Densities of diamond and graphite are `3.5` and `2.3 g mL^(-1)`, respectively. The increase of pressure on the equilibrium `C_("diamond") hArr C_("graphite")`

Densities of diamond and graphite are 3 and 2 g mL^(-1) , respectively. The increase of pressure on the equilibrium C_("diamond") hArr C_("graphite")

Is entropy increases for-the following change? Explain :C(diamond) rarr C(graphite)

The enthalpies of combustion of diamond and graphite are - 395.4 kJ and - 393.5 kJ respectively calculate the enthalpy of transformation from diamond to graphite.

The enthalpies of combustion of diamond and graphite are - 395.4 kJ and - 393.5 kJ respectively calculate the enthalpy of transformation from diamond to graphite.

Densities of diamond and graphite are 3.5g//mL and 2.3g//mL Delta_(r)H=-1.9kJ//mol Favourable conditions for formation of diamond are:

Hybridised states of carbon in graphite and diamond are respectively

" Hybridized states of carbon in graphite and diamond are respectively "

Densities of diamond and graphite are (3.5g)/(mL) and (2.3g)/(mL) . C (diamond) equilibrium C (graphite) Delta_(7)H=-1.9(kJ)/("mole") Favourable conditions for formation of diamond are: