Question

question 1

1. what is the amount of moles present in ( 1.27 cdot 10^{29} ) atoms potassium?

mol k

Explanation:

Step1: Recall Avogadro's number

Avogadro's number \(N_A = 6.022\times 10^{23}\text{ atoms/mol}\). The formula to find the number of moles \(n\) is \(n=\frac{N}{N_A}\), where \(N\) is the number of atoms.

Step2: Substitute the values

Given \(N = 1.27\times 10^{29}\text{ atoms}\) and \(N_A=6.022\times 10^{23}\text{ atoms/mol}\). Then \(n=\frac{1.27\times 10^{29}}{6.022\times 10^{23}}\).
Using the rule of exponents \(\frac{a^m}{a^n}=a^{m - n}\), we have \(n=\frac{1.27}{6.022}\times10^{29- 23}\).
\(\frac{1.27}{6.022}\approx0.211\), and \(29 - 23=6\). So \(n = 0.211\times10^{6}\).
In scientific notation \(n=2.11\times 10^{5}\text{ mol}\).

Answer:

\(2.11\times 10^{5}\text{ mol K}\)