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Elina [12.6K]
2 years ago
10

Can someone answer 1-10 for me or the one you know I’ll give you brainlest!!!

Chemistry
1 answer:
Reika [66]2 years ago
5 0
1. Coefficient
2. Subscript
3. Products
4. Reactants
5. Balance
6. Combustion
7. Decomposition
8. Single replacement
9. Double replacement
10. Synthesis
I’m sorry if I’m wrong
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A technician needs 500.0mL of a 0.500 M MgCl2 solution for some lab procedures. The stock bottle is labeled 4.00 M MgCl2. What v
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Answer:

Explanation:

To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;

    Number of moles  = molarity x volume

Parameters given;

        volume of solution = 500mL  = 0.5L

          molarity of solution = 0.5M

   Number of moles  =  0.5 x 0.5  = 0.25moles

Now to know the volume stock to take;

           Volume of stock = \frac{number of moles }{molarity}

  molarity of stock  = 4M

                 volume  = \frac{0.25}{4}   = 0.0625L or 62.5mL

4 0
3 years ago
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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