Class 11th

Classification of Elements is Necessary because it helps in understanding and remembering elements' properties better, you can understand how elements react or behave, and can guess properties of unidentified or newly discovered elements.

Yes, one can see the atoms, but not with the naked eye. One needs to use specialised microscopes to see the atoms. To see the atoms, one can use electron microscopes such as electron ptychography and TEM, or they can also use Scanning Tunneling Microscope (STM).

Helium (He) is the smallest atom. It has a smaller physical size as the protons in Helium create a stronger nuclear charge to pull its electrons closer to nucleus. The hydrogen is the lightest element.

In molecules, atoms are held together by the chemical bonds, that are strong attractive forces. The main two types of chemical bonds include ionic bond and covalent bond.

The following are the steps one must follow to draw the structure of an atom: First, find out the atomic number of the element on the periodic table, then draw a small central circle called the nucleus that includes the protons and neutrons. Next, there will be concentric circles around the nucleus for electron shells. Next, fill these shells with dots or crosses for electrons. One can fill around a maximum of 2 electrons in the first shell and a maximum of 8 in the second shell, and so on.

It talks about the atom. According to the Greek philosophers earlier, the atoms were the smallest part of matter that cannot be further divisible. Later, Dalton's atomic theory suggested that atom is the ultimate particle of matter. This theory was challenged later on by experiments done towards the end of the 19th and beginning of the 20th century. These experiments found that atoms are made up of electrons, protons and neutrons.

Kindly consider the following figure

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

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

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

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

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

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

$-742.24+1.238=-741.002\approx -741kJ/mol$

So, magnitude of $\Delta H$ = 741 kJ/mol.

(1) $C{r}^{+1}=4s^{0}3d^{5}M{n}^{+2}=4s^{0}3d^5$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(2) $N{i}^{+2}=4s^{0}3d^{8}C{u}^{+}=4s^{0}3d^{10}$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(3) $F{e}^{+2}=4s^{0}3d^{6}C{o}^{+}=4s^{1}3d^7$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(4) $V^{+2}=4s^{0}3d^{3}C{r}^{+}=4s^{o}3d^5$