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Marina CMI [18]
3 years ago
8

A 13.58 g sample of a compound contains 8.67 g of iron, Fe, 1.60 g of phosphorus, P, and oxygen, O. Calculate the empirical

Chemistry
1 answer:
antiseptic1488 [7]3 years ago
6 0

Answer:

Fe_3PO_4

Explanation:

To do this, we find the moles of each element. We get around 0.155 moles of Fe, 0.051 moles of P, and 0.206 moles of O. We then divide each one by the smallest one (which is 0.051 moles of P). We then get 3 for Fe, 1 for P, and 4 for O. This correlates to the empirical formula of the compound.

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Intermolecular forces exist between what?
Paraphin [41]

Answer:

Intramolecular forces are the forces that hold atoms together within a molecule. Intermolecular forces are forces that exist between molecules.

Explanation:

6 0
3 years ago
The property of a substance stays ______________ no matter how the substance is used.
Irina18 [472]

Any substance changes to another substance that means the change of the physical property. Like water () has different state which changes as the temperature changes. It remain as liquid in the room temperature, in solid form at or below 0°C and vapor phase on or above 100°C. But in all the stage or phase of the substance the composition of the water i.e.  remains. Thus the chemical property remains fixed when a substance change to other substance.  

3 0
3 years ago
¿Cuáles son las características de los materiales de laboratorio? Por ejemplo: exactitud, resistencia a la temperatura, etc.
boyakko [2]

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7 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!        

7 0
2 years ago
According to early chemists, which substances were classified as elements?
ladessa [460]
The best answer to your question would be B:those who could not be broken down by physical means.
6 0
3 years ago
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