To evaluate the limit \( \lim_{n \to \infty} \frac{17^7 + 27^7 + \dots + n^{77}}{n^{78}} \), we start by analyzing the series in the numerator. Consider the general term \( k^{77} \) where \( k \) ranges from 17 to \( n \). The numerator can be rewritten as a sum:
\( \sum_{k=17}^{n} k^{77} \).
The key is to approximate this sum using integral calculus for large \( n \). The sum \( \sum_{k=17}^{n} k^{77} \) can be closely approximated by the integral:
\( \int_{17}^{n} x^{77} \, dx \).
Calculating the integral, we have:
\( \int x^{77} \, dx = \frac{x^{78}}{78} + C \).
Evaluating at bounds:
\( \int_{17}^{n} x^{77} \, dx = \left[\frac{x^{78}}{78}\right]_{17}^{n} = \frac{n^{78}}{78} - \frac{17^{78}}{78} \).
The expression for the limit becomes:
\( \lim_{n \to \infty} \frac{\frac{n^{78}}{78} - \frac{17^{78}}{78}}{n^{78}} = \lim_{n \to \infty} \left(\frac{1}{78} - \frac{17^{78}}{78n^{78}}\right) \).
As \( n \to \infty \), the term \( \frac{17^{78}}{78n^{78}} \to 0 \). Hence, the limit simplifies to:
\( \frac{1}{78} \).
Therefore, the value of the limit is \( \frac{1}{78} \).
Which of the following are ambident nucleophiles?
[A.] CN$^{\,-}$
[B.] CH$_{3}$COO$^{\,-}$
[C.] NO$_{2}^{\,-}$
[D.] CH$_{3}$O$^{\,-}$
[E.] NH$_{3}$
Identify the anomers from the following.

The standard Gibbs free energy change \( \Delta G^\circ \) of a cell reaction is \(-301 { kJ/mol}\). What is \( E^\circ \) in volts?
(Given: \( F = 96500 { C/mol}\), \( n = 2 \))