Answer:
ΔH°f C₂H₅O₂N(s) = -537.2kJ
Explanation:
Based on the reaction:
4 C₂H₅O₂N(s) + 9O₂(g) → 8CO₂(g) + 10H₂O(l) + 2N₂(g)
ΔHrxn = ΔH°f products - ΔH°f reactants.
As:
ΔH°fO₂(g) = 0
ΔH°fCO₂(g) = -393.5kJ/mol
ΔH°fH₂O(l) = -285.8kJ/mol
ΔH°fN₂(g) = 0
The ΔHrxn is:
ΔHrxn = (8×-393.5kJ/mol + 10×-285.8kJ/mol) - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol
-6006kJ/mol - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol
-4×ΔH°fC₂H₅O₂N(s) = 2149kJ/mol
ΔH°fC₂H₅O₂N(s) = 2149kJ/mol / -4
<h3>ΔH°f C₂H₅O₂N(s) = -537.2kJ</h3>
<span>MgSO4 and Sr(NO3)2 are both soluble, so the ionic equation would look like this.
Mg+2 + SO4-2 + Sr+2 + 2NO3-1 ---> Mg+2 + 2NO3-2 + SrSO4
SrSO4 is insoluble. Now Mg and 2NO3 are on both sides, so cross them out, the net ionic equation is
SO4-2 + Sr+2 ----> SrSO4</span>
There are 1.078 x 10²³ molecules
<h3>Further explanation</h3>
Given
4 dm³ = 4 L Nitrogen gas
Required
Number of molecules
Solution
Assumptions on STP (1 atm, 273 K), 1 mol gas = 22.4 L, so for 4 L :
mol = 4 : 22.4
mol = 0.179
1 mol = 6.02 x 10²³ particles(molecules, atoms)
For 0.179 :
= 0.179 x 6.02 x 10²³
= 1.078 x 10²³
Answer:
Its true I think but I may be wrong lol