Answer:
-3.72 (or -3.70 depending on what values you used)
Explanation:
First, use the molar mass of Cl2 convert the mass of Cl2 to moles.
1.48 g Cl2(1 mol70.906 g)=0.02087 mol Cl2
Note that we are given ΔH=−886kJ. This refers to the enthalpy change associated with the reaction of 5mol of Cl2 by the balanced equation shown below.
2P+5Cl2⟶2PCl5ΔH=−886kJ
Therefore, to determine the enthalpy change associated with the reaction of 1.48gCl2, divide ΔH by 5molCl2 to determine the enthalpy change per mole of Cl2, then multiply by 0.02087 mol Cl2. (note: if you round up here to .021 mol of Cl2 you will get the final answer of -3.72 later)
0.02087 mol Cl2(−886 kJ5 mol Cl2)=−3.698 kJ
Rounding the answer should to three significant figures, we find that the enthalpy change associated with the reaction of 1.48gCl2 is −3.70 kJ.
Notice that coefficients in stoichiometric equations (indicating numbers of moles) are exact, so they do not constrain the number of significant figures.
Answer:
Addition reactions are thermodynamically favored at low temperatures.
Explanation:
Compared to substitutions or eliminations, addition reactions do not require to break as many bonds as them, as such, they do not require such a high input of energy (ie. temperature) in order to take place.
This is why if there's a high temperature, the reactions that require more energy -like substitutions or eliminations- will be more thermodinamically favored than the reactions that require less energy -like additions-, and viceversa.
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1) divide each percentage by the relative atomic mass of the element
2) divide all results by the smallest number
3)multiply by a whole number to get the simplest whole number ratio (if necessary)
that is to say:
Na S O
32.37÷23 22.58÷32 45.05÷16
= 1.407 = 0.7056 = 2.816 (to 4 significant figures)
the smallest number here is 0.7056 so:
1.407÷0.7056 0.7056÷0.7056 2.816÷0.7056
=1.99 approx.2 = 1 3.99 approx. 4
here there is no need to carry out step 3 as ratio obtained is already a simplest whole number ratio
so empirical formula is: Na₂SO₄