Question

using correct scientific notation, how many atoms of k are in 3.0 moles of potassium? 3.0 moles k 6.02 x 10^{23} atoms k 1 mole k 1.0 mole k 18 x 10^{23} atoms k 1.8 x 10^{24} atoms k 6.02 x 10^{23} atoms k

Explanation:

Step1: Set up the conversion

We know that \(1\) mole of any substance contains \(6.02\times10^{23}\) particles (Avogadro's number). For potassium (\(K\)), if we have \(n = 3.0\) moles of \(K\), we use the conversion factor \(\frac{6.02\times 10^{23}\text{ atoms }K}{1\text{ mole }K}\).
The formula to find the number of atoms \(N\) is \(N=n\times N_A\), where \(n\) is the number of moles and \(N_A = 6.02\times10^{23}\text{ atoms/mol}\).
Substituting \(n = 3.0\) moles into the formula: \(N=3.0\text{ moles}\times\frac{6.02\times 10^{23}\text{ atoms }K}{1\text{ mole }K}\).

Step2: Perform the multiplication

Multiply the numerical parts: \(3.0\times6.02 = 18.06\).
When multiplying numbers in scientific - notation (\(a\times10^{b}\times c\times 10^{d}=a\times c\times10^{b + d}\)), here \(b = 0\) (since \(3.0=3.0\times10^{0}\)) and \(d = 23\). So \(N=18.06\times10^{23}\text{ atoms}\).
But in scientific notation, \(a\) should be in the range \(1\leqslant a<10\). So we rewrite \(18.06\times10^{23}\) as \(1.806\times10^{24}\text{ atoms}\approx1.8\times10^{24}\text{ atoms}\) (rounded to two significant figures as \(3.0\) has two significant figures).

Answer:

\(1.8\times 10^{24}\text{ atoms }K\)