1 ग्राम `=6.023 xx 10^(23)`

The total number of protons in 10 g of calcium carbonate is (N_(0) = 6.023 xx 10^(23)) :-

The total number of protons in 10 g of calcium carbonate is (N_(0) = 6.023 xx 10^(23)) :-

The number of significant figures in the final answer of : (( 16.8 -14.2) ( 6.023 xx 10^(23)))/(2.76) is :

A metal having atomic mass 60.23 gm//mole crystallises in ABCABC close packing. Calculate the density of each metal atom if edge length is 10Å . [Given : N_(A) = 6.023 xx 10^(23)]

A body centred cubic element of density, 10.3 g cm^(-3) has a cell edge of 314 pm. Calculate the atomic mass of the element. ( N_A = 6.023 xx 10^(23) mol^(-1) )