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Natalija [7]
3 years ago
11

Consider the reaction: M + 2HCl → MCl2 + H2 When 0.85 mol of the metal, M, reacted with an aqueous HCl solution (the HCl is in e

xcess), the temperature of the solution rose because the reaction produced 4035 J of heat. What is ∆H in kJ per mol of M for this reaction? (Hint: Is this reaction exothermic or endothermic?) It is possible that your answer could be either positive or negative. If it is negative, you must include the "minus" sign. Enter your answer as a decimal number.
Chemistry
1 answer:
Sholpan [36]3 years ago
6 0

Answer : The enthalpy change of the reaction is -4.747 kJ/mol and this reaction is exothermic.

Explanation :

First we have to calculate the enthalpy change of the reaction by using the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = -4035 J

n = number of moles of metal M = 0.85 moles

\Delta H_{rxn} = enthalpy change of the reaction

Now put all the given values in above equation, we get:

\Delta H_{rxn}=\frac{-4035J}{0.85mol}=-4747.0J/mol=-4.747kJ/mol

Conversion used:   (1 kJ = 1000 J)

Sign convention of heat:

When heat is absorbed then the sign of heat is taken to be positive and is written on the reactant side and is considered as an endothermic reaction.

When heat is released then the sign of heat is taken to be negative and is written on the product side and is considered as an exothermic reaction.

Thus, the chemical reaction will be:

M+2HCl\rightarrow MCl_2+H_2+4.477kJ

Hence, the enthalpy change of the reaction is -4.747 kJ/mol and this reaction is exothermic.

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The wind energy and solar energy have their own benefits when producing electricity but one of their major drawbacks is that both these energy form depends on the environmental conditions and cannot be stored.

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3 years ago
How many atoms are in 5.00g of carbon
melisa1 [442]

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2.51 x 1023 atoms C

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MW C = 12.011

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A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l
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Explanation:

From first source, kinetic energy (K.E_{1}) ejected is 1 eV and wavelength of light is \lambda.

From second source, kinetic energy (K.E_{2}) ejected is 4 eV and wavelength of light is \frac{\lambda}{2}.

Relation between work function, wavelength, and kinetic energy is as follows.

                   K.E = \frac{hc}{\lambda} - \phi

where,        h = Plank's constant = 6.63 \times 10^{-34} J.s

                   c = speed of light = 3 \times 10^{8} m/s

Also, it is known that 1 eV = 1.6 \times 10^{-19} J

Therefore, substituting the values in the above formula as follows.

  • From first source,

                      K.E_{1} = \frac{hc}{\lambda} - \phi  

            1 eV = 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi    

    1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi        ........... (1)

  • From second source,

                  K.E_{2} = \frac{hc}{\lambda} - \phi  

          4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi        

                 6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi        ........... (2)        

Now, divide equation (2) by 2. Therefore, it will become

       {6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}

                3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}   ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

                 1.6 \times 10^{-19} = \frac{\phi}{2}

                    \phi = 3.2 \times 10^{-19}

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Thus, we can conclude that work function of the metal is 2 eV.

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