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serg [7]
3 years ago
5

An opening volumetric cylinder filled with water at volume 0.003m after period of

Chemistry
1 answer:
raketka [301]3 years ago
7 0

Answer:

= - 303.975 Joules

Explanation:

Given that:

The volume of the cylinder = 0.003 m^3

Now, to liters:

The volume = 0.003 *1000 L

= 3 L

For an open cylinder, we mean the atmospheric pressure is at standard condition.

Thus, the workdone = -PΔV

= - 1 atm (3 L)

= - 3 L atm

1 L atm to Joules

= 101.325 Joules

Therefore;

=- 3* 101.325 Joules

= - 303.975 Joules

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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
3 years ago
Trans fats: Think about how the prefix trans- is used in naming alkenes.
wel
In an alkene, cis and trans isomers are possible because the double band is rigid, cannot rotate, has groups attached to the carbons of the double bond that are fixed relative to each other, and only occurs with double bonds-possibility that molecule will have different geometries; two different molecules with slightly different properties. 
-Trans-2 ends of chain across the double bond.
While naming Cis-Trans isomers the prefix cis or trans are placed in front of the alkene name when there are cis-trans isomers. 
4 0
3 years ago
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Why the is ratio of magnesium ions to chloride ions in MgCl2 is 1:2?
Monica [59]
You should read up on Proust's law, better known as the Law of Definite Proportions. This is a chemical law that defines your question more generally, on why the ratio of elements and ions are always fixed.

Basically, this compound Magnesium(II) Chloride is MgCl2 because it has the same number of protons, neutrons, and electrons all the way. This defines the properties of the compound or atom. 
5 0
3 years ago
Please help fast, I will give brainliest.
frez [133]

Answer:

The reaction is favorable at all temperatures

Explanation:

Since G = H - TS, -H and +S would result in G = -H -TS, which will always be negative.

7 0
3 years ago
What mass of H₂ is needed to react with 8.75 g of O₂ according to the following equation: O2(g) + H2(g) → H₂O(g)?
FromTheMoon [43]

Explanation:

For reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

The given balanced equation has been:

\rm O_2\;+\;2\;H_2\;\rightarrow\;H_2OO2+2H2→H2O

From the equation, 1 mole of oxygen reacts with 2 mole of hydrogen to give 1 mole of water.

The mass of oxygen has been: 8.75 g,

Moles = \rm \dfrac{weight}{molecular\;weight}molecularweightweight

Moles of oxygen = \rm \dfrac{8.75}{32}328.75

Moles of oxygen = 0.27 mol

Since,

1 mole Oxygen = 2 mole hydrogen

0.21 mol oxygen = 0.54 mol hydrogen

Mass of hydrogen = moles \times× molecular weight

Mass of hydrogen = 0.54 \times× 2

Mass of hydrogen = 1.08 grams.

Thus, for reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

6 0
2 years ago
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