Explanation:
The chemical equation is as follows.
And, the given enthalpy is as follows.
; 
= 102.5 kJ
Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol
Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.
102.5 = ![[(\frac{1}{2})x + 498] - [(2)(243)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29x%20%2B%20498%5D%20-%20%5B%282%29%28243%29%5D)
102.5 = 
102.5 - 12 = 
x = 181 kJ
Now, total bond enthalpy of per mole of ClO is calculated as follows.
x = ![[(\frac{1}{2})181 + (\frac{1}{2})498] - 243](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29181%20%2B%20%28%5Cfrac%7B1%7D%7B2%7D%29498%5D%20-%20243)
= 339.5 - 243
= 96.5 kJ
Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.
Answer:
44 g oxygen are needed.
Explanation:
Given data:
Mass of oxygen needed = ?
Mass of ammonia = 18.2 g
Solution:
Chemical equation:
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will calculate the number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 18.2 g/ 17 g/mol
Number of moles = 1.1 mol
Now we will compare the moles of ammonia with oxygen from balance chemical equation.
NH₃ : O₂
4 : 5
1.1 : 5/4×1.1 = 1.375 mol
Mass of oxygen needed:
Mass = number of moles × molar mass
Mass = 1.375 mol × 32 g/mol
Mass = 44 g
Answer:
Na = 32.4% , % S = 22.6% and %O = 45.0%
Explanation:
% Na = 3.4/10.5. × 100%
= 32.4%
%S = 2.37/10.5 × 100%
= 22.6%
% O= 4.73/10.5 × 100%
= 45.0%xplanation:
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