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3 years ago

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Chemistry

1 answer:

stepan [7]3 years ago

5 0

Answer: atom

Explanation:

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vlabodo [156]

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6 0

3 years ago

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution for concentrations up to appro

Hatshy [7]

Answer:

The unit cell edge length for the alloy is 0.288 nm

Explanation:

Given;

concentration of vanadium, Cv = 8 wt%

concentration of iron, Cfe = 92 wt%

density of vanadium = 6.11 g/cm³

density of iron = 7.86 g/cm³

atomic weight of vanadium, Av = 50.94 g/mol

atomic weight of iron, Afe= 55.85 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_v}{\rho _v} + \frac{C__{Fe}}{\rho _{Fe}} }$

$\rho _{Avg.} = \frac{100}{\frac{8}{6.11} + \frac{92}{7.86} } = 7.68 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _v} + \frac{C__{Fe}}{A _{Fe}} }$

$A _{Avg.} = \frac{100}{\frac{8}{50.94} + \frac{92}{55.85} } = 55.42 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a BCC crystal structure, there are 2 atoms per unit cell; n = 2

$V_c=\frac{2*55.42}{ 7.68*6.022*10^{23}} = 2.397*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (2.397*10^{-23}}){^\frac{1}{3}}\\\\a = 2.88 *10^{-8} \ cm$ = 0.288 nm

Therefore, the unit cell edge length for the alloy is 0.288 nm

4 0

3 years ago

What term describes matter that is a substance made of different kinds of atoms bonded together?

Afina-wow [57]

Compounds or molecules

5 0

3 years ago

Read 2 more answers

How many moles are in 0.0723 atoms of Iron. Fe?-

Vlada [557]

Answer:

0.012× 10⁻²³ moles of iron

Explanation:

Given data:

Number of moles of iron = ?

Number of iron atoms = 0.0723 atom

Solution:

Avogadro number:

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

0.0723 atom × 1 mol / 6.022 × 10²³ atoms

0.012× 10⁻²³ mol

8 0

3 years ago

at what height is an object that has a mass of 16kg, if it’s gravitational potential energy is 7500J?

dolphi86 [110]

Answer:

46.875

Explanation:

P.E=MGH

7500=16*10*h

7500=160h

h=7500/160

h=46.875

5 0

3 years ago