Class 11th

The four key contact forces that you will read in Class 11 Physics are normal force, friction, tension, and spring force. Each of these forces needs physical contact and follows Newton's Third Law. They come into play during motion, compression, stretching, or when objects rest on surfaces. These contact forces help in drawing the free-body diagrams (FBDs) and solving problems related to the laws of motion.

An object usually experiences four primary forces. First is the gravitational force (or weight) pulling it downward. Surfaces exert normal force, while frictional force works against motion. Then the applied or tension force starts any movement. These forces can either balance each other out or be unbalanced. 

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.