Respuesta :
Answer:
a)0.4222 moles of the compound are present.
b)[tex]2.542\times 10^{23} ions[/tex] of iron(III)
c)[tex]7.6274\times 10^{23} ions[/tex] of bromide.
Explanation:
Moles = [tex]\frac{\text[Mass of compound}}{\text{Molar mas of compound}}[/tex]
a)Moles of iron (III) bromide =[tex]\frac{125 g}{296 g/mol}=0.4222 mol[/tex]
0.4222 moles of the compound are present.
Molecules of of iron (III) bromide in 0.4222 moles
1 mole = [tex]6.022\times 10^{23}[/tex] molecules
In , 0.4222 moles:
[tex]6.022\times 10^{23}\times 0.4222 molecules[/tex]
[tex]=2.542\times 10^{23} molecules[/tex] of iron (III) bromide
b)1 molecule of iron (III) bromide contain 1 iron(III) ion.Then [tex]2.542\times 10^{23} molecules[/tex] will contain:
[tex]1\times 2.542\times 10^{23} molecule= 2.542\times 10^{23} ions[/tex] of iron(III)
c) 1 molecule of iron (III) bromide contain 3 bromide ion.Then [tex]2.542\times 10^{23} molecules[/tex] will contain:
[tex]3\times 2.542\times 10^{23} molecule= 7.6274\times 10^{23} ions[/tex] of bromide.
