1 g molecule of `V_(2)O_(5)` contains :

1 gm of molecule of V_2 O_5 contains

What is the numner of molecules of CO_(2) which contain 8g of O_(2) ?

What is the number of molecules of CO_2 contains 8 g of O_2 ?

A mixture contains O_(2) and N_(2) in the proportion of 1:4 by weight. What will be the ratio of the number of molecules of O_(2) and N_(2) in the mixture?

1 g equivalent of V_2O_5 in the given reaction is equal to V_2O_5 + Zn → ZnO + V (Given at. wt. of V = A)

1 g equivalent of V_2O_5 in the given reaction is equal to V_2O_5 + Zn rarr ZnO + V (Glven at wt. of V = A)

Mole conept: Calculate the follwing in 48g of oxygen gas (diaxygen) at room temperature (25^(@) C) : (i) moles of O_(2) , (Ii) number of O_(2) molecules, (iii) number of O atoms, and (iv) number of electrons. Strategy: The molecular mass of dioxygen (O_(2)) is 32 mu Thus, its molar mass is 32.0g mol^(-1) , Now, use the needed information: (a) one mole of O_(2) contains Avogadro's number of O_(2) molecules, (b) one O_(2) molecules contains 2 oxygen atons, and (c) one oxygen atom contains 8 electrons.

Statement-I : 16g each O_(2) and O_(3) contains (N_(A))/(2) and (N_(A))/(3) atoms repectively Because Statement-II : 16g of O_(2) , and O_(3) contains same no. of atoms.

Statement-I : 16g each O_(2) and O_(3) contains (N_(A))/(2) and (N_(A))/(3) atoms repectively Because Statement-II : 16g of O_(2) , and O_(3) contains same no. of atoms.