A-31 Hallar el peso de: a. 0,75 mol-átomo de potasio; b. \( 1,2 \times 10^{15} \) átomos de carbono R. a. \( 29,32 \mathrm{~g} \) de potasio; \( \begin{array}{l}\text { b. } 2,39 \times 10^{-8} \mathrm{~g} \text { de carbono. }\end{array} \) \begin{tabular}{l} A-32 \( \begin{array}{l}\text { Determinar cuántos átomos hay en: } \\ \text { átomos de aluminio; } 0,92 \text { mol-átomo de aluminio; } \\ \text { b. } 1,5 \times 10^{23} \text { átomos de nitrógeno. }\end{array} \) \\ \hline\end{tabular}

Alright, I need to solve these two chemistry problems. Let's start with the first one: "A-31 Hallar el peso de: a. 0,75 mol-átomo de potasio; b. \( 1,2 \times 10^{15} \) átomos de carbono R. a. \( 29,32 \mathrm{~g} \) de potasio; \( \begin{array}{l}\text { b. } 2,39 \times 10^{-8} \mathrm{~g} \text { de carbono. }\end{array} \)" First, I need to understand the question. It's asking for the mass of potassium atoms and carbon atoms based on the given quantities. Let's break it down. For part a: 0,75 mol-átomo de potasio. Wait, "mol-átomo" is a bit confusing. I think it might be a typo or a specific term I'm not familiar with. Maybe it's supposed to be "mol de átomos de potasio," which means "0.75 moles of potassium atoms." That makes more sense. So, to find the mass of 0.75 moles of potassium, I need to know the molar mass of potassium. The atomic mass of potassium (K) is approximately 39.10 g/mol. So, mass = moles × molar mass. Mass = 0.75 mol × 39.10 g/mol = 29.325 g. That matches the given answer of 29.32 g, which is reasonable considering rounding. Now, part b: \( 1,2 \times 10^{15} \) átomos de carbono. To find the mass, I need to convert the number of atoms to moles first, then to grams. First, Avogadro's number is \( 6.022 \times 10^{23} \) atoms/mol. So, moles = \( \frac{1.2 \times 10^{15}}{6.022 \times 10^{23}} \). Calculating that: \( \frac{1.2}{6.022} \times 10^{15-23} = 0.1993 \times 10^{-8} \) mol = \( 1.993 \times 10^{-9} \) mol. Now, the molar mass of carbon (C) is approximately 12.01 g/mol. So, mass = moles × molar mass = \( 1.993 \times 10^{-9} \) mol × 12.01 g/mol = \( 2.39 \times 10^{-8} \) g. That matches the given answer of \( 2.39 \times 10^{-8} \) g, which is correct. Now, moving on to the second problem: "A-32 \( \begin{array}{l}\text { Determinar cuántos átomos hay en: } \\ \text { átomos de aluminio; } 0,92 \text { mol-átomo de aluminio; } \\ \text { b. } 1,5 \times 10^{23} \text { átomos de nitrógeno. }\end{array} \) \\ \hline\end{array}" This problem is asking to determine the number of atoms in given quantities of aluminum and nitrogen. For part a: "átomos de aluminio; 0,92 mol-átomo de aluminio." Again, "mol-átomo" seems like a typo. It should probably be "0.92 moles of aluminum atoms." To find the number of atoms, I'll use Avogadro's number: \( 6.022 \times 10^{23} \) atoms/mol. Number of atoms = moles × Avogadro's number = 0.92 mol × \( 6.022 \times 10^{23} \) atoms/mol = \( 5.54 \times 10^{23} \) atoms. For part b: "1,5 \times 10^{23} \text { átomos de nitrógeno.}" This is straightforward. The number of atoms is already given as \( 1.5 \times 10^{23} \) atoms of nitrogen. So, summarizing: - Part a: 0.92 moles of aluminum → \( 5.54 \times 10^{23} \) atoms - Part b: \( 1.5 \times 10^{23} \) atoms of nitrogen Everything checks out with the given information and calculations.