The digits in a properly recorded measurement are known as significant figures. It is also defined as follows. The total numbers of figures in a number including the last digit whose value is uncertain is called number of significant figures.

Suppose that the oxidation number of P in \({ HPO }_{ 3 }^{ 2- }\) be x.
Theb, 1 + x + 3 (-2) = -2
or x + 1 - 6 = -2 or x = +3

Given, \(\frac { 1 }{ 2 } { N }_{ 2 }(g)+{ \frac { 3 }{ 2 } }{ H }_{ 2 }(g)\longrightarrow { NH }_{ 3 }(g);\)
\({ \Delta }_{ f }{ H }^{ \Theta }=-91.8 \ kJ \ { mol }^{ -1 }\)
(\({ \Delta }_{ f }{ H }^{ \Theta }\) means enthalpy of formation of 1 mole of NH₃)
\(\therefore\) Enthalpy change for the formation of 2 moles of NH₃
\({ N }_{ 2 }(g)+{ 3H }_{ 2 }(g)\rightarrow { 2NH }_{ 3 }(g);\)
\({ \Delta }_{ f }{ H }^{ \Theta }=2\times -91.8=-183.6 \ kJ \ { mol }^{ -1 }\)
And for the reverse reaction. \(2{ NH }_{ 3 }(g)\longrightarrow { N }_{ 2 }(g)+{ 3H }_{ 2 }(g);{ \Delta }_{ f }{ H }^{ \Theta }=+183.6 \ kJ \ { mol }^{ -1 }\)  Hence, the value of \({ \Delta }_{ f }{ H }^{ \Theta }\) for NH₃ is +183.6 kJ mol^-1

The energy which arises due to motion of atoms or molecules in a body is known as thermal energy.It is a measure of average kinetic energy of the particles.It increases with increases with increase in temperature.

Therefore, only B_eCl₂ is linear and rest of the molecules are non-linear.

Electronic configuration of Al
In ground state \({ _{ 13 }{ Al } }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 6 },{ 3s }^{ 2 },{ 3p }_{ x }^{ 1 }\)
In excited state = \({ 1s }^{ 2 },{ 2s }^{ 2 },{ 2 }p^{ 6 },{ 3s }^{ 1 },{ 3p }_{ x }^{ 1 },{ 3p }_{ y }^{ 1 }\)
In the formation of AlCl_3,Al undergoes sp² hybridisation and it is original planar in shape. While in the formation of \({ AlCl }_{ 4 }^{ - }\) , Al undergose sp³ hybridisation.
It means empty 3p_z orbital also involced in hybridisation. Thus, the shape of \({ AlCl }_{ 4 }^{ - }\) ion is tetrahedral.

Configurations with completely filled and half-filled orbitals have extra stability. In, d-orbitals are completely filled and s-orbital is half-filled. Hence it is a more stable configuration for Cu compare to \(3d^{9}4s^{2}.\)

Frequency, \(v=\frac { c }{ \lambda } \)
\(\because \) 1 nm = 10^-9m
\(\therefore \) 580nm = 580 x 10^-9 m = 580 x 10^-7cm
\(v=\frac { 3.0\times 10^{ 8 }{ ms }^{ -1 } }{ 580\times { 10 }^{ -9 }m } =5.17\times { 10 }^{ 14 }{ s }^{ -1 }\)
\( [Velocity \ of \ light=3\times { 10 }^{ 8 }{ ms }^{ -1 }]\)
Wave number,
 \(\bar { v } =\frac { 1 }{ \lambda } =\frac { 1 }{ 580\times 10^{ -7 }cm } \)
\( =1.724\times { 10 }^{ 4 }cm^{ -1 }\)

\(2A+4B\longrightarrow 3C+4D\)
According to the given reaction, 2 moles of A rect with 4 moles of B.
Hence, 5 moles of A will react with 10 moles of
\(b\left( \frac { 5\times 4 }{ 2 } =10moles \right) \)
Limiting reagent decide the amount of product produced. According to the reaction, 4 moles of B produces 3 moles of C.
6 moles of B will produce\(\frac { 3\times 6 }{ 4 } =4.5\) moles of C.