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JulijaS [17]
3 years ago
6

In an endothermic reaction what is true of the enthalpy

Chemistry
1 answer:
kow [346]3 years ago
8 0
A. Hope this helps. :-)
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Problem PageQuestion While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synt
OverLord2011 [107]

Question: The question is incomplete. Below is the complete question and the answer;

While ethanol (CH3CH2OH is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene CH2CH2) with water vapor at elevated temperatures. A chemical engineer studying this reaction fills a 50.0 L tank at 22. °C with 24. mol of ethylene gas and 24. mol of water vapor. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 15.4 mol of ethylene gas and 15.4 mol of water vapor The engineer then adds another 12. mol of water, and allows the mixture to come to equilibrium again. Calculate the moles of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer:

Number of moles of ethanol = 11 mol

Explanation:

SEE THE ATTACHED FILE FOR THE CALCULATION

4 0
3 years ago
The O^18 :O^16 abundance ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on ea
Alex Ar [27]

Answer:less than

Explanation:

8 0
3 years ago
Why do scientists look for patterns in the world?
White raven [17]
So they can predict what happens next
4 0
3 years ago
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Explain why seeing a gas doesn not always indicate that there was a chemical change. ​
Orlov [11]
A change of state (which is a physical change) can cause gas.
7 0
2 years ago
A mixture of two compounds, A and B, was separated by extraction. After the compounds were dried, their masses were found to be:
Arlecino [84]

Answer:

The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.

Explanation:

Mass of compound A in a mixture  = 119 mg

Mass of compound A after re-crystallization = 83 mg

Percent recovery from re-crystallization :

\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100

Percent recovery of compound A:

\frac{83 mg}{119 mg}\times 100=69.74\%

Mass of compound B in a mixture  = 97 mg

Mass of compound B after re-crystallization = 79 mg

Percent recovery of compound B:

\frac{79 mg}{97 mg}\times 100=81.44\%

3 0
3 years ago
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