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shepuryov [24]
3 years ago
5

What's is the mass of 0.500 mol NH3​

Chemistry
1 answer:
Setler [38]3 years ago
4 0

Answer:

8.5155g NH3

Explanation:

the molar mass of NH3​ is 17.031 g/mol

0.5 mol NH3 x 17.031 gNH3/1 mol NH3 = 8.5155g NH3

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Can somebody answer this question for me it might be easy for you answer part 1,2,and 3
jonny [76]

Answer:

hey love! what is the question??

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3 years ago
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Which law is based on the graph that is shown below? A graph is shown with pressure on the horizontal axis and volume on the ver
DIA [1.3K]

<u>Answer:</u>

"Boyle's Law" is based on the graph that is shown below.

<u>Explanation:</u>

Boyle's law or Boyle – Mariotte law or Mariotte's law, is an experimental gas law that discusses how a gas's pressure tends to rise as the container volume start declining. This shows the relationship between pressure and volume for a fixed mass at a constant temperature, i.e., number of a gas molecules.This rule visualizes the actions of gas molecules in a confined space. This law can be understood from following equation:

p₁V₁ = p₂V₂

Above the product of the initial volume and pressure is equal to the product of the volume and pressure after a change.

5 0
3 years ago
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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
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
Explain how neutralization reactions are used in acid-base titrations.
nadya68 [22]
Acid-base tiltrations has DNA that can help in the process of neutralization. not so sure about it
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3 years ago
Please help me out its due tmr :/
lesantik [10]

Answer:

using this for points, did you get the answer because I can try and help?

Explanation:

7 0
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