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Vesna [10]
3 years ago
13

A solution contains 0.0150 m pb2 (aq) and 0.0150 m sr2 (aq). if we add so42–(aq), what will be the concentration of pb2 (aq) whe

n srso4(s) begins to precipitate
Chemistry
1 answer:
polet [3.4K]3 years ago
3 0
Ksp of SrSO4 = 3.2 x 10-7, so it will precipitate when Ksp > 3.2 x 10-7  
Same way, calculating for 0.0150 M Sr2 SrSO4 precipitates,
 when [ SO4 2-] > (3.2 x 10-7 / 0.0150) = 2.133 x 10^-05.
 Thus SrSO4 precipitates when SO4 2- > 2.133 x 10^-05 
 Now the Pb2+ at this point = 2.133 x 10^-05 x 0.0150 =3.2 x 10^-07 
 Hence Ksp for PbSO4 = 1.6 x 10-8 
 the solubility product for PbSO4 is more and the PbSO4 precipitates as
follows [Pb2+] = 1.6 x 10-8 / 2.133 x 10^-05 = 7.50 x 10^-04

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

A=6R^{2}\sqrt{3}

Here, R is radius and is related to a as follows:

R=\frac{a}{2}

Putting the value in expression for area,

A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

The value of a is 0.4961 nm

Since, 1 nm=10^{-7}cm

Thus, 0.4961 nm=4.961\times 10^{-8} cm

Putting the value,

Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}

Now, volume can be calculated as follows:

V=Area\times c

The value of c is 1.360 nm or 1.360\times 10^{-7} cm

Putting the value,

V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}

now, number of atom in unit cell can be calculated by using the following formula:

n=\frac{\rho N_{A}V_{c}}{A}

Here, A is atomic mass of Cr_{2}O_{3} is 151.99 g/mol.

Putting all the values,

n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18

Thus, there will be 18 Cr_{2}O_{3} units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of Cr^{3+} and O^{2-} is 62 pm and 140 pm respectively.

Converting them into cm:

1 pm=10^{-10}cm

Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

Putting the values,

V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758

Thus, atomic packing factor is 0.758.

6 0
3 years ago
Read 2 more answers
What volume of water would be added to 16.5 ml of a 0.0813 m solution of sodium borate in order to get a 0.0200 m s?
raketka [301]
When diluting solutions from concentrated solutions the following formula can be used 
c1v1 = c2v2 
where c1 is concentration and v1 is volume of the concentrated solution 
and c2 is concentration and v2 is volume of the diluted solution to be prepared
substituting these values 
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V = 67.1 mL
the volume of the diluted solution prepared is 67.1 mL.
the volume of water that should be added to get a final volume of 67.1 mL is (67.1 - 16.5 ) = 50.6 mL
a volume fo 50.6 mL should be added 
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Draw an ecosystem with 10 biotic and 5 abiotic factor
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I won’t draw it but I can give you 10 Biotic and 5 abiotic
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gogolik [260]

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Explain why we use 1/12 in finding atomic mass unit ?​
Nat2105 [25]

Answer:

The u (amu is the old unit name) is 1/12 of the weight of an 12C atom. The way the u is chosen ensures that all core and atom masses are multiples of 1(±0.1) u.

Explanation:

Further explanation if needed...

Carbon 12 was chosen because the chemical atomic weights based on C12 are almost identical to the chemical atomic weights based on the natural mix of oxygen. Simply because the atomic mass is defined as 1/12 of the mass of 12C. Others isotopes of carbon (13C mostly, with an abundance of 1.1% approximately) account for an average atomic mass slightly above 12.

7 0
3 years ago
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