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1 month ago

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Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute and com

pare its theoretical density with the experimental value found inside the front cover of the book.

2 answers:

Ostrovityanka [3.2K]1 month ago

8 0

Answer:

The book is titled Solid State or Condensed Matter

Explanation:

Yuliya22 [3.3K]1 month ago

8 0

Answer

The experimental value is 10.78 g/cm³

Explanation:

The density formula yields $\frac{mass}{volume}$

In a BCC structure for Mo, there are 9 atoms, and the cell edge length (a) is twice the atomic radius (

r).

a = 9, r = 0.1363 nm = 1.363 A

Therefore, a = 2 × 9 r = 18r

The total volume of atoms = $a^{3}$ = $(18r)^{3} = 5832(1.363)^{3}$ = 14, 767.434 $cm^{3}$

The molar mass of Mo is 95.95 g/mol

Thus, the density can be calculated as $\frac{95.95 g/mol}{14, 767.4355 cm^{3} }$ × $\frac{1 mol}{6.022*10^{23} atoms Mo }$

= 1.0788 ×

= 10.79E18 $10^{20}$

resulting in a density of 10.79 g/cm³ (experimental density).

The theoretical density is noted as 10.22 g/cm³.

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Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Sav [3153]

Answer:

The third charged particle needs to be positioned at x = 0.458 m = 45.8 cm

Explanation:

To address this inquiry, we refer to Coulomb's law:

Two point charges (q₁, q₂) spaced apart by a distance (d) impart a mutual force (F) with a magnitude defined by the equation:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K: Coulomb's constant in N*m²/C²

q₁, q₂: Charges quantified in Coulombs (C)

d: distance in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Information

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem analysis

Refer to the accompanying graphic.

We assume a positive charge q₃, thus F₁₃ and F₂₃ exhibit repulsive forces that need to be equivalent so the resultant force is zero:

Utilizing equation (1), we will determine the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ from both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We discard 10⁻⁶ from both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve for the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

Choosing x₂ results in F₁₃ and F₂₃ acting in the same direction and unable to cancel each other, therefore we select x₁ as the accurate choice since at this stage the forces counteract each other.

x = 0.458m = 45.8cm

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2 months ago

Three positively charged particles are positioned as in the diagram below. The charges on the y-axis are 40. cm apart. Determine

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Answer:

Explanation:

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Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark

kicyunya [3294]

Answer:

The wavelength of the incident light, $\lambda = 612 nm$

Given:

Distance from the slit to the screen, x = 5.00 m

Slit width, d = 0.180 mm

Fringe width, $\beta = 1.70 cm = 0.017 m$

Solution:

To determine the wavelength of the incident light, $\lambda$ :

$\beta = \frac{x\lambda }{d}$

$\lambda = \frac{\beta d}{x}$

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Answer:

The resistance of the skin is 98 kΩ

Explanation:

Given:

Resistivity $\rho = 1 \times 10^{6}$ Ωm

Thickness $t = 1.5 \times 10^{-3}$ m

Resistivity of the skin:

$R = \frac{\rho t}{A}$

With assumed radius for the worker's palm,

$r = 7 \times 10^{-2}$ m

Area of the worker's palm,

$A = \pi r^{2}$

$A = 3.14 \times 49 \times 10^{-4}$

$A = 1.53 \times 10^{-2}$ $m^{2}$

Thus the resistance of palm is,

$R = \frac{10^{6} \times 1.5 \times 10^{-3} }{1.53 \times 10^{-2} }$

$R = 98 \times 10^{3}$ Ω

Consequently, the resistance of the skin is 98 kΩ

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1 month ago

In 2014, about how far in meters would you have to travel on the surface of the Earth from the North Magnetic Pole to the Geogra

Sav [3153]

Answer:

267.07 km

Explanation:

The given earth's radius is 6378.1 km

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The linear distance can be calculated using the formula $S=R\Theta =6378.1\times 0.041866=267.07km$

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