how many atoms of ruthenium are in 3.23 grams of ruthenium?\nstarting amount\nfirst conversion…

how many atoms of ruthenium are in 3.23 grams of ruthenium?\nstarting amount\nfirst conversion factor\nsecond conversion factor\nfinal amount\nanswer bank\n1\n2\n3\n4\n3.23 g ru\n= 1.97×10²⁶ atoms ru\n= 1.92×10²² atoms ru\n×(\\frac{1 mol ru}{6.022×10²³ atoms ru})\n5\n6\n7\n8\n×(\\frac{6.022×10²³ atoms ru}{1 mol ru})\n×(\\frac{1 mol ru}{101.07 g ru})\n= 5.42×10⁻²² atoms ru\n= 5.31×10⁻²⁶ atoms ru\n×(\\frac{101.07 g ru}{1 mol ru})\n9

how many atoms of ruthenium are in 3.23 grams of ruthenium?\nstarting amount\nfirst conversion factor\nsecond conversion factor\nfinal amount\nanswer bank\n1\n2\n3\n4\n3.23 g ru\n= 1.97×10²⁶ atoms ru\n= 1.92×10²² atoms ru\n×(\\frac{1 mol ru}{6.022×10²³ atoms ru})\n5\n6\n7\n8\n×(\\frac{6.022×10²³ atoms ru}{1 mol ru})\n×(\\frac{1 mol ru}{101.07 g ru})\n= 5.42×10⁻²² atoms ru\n= 5.31×10⁻²⁶ atoms ru\n×(\\frac{101.07 g ru}{1 mol ru})\n9

Answer

Explanation:

Step1: Convert grams to moles

The molar mass of ruthenium (Ru) is 101.07 g/mol. We use the conversion factor $\frac{1\ mol\ Ru}{101.07\ g\ Ru}$ to convert the mass of Ru to moles. Starting with 3.23 g Ru, the amount in moles is $n = 3.23\ g\ Ru\times\frac{1\ mol\ Ru}{101.07\ g\ Ru}$.

Step2: Convert moles to atoms

We know that 1 mole of any substance contains $6.022\times 10^{23}$ atoms (Avogadro's number). So we use the conversion factor $\frac{6.022\times 10^{23}\ atoms\ Ru}{1\ mol\ Ru}$ to convert moles of Ru to atoms of Ru. The number of atoms $N=n\times\frac{6.022\times 10^{23}\ atoms\ Ru}{1\ mol\ Ru}=3.23\ g\ Ru\times\frac{1\ mol\ Ru}{101.07\ g\ Ru}\times\frac{6.022\times 10^{23}\ atoms\ Ru}{1\ mol\ Ru}$. [ \begin{align*} N&=\frac{3.23\times6.022\times 10^{23}}{101.07}\ &=\frac{19.45006\times 10^{23}}{101.07}\ &= 1.92\times 10^{22}\ atoms\ Ru \end{align*} ]

Answer:

Starting amount: 3.23 g Ru First conversion factor: $\times(\frac{1\ mol\ Ru}{101.07\ g\ Ru})$ Second conversion factor: $\times(\frac{6.022\times 10^{23}\ atoms\ Ru}{1\ mol\ Ru})$ Final amount: = 1.92×10²² atoms Ru