Answer:
![\boxed {\boxed {\sf About \ 2.999 \ moles \ of \ bromine}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%20About%20%5C%202.999%20%5C%20moles%20%5C%20of%20%5C%20bromine%7D%7D)
Explanation:
To convert from moles to molecules, we must use <u>Avogadro's Number</u>.
![6.022*10^{23}](https://tex.z-dn.net/?f=6.022%2A10%5E%7B23%7D)
This tells us the number of particles (atoms, molecules, ions, etc) in 1 mole. In this problem, the particles are molecules of bromine in 1 mole of bromine.
![6.022*10^{23} \ molecules \ Br / 1 \ mol \ Br](https://tex.z-dn.net/?f=6.022%2A10%5E%7B23%7D%20%5C%20molecules%20%5C%20Br%20%2F%201%20%5C%20mol%20%5C%20Br)
1. Convert from moles to molecules
Use Avogadro's number as a fraction or ratio.
![\frac{6.022 *10^{23} \ molecules \ Br}{1 \ mol \ Br}](https://tex.z-dn.net/?f=%5Cfrac%7B6.022%20%2A10%5E%7B23%7D%20%5C%20molecules%20%5C%20Br%7D%7B1%20%5C%20mol%20%5C%20Br%7D)
Multiply this fraction by the given number of bromine molecules.
![1.806 *10^{24} molecules \ Br *\frac{6.022 *10^{23} \ molecules \ Br}{1 \ mol \ Br}](https://tex.z-dn.net/?f=1.806%20%2A10%5E%7B24%7D%20molecules%20%5C%20Br%20%2A%5Cfrac%7B6.022%20%2A10%5E%7B23%7D%20%5C%20molecules%20%5C%20Br%7D%7B1%20%5C%20mol%20%5C%20Br%7D)
Flip the fraction so the molecules of bromine can cancel out.
![1.806 *10^{24} \ molecules \ Br* \frac{1 \ mol \ Br}{6.022 *10^{23} \ molecules \ Br}](https://tex.z-dn.net/?f=1.806%20%2A10%5E%7B24%7D%20%5C%20molecules%20%5C%20Br%2A%20%5Cfrac%7B1%20%5C%20mol%20%5C%20Br%7D%7B6.022%20%2A10%5E%7B23%7D%20%5C%20molecules%20%5C%20Br%7D)
![1.806 *10^{24} * \frac{1 \ mol \ Br}{6.022 *10^{23}}](https://tex.z-dn.net/?f=1.806%20%2A10%5E%7B24%7D%20%2A%20%5Cfrac%7B1%20%5C%20mol%20%5C%20Br%7D%7B6.022%20%2A10%5E%7B23%7D%7D)
Multiply and condense the expression into 1 fraction.
![\frac{1.806 *10^{24} \ mol \ Br}{6.022 *10^{23} }](https://tex.z-dn.net/?f=%5Cfrac%7B1.806%20%2A10%5E%7B24%7D%20%5C%20mol%20%5C%20Br%7D%7B6.022%20%2A10%5E%7B23%7D%20%7D)
Divide.
![2.999003653 \ mol \ Br](https://tex.z-dn.net/?f=2.999003653%20%5C%20mol%20%5C%20Br)
2. Round
The original measurement, 1.806*10^24 has 4 significant figures (1, 8, 0, and 6). We must round our answer to 4 sig figs. For the answer we found, that is the thousandth place.
![2.999003653 \ mol \ Br](https://tex.z-dn.net/?f=2.999003653%20%5C%20mol%20%5C%20Br)
The 0 in the ten-thousandth place tels us to leave the 9 in the thousandth place.
![\approx 2.999 \ mol \ Br](https://tex.z-dn.net/?f=%5Capprox%202.999%20%5C%20mol%20%5C%20Br)
There are about <u>2.999 moles of bromine</u> in 2.806 *10^23 molecules.