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V125BC [204]
2 years ago
8

Please help me with this science question!! 20 points ​

Chemistry
1 answer:
Volgvan2 years ago
6 0

Answer:

c the energy moved from the water to the ice cubes

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A car with a mass of 1,500 kg is traveling at a speed of 30 m/s. What force must be applied to stop the car in 3 seconds ?
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You want to determine ΔH o for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) To do so, you first determine the heat capacity
Assoli18 [71]

Answer:

(A) The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

 

(B) ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

Explanation:

Solution

 

Calculate the heat actually evolved.

                 q = mcΔt

 

Finding the mass of the reactants in grams we have.

 

Use density. (50 mL + 50 mL ) = 100 mL of solution.

 

100 mL X 1.04g/mL     = 104 grams of solution. (mass = Volume X Density)

                       

 

Find the temperature change.

 

       Δt =tfinal - tinitial = 30.4°C – 16.9°C = 13.5°C

 

    q = mcΔt

       = 104grams × 3.93J/g°C  × 13.5°C = 5.51772×103J

                                         

 

       = 5.51772 × 103 J

 

This is the heat lost in the reaction between HCl and NaOH, therefore q = -5.52 × 103 J.

 

this is an exothermic heat producing reaction.

 To calculate the total heat of the reaction or heat per mole we have

  

50.0 mL of HCl X 2.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

                            

 

The same quantity of base, 0.100 mole NaOH, was used.

The energy per unit mole is given by

  

i.e. molar enthalpy = J/mol = -5.52 × 103J / 0.100 mol

            = -5.52 × 104 J/mol

            = -55177.2 J/mol

            = -55.177 kJ/mol

 

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -55.177 kJ/mol

Heat absorbed by the calorimeter = −57.32kJ − 55.177 kJ = −2.1428KJ

The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

 

(B) For the ZnCl we have

 

Calculate the heat actually evolved.

                            q = mcΔt

 

Finding the mass of the reactants in grams we have.

 

Use density.  100 mL of solution of HCl

 

100 mL X 1.015g/mL        = 101.5 grams of solution. (mass = Volume X Density)

                       

 

Find the temperature change.

 

       Δt =tfinal - tinitial = 20.5°C – 16.8°C = 3.7 °C

 

    q = mcΔt

       = 101.5grams × 3.95J/g°C  × 3.7°C = 1483.422×103J

                                         

 

       = -1483.422×103J

 

This is the heat lost in the reaction between HCl and NaOH, therefore q = -1.483 × 103 J.

 

this is an exothermic heat producing reaction.

 To calculate the total heat of the reaction or heat per mole we have

  

100.0 mL of HCl X 1.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

                            

 

 

The energy per unit mole is given by

  

i.e. molar enthalpy = J/mol = -1.483 × 103J / 0.100 mol

                                         = -1.483 × 104 J/mol

                                         = -14834.22 J/mol

                                         = -14.834 kJ/mol

 

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -14.834 kJ/mol

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

= -14.834 kJ –(0.1587KJ/°C×3.7°C) = -15.42KJ

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

5 0
2 years ago
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