Class 11th

In Class 11 Physics, we typically talk about two main categories of forces. One is contact forces, and the second is non-contact forces. Contact forces are friction, tension, normal force, and spring force. On the other hand, non-contact forces consist of gravitational, electrostatic, and magnetic forces. These types of forces help us in understanding Newton's laws of motion, free-body diagrams (FBDs), and the concept of equilibrium.

Mass of pure carbon in coal = 0.6 * 1000gm * $\frac{60}{100}=360gm$  

Mass of carbon converted into CO = 6 $\frac{60}{100}*360=216gm\Rightarrow Mol=\frac{216}{12}=18$  mole of carbon

Mass of carbon converted into CO₂ = 360 – 216 = 144gm Þ Mole = $\frac{144}{12}=12$  mole of carbon.

C _(s) + O₂ (g) -> CO₂ (g) + 400KJ

Mole – 12 for CO₂ production, Hence total energy produced = 400 * 12 = 4800KJ

$C_{\left(s\right)}+\frac{1}{2}{O}_2\left(g\right)\underset{}{\overset{}{\to }}C{O}_{\left(g\right)}+100KJ$  

Mole = 18 for CO production, Hence energy produced = 100 * 18 = 1800KJ

Total heat = 4800 + 1800 = 6600KJ

Rb and Cs has nearly same ${\Delta }_{e.g}H=-46kJ/mole$ ${}_{}$

Ar and Kr has same ${\Delta }_{e.g.}H=+96kJ/mole$ ${}_{}$

Sequence of sub-energy level decided by the rule of  $\left(n+\mathcal{l}\right).$

$4HNO{I}_{3\left(\mathcal{l}\right)}+3KC{l}_{\left(s\right)}\underset{}{\overset{}{\to }}C{l}_{2\left(g\right)}+NOC{l}_{\left(g\right)}+2H_{2}O\left(g\right)+3KN{O}_3\left(g\right)$

$\downarrow$              

4 moles of HNO₃ produced 3 mol of KNO₃

Here mole of produced KNO₃ = $\frac{110}{101}$  

If 3 mol of KNO₃ produced by 4 moles of HNO₃

$\therefore$  1 mole of KNO₃ produced by  $\frac{4}{3}$ moles of HNO₃

and  $\frac{110}{101}$ mole of KNO₃ produced by $\frac{4*110}{3*101}$  moles of HNO₃ = 1.45 mole of HNO₃

Hence mass of HNO₃ = mole * mol.wt = 145 * 63 = 91.48 $\approx$ 91.5gm

$\Delta H=\Delta U+\Delta \left(PV\right)$

= $\Delta U+P\Delta V+V\Delta P$

= $\Delta U+P\Delta V$   (At constant pressure)

An atomic orbital is characterized by $n,\mathcal{l}$ $$ and m.

$\overrightarrow{F}=5\widehat{i}+3\widehat{j}-7\widehat{k}$

$\overrightarrow{r}=2\widehat{i}+2\widehat{j}+\widehat{k}$

$\overrightarrow{\tau }=\left|\overrightarrow{r}*\overrightarrow{F}\right|=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 3\hfill & \hfill -7\hfill \end{array}\right|\begin{array}{cc}\hfill =\widehat{i}\left(-k_1-3\right)-\widehat{j}\left(-14-5\right)+\widehat{k}\left(6-10\right)\hfill & \hfill =-17\widehat{i}+19\widehat{j}-4\widehat{k}\hfill \end{array}$  

$\frac{\Delta K}{K_i}*100=\frac{K_f-K_i}{K_i}*100$                

$=\frac{\frac{P_f^2}{2m}-\frac{P_i^2}{2m}}{\frac{P_i^2}{2m}}*100=\frac{P_f^2-P_i^2}{P_i^2}*100$               

$=\frac{{\left(1.2P_i\right)}^2-{\left(P_i\right)}^2}{P_i^2}*100=\frac{1.44P_i^2-P_i^2}{P_i^2}*100$               

 = 44%