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Chapter 2: Problem 3

(a) How many grams are there in one amu of a material? (b) Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are there in a pound-mole of a substance?

Short Answer

Expert verified
Answer: There are 1.66053906660 × 10^{-24} grams in one amu of a material, and there are 2.73159498 × 10^{26} atoms in a pound-mole of a substance.

Step by step solution

01

Convert atomic mass unit (amu) to grams

The atomic mass unit (amu) is a unit of mass used to express atomic and molecular weights. By definition, 1 amu is equal to 1/12 of the mass of a carbon-12 atom. We can use the conversion factor to convert amu to grams: 1 amu = 1.66053906660 × 10^{-24} grams
02

Calculate the number of grams in one amu of a material

We have the conversion factor from Step 1. Now we can simply use it to find the number of grams in one amu of a material: 1 amu = 1.66053906660 × 10^{-24} grams So, there are 1.66053906660 × 10^{-24} grams in one amu of a material.
03

Answer (a):

There are 1.66053906660 × 10^{-24} grams in one amu of a material.
04

Convert grams-mole to pound-mole

Now we need to find the number of atoms in a pound-mole of a substance. First, we need to convert gram-mole to pound-mole using the conversion factor: 1 pound (lb) = 453.59237 grams To convert gram-mole to pound-mole, we divide the given amount in grams by the conversion factor: 1 gram-mole = 1/453.59237 pound-mole
05

Calculate the number of atoms in a pound-mole

To find the number of atoms in a pound-mole, we use Avogadro's number, which is the number of atoms or molecules in one mole of a substance: Avogadro's number = 6.02214076 × 10^{23} Now, we can simply multiply Avogadro's number by the conversion factor from a gram-mole to a pound-mole: Number of atoms in a pound-mole = Avogadro's number × Conversion factor Number of atoms in a pound-mole = (6.02214076 × 10^{23} atoms/mole) × (453.59237 g/lb) Number of atoms in a pound-mole = 2.73159498 × 10^{26} atoms
06

Answer (b):

There are 2.73159498 × 10^{26} atoms in a pound-mole of a substance.

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Most popular questions from this chapter

The net potential energy \(E_{N}\) between two adjacent ions is sometimes represented by the expression $$E_{N}=-\frac{C}{r}+D \exp \left(-\frac{r}{\rho}\right)$$ in which \(r\) is the interionic separation and \(C\) \(D,\) and \(\rho\) are constants whose values depend on the specific material. (a) Derive an expression for the bonding energy \(E_{0}\) in terms of the equilibrium interionic separation \(r_{0}\) and the constants \(D\) and \(\rho\) using the following procedure: 1\. Differentiate \(E_{N}\) with respect to \(r\) and set the resulting expression equal to zero 2\. Solve for \(C\) in terms of \(D, \rho,\) and \(r_{0}\) 3\. Determine the expression for \(E_{0}\) by substitution for \(C\) in Equation 2.12 (b) Derive another expression for \(E_{0}\) in terms of \(r_{0}, C,\) and \(\rho\) using a procedure analogous to the one outlined in part (a).

Allowed values for the quantum numbers of electrons are as follows: \\[ \begin{aligned} n &=1,2,3, \ldots \\ l &=0,1,2,3, \ldots, n-1 \\ m_{l} &=0,\pm 1,\pm 2,\pm 3, \ldots, \pm l \\ m_{s} &=\pm \frac{1}{2} \end{aligned} \\] The relationships between \(n\) and the shell designations are noted in Table \(2.1 .\) Relative to the subshells, \(l=0\) corresponds to an \(s\) subshell \(l=1\) corresponds to a \(p\) subshell \(l=2\) corresponds to a \(d\) subshell \(l=3\) corresponds to an \(f\) subshell For the \(K\) shell, the four quantum numbers for each of the two electrons in the 1 s state in the order of \(n l m_{l} m_{s},\) are \(100\left(\frac{1}{2}\right)\) and \(100\left(-\frac{1}{2}\right) .\) Write the four quantum numbers for all of the electrons in the \(L\) and \(M\) shells, and note which correspond to the \(s, p,\) and \(d\) subshells.

Compute the percentage ionic character of the interatomic bond for each of the following compounds: \(\mathrm{MgO}, \mathrm{GaP}, \mathrm{CsF}, \mathrm{CdS},\) and \(\mathrm{FeO}\)

Silicon has three naturally-occurring isotopes: \(92.23 \%\) of \(^{28} \mathrm{Si}\), with an atomic weight of 27.9769 amu, \(4.68 \%\) of \(^{29} \mathrm{Si}\), with an atomic weight of 28.9765 amu, and \(3.09 \%\) of \(^{30} \mathrm{Si}\) with an atomic weight of 29.9738 amu. On the basis of these data, confirm that the average atomic weight of \(\mathrm{Si}\) is 28.0854 amu.

Potassium iodide (KI) exhibits predominantly ionic bonding. The \(\mathrm{K}^{+}\) and \(\mathrm{I}^{-}\) ions have electron structures that are identical to which two inert gases?
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