C5H10O5 would weight a total of 2.312 kilograms.
Answer:
Element 2
Explanation:
If we look at the model stated for element 1, it is clear that element 1 must be a noble gas. It has eight electrons in its outermost shell this implies that it has already attained a complete octet of electrons and is reluctant towards chemical reaction.
The second element belongs to group 16 since it has six electrons on its outermost shell. It is certainly more reactive than element 1 which is a noble gas.
Answer: 
kJ/mol
Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.
The standard enthalpy of formation of HCl is calculated as:
→ 
Standard Enthalpy of formation for the other compounds are:
Calcium Hydroxide: 
-1002.82 kJ/mol
Calcium chloride: -795.8 kJ/mol
Water: -285.83 kJ/mol
Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.
Calculating:
![-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]](https://tex.z-dn.net/?f=-17.2%3D%5B-795.8%2B2%28285.85%29%5D-%5B-1002.82%2B2%5CDelta%20H%5D)
So, the standard enthalpy of formation of HCl is -173.72 kJ/mol
Answer:
ΔH = 125.94kJ
Explanation:
It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:
1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ
2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ
-1/2 (1):
WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ
3/2 (2):
3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ
The sum of last both reactions:
WO3(s) + 3H2(g) → W(s) + 3H2O(g)
ΔH = 842.7kJ -716.76kJ
<h3>ΔH = 125.94kJ </h3>
Answer:
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Explanation:
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