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an element has mass no.56 and +3 oxidation state in this element no. of neutron is 30.4% more then no. of electron find the no of electron |
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Answer» Let the number of electrons in ion = xNumber of neutrons = x+0.304x = 1.304xNumber of electrons in neutral atom = x+3Number of protons in neutral atom = x+3Mass number = 56(x+3)+(1.304x)=56X = 23Number of protons = x+3 = 26Symbol of ions = Fe3+ |
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