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Chapter 3: Problem 29

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}, 23 \mathrm{~g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.

Short Answer

Expert verified
The samples can be ranked in order of increasing number of atoms as follows: 1. \(\mathrm{Na}\): \(6.022 \times 10^{23}\ \mathrm{atoms}\) 2. \(\mathrm{H_{2}O}\): \(9.033 \times 10^{23}\ \mathrm{atoms}\) 3. \(\mathrm{N_{2}}\): \(1.2 \times 10^{24}\ \mathrm{atoms}\)

Step by step solution

01

Convert \(\mathrm{H_2O}\) moles to molecules and determine the total number of atoms

Use Avogadro's number (\(6.022 \times 10^{23}\)) to convert moles to molecules: Molecules of \(\mathrm{H_2O} = 0.50\ \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{molecules/mol} = 3.011 \times 10^{23}\ \mathrm{molecules}\) Now, we can find the total number of atoms in this sample by multiplying the number of molecules by the number of atoms in one \(\mathrm{H_2O}\) molecule (2 hydrogen atoms + 1 oxygen atom = 3 atoms): Total \(\mathrm{atoms} = 3.011 \times 10^{23}\ \mathrm{molecules} \times 3\ \mathrm{atoms/molecule} = 9.033 \times 10^{23}\ \mathrm{atoms}\) Next, let's find the number of atoms in the \(\mathrm{Na}\) sample:
02

Convert mass of \(\mathrm{Na}\) to moles and determine the total number of atoms

Use the atomic weight of \(\mathrm{Na}\) from the periodic table (22.99 g/mol) to convert mass to moles: Moles of \(\mathrm{Na} = \frac{23\ \mathrm{g}}{22.99\ \mathrm{g/mol}} = 1.000\ \mathrm{mol}\) Now, use Avogadro's number to convert moles to atoms: Total \(\mathrm{atoms} = 1.000\ \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{atoms/mol} = 6.022 \times 10^{23}\ \mathrm{atoms}\) Finally, let's find the total number of atoms in the \(\mathrm{N_2}\) molecules sample:
03

Determine the total number of atoms in \(6.0 \times 10^{23} \ \mathrm{N_2}\) molecules

Since each \(\mathrm{N_{2}}\) molecule consists of 2 nitrogen atoms, we can find the total number of atoms by multiplying the number of molecules by the number of atoms in one \(\mathrm{N_{2}}\) molecule: Total \(\mathrm{atoms} = 6.0 \times 10^{23}\ \mathrm{molecules} \times 2\ \mathrm{atoms/molecule} = 1.2 \times 10^{24}\ \mathrm{atoms}\) Now, we can rank the samples in order of increasing number of atoms: 1. \(\mathrm{Na}\): \(6.022 \times 10^{23}\ \mathrm{atoms}\) 2. \(\mathrm{H_{2}O}\): \(9.033 \times 10^{23}\ \mathrm{atoms}\) 3. \(\mathrm{N_{2}}\): \(1.2 \times 10^{24}\ \mathrm{atoms}\)

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Most popular questions from this chapter

Calculate the following quantities: (a) mass, in grams, of \(0.105\) moles sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) (b) moles of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) in \(14350 \mathrm{~g}\) of this substance (c) number of molecules in \(1.0 \times 10^{-6} \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (d) number of \(\mathrm{N}\) atoms in \(0.410 \mathrm{~mol} \mathrm{NH}_{3}\)

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to \(\mathrm{NO}\) : $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(1.50 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.75 \mathrm{~g}\) of \(\mathrm{O}_{2}\) (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and of \(\mathrm{H}_{2} \mathrm{O}\) form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Several brands of antacids use \(\mathrm{Al}(\mathrm{OH})_{3}\) to react with stomach acid, which contains primarily \(\mathrm{HCl}\) $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (a) Balance this equation. (b) Calculate the number of grams of \(\mathrm{HCl}\) that can react with \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) (c) Calculate the number of grams of \(\mathrm{AlCl}_{3}\) and the number of grams of \(\mathrm{H}_{2} \mathrm{O}\) formed when \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Hydrofluoric acid, \(\mathrm{HF}(a q)\), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the \(\mathrm{HF}(a q) .\) Sodium silicate \(\left(\mathrm{Na}_{2} \mathrm{SiO}_{3}\right)\), for example, reacts as follows: \(\mathrm{Na}_{2} \mathrm{SiO}_{3}(\mathrm{~s})+8 \mathrm{HF}(a q) \longrightarrow\) $$ \mathrm{H}_{2} \mathrm{SiF}_{6}(a q)+2 \mathrm{NaF}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) How many moles of \(\mathrm{HF}\) are needed to react with \(0.300 \mathrm{~mol}\) of \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (b) How many grams of NaF form when \(0.500 \mathrm{~mol}\) of HF reacts with excess \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (c) How many grams of \(\mathrm{Na}_{2} \mathrm{SiO}_{3}\) can react with \(0.800 \mathrm{~g}\) of HF?

Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a certain number of water molecules are included in the solid structure. Its formula can be written as \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot x \mathrm{H}_{2} \mathrm{O}\), where \(x\) is the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). When a \(2.558-\mathrm{g}\) sample of washing soda is heated at \(25^{\circ} \mathrm{C}\), all the water of hydration is lost, leaving \(0.948 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). What is the value of \(x\) ?
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