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2 years ago

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A particular graduated cylinder contains 24.0 mL Br2(liquid). The density of bromine at 25 degrees C is 3.12g/cm^3. Note that in

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Answer:

$2.82x10^{23}molecules$

Explanation:

Hello.

In this case, given the volume (1cm³=1mL) and density of the bromine we are to firstly compute the mass since it will allow us to compute the representative particles:

$\rho =\frac{m}{V}$

$m=\rho *V=3.12g/cm^3*24.0cm^3\\\\m=74.88gBr_2$

Next, since the mass of one mole of diatomic bromine is 159.82 g (one bromine weights 78.91), we can next compute the moles in that sample:

$n=74.88g*\frac{1mol}{159.82g} =0.469molBr_2$

Finally, via the Avogadro's number we can compute the representative particles of bromine as follows:

$particles=0.469mol*\frac{6.022x10^{23}molecules}{1mol}\\ \\2.82x10^{23}molecules$

Best regards.

4 0

3 years ago