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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
Elnora Hickle
Wiki User
∙ 14y ago7.4x10^23 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
Wiki User
∙ 14y agoThe 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.
Wiki User
∙ 8y agoFirst, 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.
Wiki User
∙ 11y agono
Wiki User
∙ 11y ago5
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