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Ksenya-84 [330]
3 years ago
13

How can you tell where there is a steep slope on a contour map?

Chemistry
1 answer:
Katena32 [7]3 years ago
8 0

<u>Answer:</u>

Steep slopes on the contour map are identified using contour lines which are closely spaced.

<u>Explanation:</u>

Contour map also known as topographic maps are used to represent the three-dimensional portion of the earth’s surface in two-dimensional space. This map is used to represent the surface of the land such as steep, slopes, valleys. The contour maps are used in geological studies to understand the configuration of the earth's surface. Terms like map scale, vertical scale, contour lines are used on the contour map. Elevations are represented using contour lines. Contour lines placed very close to each represent the steep slopes and contour lines that are spaced farther away from each other represent the gentle slopes.

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Balance this equation, please.
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3 years ago
Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

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

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

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