Class 11th

4x – 3y + k₂ = 0

$2r=\frac{40}{5}=8\Rightarrow r=4$

${\left(x-1\right)}^2+{\left(y+2\right)}^2=16$

ΔG° = – 2.303RT log Kp = – 2.303* 2 * 300log10? ³
= – 2.303* 2 * 300 * (– 3) = 4145.4cal mol? ¹

P? = 1atm
P? = 1atm * 40/100 = 0.40atm
V? = 100 cm³
V? =?
At constant temperature, P? V? = P? V?
So V? = (P? V? ) / P? = (1atm * 100 cm³) / 0.40 atm = 250cm³
Hence, the volume of bulb B = (250 –100) cm³ = 150 cm³

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.

As 3f-orbital does not exist as value of l cannot be equal to n, hence no electron can be accommodated.

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

ATP has phosphate group, PO? ³?
So x + 4 (– 2) = – 3 or x = + 5

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