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Sladkaya [172]
3 years ago
13

Consider the balanced chemical reaction below. When the reaction was carried out, the calculated theoretical yield for carbon di

oxide was 93.7 grams, but the measured yield was 88.3 grams. What is the percent yield?
Fe2O3 + 3CO --> 2Fe + 3CO2
Chemistry
2 answers:
bekas [8.4K]3 years ago
8 0

Answer:94.23

Explanation:

Hope it helps

Arada [10]3 years ago
3 0

The percent yield represents that the efficiency of a reaction.

It represents how much times we have achieved the target yield from a particular reaction

The relation between theoretical yield, measured yield and percent yield is

percent yield = \frac{measuredyieldX100}{theoreticalyield}

for the given reaction

percent yield = \frac{88.3X100}{93.7}=94.23%

Answer: 94.23%

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Your answer is 4 protons 4 electrons and 5 neutrons

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Equal masses of substance A and substance B are at 25°C. Substance A has a lower specific heat capacity than substance B. If 50
liraira [26]
Dang bro that stuff is really hard I’m defiantly on a lower grade lol
6 0
3 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!        

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2 years ago
What will knowledge of chemistry enable you to do?
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Be a chemist. Lol. Lessen waste, make chemicals more efficient, cure diseases... and stuff :)
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3 years ago
What are some factors that affect an element's reactivity?
Alexxandr [17]
From what there’s nothing there? :)
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3 years ago
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