Question

question 3

1. how many atoms are present in 330 moles of francium?

x 10 atoms fr

Explanation:

Step1: Recall Avogadro's number

Avogadro's number \( N_A=6.022\times 10^{23}\text{ atoms/mol}\)

Step2: Use the formula \(N = n\times N_A\)

Given \(n = 330\space mol\), then \(N=330\times6.022\times 10^{23}\)

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Answer:

\(1.99\times 10^{26}\text{ atoms Fr}\) (rounded to three significant figures)