Substance A has an atomic mass number of 16 a | vedclass

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(A) Initial moles of $A$: $(n_0)_A = \frac{320}{16} = 20 	ext{ moles}$.Initial moles of $B$: $(n_0)_B = \frac{320}{32} = 10 	ext{ moles}$.Number of

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$3.38$

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(A) Initial moles of $A$: $(n_0)_A = \frac{320}{16} = 20 	ext{ moles}$.Initial moles of $B$: $(n_0)_B = \frac{320}{32} = 10 	ext{ moles}$.Number of moles of $A$ remaining after $t = 2$ days ($T_{1/2} = 1$ day): $n_A = \frac{(n_0)_A}{2^{t/T_{1/2}}} = \frac{20}{2^{2/1}} = \frac{20}{4} = 5 	ext{ moles}$.Number of moles of $B$ remaining after $t = 2$ days ($T_{1/2} = 0.5$ day): $n_B = \frac{(n_0)_B}{2^{t/T_{1/2}}} = \frac{10}{2^{2/0.5}} = \frac{10}{2^4} = \frac{10}{16} = 0.625 	ext{ moles}$.Total moles remaining: $n_{total} = 5 + 0.625 = 5.625 	ext{ moles}$.Total number of atoms: $N = n_{total} 	imes N_A = 5.625 	imes 6.023 	imes 10^{23} \approx 33.88 	imes 10^{23} = 3.388 	imes 10^{24}$.Thus,the value is approximately $3.38$.

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