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Leona [35]
3 years ago
13

A compound is found to contain 7.523% phosphorus and 92.48% iodine by weight. what is the empirical formula for this compound?

Chemistry
2 answers:
BaLLatris [955]3 years ago
7 0

Answer:

PI₃

Explanation:

The empirical formula is found using the molar ratio between phosphorus and iodine.

The phosphorus composes 7.523% by weight of the substance, or 7.523g of phosphorus per gram of the substance. The moles of phosphorus per gram of substance is calculated using the atomic weight of phosphorus (30.97 g/mol):

(7.523g P) / (1 g substance) x (mol/30.97g P) = (0.2429 mol P) / (1 g substance)

Similarly, the moles of iodine (atomic weight 126.90 g/mol) per gram of substance is found:

(92.48 g I) / (1 g substance) x (mol/126.90 g I) = (0.7288 mol I) / (1 g substance)

The molar ratio between P and I is found by dividing the two values calculated above:

(0.7288 mol I) / (1 g substance) ÷ (0.2429 mol P) / (1 g substance) = (3.00 mol I) / (mol P)

Thus, the ratio of iodine to phosphorus is 3:1, so the empirical formula (the simplest formula) is PI₃

RUDIKE [14]3 years ago
5 0
The answer is PI3
Ratio of 1:3
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Step 2:

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8 0
3 years ago
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Answer:

The reactions free energy \Delta G = -49.36 kJ

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From the question we are told that

      The pressure of (NO) is P_{NO} = 9.20 \ atm

      The  pressure of  (Cl) gas is  P_{Cl} = 9.15 \ atm

       The  pressure of nitrosly chloride (NOCl) is P_{(NOCl)} = 7.70 \ atm

The reaction is

              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

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                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

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                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

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         \Delta G^o _{NOCl} =66.1kJ/mol

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        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

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         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

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        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

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                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

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                      \Delta G = -49.36 kJ

4 0
3 years ago
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