Question

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When the following equation is balanced properly under basic conditions, what are the coefficients of the species shown? + H2B03 — 12+ B reactant, product, neither) with a coefficient of (Enter 0 Water appears in the balanced equation as a for neither.) How many electrons are transferred in this reaction?

Answer 1

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When the following equation is balanced properly under basic conditions, what are the coefficients of the species shown? 6 + 2 H2B0; — 3 12+ 2 B 2 (Enter 0 Water appears in the balanced equation as a reactant (reactant, product, neither) with a coefficient of for neither.) How many electrons are transferred in this reaction? 6 61 + 2H2B03 + 2H2O → 312 + 2B + 8OH- 61 + 2H,B0z: + 8H2O + 6e → 312 + 2B + 6e + 6H2O + 8OH-

When the following equation is balanced properly under basic conditions, what are the coefficients of the species shown? 1 s+ 2 Br - 1 Br2 + 152 (Enter 0 Water appears in the balanced equation as a neither (reactant, product, neither) with a coefficient of 0 for neither.) How many electrons are transferred in this reaction? 2 S + 2Br → Brą +52- 2Br + S + 2e → Br2 + 52- + 2e-

Sample example on how to calculate binding energy in kJ/mol

(e) 02U atomic mass = 238.05078 u number of nucleons = 238 number of protons - number of electrons = 92 number of neutrons - 238 - 92 - 146 sum of particle masses -(92 p)(1.00728 up"') + (146 n)(1.00866 u n'!)+(92 e)(0.00055 u e!) - 239.98472 u mass difference = atomic mass - sum_238.05078 u - 239.98472 -1.93394 u AE - (Am)o - (-1.93394 u)(1.66054*10-27 kg u ')(2.99792x109 ms 1)2 --2.88624x10-10 binding energy --AE --(-2.88624x10-10J) - 2.88624x10-10 Express in kJ/mol: 2.88624x10-10 6.02214*1023 atoms 1 - - 1.7381x10" kJ/mol 1000 atom mol Express in kJ/mol nucleons 1.7381*10'1 kJ/mol 238U - - 7.3031*10% kJ/mol nucleons 238 mol nucleons/mol 238U

What is the binding energy in kJ/mol La for lanthanum-138? kJ/mol 57.H+81.n—129La 1.14423E+11 The required masses (g/mol) are: H= 1.00783 : on=1.00867; 189 La = 137.90680