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tamaranim1 [39]
3 years ago
10

What is the density of an object that has a mass of 62.4grams and a volume of7.6cm^3?

Chemistry
1 answer:
Monica [59]3 years ago
5 0

Answer:

<h3>The answer is 8.2 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 62.4 g

volume = 7.6 cm³

We have

density =  \frac{62.4}{7.6}  \\  = 8.2105263157...

We have the final answer as

<h3>8.2 g/cm³</h3>

Hope this helps you

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Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu
Deffense [45]

Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

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r = \dfrac{a}{2\sqrt{2}}

where;

a = edge length.

Making "a" the subject of the formula:

a = 2 \sqrt{2} \times r

a = 2 \times 1.414 \times 135 \ pm

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

mass\  of\ atom =\dfrac{ atomic \ mass}{Avogadro  \  number}

mass\  of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4

mass of unit cell = 6.83380375 × 10⁻²² g

Density  \ of  \ unit \  cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}

Density of unit cell ( rhodium) = 12.279 g/cm³

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2 years ago
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