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love history [14]
2 years ago
11

One crystalline form of silica (SiO2) has a cubic unit cell, and from X-ray diffraction data it is known that the cell edge leng

th is 0.700 nm. If the measured density is 2.32 g/cm3, how many (a) Si4+ and (b) O2- ions are there per unit cell? The atomic weights of Si and O are, respectively, 28.09 g/mol and 16.00 g/mol.
Chemistry
1 answer:
Sav [38]2 years ago
5 0

Answer:

8 Silicon atom are present in unit cell.

16 oxygen atoms are present unit cell.

Explanation:

Number of atoms in unit cell = Z =?

Density of silica = tex]2.32 g/cm^3[/tex]

Edge length of cubic unit cell = a  = 0.700 nm = 0.700\times 10^{-7} cm

1 nm=10^{-7} cm

Molar mass of Silica  = 28.09 g/mol+16.00\times 2=60.09 g/mol

Formula used :  

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

where,

\rho = density

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

2.32 g/cm3=\frac{Z\times 60.09 g/mol}{6.022\times 10^{23} mol^{-1}\times (0.700\times 10^{-7}cm)^{3}}

Z = 8

1 silicon is 2 oxygen atoms. then 8 silicon atoms will be 16 oxygen atoms.

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2 years ago
Read 2 more answers
Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b
frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}?

(a) Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}

(b) 2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}

(c)Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}

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Answer: (d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

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Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}

Aluminum's half-equation is oxidation:

Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}

For Lead, half-equation is reduction:

Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}

Multiply first half-equation for 2 and second half-equation by 3:

2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}

3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}

Adding them:

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

The balanced redox reaction with cell notation Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)} is

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

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By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).

6 0
2 years ago
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