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vodka [1.7K]
3 years ago
13

Which statement is true about a proton and an electron?

Chemistry
1 answer:
tatyana61 [14]3 years ago
5 0

Answer: (3) They have different masses and the same charges.

Explanation:

Atoms are mainly composed of three main types of particles (in fact there are more particles that are smaller and make up these main particles): electrons, protons and neutrons.

In the specific case of  <u>electrons</u> and <u>protons</u>, <u>both have the same charge, but electrons have the opposite sign</u>. This means a proton has an electric charge of  +1 and an electron has an electric charge of  −1.

On the other hand, <u>protons have more mass than electrons.</u> In fact, thetex]mass of an electron is about approximately \frac{1}{2000} the mass of a proton, this means these negative charged particles contribute almost nothing to the total mass of an atom (unlike protons, which together with neutrons, make up virtually all of the mass of an atom).

However, each atom that is electrically neutral has the same quantity of electrons as protons.

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Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C
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The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

  • 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
  • 7O₂: 7 moles of 1 O=O bond  
  • 4CO₂: 4 moles of 2 C=O bonds  
  • 6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

\Delta H = \Delta H_{r} - \Delta H_{p}

\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol  

\Delta H = -2860 kJ/mol          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

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