Answer:
There is a production of 11.6 moles of CO₂
Explanation:
The reaction is this:
2C₂H₆(g) + 7O₂(g) ⟶ 4CO₂(g) + 6H₂O(g)
2 moles of ethane reacts with 7 moles of oxygen, to make 4 mol of dioxide and 6 moles of water vapor.
If the oxygen is in excess, we make the calculate with the ethane (limiting reactant)
2 moles of ethane produce 4 moles of dioxide
5.8 moles of ethane produce (5.8 .4)/2 = 11.6 moles
1) As can be seen from any 1H NMR chemical shift ppm tables, hydrogens which have δ values from 2ppm to 2.3ppm are hydrogens from carbon which is bonded to a carbonyl group. From this, we can conclude that our hydrogens belong to the type, but from 2 different alkyl groups because of 2 different signals.
2) So, one alkyl group is CH3 and second one can be CH or CH2.
3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2.
4) Finally, we don't have any other signals and it indicates that part of the compound which continues on CH2 is exactly the same as the first part.
The ratio remains the same, 3:2 ie 6:4
I wanna say 3,398.08 but let me know if that sounds very off
Molar mass of empirical formula = 172.8 g/mole
346.6g/172.8 g/mole = 2.00 moles
Molecular formula = C2H2Br4
Answer:
For a particular chemical reaction, the enthalpy of the reactants is -400 kJ. The enthalpy of the products is -390 kJ. The entropy of the reactants is 0.2 kJ/K. The entropy of the products is 0.3 kJ/K. The temperature of the reaction is 25oC. What can you conclude about this reaction?
It is exergonic
It is endergonic
it is a redox reaction
It is being catalyzed by an enzyme
