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question 22 (10 points) saved
which of the following contains the fewest atoms? you shouldnt need to do a calculation here.
4.0g li
4.0g rb
4.0g na
4.0g k
4.0g ca
Step1: Recall the formula for the number of moles
The formula is \(n=\frac{m}{M}\), where \(n\) is the number of moles, \(m\) is the mass, and \(M\) is the molar mass.
Step2: Calculate the number of moles for each element
- For \(Li\): \(M_{Li}=6.94\ g/mol\), \(n_{Li}=\frac{4.0\ g}{6.94\ g/mol}\approx0.58\ mol\)
- For \(Na\): \(M_{Na}=22.99\ g/mol\), \(n_{Na}=\frac{4.0\ g}{22.99\ g/mol}\approx0.17\ mol\)
- For \(K\): \(M_{K}=39.10\ g/mol\), \(n_{K}=\frac{4.0\ g}{39.10\ g/mol}\approx0.10\ mol\)
- For \(Ca\): \(M_{Ca}=40.08\ g/mol\), \(n_{Ca}=\frac{4.0\ g}{40.08\ g/mol}\approx0.10\ mol\)
- For \(Rb\): \(M_{Rb}=85.47\ g/mol\), \(n_{Rb}=\frac{4.0\ g}{85.47\ g/mol}\approx0.047\ mol\)
Since the number of atoms \(N = n\times N_{A}\) (\(N_{A}\) is Avogadro's number, \(6.022\times 10^{23}\ mol^{-1}\)), the element with the fewest moles has the fewest atoms.
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4.0g Rb