Ncert Solutions Chemistry Class 11th

As 500g of tooth-paste has = 0.2g of F?
So, 10 g of toothpaste has = (0.2g * 10? g) / 500g F = 400g of F
Hence, concentration in ppm is 400.

ΔH_rxn = ΔH_f (N? O, g) + 3ΔH_f (CO? , g) – (2ΔH_f NO? , g) – 3ΔH_f (CO, g)
= 81 + 3* (– 393) – 2 * 34 – 3 (–110)
= 81 – 1179 – 68 + 330 = – 836 kJ

One mole of water is converted to vapour at its boiling point which is 100°C and at 1 atm. For this process ΔG = 0. As phase transformation of water is an equilibrium process and at equilibrium, free energy change is always zero.

Sometimes, when maximum covalency is obtained, the halides become inert to water, thus SF? (or similarly CCl? ) is stable. This is because SF? is coordinately saturated and sterically hindered. Thus, SF? is inert to water, because of kinetic rather than thermodynamic factor.
A. PCl? + 4H? O → H? PO? + 5HCl
B. SiCl? + 4H? O → Si (OH)? + 4HCl
C. BCl? + 3H? O → B (OH)? + 3HCl

2CO (g) + O? (g) → 2CO? (g)
Δn_g = 2 – (2 + 1) = -1
ΔH = ΔE + Δn_g RT or ΔH = ΔE – 1RT
i.e. ΔH < E

The process of moving sodium and potassium ions across the cell membrane is an active transport process involving the hydrolysis of ATP to provide the necessary energy.

As Zn (OH)? , BeO, Al? O? are amphoteric, so they can react with both HCl and NaOH.

$\Delta H=-165kJ/mole$

T =?

$\Delta S=-550J{K}^{-1}$

At equilibrium $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{-165*1000}{-550}K$

$\begin{array}{l}=3*100K\\ =300K\end{array}$

i. Zn + 2NaOH → Na? ZnO? + H?
ii. 4Au + 8NaCN + O? + 2H? O → 4Na [Au (CN)? ]+ 4NaOH
iii. Cu + 4HNO? → Cu (NO? )? + 2NO? + 2H? O
(conc.)

(A) SiO? has three-dimensional network structure of Si – O bonds; while carbon dioxide consists of discrete CO? molecules. SiO? is solid, whereas CO? is a gas
(B) Because of its great affinity for oxygen, Si always occurs as the oxide, silica (SiO? ) or in the form of silicates, which are the compounds of SiO? with other metal oxides.
(C) SiO? + 2Mg → Si + 2MgO
(D) The reluctance of silicon to form pπ – pπ bonds to itself is clearly shown by the fact that silicon does not exist in graphite-like structure, but only in diamond like structure.