Colorful flowers and blueberrie looking ones and the trees
Answer:
2994 kJ
Explanation:
When one mol of ethane (C₂H₆) is combusted, 1451 kJ of heat is exchanged.
First we convert 61.9 g of C₂H₆ into moles, using its molar mass:
- 61.9 g ÷ 30 g/mol = 2.06 mol C₂H₆
Finally we <u>calculate how much heat is exchanged by the combustion of 2.06 moles of C₂H₆</u>:
- 2.06 mol * 1451 kJ/mol = 2994 kJ
To calculate percent composition, you first need to find the molar mass of C (carbon), H (hydrogen) and O (oxygen).
C is 12.01
H is 1.00
O is 16
Then multiply each by the number of atoms of each element in the formula (the number that comes after each element in the equation for example C6 means 6 carbon atoms.
C: 12.01 x 6= 72.06
H: 1x12= 12
O: 16x6= 96
Then add them up.
72.06+ 12+ 96= 180.06
Now find the percent composition of carbon.
72.06/ 180.06 x 100= 40.01%
So the answer is C 40%.
Is the following reaction spontaneous at 298 K? Answer by calculating ΔG. H2O(g) + C(s) → CO(g) + H2(g) ΔH = 131.3 kJ/mole ΔS = 134 J/mole˙K
No
