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Elina [12.6K]
3 years ago
5

Why do you think matter changes state when outside forces act upon it?

Chemistry
2 answers:
dexar [7]3 years ago
7 0
They introduce energy into the system
masya89 [10]3 years ago
3 0

Explanation:

There are three states of matter: Solid, liquid and gas. With change in pressure and temperature, matter changes its state. A solid has closely packed particles while a liquid and gas have loosely packed constituent particles. An outside force either compresses or expands the packed matter which causes it to change the state. On compression, particles come closer and gas changes to liquid and liquid to gas where as on expansion, the solid turns to liquid and liquid to gas.

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Explanation:

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particles that make up matter.

Energy Simply stated, energy is the ability to do work or cause

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8 0
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4 0
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Nadusha1986 [10]
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How does a chemical reaction change one substance into another?
sleet_krkn [62]

Answer:

Explanation:

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5 0
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
3 years ago
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