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Bess [88]
3 years ago
5

Atoms are spherical in shape. Therefore, the Pt atoms in the cube cannot fill all the available space. If only 74.0 percent of t

he space inside the cube is taken up by Pt atoms, calculate the radius in picometers of a Pt atom. The mass of a single Pt atom is 3.240 × 10−22 g. [The volume of a sphere of radius r is (4/3) πr3. The volume of a cube is l3, where l is the length of a side. Avogadro's number is 6.022 × 1023.]
Physics
1 answer:
quester [9]3 years ago
5 0

Answer:

A)6.6×10^22atoms of Pt in the cube

B)1.4×10^-8m

Explanation:

(a) Calculate the number of Pt atoms in the cube.

an edge length of platinum (Pt) = 1.0 cm.

Then Volume= 1.0 cm×1.0 cm×1.0 cm=1cm^3

Then we have volume of the cube as 1cm^3

Given:

The density Pt = 21.45 g/cm3

the mass of a single Pt atom =3.240 x 10^-22 g

Then with 1atom of the platinum element, we can calculate the number of Pt atoms in the cube as

Density of pt/mass of a single Pt atom

=(21.45 /=3.240 x 10^-22)

=6.6×10^22atoms of Pt in the cube

B)Volume of cube V=4/3πr^3

V= 4/3 ×π×r^3

V= 4.19067r^3

r^3= V/4.19067

But volume is not total volume but just 74% of it, then With 74% of the space inside the cube is taken up by Pt atoms, then we need to find 74% of volume of the cube which is 1cm^3

74/100 ×1= 0.74cm^3

Then our new volume V is 0.74cm^3

r^3=0.74/4.19067×6.620 x 10^22

r^3=2.6674×10^-24

r= 3√2.6674×10^-24

r=1.4×10^-8m

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And the distance between the two ions is

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3 years ago
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3 years ago
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Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

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Known data

q₁ = 15 μC = 15*10⁻⁶ C

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F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

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The solution of the quadratic equation is:

x₁ = 0.775m

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x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

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3 years ago
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Answer:

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