Heated mater rises and cold mater sinks
The mixture contains 62 % one isomer and 38 % the enantiomer.
Let’s say that the mixture contains 62 % of the (<em>R</em>)-isomer.
Then % (<em>S</em>) = 100 % -62 % = 38 %
ee = % (<em>R</em>) - % (<em>S</em>) = 62 % -38 % = 24 %
Answer:
4 moles of water
Explanation:
this is a combustion reaction, so the balanced equation is: 2C2H6 + 7O2 → 4CO2 + 6H2O.
the molar mass of C2H6 is 30.07g, so 40.0 g of C2H6 is 1.33 moles of C2H6.
mole ratio of H2O to C2H6 is 6/2, or 3.
1.33 moles C2H6 * 3 moles H2O/1 mole C2H6 = 4 moles H2O
As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its 
can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.
Let the three equations with given be denoted as (1), (2), (3), and the last equation (4). Let 
, 
, and 
be letters such that 
. This relationship shall hold for all chemicals involved.
There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 
shall resemble the number of 
left on the product side when the second equation is directly added to the third. Similarly
Thus
and
Verify this conclusion against a fourth species involved- 
for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.
Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.
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