Chemistry

The number of chiral carbons in open-chain aldohexose (such as glucose) is four, therefore, the number of stereoisomers = 2? = 16.

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

All monosaccharides which differ in configuration at C? and C? gives the same osazone. Since, glucose and fructose differ from each other only in configuration at C? and C? therefore, they give the same osazone. All other options given in the questions do not satisfy this condition and hence, do not from the same osazone.

The second ionization energy of K is maximum, because the second electron is removed from fully filled 3p? subshell. The second ionization energies of group 2 elements decrease down the group.
Hence, second I.E. of Ca > second I.E. of Ba.

Nitrogen, sulphur and halogens are tested in an organic compound by lassaigne's test.
The organic compound is fused with sodium metal as to convert these elements into ionisable inorganic substances.
Na + C + N → NaCN
2Na + S → Na? S
2Na + X? → 2NaX
The cyanide, sulphide or halide ions can be confirmed in aqueous solution by usual test.

${\Delta }_t=\frac{hc}{\lambda }=\frac{6.63*10^{-34}*3.08*10^8}{600*10^{-9}}$

$=\frac{6.63*3.08*10^{-17}}{600}$ J

$=765.765*10^{-21}J$

$?766*10^{-21}$ J