode ray tube. This consistent behaviour across different substances led to the conclusion that electrons are a basic constituent of all atoms.

Before 20th century, atoms were widely considered the indivisible building blocks of matter. This view goes back to Indian and Greek philosophies, as old as 400 B.C. It also became one established thought on a scientific basis by John Dalton in 1808. With his theory, several fundamental laws of chemistry were established. But those laws failed against observations like static electricity. The experimental observations made towards the end of the 19th and beginning of the 20th century definitively proved that atoms are made of subatomic particles. 

$\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow \frac{35}{300}=\frac{40}{T_2}$

$\therefore T_2=\frac{40*300}{35}=342.85K$

$T_2\left(°C\right)=69.707\approx 70°C$

  $?U=?742.24\text{?}kJ/mole?H_{298}=?$

NH₂CN (s) + $\frac{3}{2}$ O₂ (g) ® N₂ (g) + O₂ (g) + H₂O (l)

 = $?742.24+\frac{1}{2}*\frac{8.314}{1000}*298$  

$?742.24+1.238=?741.002??741kJ/mol$                  

So, magnitude of  $?H$ = 741 kJ/mol.

NaOH + Na₂CO₃

(1) When Hph is added

m.e NaOH + $\frac{1}{2}m.eN{a}_{2}C{O}_3=m.e.HCl=17.5*\frac{1}{10}=1.75$  (m.e = milli equivalents)

(2) When MeOH is added after Hph

$\frac{1}{2}meN{a}_{2}C{O}_3=m.eHCl=1.5*\frac{1}{10}=0.15$

$\begin{array}{l}\therefore m.eNaOH=1.75-0.15=1.6\\ m.eN{a}_{2}C{O}_3=0.15*2=0.3\end{array}$  

$\frac{W_{N{a}_{2}C{O}_3}}{E_{N{a}_{2}C{O}_3}}*1000=0.3$   

$Weight\mathrm{\%}ofN{a}_{2}C{O}_3=\left(\frac{\frac{0.3*53}{1000}}{0.4}\right)*100=\frac{0.3*53}{10*0.4}=\frac{15.9}{4}=3.975\mathrm{\%}\approx 4\mathrm{\%}.$               

$\begin{array}{l}\left(+6\right)\left(+2\right)\left(+3\right)\left(+6\right)\\ Cr{O_4}^{-2}+\underset{}{\underset{?}{S_2{O_3}^{-2}\to Cr{{\left(OH\right)}_4}^{-}+S{O_4}^{-2}}}\end{array}$  

O.N.C = 4

n factor of S₂O₃^-- = 8

n factor of CrO₄^-- = 3

$m.eCr{O}_4^{-2}=m.e.S_2{O}_3^{-2}$        

$V*\left(0.154*3\right)=40*\left(0.25*8\right)\therefore v=\frac{80}{0.462}$       

= 173.16 ml $\approx$ 173 ml

Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially.

After that, the slope of v-t curve is constant and positive.

After some time, velocity becomes constant and acceleration is zero.

After that, the slope of v-t curve is constant and negative.