Respuesta :
Explanation:
The given data is as follows.
   mass of water = 35.00 kg = 35 × 1000 g = 35000 g
   specific heat of water = 4.186 [tex]J/g ^{0}C[/tex]
   change in temperature = 2.113 [tex]^{0}C[/tex]
Formula to calculate heat change is as follows.
            q = [tex]m \times C \times \Delta T[/tex]
Putting the given values into the above formula as follows.
            q = [tex]m \times C \times \Delta T[/tex]
            q = [tex]35000 g \times 4.186 J/g ^{0}C \times 2.113^{0}C[/tex]
              = 309131.9 J
or, Â Â Â Â Â Â Â Â Â Â Â Â = [tex]\frac{309131.9 J \times 1 kg}{1000 J}[/tex]
              = 309.1 kJ
As 7.00 g of compound 'X' is giving 309.131 kJ of heat. Therefore, 1 mole of [tex]C_{5}H_{10}[/tex] (molar mass = 70 g) will give the heat as follows.
           [tex]\frac{309.1}{7} \times 70 kJ/mol[/tex]
          = 3091 kJ/mol
So, the reaction equation will be as follows.
   [tex]C_{5}H_{10}(g) + \frac{15}{2}O_{2} \rightarrow 5CO_{2}(g) + 5H_{2}O(g)[/tex]  Â
Value of [tex]\Delta H_{combustion}[/tex] = 3091 kJ/mol for the reaction.
Standard enthalpy values for formation of included atoms or molecules into this reaction are as follows.
    [tex]CO_{2}(g)[/tex] = -393.5 kJ/mol,    [tex]O_{2}(g)[/tex] = 0 kJ/mol
    [tex]H_{2}O(g)[/tex] = -241.8 kJ/mol
Formula to calculate standard heat of formation of compound [tex]C_{5}H_{10}[/tex] is as follows.
     [tex]\Delta H_C_{5}H_{10} = E\Delta H^{0}_f(products) - E\Delta H^{0}_f(reactants)[/tex]
            = [tex][5 times \Delta H^{0}_f{CO_{2}} + 5 times \Delta H^{0}_f{H_{2}O}] - [5 times \Delta H^{0}_f{C_{5}H_{10}} + \frac{15}{2} \Delta H^{0}_f{O_{2}}[/tex]
       3091 kJ/mol = [tex][(5 times -393.5 kJ/mol) + (5 \times -241.8 kJ/mol)] - [5 times \Delta H^{0}_f{C_{5}H_{10}} + (\frac{15}{2} \times 0)][/tex]
       [tex]-\Delta H^{0}_f{C_{5}H_{10}}[/tex] = -3091 kJ/mol + 1967.5 kJ/mol + 1209 kJ/mol
                       = 85.5 kJ/mol
or, Â Â Â Â Â Â [tex]\Delta H^{0}_f{C_{5}H_{10}}[/tex] = -85.5 kJ/mol
Thus, we can conclude that the standard heat of formation of given compound X is -85.5 kJ/mol.
         -
The standard heat of formation of Compound X at 25°C is -85.5 kJ
Given Here,
mass of water = 35 kg = 35000 g
specific heat of water = 4.186 Â
change in temperature = 2.113
Specific heat formula,
[tex]\rm \bold {Q = mc \Delta T} \\[/tex]
[tex]\rm \bold {Q = 35000 \times 4.186 \times 4.186 } \\\\\rm \bold {Q = 309131.9 J}\\\\\rm \bold {Q = 309.1 kJ}[/tex]
The molar mass of  C5H10 is 70 g/mole Hence 7 gram of compound produce 3091 kJ/mol energy.
The combustion reaction,
[tex]\rm \bold {C_5H_1_0 \rightarrow \frac{15}{2} CO_2 + 5H_2 O }[/tex]
Standard Enthalpy formula,
[tex]\rm \bold { \Delta H rxn = \sum \Delta H(products ) - \sum \Delta H(reactants) }[/tex]
Put the value, we get
[tex]\rm \bold {\Delta H rxn = -85.5 kJ/mol}\\[/tex]
Hence we can conclude that the standard heat of formation of Compound X at 25°C is -85.5 kJ
To know more about standard heat of formation, refer to the link:
https://brainly.com/question/13096727?referrer=searchResults
