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n200080 [17]
3 years ago
13

What is the displacement of the particle in the time interval 7 seconds to 8 seconds?

Chemistry
1 answer:
3241004551 [841]3 years ago
8 0

Answer:

28 is great.......... it should be snSwer

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What is the volume of a 28g object that has a physical property of .45g / m * L ?
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2 years ago
The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that
Sladkaya [172]

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Density of the He atom = 12.69 g/cm³

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From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = 1  \ atom \times  (\dfrac{1  \ mole}{  6.022 \times  10^{23}  \ atoms}) \times ( \dfrac{4.003 \ grams}{  1  \ mole})

=6.647 \times  10^{-24} \  grams

The volume can be determined as  folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = \dfrac{0.10}{2} = 0.05 nm

Converting it into cm, we have:

0.05 nm \times  \dfrac{10^{-9} \  meters}{ 1  nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}

=5 \times  10^{-9}  \ cm

Assuming that it is a sphere, the volume of a sphere is

= \dfrac{4}{3}\pi r^3

= \dfrac{4}{3}\pi  \times (5\times 10^{-9})^3

= 5.236 \times 10^{-25} \ cm^3

Finally, the density can be calcuated by using the formula :

Density = \dfrac{mass}{volume}

D =  \dfrac{6.647 \times  10^{-24} \  grams }{ 5.236 \times 10^{-25} \  cm^3}

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

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3 years ago
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