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ioda
2 years ago
13

Help pleaseeeeeeeeeee

Chemistry
1 answer:
hichkok12 [17]2 years ago
6 0

Answer:

D 1 and 3 only I am not sure

Explanation:

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2 years ago
The gas cyclobutane, C4H8(g), can be used in welding. When cyclobutane is burned in oxygen, the reaction is: C4H8(g) + 6 O2(g)4
Snowcat [4.5K]

Answer:

a

\Delta H^o _{rxn} = -2568.9 \  kJ

b

H  = 350 JK^{-1}

c

T_{max}  = 32.4 ^o C

Explanation:

From the question we are told that

 The reaction of cyclobutane and oxygen is

         C_4H_8_{(g)} + 6 O_2_{(g)} \to 4 CO_2_{(g)} + 4 H_2O_{(g)}

ΔH°f (kJ mol-1) : C4H8(g) = 27.7 ; CO2(g) = -393.5 ; H2O(g) = -241.8 ΔH° = kJ

Generally ΔH° for this reaction is mathematically represented as

      \Delta H^o _{rxn} = [[4 * \Delta H^o_f (CO_2_{(g)} ) + 4 * \Delta H^o_f(H_2O_{(g)} ] -[\Delta H^o_f (C_2H_6_{(g)} + 6 * \Delta H^o_f (O_2_{(g)}) ] ]

=>  \Delta H^o _{rxn} = [[4 * (-393.5) + 4 * (-241.8) ] -[ 27.7 + 6 * 0]

=>  \Delta H^o _{rxn} = -2568.9 \  kJ

Generally the total heat capacity of 4 mol of CO2(g) and 4 mol of H2O(g), using CCO2(g) = 37.1 J K-1 mol-1 and CH2O(g) = 33.6 J K-1 mol-1. C = J K-1 is  mathematically represented as

     H  = [ 4 * C_{CO_2_{(g)}} + 6* C_{CH_2O_{(g)}}]

=>  H  = [ 4 * 37.1 + 6* 33.6 ]

=>   H  = 350 JK^{-1}

From the question the initial temperature of reactant is  T_i  =  25^oC

Generally the enthalpy change(\Delta H^o _{rxn}) of the reaction is mathematically represented as

 |\Delta H^o _{rxn} |=  H  * (T_{max} -T_i)

  2568.9 =   350  * (T_{max} -25)  

=> \frac{2568.9 }{350}  =  T_{max} - 25

=> T_{max}  = 32.4 ^o C

   

4 0
2 years ago
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