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3 years ago

14

How many formula units are there in 103 g of NaCl?

1 answer:

charle [14.2K]3 years ago

3 0

Answer:

1.059 x 10^24 formula units

Explanation:

103/58.44 = 1.76 moles NaCl

1.76 x 6.022 x 10^23 = 1.059 x 10^24

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You are determining the density of an unknown solid. You figure out the volume by water displacement. Then you weigh the solid,

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Answer:

The calculated density will be larger

Explanation:

The calculated density will be <u>larger</u>. Because, the volume is taken accurately, by the water displacement method. But, when we the took the mass, the water was present on the unknown solid. So, the mass of that water was added to the original mass of the solid. Hence, the mass measured was larger than the original mass. We, know from the formula of density that density is directly proportional to the mass of the object.

Density = Mass/Volume

Hence, the larger measured mass means the larger value of density.

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Which of the following is a chemical property density mass volume volatility

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Explanation:

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The energy released by a nuclear fusion reaction is produced when

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Answer:

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3 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

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3 years ago