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miv72 [106K]
3 years ago
9

A small portion of a crystal lattice is sketched below. What is the name of the unit cell of this lattice? Your answer must be a

word, a very short phrase, or a standard abbreviation. Spelling counts!

Chemistry
1 answer:
Nikolay [14]3 years ago
7 0

This is an incomplete question, the given sketch is shown below.

Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)

Explanation :

Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.

There are three types of unit cell.

  • SCC (simple-centered cubic unit cell)
  • BCC (body-centered cubic unit cell)
  • FCC (face-centered cubic unit cell)

In SCC, the atoms are arranged at the corners.

Z=\frac{1}{8}\times 8=1

The number of atoms of unit cell = Z = 1

In BCC, the atoms are arranged at the corners and the body center.

Z=\frac{1}{8}\times 8+1=2

The number of atoms of unit cell = Z = 2

The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.

Z=\frac{1}{8}\times 8+\frac{1}{2}\times 6=4

The number of atoms of unit cell = Z = 4

Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)

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Consider the reaction between reactants s and o2: 2s(s)+3o2(g)→2so3(g)if a reaction vessel initially contains 7 mols and 9 mol o
nadya68 [22]

Answer: -

1 mol

Explanation: -

Number of moles of Sulphur S = 7

Number of moles of O2 = 9

The balanced chemical equation for the reaction is

2S (s)+3 O2 (g)→2SO3(g)

From the above reaction we can see that

3 mol of O2 react with 2 mol of S

9 mol of O2 will react with \frac{2 mol S x 9 mol O2}{3 mol O2}

= 6 mol of S

Unreacted S = 7 - = 1 mol.

If a reaction vessel initially contains 7 mol S and 9 mol O2

1 mole of s will be in the reaction vessel once the reactants have reacted as much as possible

6 0
3 years ago
Comparing mitosis and meiosis can you help anyone
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5 0
3 years ago
Carbon naturally occurs in two forms: diamond and graphite. Why do these two forms have very different properties?
Nookie1986 [14]
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5 0
3 years ago
A scientist measures the standard enthalpy change for the following reaction to be 67.9 kJ:
bekas [8.4K]

Answer: The standard enthalpy of formation  of H_2O(g) is  -252.1 kJ/mol.

Explanation:

The balanced chemical reaction is,

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]

Putting the values we get :

\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]

67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)]

H_f{H_2O}=-252.1kJ/mol

Thus standard enthalpy of formation  of H_2O(g) is  -252.1 kJ/mol.

3 0
3 years ago
Carbon monoxide (CO) reacts with hydrogen (H2) to form methane (CH4) and water (H2O).
Artemon [7]

Answer:

5.9x10^-2 M

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Concentration of CO, [CO] = 0.30 M

Concentration of H2, [H2] = 0.10 M

Concentration of H2O, [H2O] = 0.020 M

Equilibrium constant, K = 3.90

Concentration of CH4, [CH4] =..?

Step 2:

The balanced equation for the reaction. This is given below:

CO(g) + 3H2(g) <=> CH4(g) + H2O(g)

Step 3:

Determination of the concentration of CH4.

The expression for equilibrium constant of the above equation is given below:

K = [CH4] [H2O] / [CO] [H2]^3

3.9 = [CH4] x 0.02/ 0.3 x (0.1)^3

Cross multiply to express in linear form

[CH4] x 0.02= 3.9 x 0.3 x (0.1)^3

Divide both side by 0.02

[CH4] = 3.9 x 0.3 x (0.1)^3 /0.02

[CH4] = 5.9x10^-2 M

Therefore, the equilibrium concentration of CH4 is 5.9x10^-2 M

5 0
3 years ago
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