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Ivanshal [37]
3 years ago
10

A student visits the beach during a very hot day. She steps on the sand and jumps because it is so hot. What type of heat transf

er is demonstrated by the sand in this example?
Chemistry
2 answers:
Vlad [161]3 years ago
7 0

Answer: Conduction, because her feet are touching the hot sand.

sammy [17]3 years ago
3 0
Answer is: thermal conductuction.
Thermal conductuction is the transfer of heat<span> through physical contact. </span>Thermal conduction<span> is the transfer of heat</span> by microscopic collisions of particles. <span>Heat spontaneously flows from a hotter to a colder body.
</span>The process of heat conduction depends on four basic factors: the temperature gradient<span>, the cross section of the </span>materials<span> involved, their path length and the properties of those materials.</span>


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Calculate the radius of tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6 g/cm, and an atomic wei
Ivahew [28]

Answer:

The radius of tantalum (Ta) atom is R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm

Explanation:

From the Body-centered cubic (BBC) crystal structure we know that a unit cell length <em>a </em>and atomic radius <em>R </em>are related through

a=\frac{4R}{\sqrt{3} }

So the volume of the unit cell V_{c} is

V_{c}= a^3=(\frac{4R}{\sqrt{3} } )^3=\frac{64\sqrt{3}R^3}{9}

We can compute the theoretical density ρ through the following relationship

\rho=\frac{nA}{V_{c}N_{a}}

where

n = number of atoms associated with each unit cell

A = atomic weight

V_{c} = volume of the unit cell

N_{a} =  Avogadro’s number (6.023 \times 10^{23} atoms/mol)

From the information given:

A = 180.9 g/mol

ρ = 16.6 g/cm^3

Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.

We can use the theoretical density ρ to find the radio <em>R</em> as follows:

\rho=\frac{nA}{V_{c}N_{a}}\\\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}

Solving for <em>R</em>

\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}

Substitution for the various parameters into above equation yields

R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm

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Answer:

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