How many grams are there in 2.5 x 1023 molecules of AgNO2?

Answer 1

7.4x10^23 x ( 1 mol AgNO3/ 6.02x10^23) x (169.88g AgNO3/ 1 mol AgNO3)=
2.1x10^2g AgNO3



no of molecules=7.4x10^23
Mr of AgNO3=169.88
we can find the no of moles, therefore
in 1mol there are 6.02x10^23 molecules
7.4x10^23 molecules represent (7.4x10^23/6.02x10^23)moles


in theory when we multiply molar mass (Mr) with the no of moles, we obtain the mass of the substance in gram:
Mr x no of Mol= Mass in grams
(7.4x10^23/6.02x10^23)x169.88= 2.1x10^2

Answer 2

7.4x10^23 x ( 1 mol AgNO3/ 6.02x10^23) x (169.88g AgNO3/ 1 mol AgNO3)=
2.1x10^2g AgNO3



no of molecules=7.4x10^23
Mr of AgNO3=169.88
we can find the no of moles, therefore
in 1mol there are 6.02x10^23 molecules
7.4x10^23 molecules represent (7.4x10^23/6.02x10^23)moles


in theory when we multiply molar mass (Mr) with the no of moles, we obtain the mass of the substance in gram:
Mr x no of Mol= Mass in grams
(7.4x10^23/6.02x10^23)x169.88= 2.1x10^2

Answer 3

The molar mass of AgNO3 is 169,87....so 147 g of AgNO3 is 147/169,87=0,8654 moles. And therefor N=0,8654x6,022x10^23=5,21x10^23 molecules.

There are 5,21x10^23 molecules in 147 g of AgNO3.

Answer 4

First, divide that very large number of molecules by Avogadro's constant, which is the number of particles in 1 mole of any substance: 2.5x10e23/6.02x10e23 = 0.415 moles. Next, calculate the number of grams in mole of silver nitrite (AgNO2): Ag=108, N=14, 2(O)=32; 108+14+32=154. Finally, multiply moles by grams per mole: 0.415(154) = 63.9g.

Answer 5

no

Answer 6

5