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kkurt [141]
3 years ago
10

True or false. adding more neutrons to a neutrally charged atom will not change the charge

Chemistry
1 answer:
san4es73 [151]3 years ago
6 0
The statement is true.
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Which term is defined as the region in an atom where an electron is most likely to be located
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2 years ago
Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C
ivanzaharov [21]

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!        

7 0
2 years ago
In which cell structure does MOST of the process of cellular respiration take place?
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