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dedylja [7]
3 years ago
9

Question 16

Chemistry
2 answers:
Liono4ka [1.6K]3 years ago
4 0

Answer:

Br

Explanation:

Bromine atoms tend to gain just one electron to get to a full octet, as Bromine is in Group VII.

Thepotemich [5.8K]3 years ago
3 0
Br i’m pretty sure! :)
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A sample of an Iron Oxalato complex salt weighting 0.145 grams requires 34.88 mL of 0.015 M KMnO4 to turn the solution a very li
Amanda [17]

Answer:

5.2 ×10^-4 moles

Explanation:

Equation of reaction:

5C2O4^2- + 2MnO4^- +. 6H^+. ----------> 10CO2. +. 8H2O. + 2Mn^2+

From the information provided in the question:

Volume of potassium permanganate= 34.88ml

Concentration of potassium permanganate= 0.015M

Amount potassium permanganate= 0.015 × 34.88/1000= 5.2 ×10^-4 moles

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Does Ag metal reacts with NaOH​
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How many atoms are in 68.44 grams
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A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch
ivanzaharov [21]

Answer:

The reactions free energy \Delta G = -49.36 kJ

Explanation:

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)}

 From the reaction we can  mathematically evaluate the \Delta G^o (Standard state  free energy ) as

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

The Standard state  free energy for NO is  constant with a value  

                 \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

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

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

Now substituting this into the equation

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

                = -43 kJ/mol

The pressure constant is evaluated as

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

Substituting  values  

        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

Where R is gas constant with a value  of  R = 8.314 J / K \cdot mol

          T is temperature in K  with a given value of  T = 25+273 = 298 K

   Substituting value

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

                       = -43-6.36

                      \Delta G = -49.36 kJ

4 0
3 years ago
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