Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
Answer:
- The limiting reagent is N2O4
- 14,09g
Explanation:
- First, we adjust the reaction.
+
⇄
- Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.
We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.
Using
to form 


Using
to form 


The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.
This is the minimum measure can be formed of each product.
∴ 

Answer : The correct option is, 
Explanation :
The given element bromine belongs to the group 17 and period 4. The symbol of bromine is, Br.
The atomic number of bromine = 35
The total number of electrons present in bromine element = 35
Electronic configuration : It is defined as the arrangement of electrons around the nucleus of an atom.
Hence, the correct electronic configuration of bromine is,
