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BaLLatris [955]
3 years ago
8

Determine the molar solubility for pb3(po4)2 in pure water. ksp for pb3(po4)2 is 1.0 x 10-54.

Chemistry
1 answer:
hoa [83]3 years ago
7 0
Dissociation of Pb₃(PO₄)₂ is;
                       Pb₃(PO₄)₂(s)     ⇆      3Pb²⁺(aq)     +     2PO₄³⁻(aq)
initial                                                    -                          -
change               -X                            +3X                     +2X
Equilibrium                                          3X                        2X

Ksp            =       [Pb²⁺(aq)]³ [PO₄³⁻(aq)]²
1.0 x 10⁻⁵⁴ =         (3X)³ (2X)²
1.0 x 10⁻⁵⁴ =        108X⁵
         X      =        6.21 x 10⁻¹² M

Hence the molar solubility of Pb₃(PO₄)₂ is 6.21 x 10⁻¹² M.
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Which compound has the same empirical and molecular formula ethyne ethene ethane methane?
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Empirical formula is the simplest ratio of whole numbers of components in a compound 
molecular formula is the actual ratio of components in a compound .
the molecular formula for the compounds given are as follows
ethyne  - C₂H₂
ethene - C₂H₄
ethane - C₂H₆
methane - CH₄
the actual ratios of the elements                 simplified ratio
                     C : H                                           C : H
ethyne            2:2                                             1:1
ethene            2:4                                             1:2
ethane            2:6                                             1:3
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6 0
3 years ago
All the substances listed below are fertilizers that contribute nitrogen to the soil.
kvv77 [185]

Answer:

Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.

Explanation:

Percentage of element in compound :

=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100

(a) Urea, (NH_2)_2CO

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Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 2

N\%=\frac{2\times 14 g/mol}{60 g/mol}\times 100=46.67\%

(b) Ammonium nitrate, NH_4NO_3

Molar mass of ammonium nitrate = 80 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 2

N\%=\frac{2\times 14 g/mol}{80 g/mol}\times 100=35.00\%

(c) Nitric oxide, NO

Molar mass of nitric oxide  = 30 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 1

N\%=\frac{1\times 14 g/mol}{30 g/mol}\times 100=46.67\%

(d) Ammonia, NH_3

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Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 1

N\%=\frac{1\times 14 g/mol}{17 g/mol}\times 100=82.35\%​

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