Question

show your work...
3.90 x 10²² atoms al
\frac{1\text{ mole}}{6.02 \times 10^{23}\text{ atoms}}
problem #1
\boldsymbol{=}
\boxed{\text{rounded answer}}
\text{moles al}

Explanation:

Step1: Set up the conversion

We have \(3.90\times 10^{22}\) atoms of Al and we know that 1 mole of any substance contains \(6.02\times 10^{23}\) atoms (Avogadro's number). So we use the conversion factor \(\frac{1\space mole}{6.02\times 10^{23}\space atoms}\) to convert atoms to moles. The calculation is \((3.90\times 10^{22})\times\frac{1}{6.02\times 10^{23}}\).

Step2: Perform the calculation

First, we can rewrite the expression as \(\frac{3.90\times 10^{22}}{6.02\times 10^{23}}\). Using the rule of exponents \(\frac{a\times 10^{m}}{b\times 10^{n}}=\frac{a}{b}\times 10^{m - n}\), we have \(\frac{3.90}{6.02}\times 10^{22-23}\). Calculating \(\frac{3.90}{6.02}\approx0.648\) and \(10^{22 - 23}=10^{-1} = 0.1\). Then multiply them: \(0.648\times0.1 = 0.0648\) (or we can also calculate directly: \(\frac{3.90\times 10^{22}}{6.02\times 10^{23}}=\frac{3.90}{6.02}\times\frac{10^{22}}{10^{23}}=\frac{3.90}{6.02}\times 10^{-1}\approx0.648\times0.1 = 0.0648\), and rounding to a reasonable number of significant figures, since 3.90 has three significant figures and 6.02 has three, the result should have three significant figures. So \(0.0648\approx0.0648\) (or if we round to two decimal places for the coefficient, but more accurately, let's do the division: \(3.90\times 10^{22}\div(6.02\times 10^{23})=(3.90\div6.02)\times(10^{22}\div10^{23})\approx0.648\times0.1 = 0.0648\) moles. Rounding to three significant figures, it's approximately \(0.0648\) or \(6.48\times 10^{-2}\) moles. But let's check the calculation again: \(3.90\times10^{22}\) divided by \(6.02\times10^{23}\) is equal to \((3.90\div6.02)\times10^{22 - 23}\). \(3.90\div6.02\approx0.648\), and \(10^{-1}=0.1\), so \(0.648\times0.1 = 0.0648\) moles. If we round to two decimal places for the answer (or as per the problem's requirement, maybe two or three significant figures), but let's do the exact calculation:

\(3.90\times 10^{22}\) atoms Al \(\times\frac{1\space mole}{6.02\times 10^{23}\space atoms}=\frac{3.90\times 10^{22}}{6.02\times 10^{23}}\space moles\)

\(=\frac{3.90}{6.02}\times\frac{10^{22}}{10^{23}}\space moles\)

\(=\frac{3.90}{6.02}\times 10^{-1}\space moles\)

\(\approx0.648\times0.1\space moles\)

\(\approx0.0648\space moles\)

Rounding to three significant figures, it's \(0.0648\) moles (or \(6.48\times 10^{-2}\) moles). If we round to two decimal places, it's approximately \(0.06\) moles, but more accurately, with three significant figures, it's \(0.0648\) (or \(0.065\) if we round to three significant figures properly: \(3.90\div6.02 = 0.64784\), times \(10^{-1}\) is \(0.064784\), which rounds to \(0.0648\) or \(0.065\) depending on the rounding rule).

Answer:

\(0.0648\) (or approximately \(0.065\) or \(6.48\times 10^{-2}\)) moles Al (the exact rounded answer with three significant figures is \(0.0648\) moles, or if we consider two decimal places, \(0.06\) moles, but the precise calculation gives approximately \(0.0648\) moles)