16257063
| Question | Answer |
| Actividad 2 | 50g = No ? = N 1.000 años = t N = No · 2,72^(-0,12.094 · t) N = No · 2,72^(-0,12.094 · 1.000) N = 50 · 2,72^(-120,94) N = 1,39 · 10^(-51)g |
| ACTIVIDAD 3 | 100g = No ? = N 20.000 años = t N = No · 2,72^(-0,12.094 · t) N = No· 2,72^(-0,12.094 ·20.000) N = 100 · 2,72^(-2.418,8) N = 0g de Carbono14 |
| ACTIVIDAD 4 | 100% = No 20% = N ? = t t = {(t 1/2 )/(-ln 2)} · {ln (N/No} t = (5.760/-ln 2) · (20/100) t = 13.374,31 años han pasado |
| ACTIVIDAD 5 | 4/4 = No 1/4 = N ? = t t = {(t 1/2 )/(-ln 2)} · {ln (N/No} t = (5.760/-ln 2) ·ln {(1/4)/(4/4)} t = (5.760/-ln 2) ·ln (1/4) t = 11.520 años |
| ACTIVIDAD A) | No = ? N = ? t = 5.730 años N = No · 2,72^(-0,12.094 · t) N = No · 2,72^(-0,12.094 · 5.730) N = No · 0 N/No = 0 Pasados los años (5.730), la cantidad de Carbono 14 se va a reducir a 0. |
| ACTIVIDAD B) | 3/3 = No 1/3 = N ? = t t = {(t 1/2 )/(-ln 2)} · {ln (N/No} t = (5.760/-ln 2) ·ln {(1/3)/(3/3)} t = (5.760/-ln 2) ·ln (1/3) t = 9,129384004 |
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